Third seat loaner attempt?

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 Joined: Fri Feb 15, 2019 12:05 pm
Third seat loaner attempt?
Would you attempt a third seat loaner? The score is 4 to 3, first and third seat winning. The dealer up card is the KC...
Third seat has JC, JS, 9C, AD, QD... Does third seat order up alone? or Order up keep partner? or Pass?
If you would order up alone, would you throw the 9C on an Ace of heart lead, or throw a bar?
I ordered it up, I trumped the Ace of hearts with my 9C, the 9C got trumped by the dealer, the dealer led back an off ace, I trumped. the dealer had three trump with no diamonds, so I ended up getting set... my partner freaked out about me getting set on a loaner.... Thoughts???
Ps. we did win the game, but the set on a loaner made it exciting... lol
Third seat has JC, JS, 9C, AD, QD... Does third seat order up alone? or Order up keep partner? or Pass?
If you would order up alone, would you throw the 9C on an Ace of heart lead, or throw a bar?
I ordered it up, I trumped the Ace of hearts with my 9C, the 9C got trumped by the dealer, the dealer led back an off ace, I trumped. the dealer had three trump with no diamonds, so I ended up getting set... my partner freaked out about me getting set on a loaner.... Thoughts???
Ps. we did win the game, but the set on a loaner made it exciting... lol

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I generally call in this spot. It's just so hard to pull off a loner with JJ9 from the 3rd spot. I qualified my statement with "generally" becuz I will go alone with this hand up 43 if my partner is bad enough where I think the other team has an edge over my team. When that's the case, I go for big plays more often, trying to make luck the dominant factor, not skill.
Gotta gamble and play low with the 9C on a green lead. In fact I would say if you're not willing to gamble in this spot then you definitely should not go alone to begin with. Once you trump in with one of your bowers, the dealer only needs 1 more trump + the upcard to block your sweep.
Here's the probability the dealer has at least one more trump in addition to the upcard: 1  [(3C0 x 15C5)/18C5] = 1  (3003/8568) = 64.95%
So when you don't gamble low and instead play high with your bower you will be surrendering a 4 point sweep against the dealer's range almost 65% of the time. Can't do that. Gamble on the green lead instead as you did.
Now that said, if I'm up against two skilled players that know the key strategy of attacking 1st rd, 3rd seat loners (Dealer always discards Next if he can, and S2 always leads Next if he has it), and S2 proceeds to lead a Next card, then I'm NOT gambling low with the 9C. Now I'm trumping high with a bower and praying the dealer only has the upcard in trump and S2 doesn't have 2 trump.
The instant S2 leads a Next card there is a 67.02% chance the dealer is void in Next:
[(4C1 x 13C4) + (4C0 x 13C5)]/17C5 = (2860 + 1287)/6188 = 67.02%
If S2 is leading Next from a doubleton set (say we somehow knew one of their remaining cards was another Next card), then the probability the dealer is void in Next is 78.57%:
[(3C1 x 13C4) +(3C0 x 13C5)]/16C5 = (2145 + 1287)/4368 = 78.57%
As you can see, the 3rd seat loner defense team strategy of S2 always leading Next and the dealer always discarding Next, severely cripples 1st rd, 3rd seat loners, especially hands like JcJs9c. Against such a skilled team I would only try your loner if super desperate (down 96/97, or down big), and then I would be hoping S2 doesn't have a Next card to lead. Once S2 leads Next, then I'm trumping high and hoping to get lucky catching S4 and S2 with no more than 1 trump each, thus cleaning them out on my next lead.
Ok back to your hand, as said, I would just call, but if I did go alone, I would trump low on a green lead like the AH, and I would trump high on a Next lead against a skilled team. One more tidbit worthy of note, just alter your hand a tiny bit and I always go alone at every score except when my team is at 89:
Dealer upcard, , you're in the 3 seat with:
Having your low trump card be higher than the dealer upcard makes a significant difference. Remember I told you the dealer has almost a 65% chance of having another trump besides the upcard. Well that also means there's a 35% chance the 9C is the only trump he has. Now when you trump low on the first lead it's like you have an extra way to escape unscathed. If the dealer is not void in that lead you're obviously home free, and even if the dealer is void in that lead suit, 35% of the time he won't be able to over trump you anyways! With this configuration, I'm trumping low on every lead including a Next lead from a skilled team.
Once you chose to go alone you played the hand well. Your partner is wrong for freaking out tho. Getting set on loners happens. It's still a gambling game, and getting that coveted 4 pt sweep is so huge, one should be willing to take some risks to get it. I can tell just from your partner's reaction he doesn't go alone enough.icanplay wrote: ↑Mon Feb 03, 2020 1:52 pmI ordered it up, I trumped the Ace of hearts with my 9C, the 9C got trumped by the dealer, the dealer led back an off ace, I trumped. the dealer had three trump with no diamonds, so I ended up getting set... my partner freaked out about me getting set on a loaner.... Thoughts???
Ps. we did win the game, but the set on a loaner made it exciting... lol
Last edited by Wes (aka the legend) on Mon Feb 03, 2020 6:47 pm, edited 1 time in total.

