[fromageGB]

You are in a high level spade contract. LHO leads a nondescript heart, RHO follows with a nondescript heart and you, having all the top cards in that suit, win. In trumps you are missing just three cards, the K, the 3, and the 2. When you lead low towards the AQ in dummy, LHO plays the 2. You do not know whether he would play the 2 or the 3 from a 32 combination, but you do know he would not play the K from Kx (!) What is the chance of the the finesse winning if you play for it?

With 3 cards missing, in 100 deals you will find the following breaks : 11 are 3-0; 11 are 0-3, 39 are 2-1, 39 are 1-2. (ie LHO-RHO)

The finesse is 50% because ...
Missing the K, a finesse is 50-50.

The finesse is 11/23 (about 47.8%) because ...
Vacant Spaces tells us that LHO has 11 cards that could be the ♠K, while RHO has 12, for a total of 23 cases, so it is 11/23.

The finesse is 37/76 (about 48.7%) because ...
I have left my spectacles at home and can't tell whether it is the 2 or the 3, but I can see it is a low card. The 11 cases of 0-3 are eliminated, leaving 11 cases of 3-0, 39 cases of 1-2, and 26 cases of 1-2 (as LHO did not play the ♠K) for a total of 11+39+26=76 remaining cases. The finesse works on 11 of the first 11, none of the last 26, and the principle of Restricted Choice tells me that the finesse works on 2/3 of the 39, ie 26 of them, so the total chance of the finesse working is 11+26 = 37/76.

The finesse is 48% because ...
Of the 39 2-1 breaks where LHO originally had equal chances of ♠K3, ♠K2, and ♠32, 13 have been eliminated (as LHO did not have ♠K3) leaving 26 cases. The finessse works 50% of these 26 cases = 13 times. Of the 1-2 breaks 26 of 39 have been eliminated (as LHO has the ♠2), leaving 13 losing finesses. 0-3 has been eliminated, and 11 winning finesses of 3-0 remain. So the finesse wins in 13+11 = 24 times of 26+13+11 = 50 remaining cases, or 24/50 = 48%

Something else ...?
Please say what and why

[nige1]

IMO

• Missing K32, the finesse wins when LHO has K, K2, K3 and K32. It loses when RHO has any of those holdings. The two sets are equally likely. Hence, before FromageGB broaches the suit, his chances are 50%.
• Fromage GB stipulates that when he leads towards dummy's AQ, LHO follows with the 2. Assume that LHO will randomize his play from 32. Then the finesse wins when LHO has K2 and half the time he has K32, (11/2 + 13 = 37/2). It loses when he has 2, and half the time he has 32 (13 + 13/2 = 39/2). Hence his chances are 37/76 = about 48.7% (as in FromageGB's third case).

[manudude03]

I'll answer the above theoretical question first (which will only give you a sample probability and not the absolute probability).
I'll double all these numbers which won't affect the percentage, but makes the calculations easier.

Out of 200 cases, we would get the following possible holdings for LHO:
void: 22
2: 26
3: 26
K: 26
32: 26
K2: 26
K3: 26
K32: 22

From these, we can eliminate void, 3, K and K3 from the possible holdings taking us back to 100 possible cases leaving:
2: 26
32: 26
K2: 26
K32: 22

We apply restricted choice to 32 and K32 leaving us with:
2: 26
32: 13
K2: 26
K32: 11

We also reduce the total remaining cases by 24 (since it can't be the case say LHO had 32 and chose to play the 3). This leaves us with a probability of 37/76 or 48.7%.

However, like I said, this would only be a sample probability. For the true probability, see below:

There are 23 vacant spaces left once LHO plays the 2. There are only 4 possible holdings LHO can have now and these are (ignoring RC for now).

2: 12/23 * 11/22 = 132/506 (about 26.1%)
32: 12/23 * 11/22 = 132/506 (about 26.1%)
K2: 12/23 * 11/22 = 132/506 (about 26.1%)
K32: 11/23 * 10/22 = 110/506 (about 22.7%)

When we include RC, we half the cases when LHO has 32 and K32. This will drop 121 of the possible cases from the total. This gives the holdings new values of:

2: 132/385
32: 66/385
K2: 132/385
K32: 55/385

The finesse will be the right thing to do if LHO started with K2 or K32, so we get a total of 187/385 or about 48.57%.

[Stephen Tu]

manudude03, on 2016-June-26, 06:45, said:

However, like I said, this would only be a sample probability. For the true probability, see below:

??? I don't understand these terms "sample probability" "true probability". All methods of calculation should give you the same answer IMO or you must be doing something wrong.

Quote

There are 23 vacant spaces left once LHO plays the 2. There are only 4 possible holdings LHO can have now and these are (ignoring RC for now).

