[kenberg]

I'm wrestling with this, may I think out loud a little? Certainly I agree with LittleKid that in the asked case it is even odds when one opp plays the 2 and the other the 3.

My thoughts:

First, about restricted choice and the openeing example:
With the AKQ3 in dummy the various restricted choice arguments don't see to work when applied to the 9,T,J. The AK are played producing spots and then the Q is played. The defender with two cards remaining can play either one, even if the two are the J9, since the ten is the only card he hasn't seen. He does not even care whether declarer or his partner has the ten, it's falling. So any restricted choice argument involving inference from the third round play of the ten or jack won't work because there is an equally strong argument for the fall of the nine. Whoever holds two ot the 9,T,J could play either, and the player who held only one has his choice restricted.

The deuce, however, is different from the 7. I realize others have said this, I'm just thinking it through in my way, Take the original hand: AKQ3 in dummy with the cards splitting 4-3 on the play of the tops. The deuce must be played. If it lies with the three card holding then the rules require it, while if it lies with the four card holding then the 3 in dummy requires the deuce to be played.

So the situation is equivalent to declarer being able to ask: Which of you guys holds the deuce? They must answer. With the 7, it's not the same. The 7 is more like the jack. After three rounds of play if the 7 is with the three card holding then it must be played, but if it is with the four card holding then that player may keep the 7 or one of his other cards higher than the deuce.


So this gets us to the 4-3 odds: It's 4-3 that the person holding the deuce is the one holding four cards. It's also true that the person holding the 7 holds four cards, but they have to tell you about the deuce and they don't have to tell you about the 7.


Now the asked about example.
Missing the 2 and the 3, They have to tell you the locatin of both. If one player has one and the other has the other, then the 2 and 3 are equal, so they provide equal inference and so it's even. When one player has both then he must show them both. Surely it is easier for a person with four cards to hold both than for the person with three cards to hold bot, so the odds are in favor of the person holding both to have four cards. This is, however, one of those "inversion of probabilities" things: It is easy enough to calculate the probabilty that the player holding four cards holds the 2 and the 3, but the question is the probability of X holding four cards given that everyone followed and one player had both the 2 and the 3. This is the sort of thing Baye's theorem was invented for, and I'll get back to you after some arithmetic. Technically we need Baye's theorem in the first case as well but I think that the situation is simple enough that we can bypass it. Not in this second case, however.

[kfay]

This is definitely interesting to me.

The idea that is nagging me is this: the deuce has become a significant card, essentially, right? Doesn't that put more vacant spaces in the holder's partner's hand? Thus equalizing the probability of who holds 4 cards?

I'm dealing the 7 remaining cards in the suit, give someone the deuce, and now deal the remaining cards in the pack at random... doesn't this really make it 4:7 minus some percent?

[gnasher]

kfay, on Dec 19 2009, 03:48 PM, said:

This is definitely interesting to me.

The idea that is nagging me is this: the deuce has become a significant card, essentially, right? Doesn't that put more vacant spaces in the holder's partner's hand? Thus equalizing the probability of who holds 4 cards?

I'm dealing the 7 remaining cards in the suit, give someone the deuce, and now deal the remaining cards in the pack at random... doesn't this really make it 4:7 minus some percent?

There are 6 remaining cards, not 7.

Give the 2 to East. Now West has 13 vacant spaces and East has 12.

Take three specific high cards and give them to West. West now has 10 vacant spaces.

Take two specific high cards and give them to East. East now has 10 vacant spaces.

I have one high card left to deal. Because the vacant spaces are the same, each player has a 50% chance of getting the last high card.

So, a specific 3=3 break and a specific 4=2 break are equally likely.

There are 20 3=3 breaks and 15 4=2 breaks, so the 3=3 break occurs 20/35 or 4/7 of the time.

[kenberg]

My arithmetic sucks here, or did before I corrected it. See Cascade's/gnasher's correction below. And it's true that earlier arguments get the same answer w/o going through the complications I use. But I will leave it for whatever interest it may hold.