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Just wanna be clear here and say I have NOT made a compelling case that your loner call is wrong. All I have basically said was I would generally call. Against skilled opponents who know how to attack 3rd seat loners, I believe I have made a compelling case against going alone up 43. And I think the idea of going alone up 43 if your partner is really badtrying to make big plays and end the game as fast as possiblehas merit. A related idea to the "bad partner" adjustment is if you have little faith that your partner will lead trump on a 3rd seat call, then you might as well go alone with this hand.
I also think one could make a decent argument for going alone in this spot if their opponents are not very good. Like if S2 is the kind of player who will incorrectly lead a single ace into a loner, then bad plays like that could swing things into the "always go alone" ledger (except at 8 or 9).
Back to your actual hand, if going alone was actually a mistake, it's a marginal one at best. I mean obviously I suspect it IS a mistake, that's why I would usually just call, but I am not confident in that assessment. As always it's a math problem. And some of the crucial variables are:
1) How often will you survive the first lead if you trump low?
2) How often will going alone result in a euchre?
I also think one could make a decent argument for going alone in this spot if their opponents are not very good. Like if S2 is the kind of player who will incorrectly lead a single ace into a loner, then bad plays like that could swing things into the "always go alone" ledger (except at 8 or 9).
Back to your actual hand, if going alone was actually a mistake, it's a marginal one at best. I mean obviously I suspect it IS a mistake, that's why I would usually just call, but I am not confident in that assessment. As always it's a math problem. And some of the crucial variables are:
1) How often will you survive the first lead if you trump low?
2) How often will going alone result in a euchre?

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 Joined: Fri Feb 15, 2019 12:05 pm
thank you for the input. If we have 8 or 9, I would not go alone, for sure not! You have a valid point. My nine is the problem, and the dealer discarding greatly ups the chance of getting trumped over. Thanks for the input, you may be right. As for my partner, he would have called the same thing. He is just the type of person to think he would not have gotten set... In this case the dealer has to have three trump to set me. If I keep my partner, and the dealer has the same three trump in his hand, there is a good chance my partner does not have a trump to lead. If this is the case he is practically useless to me and we could get set anyway. I am still in limbo on this hand. If he ( my partner ) leads trump, and the dealer has three, we get our point, but we don't get two points. I am still tempted to go alone here in most cases, however, i see the limitations a little better. thanks for the great input.

