[Laocoon166]

Imagine you are playing a contract and LHO leads the ♦6. You can be 100% sure that the lead is from a singleton or doubleton and you need to guess which it is to make the hand with no other clues. Assume (from the auction say) that it is impossible for LHO to have been dealt more that 2 diamonds. Say the spots you are missing in the suit are 2, 3, 4, 6, 8.

LHO's lead is therefore from either 64, 63, 62 or 6, so it is therefore 3:1 more likely to be a doubleton? Correct? (Even a bit more considering that a priori the doubleton holdings are more likely to have been dealt than the singleton holdings).

To continue this line of thought imagine that LHO had led the ♦3 instead - either from 32 or singleton 3. Now it might seem that odds are roughly 1:1. However is there an element of Monty Hall restricted choice or something to suggest that the lead is a singleton?

i.e. There are 10 (4 + 3 + 2 + 1) possible doubletons, and 5 possible singletons. When LHO leads the 3, 9 possible doubletons are eliminated and four singletons. The 3 is therefore twice as likely to have been in the singleton group than the doubleton.

Am I making sense?

[wank]

Not really because he would often lead a non-diamond with a doubleton

[Laocoon166]

 wank, on 2016-April-02, 04:21, said:

Not really because he would often lead a non-diamond with a doubleton

Okay - assume for the purposes of this hypothetical that he would always lead the suit. Perhaps his partner had bid diamonds or something.

[gszes]

 Laocoon166, on 2016-April-02, 04:19, said:

Imagine you are playing a contract and LHO leads the ♦6. You can be 100% sure that the lead is from a singleton or doubleton and you need to guess which it is to make the hand with no other clues. Assume (from the auction say) that it is impossible for LHO to have been dealt more that 2 diamonds. Say the spots you are missing in the suit are 2, 3, 4, 6, 8.

LHO's lead is therefore from either 64, 63, 62 or 6, so it is therefore 3:1 more likely to be a doubleton? Correct? (Even a bit more considering that a priori the doubleton holdings are more likely to have been dealt than the singleton holdings).

To continue this line of thought imagine that LHO had led the ♦3 instead - either from 32 or singleton 3. Now it might seem that odds are roughly 1:1. However is there an element of Monty Hall restricted choice or something to suggest that the lead is a singleton?

i.e. There are 10 (4 + 3 + 2 + 1) possible doubletons, and 5 possible singletons. When LHO leads the 3, 9 possible doubletons are eliminated and four singletons. The 3 is therefore twice as likely to have been in the singleton group than the doubleton.

Am I making sense?

Note how you readily agree that the 6 is a 3:1 favorite vs a singleton despite the fact that it is most likely part of the doubleton group. If you merely think that way u assume the 6 is a 2:1 favorite instead of figuring the individual cases (where the result is a 3:1 favorite). You correctly assume the cases scenario with the 3 being either 1:1 (or maybe slightly less) but for some reason get into this hidden door theme that was absent when you figured the odds for the 6 (it is merely a blind spot you will correct it).

[dirtbag]

why cant he lead the 3 from 36? or the 6 from 68. i think u are assuming he will always signal hi low.

[quiddity]

 Laocoon166, on 2016-April-02, 04:19, said:

To continue this line of thought imagine that LHO had led the ♦3 instead - either from 32 or singleton 3. Now it might seem that odds are roughly 1:1. However is there an element of Monty Hall restricted choice or something to suggest that the lead is a singleton?

i.e. There are 10 (4 + 3 + 2 + 1) possible doubletons, and 5 possible singletons. When LHO leads the 3, 9 possible doubletons are eliminated and four singletons. The 3 is therefore twice as likely to have been in the singleton group than the doubleton.

Am I making sense?

No, I don't think so. Restricted choice applies when the defender has some kind of choice with one holding but not with another. Here, we assume there is no choice. The defender will always lead diamonds. The defender will always lead the 3 from 32 doubleton.

