Weekly Games 1-20
Posted: Tue Jan 21, 2020 10:51 am
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.Dlan wrote: ↑Tue Jan 21, 2020 10:51 am
Second round call of diamonds
Just wondering if others would have led the same way?
https://worldofcardgames.com/#!replayer ... %3A1%7D%5D
Wow. So basically my amazing loner gets blown to smithereens becuz Kirsten misread her hand!! Lol.Dlan wrote: ↑Tue Jan 21, 2020 11:01 am
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
The Maker does not have a spade to lead in trick 4.
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.irishwolf wrote: ↑Tue Jan 21, 2020 7:08 pmAs 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!
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 pmJD 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.
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.irishwolf wrote: ↑Tue Jan 21, 2020 7:08 pmYou 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.
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
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.Wes (aka the legend) wrote: ↑Tue Jan 21, 2020 6:25 pmHOWEVER, 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.
Anyone else wanna give this a go?Wes (aka the legend) wrote: ↑Tue Jan 21, 2020 6:25 pmPOP 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.
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.
Not so sure about that, but based on the numbers, I do think I have to change my initial conclusion below:irishwolf wrote: ↑Wed Jan 22, 2020 11:10 amLooks 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.
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.
First trick:Wes (aka the legend) wrote: ↑Tue Jan 21, 2020 6:25 pmPOP 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.
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%:
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.
You know what, let's back up here and go step by step.
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.
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.irishwolf wrote: ↑Wed Jan 22, 2020 4:35 pmWes,
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.
Your equation is wrong. There are not 16 non-trump cards there are only 15 (You're double counting the Left it appears).
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 X,X,X to make sure I don't double count those times S2 has LQ9 in trump.irishwolf wrote: ↑Thu Jan 23, 2020 10:48 amThis 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 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 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 in his hand,
and a 5.31% chance S2 has precisely with no other trump. (same issue wrong formula)
Yes you can, the other numbers = 1, so they are not necessary.
Yes, 1C1 x 17C4 represents how often S2 will have the Left in his hand, I.E. how often S2 will have 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:
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.
Or an even easier formulation:Step 1) The number of 5 card hand combinations S2 can have of
X,X
Is 3C3 x 15C2 = 105 combinations
Step 2) The number of 5 card hand combinations S2 can have of precisely
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
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
The formula I have written is correct.The full calculation for the probability of S2 having precisely 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 + 3 non-trump in his hand, same goes for , 5.31%
Hence, 5.31% + 5.31% + the probability S2 has all 3 remaining trump (LQ9) 1.23% = 11.85%
The formula for how often S2 has an unguarded Left is:
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.
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.
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.irishwolf wrote: ↑Thu Jan 23, 2020 11:41 pmAnd 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!
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.
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:irishwolf wrote: ↑Thu Jan 23, 2020 10:26 amWes,
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%
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:
Now we know it isn't that simple. The math only APPEARS that simple if one erroneously assumes dependent events are independent.
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/