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 Location: Florida
I will disagree with Wes on this hand.icanplay wrote: ↑Mon Feb 03, 2020 1:52 pmWould you attempt a third seat loaner? The score is 4 to 3, first and third seat winning. The dealer up card is the KC...
Third seat has JC, JS, 9C, AD, QD... Does third seat order up alone? or Order up keep partner? or Pass?
If you would order up alone, would you throw the 9C on an Ace of heart lead, or throw a bar?
I ordered it up, I trumped the Ace of hearts with my 9C, the 9C got trumped by the dealer, the dealer led back an off ace, I trumped. the dealer had three trump with no diamonds, so I ended up getting set... my partner freaked out about me getting set on a loaner.... Thoughts???
Ps. we did win the game, but the set on a loaner made it exciting... lol
My BPS Analysis
0.00 S1 R1
1.00 Jc
0.75 Js
0.25 9c
0.75 3 trump, 2 Bowers
0.50 Ad
0.75 2 Voids
4.00 Hand strength
.25 Value Kc  Value 9c (min up card)
3.75 vs 3.75 Order Alone, also 3.75 points implies 95% success
BPS offers some important information. First let’s look at ordering Alone. If I have 3.75 points and a 95% expectation, I will go alone unless there are 4 stoppers. There are no stoppers. There are only parlays, 3 trump and a protected Kd. I see a 50% chance at 4 points. So I am going alone.
Second, if my hand has 3.75, what does that tell me about the other 3 hands? There are about 6.25 total BPS points in the 4 hands. So 2.5 points are divvied among the 3 other hands. On a likely distribution, no hand will have more than 1.25 points. In this hand S4 has:
0.50 Ac
0.50 Kc
0.25 Tc
0.50 3 trump
0.50 A?
2.25 points out of 2.50?!, 2% probability, Crazy good hands will always win.
The same situation apply to S3’s partner, S1. S1 is likely to have a very weak hand. Why? The cards are in my hand. Next is especially weak since S3 has both Bowers. BPS’ 3.75 points translates into a generic 3 strongly expected tricks. IMO, Wes is failing to take into account the extremely low expectation of the S1 hand.

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You may be right, but not against a skilled team that attacks 3rd seat loners with the cooperative strategy of S2 always leading Next when he has it and S4 always discarding Next when possible.
As mentioned before, when you're up against a skilled team, the instant S2 leads a Next card, there is a 67.02% chance S4 is void in Next (Edit: I should say at least a 67.02% chance, since the odds of S4 being void in Next increase significantly with each additional Next card S2 may have). Of course then the question is how often will S2 have a Next card to lead, I.E. how often S2 will have at least 1 Next card:
1  [(5C0 x 13C5)/18C5] = 1  1287/8568 = 84.98%
That answer isn't precisely accurate becuz there are still some S2 calling hands in that range that I didn't tease out, hands like 3 trump (AcQcTc) or possibly two trump + an off ace, but it's close enough. So almost 85% of the time S3 is gonna be in a situation where if he trumps low he will only survive the first lead around 33% of the time. What about the other 15% of the time when S2 doesn't have a Next card to lead? Well then S2 will generally be leading his dirtiest green suit (exceptions will occur when S2 has a single Ace). It's hard to precisely quantify that. Either way, "a 50% chance at 4 points" is never happening vs a skilled team.
What matters most about S1's range is how often will he have a trump to lead us if we just call. The answer to that question is:Richardb02 wrote: ↑Tue Feb 04, 2020 8:26 pmThe same situation apply to S3’s partner, S1. S1 is likely to have a very weak hand. Why? The cards are in my hand. Next is especially weak since S3 has both Bowers. BPS’ 3.75 points translates into a generic 3 strongly expected tricks. IMO, Wes is failing to take into account the extremely low expectation of the S1 hand.
[(3C1 x 15C4) + (3C2 x 15C3)]/18C5 = (4,095 + 1365)/8568 = 63.73%
There's also 22 combos of X,X hands where S1 would pass in the first round either becuz he has reverse next blocked or a place to go in the 2nd round. Including that number has a negligible effect. It changes the percentage from 63.73% to 63.98%.
So basically S1 will have a trump to lead almost 64% of the time in this spot. That's not too shabby.

