[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]

FM75, on 2013-September-29, 16:05, said:

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]

karras166, on 2013-September-30, 05:06, said:

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]

Endymion77, on 2013-September-30, 06:17, said:

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.

This post has been edited by karras166: 2013-September-30, 08:42

[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).

[fromageGB]

helene_t, on 2013-September-30, 04:30, said:

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.

So you have demonstrated that the theory of vacant spaces says nothing in this situation. A non sequitur, as that is not the argument. The theory of vacant points does say something, and is as in the book. As Kenberg says, if you have no reason to think your specific card is one one side or the other, go with the theory. Why not?

The theory also applies to twos. If one hand shows up with a lot of twos in the play, if I was really bothered about it I would place the remaining two in the other hand. It hasn't caused a problem so far !

[helene_t]

fromageGB, on 2013-September-30, 09:04, said:

So you have demonstrated that the theory of vacant spaces says nothing in this situation. A non sequitur, as that is not the argument. The theory of vacant points does say something, and is as in the book.

Huh? Are you saying that if all we know is that LHO has ♣A then he is more likely to hold ♦Q than to hold ♦A? Because that is the argument in the book and that is wrong.

There is no such thing as a theory of vacant points, given that we have no information from the bidding.

If we were told that LHO passed as a dealer then Kerberg's argument apply. But OP said that we don't have any information from the bidding.

Quote

The theory also applies to twos.

What about Fibonnaci numbers? On the first two tricks LHO plays a 3 and an 8 while RHO plays a 4 and a 7. So I should play RHO for a 5 because he hasn't shown any Fibonacci numbers so far? OTOH, maybe I should play LHO for the five because he hasn't shown any numbers so far that don't contain a "t" in their English spelling?

[Endymion77]

fromageGB, on 2013-September-30, 09:04, said:

So you have demonstrated that the theory of vacant spaces says nothing in this situation. A non sequitur, as that is not the argument. The theory of vacant points does say something, and is as in the book. As Kenberg says, if you have no reason to think your specific card is one one side or the other, go with the theory. Why not?

There's no "theory of vacant hcp's", because from a probabalistic point of view there's no difference between a Q and A (or a 2 and an A for that matter) - they're equally likely to be dealt, and "high card points" is an arbitrary assignment made by the observer. The "theory of vacant hcp's" can only exist if we have extra information related to high card strength, e.g. someone bid when he could've passed with different cards, or passed when he could've bid.

[fromageGB]

No, Fibonacci numbers don't count. I've just invented the theory of vacant points, and if you want a theory of Fibonacci numbers that would be different, even though (perhaps) just as valid!