[cherdano]

 GrahamJson, on 2017-June-02, 08:25, said:

The situation we are catering four is when west has three cards and east two. In that case the odds are three to two that the king is with west.

Nope. The situation we are catering for is when West has 3 cards, including the 7.
The point is - the opponents have to reveal who has the 7.

[GrahamJson]

 cherdano, on 2017-June-02, 10:02, said:

Nope. The situation we are catering for is when West has 3 cards, including the 7.
The point is - the opponents have to reveal who has the 7.

Suppose S had played 9 and J, and N the 7, what difference does that make? Now the 10 and the K are outstanding and S could have either. Is it even money that the king is with north and will fall?

Put it another way. The initial odds, assuming a 3-2 break, are 3/2 that the king is with south. Now what cards played by north and south will keep those odds at 3/2? You will always end up with the king and one other card outstanding, either of which could be with south, or north. Ok, they can't play J, 10 and 9 on the first two tricks, the 7 will have to be played by north or south so does it matter who plays it?

[cherdano]

 GrahamJson, on 2017-June-02, 10:16, said:

Suppose S had played 9 and J, and N the 7, what difference does that make? Now the 10 and the K are outstanding and S could have either. Is it even money that the king is with north and will fall?

This is my last try.

If S plays J and 9 and N plays the 7, we are comparing Kj9-T7 versus JT9-K7. But in the former case, all plays were forced, whereas in the latter case S had 3 choices (play T9 or J9 or JT). So now it is actually 75% that S has the King.

You cannot treat JT97 all as equal.

[MrAce]

 GrahamJson, on 2017-June-02, 09:59, said:

That makes no difference.

Take another situation. Suppose it was AQxxx opposite xxxxx. The odds are 2/1 in favour of the finesse. Does this change when you lead up to the AQ and see the card that LHO plays? The king could be the unknown card on your left or the unknown card on your right, but it is not even money.

When you hold AQxxx vs xxxxx, in order to make 5 tricks you need 2-1 split. As you said the finesse wins when KJ or KT onside (13% X 2 =26) vs playing the A which wins only when JT-K (13 % only)
In our specific case, when you play the 2nd ♥ and see S with 7 and J there are only 2 possibilities left that matters (excluding 5-0 and 4-1 breaks since we both fail on them) As oppose to your example where there are 3 possibilities which matters)
At this point S has either JT7 or KJ7 and these holdings have EXACTLY the same odds. Can you see the difference now between the example you gave and the one we have in hand?

[GrahamJson]

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.

[cherdano]

 GrahamJson, on 2017-June-02, 15:02, said:

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.