2: 12/23 * 11/22 = 132/506 (about 26.1%)

This can't be right. You should be getting 26% exactly.
If we assign LHO the deuce, there are exactly 25c12 ways to deal LHO the other cards, 5200300 total possible deals including the deuce.
For holding the deuce alone, there are 23c12 ways = 1352078 to deal the non-trumps.
That's 1352078/5200300 = 26% exactly.

I guess if you want to do it using vacant spaces then it's 13/25 that RHO gets dealt the K once LHO followed with the deuce and 12/24 that RHO gets the 3 if also dealt the K.
13/25 * 12/24 = 26%

[fromageGB]

FAO Nigel and Wayne

I don't get this "half"
Nigel :
> Then the finesse wins when LHO has K2 and half the time he has K32, (13 + 11/2 = 37/2)
Does it not win EVERY time LHO has K32? So wins (13+11 = 24)?
> It loses when he has 2, and half the time he has 32 (13 + 13/2 = 39/2).
Does it not lose EVERY time he has 32? So loses (13+13 = 26)?
and therefore the finesse chances are 24/(24+26) = 24/50 ?
Why the half?

Wayne :
> We apply restricted choice to 32 and K32 leaving us with: ...halving the values for 32 and K32

Yes, of course if LHO started with 32, half the time he may play the 2 and half the time he may play the 3, but we know (if we brought our glasses with us) that he played the 2. Why the half?

I see in https://en.wikipedia...stricted_choice , talking about a scenario missing 4 cards, the KQ32, it says that "If East would win the first trick with the king or queen uniformly at random from ♠KQ, then that original lie 32 and KQ would reach this stage {2nd round and again LHO plays low} half the time and would take the other fork in the road half the time" but that does't help me. We KNOW which fork we are on, and we know the percentages of times the hands occur. You could equally argue that we are looking at the fork where LHO played the 3 first, and he could have taken the other fork. I don't accept this as an explanation.

Could someone who was formerly dubious about restricted choice initially but has seen the light, please point me to whatever it was that explained it to them?

[fromageGB]

Simulations of the missing KQ32 where LHO playes low twice, and we are at a "finesse or drop", would help to demonstate it, if not explain it, because there I would say the finesse is 6.72/(6.72+6.22) (percentage lies of any particular 2-2 and 3-1 breaks) = 52%, whereas restricted choice would (I believe) give a significantly different answer (but I don't know what - 68%?). Experience indicates to me that this might be the case, but can anyone refer me to a simulation?

[Stephen Tu]

fromageGB, on 2016-June-26, 09:15, said:

Yes, of course if LHO started with 32, half the time he may play the 2 and half the time he may play the 3, but we know (if we brought our glasses with us) that he played the 2. Why the half?

Because you only get to count the times you see him play the deuce. You are excluding the times you saw him play the 3.

Let's say we play 200 of these deals. LHO will be dealt the K2 26 times. He'll be dealt the "32" 26 times. He'll be dealt K32 22 times. But he only BOTH got dealt "32" AND chose to play the deuce half of that.

So if you brought your glasses, you see him follow 2 from K2 26 times, follow 2 from "32" 13 times, and follow 2 from K32 11 times. You see him follow with the 3 from "32" 13 times, and the 3 from K32 11 times. You see him follow with the 3 from K3 26 times.

If you didn't bring your glasses, then you see a low card from Kx 52 times, low card from "32" 26 times, low card from K32 22 times.

The half is correcting for the *relative frequency* of which you see the 2 chosen from 32/K32 vs. you see the 2 chosen from K2.

[Stephen Tu]

fromageGB, on 2016-June-26, 09:15, said:

I see in https://en.wikipedia...stricted_choice , talking about a scenario missing 4 cards, the KQ32, it says that "If East would win the first trick with the king or queen uniformly at random from ♠KQ, then that original lie 32 and KQ would reach this stage {2nd round and again LHO plays low} half the time and would take the other fork in the road half the time" but that does't help me. We KNOW which fork we are on, and we know the percentages of times the hands occur. You could equally argue that we are looking at the fork where LHO played the 3 first, and he could have taken the other fork. I don't accept this as an explanation.

So ajt98 vx. xxxx and we hooked the first time and we lost. Should we hook again?

There are at least two ways to look at it:
1. playing for the drop in this scenario picks up exactly one combo. KQ tight with East. Out of all the original possibilities, this accounts for 1/6 of the original 40.7% 2-2 breaks or ~6.8%
Playing for the finesse picks up both stiff honors offside. K stiff with East and Q stiff with East. It's ~24.87% that East started with a stiff. half of those will be stiff honor so ~12.4%


2. The "we get to look at which card East played scenario, and he played the Q". Again, East is only going to play the Q from KQ half the time, so only 3.4% of the original deals. West is going to play the 3 and deuce in some order 100% of the time from K32, so we still count all of those (half of the stiff honors from calculation above, since we saw that the stiff was the Q if it was a stiff), 6.2%.
The success ratios remain the same, we are just looking at half the cases because we are looking at specifically one particular fork, where we saw the Q being played.

