HCP Splits

[karras166]

So this is the situation: you (South) are in a contract of 4♥, say, and your contract hinges on a two-way guess for the diamond queen, the only card whose position you are unsure of.

 

So far West has shown up with 9HCP and East 3HCP. Does this fact alone suggest playing East for the Q♦ (i.e. excluding any possible inferences from the bidding)?

 

Before today I would have thought that it was still a 50-50 chance but a bridge book I own suggests that in this situation you should play for the most even HCP split and therefore put East with the Q.

 

I appreciate that before the hand starts the HCP are likely to split evenly but I thought this wouldn't be the case a posteriori. The phrases "Bayes probability" and "Monty Hall" are floating around my head, but I am not a mathematician. Can someone help me on this point?

 

Thanks.

[ArtK78]

There is no reason to assume that the ♦Q is in one hand or the other based solely on the division of HCP between the opponents' hands. However, if the division of the HCPs between the opponents' hands would have affected the bidding, then there is a reason for putting it in one hand or the other.

 

For example, you say that the HCP are divided 9/3 and the ♦Q is not yet accounted for. If putting the ♦Q with the 9 HCP hand would have given that player a bid somewhere along the way, and he did not bid, then there is a reason for putting it in the other hand.

 

Otherwise, you are better off focusing on the distribution of suits rather than the distribution of HCP.

[1eyedjack]

I suppose that if West has shown up with 1 x King, 1 x Queen and 4 x Jacks and that not voluntarily, and East has shown up with 1 x King, and that not voluntarily, then there possibly might arguably be more vacant spaces for East to hold the D:Q. On the other hand if West has shown up with 1 x Ace, 1 x King and 1 x Q, and East has shown up with 3 x Jacks, then they would be all square,

 

Personally I distrust this analysis. Furthermore, we like to think of these scenarios as if there were no other information available, but that seldom is the case.

[karras166]

There is no reason to assume that the ♦Q is in one hand or the other based solely on the division of HCP between the opponents' hands. However, if the division of the HCPs between the opponents' hands would have affected the bidding, then there is a reason for putting it in one hand or the other.

 

Thanks. I appreciate your point about inferences from the bidding but I am deliberately ignoring that aspect for now. Am I to take it from your answer that you think my book is wrong regarding HCP split?

 

 

<snip>

 

Furthermore, we like to think of these scenarios as if there were no other information available, but that seldom is the case.

 

Yes, I understand that I am couching the question in what might perhaps seem like an unrealistic scenario but I am just trying to sort out this particular issue in my head.

 

Another example for you - simpler (and with no vacant spaces implications), but I think it is much the same and it is troubling me. This time you are playing a contract and there are 2 aces unaccounted for. My book seems to suggest that upon finding one Ace with RHO you should place the other Ace with LHO because they are more likely to be split between the two defenders. I don't think this follows - sure at the beginning of the hand split aces was the most likely scenario but does that still hold true after discovering the position of one of the aces?

 

Hopefully this clarifies the question in my original post.

[ArtK78]

The only reason that a particular card (any particular card, not just the other ace) is more likely to be in the hand without the known ace is that the hand without the known ace has 13 other cards compared to the hand with the known ace which has 12 other cards. This is, of course, the vacant space argument. It is not because aces are likely to be evenly distributed between the unknown hands, all other things being equal. Aside from the fact that all other things are not likely to be equal, there is no validity to the claim that aces or other honor cards are likely to be evenly divided between the two unknown hands (other than the vacant spaces argument).

[Endymion77]

deleted, errors Edited September 29, 2013 by Endymion77

[karras166]

Ok, that all makes sense - just goes to show you can't believe everything you read in books.

 

The author started with the perfectly reasonable "when you are missing two aces, the chances are the same defender won't have both" and then drew these erroneous conclusions from that. I was sure I'd missed something.

[fromageGB]

I'm not convinced there's nothing in it, and in the absence of any contrary indication I would go with the idea. At least if you follow it regularly it will eventually lead to an empirical acceptance or rejection.

 

I can't help thinking that there must be something analogous to the theory of vacant spaces: the theory of vacant points, or vacant values. When there are 15 points missing, and one hand has shown up with 8 and the other nothing, and neither hand bid, then there are 3 vacant points in one hand A, and 11 in the other hand B, with 7 points to be distributed. Hand B is more likely to have the next missing points in the ratio of 11:3. ?

