karras166

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.

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

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

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.

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? 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.

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.