The remaining possible holdings keep their a priori probabilities relative to each other. So if you thought a priori that singleton 3 and doubleton 32 were equally likely, they remain equally likely after the other holdings have been eliminated.

[PhantomSac]

Yeah its not restricted choice since they will lead honestly.

[jmcilkley]

Would he lead from a doubleton? It's often a poor lead, seeing up declarer's second suit. So I would assume it's a singleton.

[kenberg]

I will look at the lead of the 3.

The stipulation was that we should assume that lho holds either one or two diamonds and that he would, for whatever the reason, surely lead a diamond. I think we can include the unstated assumption that he leads the top of a doubleton.

You might say that his choices are "very restricted". Whether he has the singleton 3 or the doubleton 32, the conditions of the problem absolutely force the lead of the 3.

Now we consider all hands consistent with the stipulations. As I get it, we must place the 3 on our left, all of the diamonds that we cannot see, except for the deuce, on our right, and then see what the chances are that the deuce is in one hand or the other.

That is: The lead of the 3 determines that the hand must be one of a certain exactly prescribed set of hands. When the hand is of this type the 3 will always be led, when the hand is not of that type the 3 will never be led.

We are told which higher spots are unseen, but I think we need to count all of the higher missing diamonds. Let us say that there are 5 diamonds higher than the 3 that, in addition to the deuce, we cannot see. They all lie to our right. That leaves 8 spaces to put the 2 on our right, 12 places to put the 2 on our left. So the odds are 3 to 2 that the deuce is on our left.

But wait. Maybe he holds the Q53? No. The problem stipulates that he does not. But how about Q3? As I understand the problem, that also is ruled out. The lead of the 3, as I get it, determines that the 3 is on our left (obviously) and also that all diamonds higher than the 3 are on our right. And, importantly, nothing else about the hands can be deduced from the lead of the 3. So we simply do empty spaces to compute the odds.

I think that what I say above is correct, at least it sounds right to me, but notice how heavily I lean on assumptions that are unlikely to be true in practice. For example, it was suggested that we know he would lead a diamond because partner overcalled. Did he overcall at the 2 level vulnerable? If the opponents have the 2 and the 3, and then 5 more unseen diamonds, I think the guy on my right has, or very likely has, 6 of them, making the 3 a stiff. Yes, people overcall even vul at the 2 level on five baggers, but they would need more hcps to do it, or at least I would. We can easily think of a dozen variants like this.


So my claim is that if we take the stipulations as being as I understand them, and as absolutely forcing on our analysis, then we place all diamonds higher than the 3 on our right and apply empty spaces to figure where the 2 is. But I also claim this would very seldom be the best way to proceed.


Mathematics is very useful. But care is needed in making the assumptions explicit, and then care is needed in making sure that the assumptions are not so stringent that they destroy the usefulness of the conclusion.

[PhantomSac]

You are forgetting that you are more likely to be dealt 32 than stiff 3, even with empty spaces. One specific 2-3 break is more likely than one specific 1-4 break.

Another way to look at it is that RHO is less likely to have the 2 for empty spaces reasons, he is known to have the 864 of diamonds.

[jogs]

 Laocoon166, on 2016-April-02, 04:19, said:

Imagine you are playing a contract and LHO leads the ♦6. You can be 100% sure that the lead is from a singleton or doubleton and you need to guess which it is to make the hand with no other clues. Assume (from the auction say) that it is impossible for LHO to have been dealt more that 2 diamonds. Say the spots you are missing in the suit are 2, 3, 4, 6, 8.

LHO's lead is therefore from either 64, 63, 62 or 6, so it is therefore 3:1 more likely to be a doubleton? Correct? (Even a bit more considering that a priori the doubleton holdings are more likely to have been dealt than the singleton holdings).