 

Endymion's argument of equal likelihood when there are 2 aces missing because there are only 4 initial possibilities is only valid if each of the 4 possibilities is equally likely. Once you assign different probabilities for those 4, then the conclusion is different.

 

In the extreme case, where hand A has shown 8 points and failed to open, but B has shown one ace, then I think Endymion might accept that A will not have the other ace. He therefore probably assigns a step function to the distribution of high card values, flat likelihood all the way to a sudden impossibility. I would think a more gradual transition of probabilities might apply.

 

Certainly if you are playing 7NT undoubled with 2 aces missing, and the person on lead shows up with one of them, I think it is fair to assume his partner has the other. Yes, we did have a case of someone being in 7NT undoubled minus one ace, and the leader eventually turned up with the ace. When asked why he didn't double, he said "Well, it was against you (respected senior player he was in awe of) and I thought you'd make it".

[helene_t]

I think I know that book - card play techniques by Mollo ? Great book but unfortunately the authors don't grasp probabilities .

[Endymion77]

deleted, errors

[Endymion77]

Looks like the math in the book is correct after all. If you deal just 4 cards to the 2 players: 2 aces (A1 and A2) and 2 pips (x1 and x2), there are 24 possible distributions:

 

P1 P2

-----

A1 A2

x1 x2

 

A1 A2

x2 x1

 

A1 x1

A2 x2

 

A1 x1

x2 A2

 

A1 x2

A2 x1

 

A1 x2

x1 A2

 

A2 A1

x1 x2

 

A2 A1

x2 x1

 

A2 x1

A1 x2

 

A2 x1

x2 A2

 

A2 x2

A1 x1

 

A2 x2

x1 A1

 

x1 A1

A2 x2

 

x1 A1

x2 A2

 

x1 A2

A1 x2

 

x1 A2

x2 A1

 

x1 x2

A1 A2

 

x1 x2

A2 A1

 

x2 A1

x1 A2

 

x2 A1

A2 x1

 

x2 A2

x1 A1

 

x2 A2

A1 x1

 

x2 x1

A1 A2

 

x2 x1

A2 A1

 

In 12 of those cases, A1 is dealt to the first player (the one who showed up with the ace). However, once he shows with the ace, 8 times his other card is x1 or x2 and only 4 times his other card is A2. So it's more likely that the ace is in the other player.

 

The distributions I listed above:

 

Opp 1 - Opp 2

x - A1 A2

A1 - A2

A2 - A1

A1 A2 - x

 

aren't equally likely. Here's how often they happen:

 

Case 1: 4 times

Case 2: 8 times

Case 3: 8 times

Case 4: 4 times

 

Once it's between case 2 or case 4, case 2 is twice as likely (in the case of 4 cards only of course, when 13 cards are dealt the difference will be much more subtle but would still be there). So I hope I didn't confuse anyone. :)

[karras166]

I think I know that book - card play techniques by Mollo ? Great book but unfortunately the authors don't grasp probabilities .

 

That's right Helene. There are certainly a few errors it seems but a thoroughly enjoyable read.

 

Looks like the math in the book is correct after all.

In 12 of those cases, A1 is dealt to the first player (the one who showed up with the ace). However, once he shows with the ace, 8 times his other card is x1 or x2 and only 4 times his other card is A2. So it's more likely that the ace is in the other player.

 

It's this point that reminded me of the Monty Hall problem, and exactly the point I was trying to understand - thanks for doing the analysis Endymion. On the face of it, it seems that the remaining ace is just as likely to be in either hand but I suppose the starting probabilities are still valid. I guess this analysis can be extended to HCP split as well as in my first example. But as others say above often there will be a known side suit to take into account.

[fromageGB]

As predicted. The theory of vacant points wins the day.

Nothing to do with the monty hall problem, though. That would entail a defender showing you a missing K and saying "still fancy the finesse?"

[FM75]

If you truly wanted an answer to the question of where the missing card is, without taking into account the bidding or play to that point other than the card discovered, then you might as well have framed it where the 2 was missing.

 

That should make it clear that the vacant points argument is specious, without using information that would take into account the presence of the cards.

 

Added:

The other thing here is that the likelihood of the result depends on how many tricks have been played. At trick 13, the missing card is certain to turn up in the other hand. At trick 12 it is 2-1 to be in the other hand unless it would be limited by distribution at that point.

