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Dlan
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Post by Dlan » Tue Jan 21, 2020 10:51 am

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Second round call of diamonds

Just wondering if others would have led the same way?

https://worldofcardgames.com/#!replayer ... %3A1%7D%5D



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Post by Dlan » Tue Jan 21, 2020 11:01 am

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I see this a lot where players don't seem to recognize the jack as a possible trump

https://worldofcardgames.com/#!replayer ... %3A1%7D%5D

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Post by irishwolf » Tue Jan 21, 2020 5:28 pm

Well, answer your own reason for leading an off suit when you have such a strong hand? Why do you think that is better than leading the Right? What are you trying to accomplish and why do you think it a better strategy?

~Irishwolf

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Post by irishwolf » Tue Jan 21, 2020 6:20 pm

In hand 2, it is difficult for me to see the TD would not be picked up - biddable hand for sure!

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Post by Wes (aka the legend) » Tue Jan 21, 2020 6:25 pm

Dlan wrote:
Tue Jan 21, 2020 10:51 am
Image
Second round call of diamonds

Just wondering if others would have led the same way?

https://worldofcardgames.com/#!replayer ... %3A1%7D%5D
Me and Edward have talked about this spot before. I used to be a "gotta lead the Right outta this configuration" type of guy. IIRC I've given advice on this forum saying as much. Edward believed leading a garbage card out of this configuration was the best play. Intuitively, I thought he was wrong, but just for sh*#s and giggles, I tested out Edward's way for a few thousand games, and maybe my brain is playing tricks on me but I felt like my euchre rate in this spot dropped significantly when I led that garbage card.

Now the idea behind leading the Right is pretty clear and simple. Right away, you get a chance to possibly clear out 2 enemy trump with one lead. That's pretty cool. The problem is, once you take this line and the Left doesn't come out, you've usually put your team in a position where your partner HAS to get one trick or you're euchred. Whereas if you follow Edward's advice, and lead that garbage card with this hand, you don't necessarily need your partner's help at all! It's like you're basically setting up an end play to get 3 tricks on your first garbage suit lead. If the 2 seat wins the first trick, you're now last to act and can trump in low on the next lead without fear of getting overtrumped, then you lead another garbage card. If 2 seat takes that trick you're home free as you're guaranteed the last 2 tricks with (Card_A-D) (Card_J-D). And if anywhere during the hand your partner takes a trick instead, you also have at least a point wrapped up.

Another way to look at it: Your hand is not as strong as it looks given that you've ordered up the enemy giving them a trump + a void. Perhaps playing this hand defensively trying to eke out a point while minimizing the chances of a euchre is the best course of action given the marginality of your holding. I think it is.

HOWEVER, there may be one BIG exception to this rule. And it only comes up when your team is down 9-8 like in your situation! Ok we get the idea of leading a garbage card. In theory this line gets euchred less and ekes out a point more often. Nobody's put this to the test but that's what I believe. BUT leading the Right will net your team more 2 point marches. At 0-0, I'm leading a garbage card, but down 9-8 I'm going for the gusto and leading the Right bower maximizing my team's chances of getting 2 points and winning the game right now. And sure, this line will get euchred more than playing it safe and leading a garbage card, but it's not THAT big of a difference. Your team is still gonna score a point most of the time no matter what you do. So down 9-8 I'm going for 2!! F*#@ playing for overtime!!

Overall I think you played the hand well Dlan. Your line is reasonabe as there's no hard data to prove you wrong or right. Even, my so-called exception to the rule "Lead the right when down 9-8" has no hard data behind it.

POP QUIZ for everyone: If Dlan leads the Right down 9-8, does his team get 1 point, 2 points, or euchred? Assume everyone plays well.

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Post by Wes (aka the legend) » Tue Jan 21, 2020 6:37 pm

Dlan wrote:
Tue Jan 21, 2020 11:01 am
Image

I see this a lot where players don't seem to recognize the jack as a possible trump

https://worldofcardgames.com/#!replayer ... %3A1%7D%5D
Wow. So basically my amazing loner gets blown to smithereens becuz Kirsten misread her hand!! Lol.

Note: It was the (Card_10-D) upard and the dealer had:

(Card_A-H) (Card_J-H) (Card_9-D) (Card_J-C) (Card_10-C)

Change her hand very slightly to this:

(Card_A-H) (Card_J-H) (Card_9-D) (Card_J-C) (Card_10-S)

And it wouldn't be clear at all that she misread her hand becuz she now has a nice euchre hand if she passes with a Next call in trouble.

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Post by irishwolf » Tue Jan 21, 2020 6:37 pm

Not sure why the low spade is led on trick 3? Your partner is of NO help. Keep leading trump, you just might force the opponents to slough a spade.

In hand 2, it is difficult for me to see the TD would not be picked up - biddable hand for sure!

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Post by irishwolf » Tue Jan 21, 2020 7:08 pm

We haven't heard from DON? I am curious his comments?

As to your statement, I would take a different route. (WES said) "Now the idea behind leading the Right is pretty clear and simple. Right away, you get a chance to possibly clear out 2 enemy trump with one lead. That's pretty cool. The problem is, once you take this line and the Left doesn't come out, you've usually put your team in a position where your partner HAS to get one trick or you're euchred. Whereas if you follow Edward's advice, and lead that garbage card with this hand, you don't necessarily need your partner's help at all! It's like you're basically setting up an end play to get 3 tricks on your first garbage suit lead. If the 2 seat wins the first trick, you're now last to act and can trump in low on the next lead without fear of getting overtrumped, then you lead another garbage card. If 2 seat takes that trick you're home free as you're guaranteed the last 2 tricks with (Card_A-D) (Card_J-D). And if anywhere during the hand your partner takes a trick instead, you also have at least a point wrapped up"

If the Left is guarded, you don't have 3 tricks (2.5 to 2.75 tricks) and it will be guarded 27% by the dealer and 20% by S2. Almost half the time. More problems can surface by not leading the JD. Almost like that problem, Dealer wins the the first trick with an ace, now puts you in a squeeze. So about 27% you get over trumped and 35% S3 has no trump. Not where I want to go!

JD AD KD can only be overtaken with the guarded left if you lead the Right. I contend that this hand is strong enough that S1 should be leading the JD first. What you will do is (statistically) remove that one trump from S2 (47%) you will take him out of play with trump and sometimes the Left. It does not matter to take your partner's Left. And you will reduce the dealer to just his JH if he has it and take the JH if not guarded by S2.

In the hand as it played out, S1 is euchred on trick 3 if S2 has it instead of S3. Or if the dealer has it guarded. That's how easy it is to euchre the maker. Not leading the JD, you just increased you chance to get euchred.

You don't need your partner for trump, you need him for ACES, and this is what justifies the JD lead. The strategy is that now you get two bites of the apple looking for your partner to have one of those two aces. Even if S2 wins the first off suit lead, now S3 gets a chance to win the next lead, be it a different suit.

If the dealer has the Left + 2, you still have a shot if S3 has an off suit ace. This is statistically the best way to make your point. And sometimes if S3 has the ace on trick 2, you can also get a sweep and puts you behind the dealer.

With a weaker order JD + 2 lower trumps, different story. You need S3 to have a trump and/or ace. But to each his own.

That's my comment!

~Irishwolf

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Post by Dlan » Tue Jan 21, 2020 7:59 pm

I had thought I posted this, but I may have forgot to hit submit. Anyway Irishwolf, here's my story and I'm sticking to it....

This type of hand seems to come up a lot. To be honest, I was trying something different. I don’t really track points vs euchres, but leading the right hasn’t worked out well. The other thing was, I had next deal and a point would put our team in a good position to win.

