[kgr]

I messed up in the other post, let me try again:

---

contract 6♠.
LHO leads ♥x, RHO takes with the ACE and return a ♥ for LHO to ruff.
LHO plays ♣ and you take ♣A, (♣K, ♦A, ♦K all follow) ♠A,
You play small ♠ and LHO follows small.
Probability that LHO has ♠Q?
Vacant Spaces:
13-1(H)-3(S):13-4(H)-1(S)=9:8=0.529412
Also take other suits into account (HOW WRONG IS IT?):
Played ♣A: 13-1(H)-3(S)-1(C ):13-4(H)-1(S)-1(C )=8:7=0.533333
Played ♣A, ♣K: 13-1(H)-3(S)-2(C ):13-4(H)-1(S)-2(C )=7:6=0.538462
Played ♣A, ♣K, ♦A: 13-1(H)-3(S)-2(C )-1(D):13-4(H)-1(S)-2(C )-1(D)=6:5=0.545455
Played ♣A, ♣K, ♦A, ♦K: 13-1(H)-3(S)-2(C )-2(D):13-4(H)-1(S)-2(C )-2(D)=5:4=0.555556
What is the real probability of these last? Is it closer to 0.529412 or closer to the wrongly calculated?

[MFA]

Vacant spaces is a rough model that can't handle partial information. Either cards are counted or they are not.

It is usual much more accurate not to count in partial counts in the suits than to do so. I would expect 9:8 to be closest to the right answer of the numbers you present.

If we cash one club, all we really get to know is that the suit is not 8-0 or 0-8. A very insignificant piece of information. And so on when we cash a second club and one or two diamonds.

[inquiry]

Let me answer this with a question.

See my Risk slam bonus question in general bridge discussion. The ending is...

East having lead a heart and west having followed with a low heart....

So using your example, we know now that South had 5 hearts and is down to just one heart, either the Queen or a small and north who started with 2 hearts is down to just one heart, either the Queen or a small.

If we calculate based on known cards, as you did, there is an equal chance north has the biggest heart as there is south has the biggest heart

---

So my question to you, does this sound right? If the answer to that is no, perhaps you can answer your own question above.


(and if you think the odds are the same that north has the Q and South in this ending, look up "monte hall problem" or something similiar on the web -- might should be wayne brady problem these days)

[ceeb]

Probability that LHO has the ♠Q

4:3 from the ♥ distribution alone
9:8 (4% less) given the further information that the ♠ pips are 3=1 -- per vacant spaces. This estimate technically
Increases by 1/7400 knowing no club void.
Increases by 1/360 knowing no club void or singleton.
Cashing the ♦A tells you nothing further.
Increases by 1/130 knowing no minor shortage.

Summary: The number 0.537 from accurate calculation is over twice as close to the approximate 9:8 than it is to the "wrong answer" 5:4.

[kgr]

inquiry, on Oct 2 2010, 05:30 PM, said:

Let me answer this with a question.

See my Risk slam bonus question in general bridge discussion. The ending is...

East having lead a heart and west having followed with a low heart....

So using your example, we know now that South had 5 hearts and is down to just one heart, either the Queen or a small and north who started with 2 hearts is down to just one heart, either the Queen or a small.

If we calculate based on known cards, as you did, there is an equal chance north has the biggest heart as there is south has the biggest heart

---

So my question to you, does this sound right? If the answer to that is no, perhaps you can answer your own question above.


(and if you think the odds are the same that north has the Q and South in this ending, look up "monte hall problem" or something similiar on the web -- might should be wayne brady problem these days)

Ben,
My feeling says that South is more likely to have ♥Q because he had more Hearts initially.
But I don't see how this relates to the Monte Hall problem and even less how it relates to my question.

[kgr]

ceeb, on Oct 3 2010, 12:17 AM, said:

Probability that LHO has the ♠Q

4:3 from the ♥ distribution alone
9:8 (4% less) given the further information that the ♠ pips are 3=1 -- per vacant spaces. This estimate technically
Increases by 1/7400 knowing no club void.
Increases by 1/360 knowing no club void or singleton.
Cashing the ♦A tells you nothing further.
Increases by 1/130 knowing no minor shortage.

Summary: The number 0.537 from accurate calculation is over twice as close to the approximate 9:8 than it is to the "wrong answer" 5:4.

Thank you Ceeb. I'm convinced now.
- I guess that counting discards would give an even worse result.
BTW: How did you come to 1/7400 and 1/360 in:
...Increases by 1/7400 knowing no club void.
Increases by 1/360 knowing no club void or singleton....
Do you have a program for it or is there some logic to come to these figures?

[Cascade]

kgr, on Oct 3 2010, 07:03 PM, said:

inquiry, on Oct 2 2010, 05:30 PM, said:

Let me answer this with a question.