[kenberg]

I have seen arguemnts of this general nature. Let me describe a variant that I think is correct and then say why it does not apply in the current case.

 

Suppose lho shows up with 9 points more or less right away. Perhaps he starts wit the AKQ of clubs and you ruff the third round. Suppose it seems right to draw trump right away, you are missing the Q of trump, and there is another K out there somewhere, not in trump. Say it seems right to draw trump right way. OK, lho probably cannot have both the trump Q and the unseen K. Certainly he cannot have it if he was dealer, since he would have opened the bidding. So he can have the K w/o the Q, the Q w/o the K, or neither. All equally likely so the odds are 2 to 1 the Q is with rho.

 

That reasoning seems sound.

 

 

But if you have time to first locate the K with rho then, as near as I can see, we are back to 50-50.

 

 

Perhaps it makes sense to simply follow the advice of playing for the Q to be in the weaker hand. If it's 50-50, well, it's 50-50. And sometimes there will be an analysis, if you think long enough, to make it better than 50-50 that the Q LIES WITH rho. It would be unusual for lho to show up with 9 and for an analysis to show the Q is still more, rather than equally, likely to be with him. A corollary would be that after lho shows 9 points and you get the lead, you may as well play for the Q to be with rho w/o bothering to find the location of the missing K. If the K lies with lho then it is somewhere between highly likely and certain that the Q lies with rho. If the K lies with rho it's a coin flip. So just play rho for the Q.

[helene_t]

Looks like the math in the book is correct after all. If you deal just 4 cards to the 2 players: 2 aces (A1 and A2) and 2 pips (x1 and x2), there are 24 possible distributions:

Yes if you only give 2 cards to each player then the empty spaces argument would be strong. You can make it even stronger by giving just one card to each player - now knowing that LHO has one of the cards makes it a 100% bet that RHO has the other card!

 

But you give 13 cards to each player. Knowing that LHO has one of the aces makes it 12/25 that he also has the other ace.

 

In practice you will always know a lot more of course. But the OP's question was about the hypothetical situation that all we know is that LHO has ♠A. And the the question: what is the probability that LHO also has ♥A?

[Endymion77]

Yes if you only give 2 cards to each player then the empty spaces argument would be strong. You can make it even stronger by giving just one card to each player - now knowing that LHO has one of the cards makes it a 100% bet that RHO has the other card!

 

Yes, as I mentioned above "Once it's between case 2 or case 4, case 2 is twice as likely (in the case of 4 cards only of course, when 13 cards are dealt the difference will be much more subtle but would still be there)."

 

The number of unknown cards obviously depends on which trick the situation arises, it might easily happen when there are only 4 cards left in the defenders, or 6, or 26 (if it happens at trick 1).

 

But the OP's question was about the hypothetical situation that all we know is that LHO has ♠A. And the the question: what is the probability that LHO also has ♥A?

 

And the answer is that it depends on the number of remaining cards in the opponents but RHO is always more likely to have it.

[helene_t]

FWIW the example that the book gives was

 

xx

 

opposite

 

KJx.

 

They suggest making a discovery play to see who has the ace in a different suit. Now if the player you finesse against had the ace in the other suit, you play him for the queen (by playing low to the Jack). Otherwise, you play him for the Ace (by playing low to the Ace).

 

This is clearly wrong since even if there is an empty space argument giving the player you finesse against an above-50% chance of holding the ace, the same would apply to the queen.

[Endymion77]

Yes, this example is clearly wrong.

[karras166]

In 12 of those cases, A1 is dealt to the first player (the one who showed up with the ace). However, once he shows with the ace, 8 times his other card is x1 or x2 and only 4 times his other card is A2. So it's more likely that the ace is in the other player.

 

But surely in this four (relevant) card ending which you describe once RHO (say) has shown up with A1 LHO will have shown up with either x1 or x2 and so we are back to 50:50. So from what I can see this argument would only be relevant mid-trick which isn't of much use at all.

 

 

Nothing to do with the monty hall problem, though. That would entail a defender showing you a missing K and saying "still fancy the finesse?"

 

I suppose what reminded me of Monty Hall was that you start off with a statement "it's unlikely both aces are in the same hand" and when one Monty/your opps reveal the position of one of the aces, does the original probability still apply like in Monty Hall. (Again let me stress I am not a mathematician.)

 

 

And the answer is that it depends on the number of remaining cards in the opponents but RHO is always more likely to have it.