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Wes (aka the legend) wrote: ↑Mon Feb 03, 2020 5:30 pmThe instant S2 leads a Next card there is a 67.02% chance the dealer is void in Next (Edit: I should've said "at least" a 67.02% chance):
[(4C1 x 13C4) + (4C0 x 13C5)]/17C5 = (2860 + 1287)/6188 = 67.02%
If S2 is leading Next from a doubleton set (say we somehow knew one of their remaining cards was another Next card), then the probability the dealer is void in Next is 78.57%:
[(3C1 x 13C4) +(3C0 x 13C5)]/16C5 = (2145 + 1287)/4368 = 78.57%
Ok, I need to fix some things. I apologize ahead of time as now I'm totally veering off course. I said above that once S2 leads Next there is at least a 67.02% chance S4 is void in Next if said team was deploying the aforementioned cooperative strategy of attacking 1st rd, 3rd seat loners. But that 67.02% is assuming S2 only has 1 Next card, just as the 78.57% number was assuming S2 had 2 next cards. There's a better way to break this down, to just one number. I.E. once S2 leads Next what is the overall probability S4 is void in Next? Ok here we go.Wes (aka the legend) wrote: ↑Mon Feb 03, 2020 5:30 pmAs mentioned before, when you're up against a skilled team, the instant S2 leads a Next card, there is a 67.02% chance S4 is void in Next (Edit: I should say at least a 67.02% chance, since the odds of S4 being void in Next increase significantly with each additional Next card S2 may have). Of course then the question is how often will S2 have a Next card to lead, I.E. how often S2 will have at least 1 Next card:
1  [(5C0 x 13C5)/18C5] = 1  1287/8568 = 84.98%
That answer isn't precisely accurate becuz there are still some S2 calling hands in that range that I didn't tease out, hands like 3 trump (AcQcTc) or possibly two trump + an off ace, but it's close enough. So almost 85% of the time S3 is gonna be in a situation where if he trumps low he will only survive the first lead around 33% of the time.
First break down the probabilities S2 has 1 Next card, 2 Next cards, 3, etc:
1 Next card: (5C1 x 13C4)/18C5 = 3575/8568 = 41.73%
2 Next cards: (5C2 x 13C3)/18C5 = 2,860/8568 = 33.38%
3 Next cards: (5C3 x 13C2)/18C5 = 780/8568 = 9.10%
4 Next cards: (5C4 x 13C1)/18C5 = 65/8568 = .76%
5 Next Cards: (5C5 x 13C0)/18C5 = 1/8568 = .01%
Remember I said S2 will have at least 1 Next card 84.98% of the time. If we add up all those percentages that's the number we should get and indeed 41.73 + 33.38 + 9.10 + .76 + .01 = 84.98%
Ok Next step, figure out the corresponding Void percentages of S4:
When S2 has 1 Next card S4 is void this percentage of time:
[(4C0 x 13C5) + (4C1 x 13C4)]/17C5 = (1,287 + 2860)/6188 = 67.02%
When S2 has 2 Next cards S4 is void this percentage of time:
[(3C0 x 13C5) + (3C1 x 13C4)]/16C5 = (1,287 + 2145)/4368 = 78.57%
When S2 has 3 Next Cards S4 is void this percentage of the time:
[(2C0 x 13C5) + (2C1 x 13C4)/15C5 = (1,287 + 1430)/3003 = 90.48%
When S2 has 4 Next cards S4 will be void in Next 100% of the time given that there is only 1 Next card left, and S4 is always voiding himself in Next, even if the AS is his only Next card. When S2 has 5 Next cards then obviously S4 is void in Next 100% of the time since there are no Next cards left.
Ok so the magical question is given that S2 leads Next what is the overall probability S4 will be void. To get that answer we have to add up all the conditional probabilities which looks like this:
(.4173/.8498)(.6702) + (.3338/.8498)(.7857) + (.0910/.8498)(.9048) + (.0076/.8498)(1) + (.0001/.8498)(1) = 74.37%
Ok so there it is, if one goes alone from the 3rd seat, 1st round with JcJs9cAdQd against a skilled team, and S2 leads next, the overall probability S4 is void in Next is 74.37% of the time.
Note: that number will be a tad off becuz as I said in another post there are some hands S2 would've called with that we have in his passing range. But we're talking about a very small number of hand combos here. There are only 105 combos of AcQcTcXX that S2 could have called with, I.E. 1.23% of his range (as most people don't call with two low trump + an off ace from that spot), so overlooking this hand should have a minimal confounding effect. So the 74.37% number is a great approximation.
Ok why did I do all this? IDK mostly cuz I'm anal and bored. But also becuz I was always curious at how well the cooperative strategy of S2 always leading Next and S4 always discarding Next worked out. Well it turns out it's pretty damn effective, putting a S3 JcJs9cAdQd loner in a tight squeeze around 74% of the time!

