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play for 1 loser


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Maybe I've made the mistake of treating the 7 as equal to the J, 10 and 9 thinking that they could be played at random. However obviously if the 7 isn't played by north or south then the 8 is promoted. So, if the odds in the given situation are evens then it follows that if north had played the 7 on the first trick then the odds are strongly on the king being with south in order for the a priori odds of 3 to 2 to be maintained.

Basically, the situation is the following. We have to assume the suit splits 3=2. But we also get to ask a question - namely, "Who has the 7?" If South has the 7, the remaining cards split 2-2, and it's 50-50. If North has the 7, the remaining cards split 3-1, so it's 75% South has the king.

 

Obviously, all this ignores that if North has J9, he does not know in the first round that J9 are equal. See post #12 for that.

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