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Post by irishwolf » Tue Jan 21, 2020 8:09 pm

POP QUIZ for everyone: If Dlan leads the Right down 9-8, does his team get 1 point, 2 points, or euchred? Assume everyone plays well.

You will win four tricks! Lead Right, then the other two as dealer could have 9,10,Q. Assuming you lead spade on trick 4.

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Post by Wes (aka the legend) » Tue Jan 21, 2020 9:12 pm

irishwolf wrote:
Tue Jan 21, 2020 8:09 pm
POP QUIZ for everyone: If Dlan leads the Right down 9-8, does his team get 1 point, 2 points, or euchred? Assume everyone plays well.

You will win four tricks! Lead Right, then the other two as dealer could have 9,10,Q. Assuming you lead spade on trick 4.
The Maker does not have a spade to lead in trick 4.

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Post by Wes (aka the legend) » Tue Jan 21, 2020 9:59 pm

irishwolf wrote:
Tue Jan 21, 2020 7:08 pm
As to your statement, I would take a different route.

If the Left is guarded, you don't have 3 tricks (2.5 to 2.75 tricks) and it will be guarded 27% by the dealer and 20% by S2. Almost half the time. More problems can surface by not leading the JD. Almost like that problem, Dealer wins the the first trick with an ace, now puts you in a squeeze. So about 27% you get over trumped and 35% S3 has no trump. Not where I want to go!
If the dealer wins the first trick with an off ace and puts you in a squeeze, you have a choice to make with this specific holding.

1) You could throw off with your other garbage card giving your partner a chance to take this trick. If your partner takes the trick your team is home free, and if the 2 seat takes the trick you still have the lead in the right spot. The problem with this line tho is when the 2 seat takes the 2nd trick, your team is royally screwed every time the dealer has a guarded left becuz when you invariably get the lead on the 4th trick the dealer will now have you cornered and there's no escape.

2) Instead of throwing off on the dealer's double lead you trump in with your KD gambling that the 2 seat does not have the Left, and then you lead your garbage card setting up a an end play with your AdJd. If the 2 seat takes the 3rd trick, bam! Your team is home free as your end play will now be successful and take the last 2 tricks. If your partner takes the 3rd trick, then your team is home free again and scores a point. If the dealer takes the 3rd trick, your team still scores a point every time as long as the 2 seat does not hold the Left. Notice that when you run this play, every time the dealer has a guarded Left you render him completely impotent. You strip it every time. You only get screwed when the Left is in the 2 seat's hand. If it's in your partner's hand then ofc your team already has a point locked up.

I think the safest and best route is #2. Trump in on the 2nd trick and lead the garbage card.
irishwolf wrote:
Tue Jan 21, 2020 7:08 pm
JD AD KD can only be overtaken with the guarded left if you lead the Right. I contend that this hand is strong enough that S1 should be leading the JD first. What you will do is (statistically) remove that one trump from S2 (47%) you will take him out of play with trump and sometimes the Left. It does not matter to take your partner's Left. And you will reduce the dealer to just his JH if he has it and take the JH if not guarded by S2.

In the hand as it played out, S1 is euchred on trick 3 if S2 has it instead of S3. Or if the dealer has it guarded. That's how easy it is to euchre the maker. Not leading the JD, you just increased you chance to get euchred.
This claim I disagree with, although I can't refute it. By leading a garbage card it is very hard for this hand to get euchred if played well. A parlay basically has to occur. 1) Seat 2 has to have the Left and 2) Seat 2 has to be able to use it at the right time which won't always be the case becuz he may have to follow suit.
irishwolf wrote:
Tue Jan 21, 2020 7:08 pm
You don't need your partner for trump, you need him for ACES, and this is what justifies the JD lead. The strategy is that now you get two bites of the apple looking for your partner to have one of those two aces. Even if S2 wins the first off suit lead, now S3 gets a chance to win the next lead, be it a different suit.
Right of course. That's the idea behind leading the Right. We get to potentially clear out two enemy trump with one lead and if our partner has 1 or more off suit Aces, our trump lead helps promote them. We're certainly not banking on our partner to have 2 trump in this spot.

What's nice about playing the hand the way Dlan did, is that we have a good chance to score a point EVEN IF our partner has no off aces to help. And sometimes when our partner has 1 trump and a void somewhere, Dlan's line will enable him to help in that way. An option that's gone once we lead the Right.

Also, those times the dealer has 3 trump, Dlan's line is way more likely to escape with a point than leading the Right. That's important too, and it's for the same reason that Dlan's line will often strip the dealer of a guarded Left, if played right.
irishwolf wrote:
Tue Jan 21, 2020 7:08 pm
If the dealer has the Left + 2, you still have a shot if S3 has an off suit ace. This is statistically the best way to make your point. And sometimes if S3 has the ace on trick 2, you can also get a sweep and puts you behind the dealer.


I mean, yeah, that's obviously your claim, but no back of the envelope math can sway me on this. I need hard data on this one. Without hard data, like actually simulating this spot for however many hands necessary, I think we're all forced to be agnostic on which line is best. I will advocate for leading a garbage card here but there's no F-ing way I would bet my life I'm right. I'm definitely unsure, but that's the side I'm betting on until we get real evidence.
irishwolf wrote:
Tue Jan 21, 2020 7:08 pm
With a weaker order JD + 2 lower trumps, different story. You need S3 to have a trump and/or ace. But to each his own.

That's my comment!

~Irishwolf
I play it the same way with a weaker Right + 2 order, but the line of leading an offsuit garbage card is probably most ideal when you specifically hold

(Card_J-D) (Card_A-D) (Card_K-D)

becuz it's so much easier to set up an end play and eke out 3 tricks with this hand becuz Seat 2 has to have that one key card (the Left) to overtrump you and screw up your play. Whereas with a weaker Right + 2, there could be multiple trump cards Seat 2 could have that could ruin your day.
Last edited by Wes (aka the legend) on Wed Jan 22, 2020 1:20 pm, edited 1 time in total.

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Post by Wes (aka the legend) » Wed Jan 22, 2020 12:14 am

Wes (aka the legend) wrote:
Tue Jan 21, 2020 6:25 pm
HOWEVER, there may be one BIG exception to this rule. And it only comes up when your team is down 9-8 like in your situation! Ok we get the idea of leading a garbage card. In theory this line gets euchred less and ekes out a point more often. Nobody's put this to the test but that's what I believe. BUT leading the Right will net your team more 2 point marches. At 0-0, I'm leading a garbage card, but down 9-8 I'm going for the gusto and leading the Right bower maximizing my team's chances of getting 2 points and winning the game right now. And sure, this line will get euchred more than playing it safe and leading a garbage card, but it's not THAT big of a difference. Your team is still gonna score a point most of the time no matter what you do. So down 9-8 I'm going for 2!! F*#@ playing for overtime!!

Overall I think you played the hand well Dlan. Your line is reasonabe as there's no hard data to prove you wrong or right. Even, my so-called exception to the rule "Lead the right when down 9-8" has no hard data behind it.
Giving this more thought, logically speaking there can't be an "exception to the rule". If leading a garbage card has a higher EV than leading the Right, then that's the correct strategy no matter what the score. My bifurcation in strategy (Lead a garbage card at every score except lead the Right down 9-8) betrays the fact that I don't really know which strategy has a higher EV, which I've already admitted I don't.

So my overall thinking on this is: I'm leading a garbage card from this hand configuration becuz I suspect this line has a higher EV than leading the Right, but becuz I suspect it's close and I could be wrong, I'm erring on the side of maximizing my 2 pt chances down 9-8.