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OK Wes, I can see that I failed to properly account for the “Lead Next” convention. So lets say S3 gets 4 points 26% of the time. In fact let’s be conservative and assume that S3 earns 4 points, 20% of the time. The convention only helps in stopping trump. It has minimal impact on the euchre rate.
So my EV for ordering is:
4 x 20% + 1 x 75%  2 x 5% = 0.80+.75 0.10 = 1.45
That is strong.
S3’s hand is so strong I don’t want to even conjecture about S4 ordering and getting euchred or S1 ordering Next. I suggest that 1.25 points in my hand (yes this hand has a 1.45 EV), is better than “maybe” 1.25 points dealing with not just unknowns, but low probability unknowns.
I agree with your aversion to ordering from S3 with “average” hands, but this hand contains 3 very likely tricks. That makes this a very strong hand, which dominates the other hands so strongly that the strategy should be changed.
So my EV for ordering is:
4 x 20% + 1 x 75%  2 x 5% = 0.80+.75 0.10 = 1.45
That is strong.
S3’s hand is so strong I don’t want to even conjecture about S4 ordering and getting euchred or S1 ordering Next. I suggest that 1.25 points in my hand (yes this hand has a 1.45 EV), is better than “maybe” 1.25 points dealing with not just unknowns, but low probability unknowns.
I agree with your aversion to ordering from S3 with “average” hands, but this hand contains 3 very likely tricks. That makes this a very strong hand, which dominates the other hands so strongly that the strategy should be changed.

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I going to disagree with the math and conclusions drawn by Wes. Way too optimistic at 74%:
" Ok so there it is, if one goes alone from the 3rd seat, 1st round with JcJs9cAdQd against a skilled team, and S2 leads next, the overall probability S4 is void in Next is 74.37% of the time."
So looking at all the combination that S2 has with next in combination of those relevant with the dealer having 1 or none. Dealer having more than one is not relevant.
S2/S4 HAS NEXT CARD RELEVANT COMBINATIONS (5 UNKNOWN NEXT CARDS)
0/0 .15 .15 = .023 (S2 VOID – SO NOT ADDITIVE TO THE TOTAL)
1/0 .42 .15 = .063
1/1 .42 .42 = .176
2/0 .33 .15 = .050
2/1 .33 .42 = .139
3/0 .09 .15 = .014
3/1 .09 .15 = .038
TOTAL ~0.483 (MAY BE OFF A FRACTION DUE TO ROUNDING!)
Of course a good Ploy and this ploy came from discussions as worked out from Euchre Science about 12 years ago.
As to trumping with the 9C to the AH, of course the correct play. However, if next had been led, and you know the opponents know the ploy you have to trump with a Jack and back off your Loner Attempt. Now lead your AD while holding your other Jack and 9C.
~Irishwolf
" Ok so there it is, if one goes alone from the 3rd seat, 1st round with JcJs9cAdQd against a skilled team, and S2 leads next, the overall probability S4 is void in Next is 74.37% of the time."
So looking at all the combination that S2 has with next in combination of those relevant with the dealer having 1 or none. Dealer having more than one is not relevant.
S2/S4 HAS NEXT CARD RELEVANT COMBINATIONS (5 UNKNOWN NEXT CARDS)
0/0 .15 .15 = .023 (S2 VOID – SO NOT ADDITIVE TO THE TOTAL)
1/0 .42 .15 = .063
1/1 .42 .42 = .176
2/0 .33 .15 = .050
2/1 .33 .42 = .139
3/0 .09 .15 = .014
3/1 .09 .15 = .038
TOTAL ~0.483 (MAY BE OFF A FRACTION DUE TO ROUNDING!)
Of course a good Ploy and this ploy came from discussions as worked out from Euchre Science about 12 years ago.
As to trumping with the 9C to the AH, of course the correct play. However, if next had been led, and you know the opponents know the ploy you have to trump with a Jack and back off your Loner Attempt. Now lead your AD while holding your other Jack and 9C.
~Irishwolf