 

Again doesn't this only follow if you haven't seen the other opp follow with a small x?

 

 

FWIW the example that the book gives was

 

xx

 

opposite

 

KJx.

 

For completeness' sake I may as well post the whole problem as I think the authors have made not only a maths error but a bridge error too.

 

[hv=pc=n&s=sk32hak5432d32c32&n=sa4hqj76dkj4ck654]133|200[/hv]

 

So as Helene says the author suggests that an expert would locate the A♣ before playing on diamonds, playing whoever showed up with the ♣A not to also have the ♦A - apparently fallacious logic as discussed above. However it looks to me like the hand should be played by ruffing out spades, drawing trumps and playing to the JD endplaying East if he wins with the Q.

[helene_t]

I was a bit annoyed about this example when I read the book because it is actually an interesting hand so it's a shame that the "solution" is so flawed.

 

The endplay works if RHO has ♣A and ♦AQ, and not three trumps. Of course it also works if LHO has ♣A but in that case you are always safe. Alternatively you could play a low diamond immediately, hoping that LHO may step up with the ace if you play the suit before too much information is available.

[sfi]

If you truly wanted an answer to the question of where the missing card is, without taking into account the bidding or play to that point other than the card discovered, then you might as well have framed it where the 2 was missing.

 

That should make it clear that the vacant points argument is specious, without using information that would take into account the presence of the cards.

 

Added:

The other thing here is that the likelihood of the result depends on how many tricks have been played. At trick 13, the missing card is certain to turn up in the other hand. At trick 12 it is 2-1 to be in the other hand unless it would be limited by distribution at that point.

 

Actually, playing for the hand not holding a specific 2 to hold the important card works just as well. However, if you're going to do this you have to decide on your "discovery card" at the beginning of the hand for it to matter. Otherwise you're acting on information you already know.

 

And it really is a vacant spaces calculation. The hand with the previously chosen card only has a 48% chance to hold any other specific card if that is all the information you have on the hand. The only reason people suggest the aces are split is because people notice aces more than twos.

 

Of course it's rarely all the information you have on a hand so while true, it's not terribly valuable.

[Endymion77]

But surely in this four (relevant) card ending which you describe once RHO (say) has shown up with A1 LHO will have shown up with either x1 or x2 and so we are back to 50:50. So from what I can see this argument would only be relevant mid-trick which isn't of much use at all.

 

Not really, because the small card is irrelevant. You already know that he has at least one small card so which exactly small card he shows you doesn't matter (just like it doesn't matter which exactly door Monty will open to show you a booby prize, you already know that there's at least 1 remaining door with a booby prize).

[karras166]

Not really, because the small card is irrelevant. You already know that he has at least one small card so which exactly small card he shows you doesn't matter (just like it doesn't matter which exactly door Monty will open to show you a booby prize, you already know that there's at least 1 remaining door with a booby prize).

 

You say "8 times his other card is x1 or x2 and only 4 times his other card is A2. So it's more likely that the ace is in the other player." But when x1 or x2 is found with the other player that eliminates 4 possibilites so we are back to the first player having either A1 and x1 or A1 and A2. Sorry if I am missing something.

Edited September 30, 2013 by karras166

[Endymion77]

He either has x1/A2 or x2/A2 or x1/x2. If he has x1/A2 he plays x1 and if he has x2/A2 he plays x2 and if he has x1/x2 he plays a random card (50% chance to play x1 or x2). Right? So there's a 50% chance he'll show x1 and 50% chance he'll show x2 regardless of his second card. But 2 times out of 3, his second card is A2 and not the other "x" card, and it doesn't matter which "x" card he shows you - x1 or x2 (his choice might or might not have been restricted, but it's 50-50 to play either x1 or x2 at trick 1 assuming he never discards the missing ace of course).

 

Just like Monty Hall's doors, let's say you select door A. Then there are 3 possibilities:

 

- prize is behind door A. Then he opens door B or door C randomly (50% B, 50% C).

- prize is behind door B. Then he opens door C.

- prize is behind door C. Then he opens door B.

 

So there's a 50% chance he'll open door B, and 50% chance he'll open door C. But it doesn't matter which door he opens, you should switch (it doesn't matter which small card the defender plays, you should "switch" ie look for the A in his hand because he had 2 places to have it while the defender with the other ace had 1 place to have it).