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Post by Wes (aka the legend) » Wed Jan 22, 2020 12:18 am

Wes (aka the legend) wrote:
Tue Jan 21, 2020 6:25 pm
POP QUIZ for everyone: If Dlan leads the Right down 9-8, does his team get 1 point, 2 points, or euchred? Assume everyone plays well.
Anyone else wanna give this a go?

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Post by Tbolt65 » Wed Jan 22, 2020 2:05 am

I look at is this way. I ask myself How am I going to allow my partner to get a trick? Whats the best way he/she "Might" take one for me or two for that matter for a possible march or point? Will they take it with a trump, or an off suit ace or other? The main goal is to not get euchred. On a next call, for round two I tend to favor leading trump of some kind and if I take it I lead then the turned down suit. How ever depending on the score, players, card distributation and If I may need to play it safe. I will then lead my garbage card and try to hit my partners void or Ace. If the other team takes the trick then fine. Then I essentially set up what Wes talks about. Although I do leave myself exposed. That can happen at times. Also have to take into consideration that the opponents may not play well and may hold onto their cards, and not play optimally. Not with just this scenario but in all. So many times where one shoulda got euchred or only one point the opponents falter and make a key mistake. That is also another factor to consider. Is how well your opponents play while in the middle of playing this hand.

So basically I don't always play it one way either, but in the end I will favor playing it safe and getting points. I hate giving up two points.

I am fine with Dlan opening lead here.

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Post by Wes (aka the legend) » Wed Jan 22, 2020 3:10 am

irishwolf wrote:
Tue Jan 21, 2020 7:08 pm
If the Left is guarded, you don't have 3 tricks (2.5 to 2.75 tricks) and it will be guarded 27% by the dealer and 20% by S2.
BTW your probability that the dealer is guarded is close enough (27.78% is the number I'm getting), but I believe you're wrong about the probability of S2 being guarded. 20% is way too high. The probability that Seat 2 is guarded = the probability Seat 2 has the Left multiplied by the probability Seat 2 has at least 1 more trump in addition to the Left.

The probability Seat 2 has the Left:

1 -[(17/18)(16/17)(15/16)(14/15)(13/14)] = 27.78%

The probability Seat 2 has at least 1 more trump in addition to the Left:

1- [(15/17)(14/16)(13/15)(12/14)] = 42.65%

Thus the probability that Seat 2 is guarded: 27.78% x 42.65% = 11.85%

Solving this using factorials:

The probability Seat 2 has precisely (Card_J-H) (Card_Q-D):

[(1c1 x 1c1 x 16c3)/18c5] - those times S2 has JhQd9d [(3c3 x 15c2)/18c5)]

This translates to (560/8568) - (105/8568) = 5.31%

The probability that Seat 2 has precisely (Card_J-H) (Card_9-D):

Same calculations and answer as above for JHQD, 5.31%

The probability Seat 2 has all three remaining trump:

(Card_J-H) (Card_Q-D) (Card_9-D)

[(3c3 x 15c2)/18c5] which translates to 105/8568 = 1.23%

Add all those probabilities together. 5.31% + 5.31% + 1.23% = 11.85%

Thus the probability that Seat 2 is guarded is 11.85%

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Post by irishwolf » Wed Jan 22, 2020 10:55 am

You are mistaken on your calculations:

Nit pitting on the is it 27 or 27.78%. If there are two unknown trumps (QD/9D) there is a 50% he has one of those trumps (42% either) and a 6.5% chance he has both.


BTW your probability that the dealer is guarded is close enough (27.78% is the number I'm getting), but I believe you're wrong about the probability of S2 being guarded. 20% is way too high. The probability that Seat 2 is guarded = the probability Seat 2 has the Left multiplied by the probability Seat 2 has at least 1 more trump in addition to the Left.

The probability Seat 2 has the Left:

1 -[(17/18)(16/17)(15/16)(14/15)(13/14)] = 27.78%

The probability Seat 2 has at least 1 more trump in addition to the Left:

1- [(15/17)(14/16)(13/15)(12/14)] = 42.65%

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Post by irishwolf » Wed Jan 22, 2020 11:10 am

Looks like you have made a case for S1 to lead the JD? If ther 9s is led and the dealer has AS/Xs now S2 to can over the KD OR AD.

Thus the probability that Seat 2 is guarded is 11.85%

The biggest vulnerability for leading the JD is when the dealer has the Left and the AS. Small percentage and even becomes an issue if the 9S or JD is led.

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Post by Tbolt65 » Wed Jan 22, 2020 1:06 pm

I forgot to mention in my previous post. If the 2nd round declaration of Trump is not next and it's the same hand make up as shown but just a different color. I am almost always leading my garbage card. With the current hand configuration.

But as the variables changes so does my line of play.

Tbolt65
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Post by Wes (aka the legend) » Wed Jan 22, 2020 1:07 pm

irishwolf wrote:
Wed Jan 22, 2020 11:10 am
Looks like you have made a case for S1 to lead the JD? If ther 9s is led and the dealer has AS/Xs now S2 to can over the KD OR AD.

Thus the probability that Seat 2 is guarded is 11.85%

The biggest vulnerability for leading the JD is when the dealer has the Left and the AS. Small percentage and even becomes an issue if the 9S or JD is led.
Not so sure about that, but based on the numbers, I do think I have to change my initial conclusion below:
If the dealer wins the first trick with an off ace and puts you in a squeeze, you have a choice to make with this specific holding.

1) You could throw off with your other garbage card giving your partner a chance to take this trick. If your partner takes the trick your team is home free, and if the 2 seat takes the trick you still have the lead in the right spot. The problem with this line tho is when the 2 seat takes the 2nd trick, your team is royally screwed every time the dealer has a guarded left becuz when you invariably get the lead on the 4th trick the dealer will now have you cornered and there's no escape.

2) Instead of throwing off on the dealer's double lead you trump in with your KD gambling that the 2 seat does not have the Left, and then you lead your garbage card setting up a an end play with your AdJd. If the 2 seat takes the 3rd trick, bam! Your team is home free as your end play will now be successful and take the last 2 tricks. If your partner takes the 3rd trick, then your team is home free again and scores a point. If the dealer takes the 3rd trick, your team still scores a point every time as long as the 2 seat does not hold the Left. Notice that when you run this play, every time the dealer has a guarded Left you render him completely impotent. You strip it every time. You only get screwed when the Left is in the 2 seat's hand. If it's in your partner's hand then ofc your team already has a point locked up.

I think the safest and best route is #2. Trump in on the 2nd trick and lead the garbage card.

I thought we could gamble that S2 doesn't have the left and trump in even on a double lead, but now seeing that there's a 27.78% chance that Seat 2 has the Left, we HAVE to play off those times the dealer takes the first trick on our offsuit lead and then puts us in a squeeze double leading the same suit. 27.78% is just too damn high to gamble in that spot. After we throw off our other garbage suit, if S2 trumps the double lead with the Left we are home free, and if Seat 2 has Left + 1 and trumps low we are home free also. If S2 plays off in this spot, hopefully our partner has 1 trump to trump in or the boss card of that suit if the dealer is not double leading a boss card, if so we score a point that way too.

Now, if the dealer takes the first trick (by trumping in or with an off suit boss card) and then leads a non-trump fresh suit that we are void in, I.E. spades in this example, THEN we should trump in right away, and lead a garbage suit imo. It's important we get a trick in, set up our end play and don't have the lead on trick 4. That way we can strip the dealer of a guarded Left every time. For S2 to screw us in this specific spot a parlay must occur: S2 must 1) be void in the fresh suit and 2) have the Left. I'll take my chances now.