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CORRECTION:
S2/S4 HAS NEXT CARD RELEVANT COMBINATIONS (5 UNKNOWN NEXT CARDS)
First number is S2 having Next card, while S4 has one or no next card.
0/0 .15 .15 = .023 (S2 VOID – SO NOT ADDITIVE TO THE TOTAL)
1/0 .42 .15 = .063
1/1 .42 .42 = .176
2/0 .33 .15 = .050
2/1 .33 .42 = .139
3/0 .09 .15 = .014
3/1 .09 .42 = .038
TOTAL ~0.518 (MAY BE OFF A FRACTION DUE TO ROUNDING!)
~Irishwolf
S2/S4 HAS NEXT CARD RELEVANT COMBINATIONS (5 UNKNOWN NEXT CARDS)
First number is S2 having Next card, while S4 has one or no next card.
0/0 .15 .15 = .023 (S2 VOID – SO NOT ADDITIVE TO THE TOTAL)
1/0 .42 .15 = .063
1/1 .42 .42 = .176
2/0 .33 .15 = .050
2/1 .33 .42 = .139
3/0 .09 .15 = .014
3/1 .09 .42 = .038
TOTAL ~0.518 (MAY BE OFF A FRACTION DUE TO ROUNDING!)
~Irishwolf

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Before I respond to your post I gotta fix a minor error I made that will be relevant to this spot.Richardb02 wrote: ↑Wed Feb 05, 2020 8:49 pmOK Wes, I can see that I failed to properly account for the “Lead Next” convention. So lets say S3 gets 4 points 26% of the time. In fact let’s be conservative and assume that S3 earns 4 points, 20% of the time. The convention only helps in stopping trump. It has minimal impact on the euchre rate.
So my EV for ordering is:
4 x 20% + 1 x 75%  2 x 5% = 0.80+.75 0.10 = 1.45
That is strong.
I didn't probably account for the AcQcTcXX part of S1's range so the numbers are slightly off. Here's the correct equation:Wes (aka the legend) wrote: ↑Tue Feb 04, 2020 9:30 pmWhat matters most about S1's range is how often will he have a trump to lead us if we just call. The answer to that question is:
[(3C1 x 15C4) + (3C2 x 15C3)]/18C5 = (4,095 + 1365)/8568 = 63.73%
There's also 22 combos of X,X hands where S1 would pass in the first round either becuz he has reverse next blocked or a place to go in the 2nd round. Including that number has a negligible effect. It changes the percentage from 63.73% to 63.98%.
So basically S1 will have a trump to lead almost 64% of the time in this spot. That's not too shabby.
[(3C1 x 15C4) + (3C2 x 15C3) + 22]/(18C5  83) = (4,095 + 1365 + 22)/8485 = 64.61%
The "22" number represents the combos of AcQcTc that S1 still has after he passes. The "83" number represents the 83 combos of AcQcTc that cannot be in S1's passing range becuz he would've called with those hands. There are a total of 105 combos of AcQcTcXX and now all (22 + 83) are properly accounted for. Ok moving on.
Ok Richard, you've made half an argument. Let's go with your EO of going alone = 1.45. Now you need to calculate the EO of just calling and compare your EO of going alone to that number. Assume S1 is competent and always leads trump to your 3rd seat calls or leads an off ace if he has no trump. And assume my math is accurate when I say S1 will have a trump to lead 64.61% of the time. Ok the model for the EO of calling would look like this:
.6461[(2pts x Z%) + (1pt x Z%) + (2pts x Z%)] + .3539[(2 pts x Z%) + (1pt x Z%) + (2pts x Z%) = EO
Basically guesstimate what the EO of calling is the 64.61% of the time S1 leads trump, then guesstimate what the EO of calling is the 35.39% he has no trump to lead. Then add those two numbers together to get the total EO. I'm gonna let you fill those numbers in with your best guesses and then you can post your EO of calling. I already know my best guesses but I'd rather work with your bias than mine.
Some other percentages that may help hone your guesses:
To reiterate, the probability S1 has no trump is:
(3C0 x 15C5)/(18C5  83) = 35.39%
The probability S1 has no trump and no off aces:
(3C0 x 2C0 x 13C5)/(18C5  83) = 1287/8485 = 15.17%
The probability S1 has no trump but at least 1 off Ace to lead =
[(3C0 x 2C1 x 13C4)/(18C5  83) + (3C0 x 2C2 x 13C3)/(18C5  83)] = 20.22%
Which means when S1 has no trump he will have an off Ace to lead 20.22/35.39 = 57.13% of the time.