I like to think that math makes intuitive sense but when dealing with probabilities this often needs to be modified to "intuitive sense after you think about it right".

Suppose you have two random events A and B that may or may not occur. There is a certain probability either will occur, that both will occur, that neither will occur. There is also the probability that A will occur given that B occurred, and vice versa.



Now I am going to suggest that we can get away with saying that we need only consider hands where the suit turns out to split 4-3, one direction or the other. Everything in probability is a little risky, but I think this is kosher.

Also I want to concentrate on E being the guy with the 2 in the first example or the 23 in the second.

It may be a little easier to calculate the probability that E holds three cards. This will be 1 minus the probability that he holds 4.

I will take the second example.

The event A is E holds three cards

The event B is E holds the 2 and the 3.


Here is the key item: Let A & B mean that both A and B occur. Let P(A) be the probability that A occurs. Let P(A|B) be the probability that A will occur given that B occurs. This means you watch for all of the times when B holds the 2 and the 3, and see what percentage of the time he holds three cards (where we look only at the deals where he holds either 3 or 4).


Principle: P(A&B)=P(B) X P(A|B). This is not so crazy. Without worrying about the specifics of what A and B are, suppose that B occurs 2/3 of the time and of the cases where B occurs, A also occurs 1/4 of the time. The principle says that A and B occur together (2/3) X (1/4)= 1/6 of the time.



If you are with me so far, then you will probably also accept that P(A&B)=P(A) X P(B|A) .

So

P(B) X P(A|B)=P(A) X P(B|A)


We would like to calculate P(A|B), the probability that E holds three cards given that he holds the 2 and the 3. If we can calculate the other probabilities, then we have this one as well from the equation.


The others come from combinatorics. Let em do the simple case first, the one where the tops are played from AKQ3 and E shows the deuce. This time B is the event that E holds the deuce.

P(B) is the probability that E holds the deuce. We are talking of a priori probabilities here, or almost a priori meaning that we are assuming the suit splits 4-3 one way or the other but we are assuming nothing more. So P(B)=1/2. Either player is equally likely to hold the deuce.



Similarly, P(A)=1/2. One player holds 3, the other 4, a priori it's fifty-fifty which way. So P(A|B)=P(B|A), and P (B|A) is the probability that E holds the 2 given that he holds three cards. That's 3/7, as advertised


Now to the main event. B is the event that E holds the 2 and the 3.

P(A) is still 1/2, but the other tow probabilities require a little more effort.


P(B) is the probability that E holds the 2 and the 3. Again, we are assuming that the suit is split 4-3 one way or the other. The probability that E holds 3 is the same as the probability that he holds 4, and so the probabilty that E holds both the 2 and the 3 is the average of the prob of him holding the 2 and 3 when holding three cards and the prob of him holding the 2 and the 3 when holding 4 cards.

Correcting my previous post, agreeing with Cascade that 4+3 is actually 7, not 6, we get:

From combinatorics, P(E holds 2&3 | E holds three cards) is 5 choose 1 divided by 7 choose 3 (the number of three card holdings including the 2 and the 3 divided by the number of three card holdings). This is 5 divided by 35, or 1/7.

P(E holds 2&3 | E holds four cards) is 5 choose 2 divided by 7 choose 4. This is 10 divided by 35, or 2/7.


P(B) is the average of 1/7 and 2/7, namely 3/14 You still there? The hardest part is over.



P(B|A) is the probability of E holding the 2 and 3 given that E holds three cards. We did this above, the answer was 1/7.





So we have agreed that

P(B) X P(A|B)=P(A) X P(B|A)

from which (3/14) X P(A|B)=(1/2) X (1/7)=1/14

So P(A|B)=(1/14)/(3/14)=1/3.

When the suit splits 3-4 and one hand holds the 2 and 3, the probability that this hand has length three is 1/3, and so the probability it has length four is 2/3.