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Post by Wes (aka the legend) » Wed Jan 22, 2020 2:07 pm

Wes (aka the legend) wrote:
Tue Jan 21, 2020 6:25 pm
POP QUIZ for everyone: If Dlan leads the Right down 9-8, does his team get 1 point, 2 points, or euchred? Assume everyone plays well.
First trick:

S1 leads the (Card_J-D)
S2 plays the (Card_J-C)
S3 plays the (Card_J-H)
S4 plays the (Card_10-D)

Good guys take the first trick with the (Card_J-D)

2nd trick:

Now that the Left has been exposed, and the 9D/QD is still in the wild, the Maker must lead trump again.

S1 leads the (Card_A-D)
S2 plays the (Card_Q-S)
S3 plays the (Card_9-S)
S4 plays the (Card_9-D)

Good guys take the 2nd trick with the (Card_A-D) .

Notes: S2 correctly plays the QS preserving his guarded King in clubs as long as possible. When S3 throws off the 9S this is a signal to his partner that he DOES NOT have spades covered. In this spot S3's job is to preserve his suited ace for the last 2 tricks if he has one. When S3 plays the 9S on the 2nd trick lead he is telling his partner he does not have As9s becuz if he did the last thing he would do is break that up. S3 also correctly holds unto his KH at this stage just in case the AH is buried.

3rd trick:

Becuz the QD is still out in the wild, the maker must lead his last boss trump.

S1 leads the (Card_K-D)
S2 plays the (Card_Q-C)
S3 plays the (Card_K-H)
S4 plays the (Card_10-S)

Good guys take the 3rd trick with the (Card_K-D) .

Notes: S2 is forced to unguard his KC to keep his AS. S3 plays the KH which emphatically tells his partner "I DO NOT HAVE HEARTS COVERED" becuz if S3 had hearts covered he would NEVER play the KH. If S3 had hearts covered with the KH then he would have to have AHKH and in that case he would always play the AH first letting his partner know he does have hearts covered. That's why when S3 playes the KH first we know without a shadow of a doubt S3 is not covering Hearts.

4th trick:

The maker has the lead with a garbage heart and a garbage club left. Knowing that his partner is never covering hearts, the maker has an easy decision to lead the club. If the maker had a garbage club and a garbage spade left, he would also have an easy decision to lead a club since his partner threw off a spade earlier in the hand signaling he's not covering spades. And if the Maker had 2 bad choices in the end with a garbage spade and a garbage heart, he should choose to lead the spade becuz his partner playing the KH is a stronger signal that he's not covering a suit then playing the 9S earlier in the hand.

S1 leads the (Card_10-C)
S2 plays the (Card_K-C)
S3 plays the (Card_A-C)
S4 plays the (Card_K-S)

Good guys take the 4th trick with the (Card_A-C) .

Notes: Seat 4 correctly throws away his non-boss spade hoping in vain his boss heart takes the last trick.

5th trick:

S3 plays the (Card_9-C)
S4 plays the (Card_A-H)
S1 plays the (Card_10-H)
S2 plays the (Card_A-S)

Good guys take the 5th trick with the boss (Card_9-C).

That's all 5 tricks for the game winning march!

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Post by Tbolt65 » Wed Jan 22, 2020 3:20 pm

Good break down on reading the hand as it is played out Wes.

Tbolt65
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Wes (aka the legend)
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Post by Wes (aka the legend) » Wed Jan 22, 2020 3:35 pm

irishwolf wrote:
Wed Jan 22, 2020 10:55 am
You are mistaken on your calculations:

Nit pitting on the is it 27 or 27.78%. If there are two unknown trumps (QD/9D) there is a 50% he has one of those trumps (42% either) and a 6.5% chance he has both.
I stand by my calculations. I think you're wrong here. If S2 has the Left, then there are only 4 other slots he can have the QD or the 9D or both. This isn't that hard. Once we know the probability of S2 having the Left is 27.78%:

1 -[(17/18)(16/17)(15/16)(14/15)(13/14)] = 27.78%

or using factorials:

(1c1 x 17c4)/18c5 = 27.78%

Then all we have to do is figure out the probability S2 has "At least one" more trump in addition to the Left. Mathematically "At least 1": = Total - None, which translates to:

1 - [(15/17)(14/16)(13/15)(12/14)] = 42.65%

Then we just multiply the probabilities together 27.78% x 42.65% = 11.85%

That's how often S2 will be guarded in this spot.

Wes (aka the legend)
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Post by Wes (aka the legend) » Wed Jan 22, 2020 4:13 pm

irishwolf wrote:
Wed Jan 22, 2020 10:55 am
You are mistaken on your calculations:

If there are two unknown trumps (QD/9D) there is a 50% he has one of those trumps (42% either) and a 6.5% chance he has both.
Once we know S2 has the Left, there is not a 6.5% chance he also has both the QD and the 9D in his hand. It's a 1.23% chance.

The formula for that calculation:

[(3c3 x 15c2)/18c5] = 105/8568 = 1.23%

Which is simply the probability S2 has all 3 remaining trump (LQ9) in his 5 card hand.

There is a 6.54% chance S2 has (Card_J-H) (Card_Q-D) in his hand,

and there's a 5.31% chance S2 has precisely (Card_J-H) (Card_Q-D) with no other trump.

And there is a 6.54% chance S2 has (Card_J-H) (Card_9-D) in his hand,

and a 5.31% chance S2 has precisely (Card_J-H) (Card_9-D) with no other trump.

And as said above, there is a 1.23% S2 has all 3 remaining trump in his hand:

(Card_J-H) (Card_Q-D) (Card_9-D)

5.31% + 5.31% + 1.23% = 11.85% chance S2 is guarded.
Last edited by Wes (aka the legend) on Wed Jan 22, 2020 11:16 pm, edited 1 time in total.

irishwolf
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Post by irishwolf » Wed Jan 22, 2020 4:35 pm

AS I SAID WRONG CALCULATION. You don't know what you are doing.

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Post by irishwolf » Wed Jan 22, 2020 4:55 pm

For independent events, the probability of both occurring is the product of the probabilities of the individual events: Pr(AandB)=Pr(A∩B)=Pr(A)×Pr(B). Example: if you flip a coin twice, the probability of heads both times is: 1/2×1/2=1/4.

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Post by Wes (aka the legend) » Wed Jan 22, 2020 5:25 pm

irishwolf wrote:
Wed Jan 22, 2020 4:35 pm
AS I SAID WRONG CALCULATION. You don't know what you are doing.
I am calculating the probability that S2 has a guarded Left which is 11.85%. Substantiate your claim. Show me where my work is wrong.

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Post by Wes (aka the legend) » Wed Jan 22, 2020 6:21 pm

irishwolf wrote:
Wed Jan 22, 2020 4:35 pm
AS I SAID WRONG CALCULATION. You don't know what you are doing.
You know what, let's back up here and go step by step.

1) Do you agree that there are 6 exposed cards here (the upcard + the 5 cards in S1's hands) and thus 18 unexposed cards?

2) Do you then agree that there are 18C5 possible hand combos that S2 can have? Which translates to 8,568 combos.

Are you with me so far?

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Post by Wes (aka the legend) » Thu Jan 23, 2020 1:16 am

irishwolf wrote:
Wed Jan 22, 2020 4:35 pm
AS I SAID WRONG CALCULATION. You don't know what you are doing.
BTW for those interested (something tells me no one is lol) there's actually another way you can solve the probability problem of how often S2 is guarded.
You can solve it indirectly by figuring out how often S2 is NOT guarded and then subtract that number from 1.