Where you are going wrong is counting all instances where a player played from equals, instead of excluding the instances where he would have chosen to play a different card. We are assuming that a player would play randomly from equals, not always the lowest.

[fromageGB]

Hey, thanks Stephen. I think I'm getting there ...

[manudude03]

Stephen Tu, on 2016-June-26, 09:04, said:

??? I don't understand these terms "sample probability" "true probability". All methods of calculation should give you the same answer IMO or you must be doing something wrong.



This can't be right. You should be getting 26% exactly.
If we assign LHO the deuce, there are exactly 25c12 ways to deal LHO the other cards, 5200300 total possible deals including the deuce.
For holding the deuce alone, there are 23c12 ways = 1352078 to deal the non-trumps.
That's 1352078/5200300 = 26% exactly.

I guess if you want to do it using vacant spaces then it's 13/25 that RHO gets dealt the K once LHO followed with the deuce and 12/24 that RHO gets the 3 if also dealt the K.
13/25 * 12/24 = 26%

You're forgetting about what happened at trick 1. Both sides played non-descript cards. They may be non-descript, but they were cards.

[rmnka447]

After LHO plays the 2, the universe of possible card holdings are

LHO RHO
2 K3
32 K
K2 3
K32 -

You can throw out LHO 2 RHO K3 because no matter what you do, you're losing a trick.

There are 3 combinations of cards breaking 2-1, so out of the 39 times the cards break 2-1 in a hundred deals, any particular combination occurs 13 times.
There is only 1 combination where the cards break 3-0, so we would expect that to occur 11 times in a hundred deals.

If LHO would always play 2 from any combination that has the 3-2 in it, then taking the finesse is right in 24 deals and wrong in 13 deals, roughly 2-1 odds for finessing.

If LHO would vary what is played from the 3-2 in any combination, then the frequency of those combinations is reduced. Let's assume LHO is as likely to play either the 3 or 2 when holding both. Then the likelihood of the finesse working is 13 deals for K2 - 3 and 5.5 (11x0.5) for K32 - which equals 18.5 deals in a hundred. The likelihood of the finesse failing is 6.5 (13x0.5) in a hundred deals. Overall, the odds are a little under 3-1 for finessing.

So, when finessing or not can affect the results, the odds favor finessing.

[manudude03]

Stephen Tu, on 2016-June-26, 09:04, said:

??? I don't understand these terms "sample probability" "true probability". All methods of calculation should give you the same answer IMO or you must be doing something wrong.



This can't be right. You should be getting 26% exactly.
If we assign LHO the deuce, there are exactly 25c12 ways to deal LHO the other cards, 5200300 total possible deals including the deuce.
For holding the deuce alone, there are 23c12 ways = 1352078 to deal the non-trumps.
That's 1352078/5200300 = 26% exactly.

I guess if you want to do it using vacant spaces then it's 13/25 that RHO gets dealt the K once LHO followed with the deuce and 12/24 that RHO gets the 3 if also dealt the K.
13/25 * 12/24 = 26%

It would be 26% if there hadn't been a trick beforehand. Since there has been, the appropriate numbers are 23c11 and 21c11 which would be 352716/1352078 which is just under 26.1%.

[Stephen Tu]

manudude03, on 2016-June-26, 13:22, said:

It would be 26% if there hadn't been a trick beforehand. Since there has been, the appropriate numbers are 23c11 and 21c11 which would be 352716/1352078 which is just under 26.1%.

Is this really the right way to calculate for that previous trick, just decrease vacant spaces by 1? I kind of feel like this is wrong somehow but can't really explain it. To me if they led say side suit which was say JT opposite AKQ all we get to do is eliminate the 8-0 breaks and the odds aren't affected by nearly that much.

Maybe someone with a deeper understanding can explain which one of us is wrong.

[SteveMoe]

Stephen Tu, on 2016-June-26, 17:26, said:

Is this really the right way to calculate for that previous trick, just decrease vacant spaces by 1? I kind of feel like this is wrong somehow but can't really explain it. To me if they led say side suit which was say JT opposite AKQ all we get to do is eliminate the 8-0 breaks and the odds aren't affected by nearly that much.

Maybe someone with a deeper understanding can explain which one of us is wrong.

Stephen, I agree with you. The only thing first trick told us was everyone played a card to the trick, so that suit did not break 0 for something. The play of that card provided no useful information as to vacant spaces. Thus the 12/13 ratio still applies at trick two.