[hrussin]

I have been reading Root's How to Play a Bridge Hand, and I am confused by his description of this hand (p10, in the section "To Finesse, or to Play for the Drop"):

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Root writes: "You cash the ace and king and the jack does not fall. Then you lead toward dummy's queen-ten and West follows with a low card. Since the opponents began with only six cards, the odds are slightly in favor that East has the jack; so playing the queen offers a better chance than the finesse."

Why are the odds slightly in favor that East has the jack?

Is it because, if you West plays low, there are now 5 outstanding cards, of which one must be the jack? But, not knowing the distribution, how can we say the odds are still over 50% for East to have the jack?

[apollo1201]

Suits breaking 4-2 is more common than 3-3. However, now you know West can follow 3 times, the a priori 4-2 split odds have to be revised: it can only be 4W+2E and not 2W+4E. And 3-3 is more likely than J 4th in W’s hand and low doubleton to E.
The same principle says for the Q to play the drop w/ 9 cards (the famous nine never).
But in both cases, finesse is only slightly inferior in terms of odds of success. Other information on opponents’ hands (bidding, known lengths or shortages in other suits) might make you decide to play differently.

[steve2005]

This is a combination where you are likely to have more information than others. But is still likely to be close.

[Stephen Tu]

Basically you are comparing how often the suit is dealt:
xxx Jxx, where you have to play for the drop
vs.
Jxxx xx, where you need to finesse.

xxx Jxx is half of the 3-3 splits, which is 0.5 * 35.528% or 17.764% of all possible deals a priori (many of which have now been eliminated with both players following twice and LHO now following a third time). Why this number? There are 26c13 = 26!/(13!*13!) = 10400600 total ways to deal the 26 cards we don't see to the opponents. If the suit is split xxx jxx There are 5c2 ways to deal the small spades that are not the J to the opponents. Also given 3-3 spades there are 20c10 ways to deal the 20 non-spade cards to the opponents. So overall xxx jxx is:
(5c2*20c10)/26c13= 10*184756/10400600 = 17.764%

Jxxx xx, there are again 5c2 ways to deal the small spades. But now it's 20c9 ways to deal the other cards, so the formula is:
(5c2*20c9)/26c13 = 10*167960/10400600 = 16.149%
This is equal to 2/3 of the half of 4-2 splits which split with LHO having 4, which is 2/3 * 1/2 * 48.447% = 16.149%; J is in the 4 cd holding 4/6 = 2/3 of the time.

Another way to think of it is a "vacant spaces" argument. After LHO follows 3 times low, there are 10 slots for LHO to have the J, but 11 slots for RHO to have the J. So it should be 11:10 that RHO has it. And indeed 11/10 = 1.1:1, 17.764:16.149 = 1.1:1. So basically the drop will work 11/21 or 52.38% at the decision point, having eliminated all but two possibilities.

Note that this only holds if you have no other information about the hand. If RHO in the bidding or the play shows up with a longer side suit than LHO, not balanced out by a countering imbalance in another, then this will tilt the odds back in favor of the finesse.

[hrussin]

Stephen Tu, on 2018-February-19, 18:51, said:

Basically you are comparing how often the suit is dealt:
xxx Jxx, where you have to play for the drop
vs.
Jxxx xx, where you need to finesse.

xxx Jxx is half of the 3-3 splits, which is 0.5 * 35.528% or 17.764% of all possible deals a priori (many of which have now been eliminated with both players following twice and LHO now following a third time). Why this number? There are 26c13 = 26!/(13!*13!) = 10400600 total ways to deal the 26 cards we don't see to the opponents. If the suit is split xxx jxx There are 5c2 ways to deal the small spades that are not the J to the opponents. Also given 3-3 spades there are 20c10 ways to deal the 20 non-spade cards to the opponents. So overall xxx jxx is:
(5c2*20c10)/26c13= 10*184756/10400600 = 17.764%

Jxxx xx, there are again 5c2 ways to deal the small spades. But now it's 20c9 ways to deal the other cards, so the formula is:
(5c2*20c9)/26c13 = 10*167960/10400600 = 16.149%
This is equal to 2/3 of the half of 4-2 splits which split with LHO having 4, which is 2/3 * 1/2 * 48.447% = 16.149%; J is in the 4 cd holding 4/6 = 2/3 of the time.

Another way to think of it is a "vacant spaces" argument. After LHO follows 3 times low, there are 10 slots for LHO to have the J, but 11 slots for RHO to have the J. So it should be 11:10 that RHO has it. And indeed 11/10 = 1.1:1, 17.764:16.149 = 1.1:1. So basically the drop will work 11/21 or 52.38% at the decision point, having eliminated all but two possibilities.

Note that this only holds if you have no other information about the hand. If RHO in the bidding or the play shows up with a longer side suit than LHO, not balanced out by a countering imbalance in another, then this will tilt the odds back in favor of the finesse.

Thanks for the detailed response! Very helpful.