To do this we need to do the following steps:

1) Figure out the probability Seat 2 does NOT have the Left in his hand.

2) Then we need to figure out the probability Seat 2 ONLY has the Left in his hand with four other non-trump cards.

3) Add those two probabilities together.

4) Then subtract that total probability from 1 and then we're done.

Ok let's go!

Step 1: The probability Seat 2 does NOT have the Left in his hand is:

17C5/18C5 = 6,188/8,568 = .7222 or 72.22%

Step 2: The probablity Seat 2 ONLY has the Left in his hand with four other non-trump cards:

[(1C1 x 15C4)/18C5] = 1365/8,568 = .1593 or 15.93%

Step 3: Add those probabilities together, .7222 + .1593 = .8815 or 88.15%.

So Seat 2 will NOT be guarded 88.15% of the time.

Step 4: 1 - .8815 = .1185 or 11.85%.

Therefore, Seat 2 will be guarded 11.85% of the time, same answer as before.

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Post by irishwolf » Thu Jan 23, 2020 10:26 am

Wes,

You are correct for the following: down to Step 2 below. Issue is that events occurring independently occurring at the same time is the product of (multiplied together not added or subtracted the way you are doing).

Look at it this way: What the chance it will Rain today, weather man says 70% (so that is .70). What is the chance I forget my umbrella (say 50%). Those are two independent events. So the chance both occur together is .70 x .50 = .35 or 35% chance it will rain and I will forget my umbrella.

So to the Left that is an independent event = 27.8% chance S2 has it.
Now what is the chance that two Diamonds will not be dealt at the same time the JH is dealt? 2c0 16c5 / 18c5 = 14.2%
or you can get it this way as well: (2c2 16c3 18c5 + 2c1 16c4 18c5 {chance has 1 = 42.5 + has both = 6.5 = 49.1%}) x 27.8
____________________________________

To do this we need to do the following steps:

1) Figure out the probability Seat 2 does NOT have the Left in his hand.

2) Then we need to figure out the probability Seat 2 ONLY has the Left in his hand with four other non-trump cards.

3) Add those two probabilities together.

4) Then subtract that total probability from 1 and then we're done.

Ok let's go!

Step 1: The probability Seat 2 does NOT have the Left in his hand is:

17C5/18C5 = 6,188/8,568 = .7222 or 72.22% (or 1 - 27.8) correct

THEN YOU GO OFF THE CHART WITH THIS: (NOT CORRECT STEPS 2 & 4)

Step 2: The probablity Seat 2 ONLY has the Left in his hand with four other non-trump cards:

[(1C1 x 15C4)/18C5] = 1365/8,568 = .1593 or 15.93% (WRONG BECAUSE YOU DISREGARD THE CHANCE OF NOT HAVING EITHER QD OR 9D IN THIS CALCULATION!)

There are 2 Diamonds, trying to find probability has no Diamonds which is
2c0 = 1 x (16c5 = 4368) / 8568 = 51% (rounded) x 27.8 = 14.2% has the left and not the QD or 9D. Its the product.

Step 3: Add those probabilities together, .7222 + .1593 = .8815 or 88.15%.

Events occurring at the same time is the product of those event occurring independently. You can't add these together.)

So Seat 2 will NOT be guarded 88.15% of the time.

Step 4: 1 - .8815 = .1185 or 11.85%. (WRONG)

Therefore, Seat 2 will be guarded 11.85% of the time, same answer as before.

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Post by irishwolf » Thu Jan 23, 2020 10:48 am

This wrong and not how to solve this in this manner:

______________

"You are mixing events, treat as independent! 15X14X13 / 3X2X1 = 455 / 8568 = 5.3% for having JH QD"

is a 6.54% chance S2 has (Card_J-H) (Card_Q-D) in his hand, (not sure why you are doing this but it is correct or 27.8 x 27.8 - two independent events)

and there's a 5.31% chance S2 has precisely (Card_J-H) (Card_Q-D) with no other trump. (No clue what you are doing - this wrong using the wrong formula)

And there is a 6.54% chance S2 has (Card_J-H) (Card_9-D) in his hand,

and a 5.31% chance S2 has precisely (Card_J-H) (Card_9-D) with no other trump. (same issue wrong formula)
________________________

So this is 15c3 / 18c5 - where is the rest of the calculation? Can't get there like this.

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Post by irishwolf » Thu Jan 23, 2020 10:59 am

I see your mistake:

Step 2: The probability Seat 2 ONLY has the Left in his hand with four other non-trump cards:

[(1C1 x 15C4)/18C5] = 1365/8,568 = .1593 or 15.93% THERE ARE 18 UNKNOWN CARDS AND HERE YOU USE 16 CARDS. NOT SURE WHY? 1C1 17C4 / 8568 = 2380 / 8568 = 27.8 AND 1.00 - 27.8 = 72.2% LEFT BY ITSELF BUT THIS SAYS NOTHING ABOUT THE QD & 9D WHICH IS STILL IN THE 17 UNKNOWN.

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Post by Wes (aka the legend) » Thu Jan 23, 2020 2:24 pm

irishwolf wrote:
Wed Jan 22, 2020 4:35 pm
AS I SAID WRONG CALCULATION. You don't know what you are doing.
irishwolf wrote:
Wed Jan 22, 2020 4:55 pm
For independent events, the probability of both occurring is the product of the probabilities of the individual events: Pr(AandB)=Pr(A∩B)=Pr(A)×Pr(B). Example: if you flip a coin twice, the probability of heads both times is: 1/2×1/2=1/4.
irishwolf wrote:
Wed Jan 22, 2020 4:35 pm
Wes,

You are correct for the following: down to Step 2 below. Issue is that events occurring independently occurring at the same time is the product of (multiplied together not added or subtracted the way you are doing).

Look at it this way: What the chance it will Rain today, weather man says 70% (so that is .70). What is the chance I forget my umbrella (say 50%). Those are two independent events. So the chance both occur together is .70 x .50 = .35 or 35% chance it will rain and I will forget my umbrella.
Ok, I think what's really going on here is I am doing calculations that you don't really know how to do. All this math stuff seems to be a bit over your head. So let me help you out now. After all, that's what I'm here for. When you're trying to figure out how many 5 card hand combinations Seat 2 can have with a guarded Left, you don't multiply all possibilities. You add them up.

So for example. When Seat 2 is guarded there are only 3 different hand combo possibilities:

1) Those times S2 has (Card_J-H) (Card_Q-D) (Card_9-D) X,X

2) Those times S2 has precisely (Card_J-H) (Card_Q-D) X,X,X

3) Those times S2 has precisely (Card_J-H) (Card_9-D) X,X,X


Step 1) The number of 5 card hand combinations S2 can have of

(Card_J-H) (Card_Q-D) (Card_9-D) X,X

Is 3C3 x 15C2 = 105 combinations

Step 2) The number of 5 card hand combinations S2 can have of precisely

(Card_J-H) (Card_Q-D) X,X,X

(1C1 x 1C1 x 16C3) - (3C3 x 15C2) = 455 combinations

Note: I subtracted out (3C3 x 15C2) cuz I don't wanna double count those times S2 has LQ9 in trump.

Step 3) The number of 5 card hand combinations S2 can have precisely

(Card_J-H) (Card_9-D) X,X,X

(1C1 x 1C1 x 16C3) - (3C3 x 15C2) = 455 combinations

Step 4) Add up all the 5 card hand combinations:

105 + 455 + 455 = 1,015

Step 5) Then, to get the probability of S2 being guarded, divide that number (1,015) by the total amount of possible 5 card hand combinations. Since there are 18 unexposed cards, the total hand combos possible = 18C5 = 8568 possible combinations.