[gnasher]

kenberg, on Dec 19 2009, 05:13 PM, said:

[complicated stuff]

I confess that I haven't tried to follow this in detail, but I think you failed at the final hurdle:

Quote

P(A) is still 1/2
...
P(B) is the average of 1/5 and 2/5, namely 3/10
...
P(B|A) is the probability of holding the 2 and 3 given the holding of three cards. We did this above, the answer was 1/5.

So we have

P(B) X P(A|B)=P(A) X P(B|A)

from which (2/5) X P(A|B)=(1/2) X (3/10)

I think you have
  3/10 x P(A|B) = 1/2 x 1/5
so
  P(A|B) = 1/3
That is, the player who produces the 2 and the 3 will hold four cards 2 times out of 3.

Which is, of course, the same result as these:

MarkDean, on Dec 17 2009, 08:00 AM, said:

there is a 2/3 chance the hand with the 2 and 3 have the last card.

gnasher, on Dec 17 2009, 11:28 AM, said:

it's 2:1 that they're 2=3.

[Cascade]

kenberg, on Dec 20 2009, 05:13 AM, said:

P( is the probability that E holds the 2 and the 3. Again, we are assuming that the suit is split 4-3 one way or the other. The probability that E holds 3 is the same as the probability that he holds 4, and so the probabilty that E holds both the 2 and the 3 is the average of the prob of him holding the 2 and 3 when holding three cards and the prob of him holding the 2 and the 3 when holding 4 cards.

From combinatorics (I can explain if needed), P(holds 2&3 | holds three cards) is 4 choose 1 divided by 6 choose 3 (the number of three card holdings including the 2 and the 3 divided by the number of three card holdings. This is 4 divided by 20, or 1/5.

P(holds 2&3| holds four cards) is 4 choose 2 divided by 6 choose 4. This is 6 divided by 15, or 2/5.

In a 4=3 split there are seven cards not six.

P(2&3 | 3 cards) = 5C1/7C3 = 1/7

P(2&3 | 4 cards) = 5C2/7C4 = 2/7

P(2&3) = 3/14

P(2&3) * P(3 cards | 2&3) = P(3 cards) * P(2&3 | 3 cards) becomes

3/14 * P(3 cards | 2&3) = 1/2 * 1/7

so that P(3 cards | 2&3) = 1/3

[kenberg]

Yep. Glad I asked for a check! 4+3=7. I once knew that.

Very embarrassing because I had done it quickly before and came up with 2/3 and 1/3. When I got 5/8 and 3/8 this time I just figured that I had made a mistake before.

Thanks

[aguahombre]

Can think of several applications for this knowledge. Thanks all, for a great topic which I have been too mathematically impaired to figure out over the years.

[Winstonm]

How does opponents' bidding affect this analysis?

[Phil]

Ira Chorush had presented a similar problem to Gnome a few years ago, although I think the suit combo was AKQ42 opposite the 5 (I thought). Maybe he remembers better than I do.

[Cascade]

There is a cute trick to calculate the probabilities:

If we consider a situation like

AKQx

xx

With defenders playing

West: BBs East: Bss or similar where B = a big card and s = a small card.

And there is one big card to be played.

Then if we say that west has played W big cards and east has played E big cards and that there W + E + 1 big cards altogether.

So the big cards are divided either W + 1 : E or W : E + 1

The number of these combinations are respectively

C(W + E + 1 , W + 1) = C(W + E + 1, E) and

C(W + E + 1 , W) = C(W + E + 1, E + 1)

The ratio of these two numbers are the odds of west having the missing card.

C(W + E + 1 , W + 1) : C(W + E + 1, E + 1)

It is reasonably simple high school combinatorics but this simplifies to

E + 1 : W + 1

In plain language, add one to the number of big cards played by east and add one to the number of big cards played by west and there you have it.

Here are all of the cases:

1.

AKQ9

87

Bss sss

W=1 E=0 Odds of west having the outstanding card are 0+1:1+2 = 1:2

P(W) = 1/3 P(E) = 2/3

2.

AKQ8

76

a/ Bss Bss odds 2:2 = 1:1

P(W) = P(E)= 1/2

b/ BBs sss odds 1:3

P(W) = 1/4 P(E) = 3/4

3.