To continue this line of thought imagine that LHO had led the ♦3 instead - either from 32 or singleton 3. Now it might seem that odds are roughly 1:1. However is there an element of Monty Hall restricted choice or something to suggest that the lead is a singleton?

i.e. There are 10 (4 + 3 + 2 + 1) possible doubletons, and 5 possible singletons. When LHO leads the 3, 9 possible doubletons are eliminated and four singletons. The 3 is therefore twice as likely to have been in the singleton group than the doubleton.

Am I making sense?

You can not be 100% sure it is a singleton or doubleton. Also the probability of singletons and doubletons are not equal.

[kenberg]

 PhantomSac, on 2016-April-04, 10:25, said:

You are forgetting that you are more likely to be dealt 32 than stiff 3, even with empty spaces. One specific 2-3 break is more likely than one specific 1-4 break.

Another way to look at it is that RHO is less likely to have the 2 for empty spaces reasons, he is known to have the 864 of diamonds.

I am not so sure I have forgotten anything here. I agree that the 2 is more likely to be on the left, so if we disagree the question is what are the odds. OP said the 86432 are the unseen spots but I was assuming maybe we are missing some honors as well. If we are not missing honors, no one much cares about where the deuce is because they are about to cash their honors.

Suppose, to take an extreme case, that you (the guesser) have no diamonds at all in your hand and no diamonds in the dummy. Suppose the 3 is led, and the conditions of the problem are in force. This means that rho holds either the 4,5,...,A of diamonds and two non diamonds or else he holds the 2,4,5,...,A of diamonds and one non-diamond. The opponents hold all 13 diamonds and 13 known non-diamonds. There are (13 times 12)/2 =78 hands that rho might hold if he holds two non-diamonds and 13 possible hands if he holds 1 non-diamond. So the odds are 78 to 13, or 6 to 1, that the deuce is on the left. Counting empty spaces, after the 4 through K are placed on the right and the 3 on the left, we find 12 empty spaces on the left to place the deuce, 2 empty spaces on the right to place the deuce, giving the same 6 to 1 odds.

It all depends on exactly what is being asked.

A game: You are playing with the devil for your soul. Your hand and dummy are dealt first and will not change. Between your hand and dummy you see AQT975. The KJ86432 of diamonds are unseen. The devil now repeatedly shuffles and deals the remaining cards, 13 to your left, 13 to your right. If lho is dealt a hand with either the stiff 3 or the doubleton 32 of diamonds, a bell rings. If the bell does not ring then the devil re-shuffles and re-deals. He is an honest devil so the shuffle is random.
When the bell rings, you are to guess the location of the deuce.
It seems to me we are asking: What is the conditional probability that the 2 is on our left, given that the 3 is on our left and the KJ864 is on our right? For this it seems empty spaces should work.
More formally, we just count hands.
The opponents hold 7 diamonds so there are 19 non-diamonds to be distributed.
If the KJ8642 are on the right we must place 7 more cards there from the 19 non-diamonds.
If the KJ864, but not the 2, are on the right we must place 8 more cards there from the 19 non-diamonds.
The first can be done in Binomial(19,7) ways and the second can be done in Binomial(19,8) ways. If we divide Binomial(19,8) by Binomial(19,7) we get 8/12 or 2/3.
If we count empty spaes after placing the 3 on the left and the KJ864 on the right we get 2 on the left and 8 on the right, for the same result. We get 3 to 2 odds, or a probability of 3/(3+2) = 0.6, that the deuce is on the left.

This is the conditional probability after the 3 is led, providing we agree that the 3 would be led if and only if lho held either the stiff 3 or the doubleton 32. We are to use only the fact that lho holds either the stiff 3 or the doubleton 32 and he will lead one of them We are not to consider how this mighty come about.

The problem, as I am interpreting it, is very artificial. I see it as: Given that lho holds the diamond 3 and no diamond higher than the 3, what is the probability that he holds the 2? Phrased this way, we can forget about the lead since the stipulation is that he must lead the 3 whether he has the deuce or not. We are not to take into account how we know he has at most two diamonds nor why we know he must lead a diamond.