1015/8568 = 11.85%

QED
irishwolf wrote:
Wed Jan 22, 2020 4:35 pm
So to the Left that is an independent event = 27.8% chance S2 has it.
Now what is the chance that two Diamonds will not be dealt at the same time the JH is dealt? 2c0 16c5 / 18c5 = 14.2%
Your equation is wrong. There are not 16 non-trump cards there are only 15 (You're double counting the Left it appears).

15C4 = 1365 combinations. 15C4/18C5 = 15.93% which is the probability of S2 having an unguarded Left bower in his hand.

Stop trying to do stuff you really don't know how to do Irishwolf. You are out of your element here.
irishwolf wrote:
Thu Jan 23, 2020 10:48 am
This wrong and not how to solve this in this manner:

______________

"You are mixing events, treat as independent! 15X14X13 / 3X2X1 = 455 / 8568 = 5.3% for having JH QD"

is a 6.54% chance S2 has (Card_J-H) (Card_Q-D) in his hand, (not sure why you are doing this but it is correct or 27.8 x 27.8 - two independent events)

and there's a 5.31% chance S2 has precisely (Card_J-H) (Card_Q-D) with no other trump. (No clue what you are doing - this wrong using the wrong formula)

And there is a 6.54% chance S2 has (Card_J-H) (Card_9-D) in his hand,

and a 5.31% chance S2 has precisely (Card_J-H) (Card_9-D) with no other trump. (same issue wrong formula)
I realize you have no clue what I'm doing Irishwolf, that much is clear. I subtracted out the probability S2 has all LQ9 in trump (1.23%) from the probability that S2 has (Card_J-H) (Card_Q-D) X,X,X to make sure I don't double count those times S2 has LQ9 in trump.

So Seat 2 will have (Card_J-H) (Card_Q-D) X,X,X, 6.54% of the time and once you subtract out those times he has (Card_J-H) (Card_Q-D) (Card_9-D) X,X, he will have PRECISELY (Card_J-H) (Card_Q-D) X,X,X, i.e with no other trump 5.31% of the time.

Do the same thing with (Card_J-H) (Card_9-D), 6.54% -1.23% = 5.31%

Then add the probabilities up: 5.31% + 5.31% + 1.23% = 11.85%

S2 is guarded 11.85% of the time.

QED
irishwolf wrote:
Thu Jan 23, 2020 10:48 am
So this is 15c3 / 18c5 - where is the rest of the calculation? Can't get there like this.
Yes you can, the other numbers = 1, so they are not necessary.

The full calculation for the probability of S2 having precisely (Card_J-H) (Card_Q-D) and 3 non trump cards =

(1C1 x 1C1 x 1C0 x 15C3)/18C5

Since 1C1 = 1, and 1C0 = 1, we can rewrite it as

15C3/18C5 which = 455/8568 = 5.31%

That's how often S2 will have PRECISELY (Card_J-H) (Card_Q-D) + 3 non-trump in his hand, same goes for (Card_J-H) (Card_9-D), 5.31%

Hence, 5.31% + 5.31% + the probability S2 has all 3 remaining trump (LQ9) 1.23% = 11.85%

QED
irishwolf wrote:
Thu Jan 23, 2020 10:59 am
I see your mistake:

Step 2: The probability Seat 2 ONLY has the Left in his hand with four other non-trump cards:

[(1C1 x 15C4)/18C5] = 1365/8,568 = .1593 or 15.93% THERE ARE 18 UNKNOWN CARDS AND HERE YOU USE 16 CARDS. NOT SURE WHY?

Again, same theme here. You're the one F***ING this up and making mistakes here. Not me.

I left out 2C0 becuz it = 1, and is therefore unnecessary. 2C0 represents the two other trump (Qd9d) and I am choosing zero since I'm trying to figure out how often S2 has the Left + 4 other non-trump cards.

So the full calculation is [(1C1 x 2C0 x 15C4)/18C5] = 1365/8568 = 15.93%

Since 1C1 = 1 and 2C0 = 1, I really only need to write: 15C4/18C5

BTW 1C1 represents the Left. There is only 1 JH and I am choosing it.

To drive this last piont home, we can use another example. Assuming 24 unexposed cards (in a euchre deck obvly) how many 5 card hand combinations are there where you will have all red cards?

The answer is 12C5. There are 12 red cards and you are choosing 5.

Sure I could also write it as 12C5 x 12C0 to account for the black cards I am not choosing but that would be unnecessary since 12C0 = 1, so I can just write

12C5, and for those wondering 12C5 = 792 hand combos.

And the probability of getting such a hand is 12C5/24C5 = 792/42,504 = 1.86%
irishwolf wrote:
Wed Jan 22, 2020 4:35 pm
1C1 17C4 / 8568 = 2380 / 8568 = 27.8 AND 1.00 - 27.8 = 72.2% LEFT BY ITSELF BUT THIS SAYS NOTHING ABOUT THE QD & 9D WHICH IS STILL IN THE 17 UNKNOWN.
Yes, 1C1 x 17C4 represents how often S2 will have the Left in his hand, I.E. how often S2 will have (Card_J-H) and 4 other cards that could be trump or non-trump. Here's the calculation you would do to figure out the number of hand combinations where S2 will have the Left + 4 non-trump cards:

1C1 x 2C0 x 15C4 = 1365

There are 1,365 five card hand combinations where S2 will have the Left + no other trump. 1365 combos out of 8568 total combos = 15.93%

Now figure out how often S2 will NOT have the Left at all:

I'll write out the whole equation so you don't get confused.

[(1C0 x 17C5)/18C5] = 6,188/8568 = 72.22%

15.93% + 72.22% = 88.15%

88.15% = the total probability S2 will NOT be guarded

Or S2 will be guarded 100% - 88.15% = 11.85% of the time

QED
Last edited by Wes (aka the legend) on Thu Jan 23, 2020 6:43 pm, edited 1 time in total.

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Post by irishwolf » Thu Jan 23, 2020 5:39 pm

You know Wes, I give up on you. You made something so easy and will not listen to common sense. JH guarded is about half the time of 27.8%. Just that simple!

YOU ARE NUTS! Go see any statistic book! You need help. I asked how you got 3c2 15c3 18c5 = 455 / 8568 = 5.3% wrong calculation and you just ignore it. The formula was right but you for got to multiply 455 by 3 (3c2) / 18c5 = 15.9%. But that is unrelated to where you need to go.

No more comments from me on this topic! I tried to help you but fruitless.

~Irishwolf

Ok, I think what's really going on here is I am doing calculations that you don't really know how to do. All this math stuff seems to be a bit over your head. So let me help you out now. After all, that's what I'm here for. When you're trying to figure out how many 5 card hand combinations Seat 2 can have with a guarded Left, you don't multiply all possibilities. You add them up.

Wes (aka the legend)
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Post by Wes (aka the legend) » Thu Jan 23, 2020 6:58 pm

irishwolf wrote:
Thu Jan 23, 2020 5:39 pm
You know Wes, I give up on you. You made something so easy and will not listen to common sense. JH guarded is about half the time of 27.8%. Just that simple!
No is it not. You have to count the number of combos S2 is guarded and then divide that number by total possible combos. I'm sorry that you seemingly don't have the ability to realize that.
irishwolf wrote:
Thu Jan 23, 2020 5:39 pm
YOU ARE NUTS! Go see any statistic book! You need help. I asked how you got 3c2 15c3 18c5 = 455 / 8568 = 5.3% wrong calculation and you just ignore it.