AKQ7

65

a/ BBs Bss odds 2:3

P(W) = 2/5 P(E) = 3/5

b/ BBB sss odds 1:4

P(W) = 1/5 P(E) = 4/5

4.

AKQ6

54

a/ BBs BBs odds 3:3 = 1:1

P(W) = P(E) = 1/2

b/ BBB Bss odds 2:4 = 1:2

P(W) = 1/3 P(E) = 2/3

5.

AKQ5

43

BBB BBs odds 3:4

P(W) = 3/7 P(E) = 4/7

[kenberg]

Very elegant and clear. Which is another way of saying that I wish I would have worked it through that way.

[kenberg]

I was thinking a little about the formula E+1 : W+1 with regard to restricted choice. It seems to apply, and it highlights a perhaps interesting point.

For example
AKT23 opposite 4567. You (North, with the high cards) play the king, the Q falls on your left, you go to the board and lead a spot, rho follows with a spot. You have seen all the small cards, how about the remaining large card. One person, say E has played one big card so E is 1, and W=0. The formula says 2:1.odds.

The key to the general analysis was that they must show you all of the small cards, however they lie. Here, when the ace is played and the Q drops, one could imagine E being cute. Not likely in this situation, since W might hold the king, but suppose instead you opened this suit showing a five card major, the first trick in a side suit was taken on the board, the deuce was led, W followed with a spot, you played the ace, and the Q fell. What to think? W, perhaps with QJx, has seen his partner's stiff spot and might very well wish to get you to spend a board entry to try for a restricted choice play. So the fall of the Q does not, in itself, produce the 2:1 odds. Those odds appear after you go to the board at trick 3, lead the suit at trick 4, and the spot appears on your right. The small cards have been shown and the formula now, but not earlier, applies. 1+1 : 0+1 gives 2:1.

A rather nice formula.

[nige1]

Cascade, on Dec 21 2009, 01:13 AM, said:

There is a cute trick to calculate the probabilities: If we consider a situation like AKQx opposite xx
With defenders playing  West: BBs East: Bss
or similar where B = a big card and s = a small card. And there is one big card to be played. Then if we say that west has played W big cards and east has played E big cards and that there W + E + 1 big cards altogether.
So the big cards are divided either W + 1 : E or W : E + 1
The number of these combinations are respectively
C(W + E + 1 , W + 1) = C(W + E + 1, E) and
C(W + E + 1 , W) = C(W + E + 1, E + 1)
The ratio of these two numbers are the odds of west having the missing card.
C(W + E + 1 , W + 1) : C(W + E + 1, E + 1)
It is reasonably simple high school combinatorics but this simplifies to
E + 1 : W + 1
In plain language, add one to the number of big cards played by east and add one to the number of big cards played by west and there you have it. Here are all of the cases:

• AKQ9 opposite 87
  Bss sss
  W=1 E=0 Odds of west having the outstanding card are 0+1:1+2 = 1:2
  P(W) = 1/3 P(E) = 2/3
  
• AKQ8 opposite 76
  
  • Bss Bss odds 2:2 = 1:1
    P(W) = P(E)= 1/2
    
  • BBs sss odds 1:3
    P(W) = 1/4 P(E) = 3/4
  
  
• AKQ7 opposite 65
  
  • BBs Bss odds 2:3
    P(W) = 2/5 P(E) = 3/5
    
  • BBB sss odds 1:4
    P(W) = 1/5 P(E) = 4/5
  
  
• AKQ6 opposite 54
  • BBs BBs odds 3:3 = 1:1
    P(W) = P(E) = 1/2
    
  • BBB Bss odds 2:4 = 1:2
    P(W) = 1/3 P(E) = 2/3 
  
  
• AKQ5 opposite 43
  BBB BBs odds 3:4
  P(W) = 3/7 P(E) = 4/7

Thank you Cascade! Brilliant!