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ONE MORE TIME: TRY THIS (SCRATCH THOSE TWO PREVIOUS POSTS)
1/0 .42 .23 = .100
1/1 .42 .48 = .200
2/0 .33 .23 = .076
2/1 .33 .48 = .158
3/0 .09 .51 = . 046
3/1 .09 .28 = .025
TOTAL 0 .605 (MAY BE OFF A FRACTION DUE TO ROUNDING!)
1/0 .42 .23 = .100
1/1 .42 .48 = .200
2/0 .33 .23 = .076
2/1 .33 .48 = .158
3/0 .09 .51 = . 046
3/1 .09 .28 = .025
TOTAL 0 .605 (MAY BE OFF A FRACTION DUE TO ROUNDING!)

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You're doing it again Irishwolf. You're multiplying dependent probabilities to get your answers. You can't do that.irishwolf wrote: ↑Wed Feb 05, 2020 11:16 pmI going to disagree with the math and conclusions drawn by Wes. Way too optimistic at 74%:
So looking at all the combination that S2 has with next in combination of those relevant with the dealer having 1 or none. Dealer having more than one is not relevant.
S2/S4 HAS NEXT CARD RELEVANT COMBINATIONS (5 UNKNOWN NEXT CARDS)
0/0 .15 .15 = .023 (S2 VOID – SO NOT ADDITIVE TO THE TOTAL)
1/0 .42 .15 = .063
1/1 .42 .42 = .176
2/0 .33 .15 = .050
2/1 .33 .42 = .139
3/0 .09 .15 = .014
3/1 .09 .15 = .038
TOTAL ~0.483 (MAY BE OFF A FRACTION DUE TO ROUNDING!)
For example, you claim S2 has an approx 42% probablity of having 1 Next card, which is correct:
(5C1 x 13C4)/18C5 = 3575/8568 = 41.73%
Then you claim that S4 has an approx 15% probability of having no trump, which is also correct:
(5C0 x 13C5)/18C5 = 1287/8568 = 15.02%
Then you multiply these probabilities together:
And you did the same for those times S2 and S4 have 1 Next card, etc:
But that's a NO NO becuz those probabilities are NOT independent. If S2 has 1 Next card that means there are 4 Next cards left in the deck, not 5, and thus that instantly changes the chances S4 is void in Next (It's no longer 15%) and it changes the chances S4 has only 1 Next card (It's no longer 42%). I.E. "the incidence of one event (S2 having a Next card) does affect the probability of the other event" (S4 being void in Next or S4 having 1 Next card he can void).
A reminder:
Once we give S2 one Next card, the proper math for figuring out how often S4 is void in Next is NO LONGER THIS: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 ... ntevents/
(5C0 x 13C5)/18C5 = 1287/8568 = 15.02%
And the probability S4 has one Next card is NO LONGER THIS:
(5C1 x 13C4)/18C5 = 3575/8568 = 41.73%
We have to account for the fact that 1 Next card is in S2's hand, and thus there are only 4 Next cards left. Here is how often S4 is void in Next once we know S2 has 1 Next card:
[(4C1 x 13C4) + (4C0 x 13C5)]/17C5 = (2860 + 1287)/6188 = 67.02%
4C1 represents those hand combos S4 has one Next card and can thus discard it to void himself in Next. 4C0 represents those hand combos S4 has no Next cards and is thus already void in Next. 17C5 acknowledges the fact that once we know S2 has a Next card then there are now 17 unseen cards left, so total combos possible = 17C5 = 6188.
My numbers are fine Irishwolf.
If you accept the math below:
...And you kinda HAVE to accept that cuz the math don't lie, then the rest of my math follows accordingly.
Last edited by Wes (aka the legend) on Thu Feb 06, 2020 1:25 am, edited 1 time in total.

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As we already established, your probability that S2 has 1 Next card is correct, approx 42%.
I see where you're getting the probability S4 is void in Next = 23%
(4C0 x 14C5)/18C5 = 2002/8568 = 23.37%
But you're making a mistake. Once S2 has 1 next card it is no longer 18C5, it's 17C5.
Here's the correct math:
The probability S4 is void in Next given that S2 has 1 Next card =
(4C0 x 13C5)/17C5 = 1287/6188 = 20.80%
Also, the probability S4 has just 1 Next card given that S2 has 1 Next card =
(4C1 x 13C4)/17C5 = 2860/6188 = 46.22%
Add those probabilities together to get the total probability S4 will be void in Next when S2 has 1 Next card. 20.80 + 46.22 = 67.02%
I assume the rest of your math has the same issue, so I'll stop there.