You misread. I did not say that. This is what I wrote:
Step 1) The number of 5 card hand combinations S2 can have of

(Card_J-H) (Card_Q-D) (Card_9-D) X,X

Is 3C3 x 15C2 = 105 combinations

Step 2) The number of 5 card hand combinations S2 can have of precisely

(Card_J-H) (Card_Q-D) X,X,X

(1C1 x 1C1 x 16C3) - (3C3 x 15C2) = 455 combinations

Note: I subtracted out (3C3 x 15C2) cuz I don't wanna double count those times S2 has LQ9 in trump.

Step 3) The number of 5 card hand combinations S2 can have precisely

(Card_J-H) (Card_9-D) X,X,X

(1C1 x 1C1 x 16C3) - (3C3 x 15C2) = 455 combinations

Step 4) Add up all the 5 card hand combinations:

105 + 455 + 455 = 1,015
Or an even easier formulation:
The full calculation for the probability of S2 having precisely (Card_J-H) (Card_Q-D) and 3 non trump cards =

(1C1 x 1C1 x 1C0 x 15C3)/18C5

Since 1C1 = 1, and 1C0 = 1, we can rewrite it as

15C3/18C5 which = 455/8568 = 5.31%

That's how often S2 will have PRECISELY (Card_J-H) (Card_Q-D) + 3 non-trump in his hand, same goes for (Card_J-H) (Card_9-D), 5.31%

Hence, 5.31% + 5.31% + the probability S2 has all 3 remaining trump (LQ9) 1.23% = 11.85%
The formula I have written is correct.
irishwolf wrote:
Thu Jan 23, 2020 5:39 pm
The formula was right but you for got to multiply 455 by 3 (3c2) / 18c5 = 15.9%. But that is unrelated to where you need to go.
The formula for how often S2 has an unguarded Left is:

(1C1 x 2C0 x 15C4)/18C5 = 1365/8568 = 15.93%

The formula for how often S2 will NOT have the Left in his hand at all is:

(1C0 x 17C5)/18C5 = 6,188/8568 = 72.22%

Thus S2 will be guarded 1 - (15.93% + 72.22%) = 11.85%

Or the direct way to solve for how often S2 has a guarded Left is:

[(1C1 x 1C1 x 1C0 x 15C3) + (1C1 x 1C1 x 1C0 x 15C3) + (3C3 x 15C2)]/18C5 =

11.85%

Which also breaks down to: [2(15C3) + 15C2]/18C5 = 11.85%

There is NOTHING you can say or do to change that. The math doesn't lie. 2+2 will = 4 no matter how badly you want it to equal 5. The fact that you don't understand how to do the math is not my problem. It's yours. Just note that I tried to help you.
Last edited by Wes (aka the legend) on Thu Jan 23, 2020 8:07 pm, edited 1 time in total.

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Post by Wes (aka the legend) » Thu Jan 23, 2020 7:25 pm

irishwolf wrote:
Thu Jan 23, 2020 5:39 pm
You know Wes, I give up on you. You made something so easy and will not listen to common sense. JH guarded is about half the time of 27.8%. Just that simple!
Wait a second WTF. Today, you're saying S2 is guarded about half the time of 27.8%. So around 13.9%. Closer to the truth, but still wrong.

Earlier itt you said this:
irishwolf wrote:
Thu Jan 23, 2020 5:39 pm
If the Left is guarded, you don't have 3 tricks (2.5 to 2.75 tricks) and it will be guarded 27% by the dealer and 20% by S2.
A number so off that it was intuitively obvious you screwed up somewhere which is what started this whole discussion. So which one is it? Approx 13.9% or approx 20%? IMO, if you're gonna be wrong it's better to be wrong in style, so I would choose the latter in your shoes.

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Post by irishwolf » Thu Jan 23, 2020 11:41 pm

Yes, you are correct, I did say 20% as that was from recall with the aid of calculating. And I admit but not as dumb as you getting euchred on the Four trump hand below. Funny that you never responded to you poor play leading the low spade on trick 3. Had you held off until trick 5 you would not have been euchred. Pretty poor play I would say.

by irishwolf » Tue Jan 21, 2020 6:37 pm

https://worldofcardgames.com/#!replayer ... %3A1%7D%5D

Not sure why the low spade is led on trick 3? Your partner is of NO help. Keep leading trump, you just might force the opponents to slough a spade.

In hand 2, it is difficult for me to see the TD would not be picked up - biddable hand for sure!

Wes (aka the legend)
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Post by Wes (aka the legend) » Fri Jan 24, 2020 3:18 am

irishwolf wrote:
Thu Jan 23, 2020 11:41 pm
And I admit but not as dumb as you getting euchred on the Four trump hand below. Funny that you never responded to you poor play leading the low spade on trick 3. Had you held off until trick 5 you would not have been euchred. Pretty poor play I would say.

by irishwolf » Tue Jan 21, 2020 6:37 pm

https://worldofcardgames.com/#!replayer ... %3A1%7D%5D

Not sure why the low spade is led on trick 3? Your partner is of NO help. Keep leading trump, you just might force the opponents to slough a spade.

In hand 2, it is difficult for me to see the TD would not be picked up - biddable hand for sure!
Wow somehow I didn't realize you were talking about my hand that post. Thanx for catching that one. Yeah I don't know what the hell I was thinking. Clear mistake. As I've told Don before, and I'll say it again to others. If I ever make a mistake or a dubious play, feel free to point that out. I wouldn't even mind if someone made a "Mistakes Wes made" thread. Would be fun to me.

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Post by Wes (aka the legend) » Fri Jan 24, 2020 4:28 am

irishwolf wrote:
Thu Jan 23, 2020 11:41 pm
Had you held off until trick 5 you would not have been euchred.
I'm actually euchred no matter how the hand plays out becuz Dlan is gonna hold onto that QS til the very end given that he already knows his clubs are worthless (I trumped clubs on the 2nd trick). But regardless, still poor play on my part. Total brain fart.

Wes (aka the legend)
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Joined: Wed Jun 13, 2018 3:03 pm

Post by Wes (aka the legend) » Fri Jan 24, 2020 11:23 pm

One last way to solve the riddle of how often Seat 2 has a guarded left in Dlan's hand:

Figure out how many Left + 1 hand combinations S2 can have:

1C1 x 2C1 x 15C3 = 910 combos

Figure out how many Left + 2 hand combos S2 can have:

3C3 x 15C2 = 105 combos

Add those combos together.

910 + 105 = 1,015

Divide those 1,015 combos by total possible combos.

Total possible combos = 18C5 = 8,568

1,015/8,568 = 11.85%

Wes (aka the legend)
Posts: 743
Joined: Wed Jun 13, 2018 3:03 pm

Post by Wes (aka the legend) » Sat Jan 25, 2020 8:14 pm

For those who may be wondering, where did Irishwolf go wrong? The key to answering that question lies in the 3 quotes below:
irishwolf wrote:
Thu Jan 23, 2020 10:48 am
This wrong and not how to solve this in this manner:

______________

"You are mixing events, treat as independent!
irishwolf wrote:
Wed Jan 22, 2020 4:55 pm
For independent events, the probability of both occurring is the product of the probabilities of the individual events: Pr(AandB)=Pr(A∩B)=Pr(A)×Pr(B). Example: if you flip a coin twice, the probability of heads both times is: 1/2×1/2=1/4.
irishwolf wrote:
Thu Jan 23, 2020 10:26 am
Wes,

You are correct for the following: down to Step 2 below. Issue is that events occurring independently occurring at the same time is the product of (multiplied together not added or subtracted the way you are doing).