[dellache]

Cascade, on Dec 21 2009, 06:13 AM, said:

There is a cute trick to calculate the probabilities:

If we consider a situation like

AKQx

xx

With defenders playing

West: BBs East: Bss or similar where B = a big card and s = a small card.

And there is one big card to be played.

Then if we say that west has played W big cards and east has played E big cards and that there W + E + 1 big cards altogether.


It is reasonably simple high school combinatorics but this simplifies to

E + 1 : W + 1

May I add my 2 pence ?

1. Alternate way to explain the AKQ3 Vs 54 problem
East plays the 2, how do we show that the chances he holds the last B-card is 4/7 using "vacant places theory for dummies" ?

Let's consider a new game (not bridge!) with the ♥-JT98762 and a Joker (that's 8 cards).
Rules :
a. I shuffle and deal in two packs of 4 cards each.
b. You look at the two packs.
c. YOU have to tell me in which pack (let's call it PACK1) the 2 of Hearts stands.
Now we have PACK1 = ???2 Versus PACK2 = ????
That's 7 vacant places remaining for the Joker. So the Joker is in the PACK2 4 times out of 7 (right ?).

Corollary : the last B-card is with the ♥2, 4 times out of 7.


2. Cascade's theorem complicated

Suppose now that we have the 2 Major suits which consist of AKQx facing xx.
See Max Rebattu's "Bol tips" problem : Here

We suppose that each players has exactly 7 major cards. Using the same notations as Cascade (ES and EH for spades and hearts respectively), it's (rather) easy to show that the probability of West having the last Heart (relative to East having the last Heart) is :
(EH+1)/(WH+1) * (WS+1)/(ES+1)

[Well if Spades are AKQ2 facing 43 (edit)... you have the single suit formula again at least].

I f you apply this to the "Bol Tips" problem, and if you suppose that the double squeeze works (East and West both have 7 major cards), then application of this formula gives you a 2:1 chance of West having the last Heart.

Mmmm... anyone for a "joker game analogy" in order to explain this with "vacant places for the dummies" ?

Cheers,

[barryallen]

Cascade, on Dec 21 2009, 01:13 AM, said:

There is a cute trick to calculate the probabilities:

If we consider a situation like

AKQx

xx

With defenders playing

West: BBs East: Bss or similar where B = a big card and s = a small card.

And there is one big card to be played.

Then if we say that west has played W big cards and east has played E big cards and that there W + E + 1 big cards altogether.

So the big cards are divided either W + 1 : E or W : E + 1

The number of these combinations are respectively

C(W + E + 1 , W + 1) = C(W + E + 1, E) and

C(W + E + 1 , W) = C(W + E + 1, E + 1)

The ratio of these two numbers are the odds of west having the missing card.

C(W + E + 1 , W + 1) : C(W + E + 1, E + 1)

It is reasonably simple high school combinatorics but this simplifies to

E + 1 : W + 1

In plain language, add one to the number of big cards played by east and add one to the number of big cards played by west and there you have it.

Here are all of the cases:

1.

AKQ9

87

Bss sss

W=1 E=0 Odds of west having the outstanding card are 0+1:1+2 = 1:2

P(W) = 1/3 P(E) = 2/3

2.

AKQ8

76

a/ Bss Bss odds 2:2 = 1:1

P(W) = P(E)= 1/2

b/ BBs sss odds 1:3

P(W) = 1/4 P(E) = 3/4

3.

AKQ7

65

a/ BBs Bss odds 2:3

P(W) = 2/5 P(E) = 3/5

b/ BBB sss odds 1:4

P(W) = 1/5 P(E) = 4/5

4.

AKQ6

54

a/ BBs BBs odds 3:3 = 1:1

P(W) = P(E) = 1/2

b/ BBB Bss odds 2:4 = 1:2

P(W) = 1/3 P(E) = 2/3

5.

AKQ5

43

BBB BBs odds 3:4

P(W) = 3/7 P(E) = 4/7

More than good enough. Cannot believe I have over looked this for so long, maybe because I thought the odds were not that great in certain circumstances.

If you cannot be bothered with the math, just go with the BOLS tip.