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I agree and will revisit.

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 Joined: Tue Apr 24, 2018 9:33 pm
I am in agreement with your 74% number. And the dealer must even discard the singleton Ace of next which many still cannot believe it is a good thing to do.
This why if 3rd seat goes alone, S3 Must trump the (Next) lead at least as high as the upcard. It's funny how popular this ploy has become in recent years.
"Well it turns out it's pretty damn effective, putting a S3 JcJs9cAdQd loner in a tight squeeze around 74% of the time!"
This why if 3rd seat goes alone, S3 Must trump the (Next) lead at least as high as the upcard. It's funny how popular this ploy has become in recent years.
"Well it turns out it's pretty damn effective, putting a S3 JcJs9cAdQd loner in a tight squeeze around 74% of the time!"

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 Joined: Wed Jun 13, 2018 3:03 pm
Yep. Also, even if I address the issue of S2's range not being entirely accurate...
...it predictably doesn't move the needle much becuz we're only talking about 1.23% of his range, but I'll redo the math anyways to account for this:Wes (aka the legend) wrote: ↑Wed Feb 05, 2020 1:04 amNote: that number will be a tad off becuz as I said in another post there are some hands S2 would've called with that we have in his passing range. But we're talking about a very small number of hand combos here. There are only 105 combos of AcQcTcXX that S2 could have called with, I.E. 1.23% of his range (as most people don't call with two low trump + an off ace from that spot), so overlooking this hand should have a minimal confounding effect. So the 74.37% number is a great approximation.
18C5 = 8,568 does't precisely represent S2's total possible range since S2 already passed, and S2 is never passing with AcQcTcXX. There are 105 combos of AcQcTcXX, (3C3 x 15C2 = 105) so S2's total possible passing range is really 18C5  105 = 8,463 combos.
Of those 105 AcQcTcXX hands, 50 of them have 1 Next card, 10 of them have 2 Next cards and 45 of them have no Next cards:
1 Next card: 3C3 x 5C1 x 10C1 = 50
2 Next cards: 3C3 x 5C2 x 10C0 = 10
No Next cards: 3C3 x 5C0 x 10C2 = 45
So back to the first step incorporating our new more precise numbers:
First break down the probabilities S2 has 1 Next card, 2 Next cards, 3, etc:
1 Next card: [(5C1 x 13C4)  50)]/(18C5105) = 3525/8463 = 41.65%
2 Next cards: [(5C2 x 13C3)  10]/(18C5  105) = 2,850/8463 = 33.68%
3 Next cards: (5C3 x 13C2)/(18C5  105) = 780/8463 = 9.22%
4 Next cards: (5C4 x 13C1)/(18C5  105) = 65/8463 = .77%
5 Next Cards: (5C5 x 13C0)/(18C5  105) = 1/8463 = .01%
Add up the probabilities and now S2 will have at least 1 Next card 85.33% of the time. We can double check this number by figuring out the probability S2 will have no Next cards: [(5C0 x 13C5)  45]/(18C5  105) = 14.68%. 14.68% + 85.33% = 100.01%, IOW 100% excluding rounding errors. These percentages adding up to 100% shows we are correct. Now we have enough information to find the overall probability of S4 being void in Next given that S2 led Next (with AcQcTc appropriately factored out of S2's range). Again, just add up the conditional probabilities with our new numbers:
(.4165/.8533)(.6702) + (.3368/.8533)(.7857) + (.0922/.8533)(.9048) + (.0077/.8533)(1) + (.0001/.8533)(1) = 74.42%
So before our final conclusion was 84.98% of the time S2 will have a Next card to lead, and when S2 leads Next S4 will be void on average 74.37% of the time. After we factor out AcQcTcXX hands from S2's range the numbers slightly change to 85.33% of the time S2 will have a Next card and S4 being void in Next once S2 leads it 74.42% of the time. Basically no significant difference. What if I go even further and additionally factor out '2 low trump + 2 off Aces' and even '2 trump + an off Ace' hands that an aggressive S2 may call with when he blocks no suits. I did that too, and to save space, nothing much changes once again. The final numbers in that last instance was S2 will have a Next card to lead 85.47% of the time, and once S2 leads Next, S4 will be void 74.53% of the time.