Look at it this way: What the chance it will Rain today, weather man says 70% (so that is .70). What is the chance I forget my umbrella (say 50%). Those are two independent events. So the chance both occur together is .70 x .50 = .35 or 35% chance it will rain and I will forget my umbrella.

So to the Left that is an independent event = 27.8% chance S2 has it.
Now what is the chance that two Diamonds will not be dealt at the same time the JH is dealt? 2c0 16c5 / 18c5 = 14.2%
Irishwolf made the common mistake of treating two events as independent when they are NOT independent. Now let's look at the last bolded paragraph. IW is trying to figure how how often S2 has an unguarded Left. Irishwolf agrees that the probability that S2 has the Left in his 5 card hand = 1C1 X 17C4/18C5 = 27.78%. No problem there. Then he calculates the probability "that two Diamonds will not be dealt at the same time the JH is dealt". Using his formula that probability would be:

2C0 x 16C5/18C5 = 50.98%

He then assumes that these two events are independent, and since "[f]or independent events, the probability of both occurring is the product of the probabilities of the individual events", IW multiplies them to get his answer:

.2778 x .5098 = 14.2% as IW stated. That's how often S2 will have an unguarded Left according to IW's math.

But again, the problem is these events are NOT independent. If I have the Left in my hand that immediately changes the probability that I will not have any more trump (Now there's only 4 slots left in my 5 card hand to not have the Qd or 9D vs 5 slots left if I didn't have the JH). Same goes for when I don't have the QD and the 9D in my hand--that changes the probability of me having the Left in my hand. These probabilities are DEPENDENT as the incidence of one event DOES affect the probability of the other event! Therefore, when you try to multiply them to get your answers you will invariably be wrong.

Now let's go back to IW's calculation:
irishwolf wrote:
Thu Jan 23, 2020 10:26 am
So to the Left that is an independent event = 27.8% chance S2 has it.
Now what is the chance that two Diamonds will not be dealt at the same time the JH is dealt? 2c0 16c5 / 18c5 = 14.2%
Remember, IW is trying to figure out how often S2 has an unguarded Left. Since the probability S2 will have the JH in his 5 card hand and the probability S2 will not have the QD and the 9D in his 5 card hand are NOT independent you can't treat them as separate events. That's what IW is doing for his combination formula for figuring out "the chance that two Diamonds will not be dealt at the same time the JH is dealt" represented as:

2c0 x 16c5

2c0 English translation = There are two other diamonds besides the JH and I am choosing zero of them.

16c5 translation = there are 16 other non-QD/9D unseen cards and I am choosing 5 of them.

Hence this hand represents all 5 card hand combos where S2 does not have the QD or the 9D, which happens 50.98% of the time. Problem is some of those hand combos will have the Left in it--as the JH has to be one of those 16 non-QD/9D unseen cards--which means if we multiply this probability (50.98%) by the probability S2 has the Left in his 5 card hand (27.78%) WE ARE NOW DOUBLE COUNTING HAND COMBOS WHERE S2 HAS THE LEFT. Double counting is what happens when you try to treat dependent events as if they're independent. And that's why IW's numbers are off.

To get the correct number, I.E. to correctly figure out how often S2 has an unguarded Left, you simply set up the correct combination function, understanding these events are not independent, so you don't end up double counting combos:

1C1 x 2C0 x 15C4

1C1 translation: There is 1 JH in the remaining 18 unseen cards and I am choosing it.

2C0 translation: There are two other trump (QD/9D) and I am choosing zero of them.

15C4 translation: After I choose the JH and don't choose the two other trump (QD/9D), there are 15 other non-trump cards left to choose from and I am choosing 4 of them, which completes my 5 card unguarded Left hand (JH + 4 non-trump).

1C1 x 2C0 x 15C4 = 1,365 combos

There are 1,365 different five card hand combinations for S2 to have an unguarded Left. To get the probability of that occurring you simply divide that number by total possible combinations which is:

Total possible combinations = 18C5 = 8,568

1365/8568 = 15.93%

S2 will have an unguarded Left 15.93% of the time, NOT 14.2% of the time. All of IW's math mistakes itt come from treating dependent events as independent.

Let's finish this off with one last IW quote:
irishwolf wrote:
Thu Jan 23, 2020 5:39 pm
You know Wes, I give up on you. You made something so easy and will not listen to common sense. JH guarded is about half the time of 27.8%. Just that simple!
Now we know it isn't that simple. The math only APPEARS that simple if one erroneously assumes dependent events are independent.
Last edited by Wes (aka the legend) on Sat Jan 25, 2020 11:46 pm, edited 1 time in total.

Wes (aka the legend)
Posts: 743
Joined: Wed Jun 13, 2018 3:03 pm

Post by Wes (aka the legend) » Sat Jan 25, 2020 10:05 pm

A fun example illustrating the perils of mistaking dependent events for independent events:

The cards have just been dealt. The upcard is the (Card_9-S) and you're in Seat 1 and you have not looked at your hand yet.

What is the probability you will have no trump in your hand in the first round?

Here's the formula:

(6C0 x 17C5)/23C5

6C0 translation: There are 6 remaining trump and I am choosing zero of them.

17C5 translation: There are 17 remaining non-trump cards and I am choosing 5 of them to fill out my 5 card non-trump hand.

23C5 translation = total possible 5 card hand combos out of 23 unseen cards.

6C0 x 17C5/23C5 = 6,188/33,649 = 18.39%

There is an 18.39% chance S1 has no trump in his hand.

Now what are the odds S1 has no off aces in his hand? Here's the formula:

(3C0 x 20C5)/23C5

3C0 translation: There are 3 off aces and I am choosing zero of them.

20C5 translation: There are 20 remaining non-off ace cards and I am choosing 5 of of them to fill out my 5 card hand.

3C0 x 20C5/23C5 = 15,504/33,649 = 46.08%

Ok so 46.08% of the time S1 will have no off aces in his hand and 18.39% of the time S1 will have no trump in his hand.

Now what is the probability S1 will have no trump and no off aces in his hand? Well that's simple right!? Just multiply the probability S1 has no trump (18.39%) by the probability S1 has no off aces (46.08%).

.1839 x .4608 = 8.47%

S1 will have no trump & no off aces 8.47% of the time. QED Bitches. Peace out!

NSFMFs!!!

The probability S1 has no trump and the probability S1 has no off aces are NOT independent! So you can't just find the product of both probabilities!

For example, if I have an (Card_A-D) in my hand that changes the chances I will have no trump in my hand. And conversely if I have the (Card_Q-S) in my hand that changes the chances I will have an off Ace in my hand.

A reminder:
In probability, two events are independent if the incidence of one event does not affect the probability of the other event. If the incidence of one event does affect the probability of the other event, then the events are dependent.

https://brilliant.org/wiki/probability- ... nt-events/

Since the incidence of having an off ace in our hand DOES AFFECT the probability of the other event, having no trump in our hand, and vice versa, these probabilities are dependent. Therefore (18.39% x 46.08% = 8.47%) CANNOT be correct. Here's the correct combination formula that takes dependence into account:

Probability S1 has no trump + no off aces:

(6C0 x 3C0 x 14C5)/23C5

6C0 = there are 6 remaining trump and you choose zero of them.

3C0 = there are 3 remaining off aces and you choose zero of them.

14C5 = there are 14 remaining cards that are neither a trump nor an off ace and you choose 5 of them, thus filling out your 5 card hand.

(6C0 x 3C0 x 14C5)/23C5 = 2,002/33,649 = 5.95%

So the correct answer is once you see the upcard, there is a 5.95% chance you will have no trump + no off aces, not 8.47%.

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