See my Risk slam bonus question in general bridge discussion. The ending is...

Dealer: ????? Vul: ???? Scoring: Unknown

♠   ♥ ? ♦ Q ♣   ♠   ♥ AJ ♦   ♣   ♠   ♥   ♦ J ♣   ♠   ♥ ? ♦   ♣

East having lead a heart and west having followed with a low heart.... So using your example, we know now that South had 5 hearts and is down to just one heart, either the Queen or a small and north who started with 2 hearts is down to just one heart, either the Queen or a small. If we calculate based on known cards, as you did, there is an equal chance north has the biggest heart as there is south has the biggest heart

So my question to you, does this sound right? If the answer to that is no, perhaps you can answer your own question above.


(and if you think the odds are the same that north has the Q and South in this ending, look up "monte hall problem" or something similiar on the web -- might should be wayne brady problem these days)

Ben,
My feeling says that South is more likely to have ♥Q because he had more Hearts initially.
But I don't see how this relates to the Monte Hall problem and even less how it relates to my question.

This is exactly what the Monte Hall problem says.

When the spectator in the Monte Hall chose one door it was more likely the prize was in one of the other two.

Therefore when the host opens one of the other doors which he can always door thus giving you no additional information the prize is still more likely to be in the other unchosen door.

If one opponent is know to have been dealt five hearts then he can always show you four non-queens. That doesnt change that he is 5:2 on to have the queen when his partner was known to start with two hearts.

[ceeb]

kgr, on Oct 3 2010, 02:09 AM, said:

ceeb, on Oct 3 2010, 12:17 AM, said:

BTW: How did you come to 1/7400 and 1/360 in:
...Increases by 1/7400 knowing no club void.
Increases by 1/360 knowing no club void or singleton....
Do you have a program for it or is there some logic to come to these figures?

Just a program. I specify opposing cards of
sq 4sx 5h 8d 8c, and a condition such as hε1 or (hε1)&(sx ε 3)
and it computes a table of of the 73 or 17 or whatever possible distributions (counting sq as a suit) and their corresponding probabilities. Easy manipulations on the table give various results.

Answers to simple problems often turn out to be nice fractions with small denominators -- which in retrospect can probably be seen logically such as via vacant spaces -- but 1/360 etc are just rounded versions of decimal numbers with a lot of arithmetic behind them.

[inquiry]

I'm sorry if my proposed question didn't help. I assume that cascade answer about Monte Hall problem solved your question about that.

My point was you can not apply vacant space all the time to all the cards played. The example showed one vacant space in left in both opponents hand but the obvious conclusion from that "fact" would be wrong. That is, the heart finessee versus teh drop are not 50% each. We can all work out the finesse is 5:2 favorite in that example.

This brings us to yoru problem. I would totally ignore the fact that both opponents followed with two small cards in both minors. Here is the rule I follow, I ignore insignificant cards in a side suit, until some one shows out, or ... if by following suit it eliminates some of the possible distributions for the key suit. That is, once one of the opponents showed enough, say clubs, or clubs and diamonds combined, that a 5-0 trump slit is elimainated, i correct the calculations for no 5-0 trump split. Etc.

So if we ignore the side suits, when you lead a low spade towards dummy (after cashing the ♠A), west follows low, there is one spade out (the queen). It is either with WEST or with EAST. We know that West has 9 vacant spaces (3 known spades, 1 known heart), and East has 8 vacant spaces (4 known hearts and 1 known spade). So it is 9:8 that the spade Queen is with West. which has us exactly back to your 52.9421% chance.

[Free]

Following Fred's rule you should finesse.

[gwnn]

Monty Hall.

[kgr]

Thanks all for the answers. This was interesting.
...and of course a last question:
If both opps still follow to a suit. What is the impact if you know for sure that they are not playing random, but allows from low to high (This is a realty against some opps). Can you then include it somehow in the vacant space theory or does it not make any difference?
Thanks,
Koen

[ceeb]

kgr, on Oct 4 2010, 06:59 AM, said:

If both opps still follow to a suit. What is the impact if you know for sure that they are not playing random, but allows from low to high (This is a realty against some opps). Can you then include it somehow in the vacant space theory or does it not make any difference?

That may be two different questions. It surely makes a difference, and somewhat you can account for it via vacant spaces.

Suppose
   ♠AKQ
♠72    ♠43
   ♠J
is revealed somehow -- perhaps through cashing 2 rounds of ♠.

Under the assumption that LHO has denied the ♠6543 and RHO denied the ♠2 it seems a good approximation to deduct 1 and 4 vacant spaces from LHO and RHO respectively.

I have a feeling that this kind of machination must be applied with care (especially worried about bias in the way the information is obtained) and tentatively. But it could also be an underexploited gold mine.