[dickiegera]

I know that it is said that with 9 in a suit you play to drop a missing Queen holding the Ace and King.

I also know that missing 4 cards in the suit that they are more likely to be 3-1 rather than 2-2.

Can someone explain the math on playing for the drop.

Thank you

[manudude03]

First of all, it is very close whether to finesse or play for the drop.

As for the maths, to deal with round numbers, a 2-2 break is 40% while a 3-1 break either way is 50%. However, the key point is that the 3-1 is either way. Once LHO follows to the second round, he can no longer have started with 1 so the appropriate odds are 40:25 for drop/finesse. Yes, you can add 12% to either option for stiff Q.

[awm]

manudude03, on 2012-July-01, 14:01, said:

First of all, it is very close whether to finesse or play for the drop.

As for the maths, to deal with round numbers, a 2-2 break is 40% while a 3-1 break either way is 50%. However, the key point is that the 3-1 is either way. Once LHO follows to the second round, he can no longer have started with 1 so the appropriate odds are 40:25 for drop/finesse. Yes, you can add 12% to either option for stiff Q.

This is not quite right and the odds are not nearly so far.

Originally there were six possible 2-2 breaks (four choose two is six) and eight possible 3-1 breaks (four singletons, and the singleton can be on either side). 2-2 break is 40.7% and 3-1 break is 49.7%, so if we divide we get that each particular 2-2 break is 40.7/6 = 6.78% and each particular 3-1 break is 49.7/8 = 6.21%. Now that we have seen three cards at the point where we make the decision, only a single 2-2 break remains possible (queen in the fourth hand) and only a single 3-1 break remains possible (queen in the second hand). Thus the odds favor the drop by 6.78:6.21 which is just a little better than even odds.

If you know something else about the hand (for example, regarding distribution in the other suits) it can easily swing things in favor of taking the finesse.

[FrancesHinden]

dickiegera, on 2012-July-01, 13:54, said:

I know that it is said that with 9 in a suit you play to drop a missing Queen holding the Ace and King.

I also know that missing 4 cards in the suit that they are more likely to be 3-1 rather than 2-2.

Can someone explain the math on playing for the drop.

Thank you

Without worrying about the exact numbers, you reconcile the apparent contraction by noticing that although they are more likely to be 3-1 either way round than 2-2, when LHO follows to the second round, he can't have a singleton so you are now comparing 3-1 with length on one particular side, with 2-2. The former is less likely.

[AlexJonson]

I like the vacant spaces approach. When opponent A showed two of the small cards and B showed one, A has eleven spaces and B has twelve, so play to drop the queen from B.

As has been Implied if you know anything about the distribution of another suit, you can adjust the 11:12 spaces and decide.

[JLOGIC]

dickiegera, on 2012-July-01, 13:54, said:

I know that it is said that with 9 in a suit you play to drop a missing Queen holding the Ace and King.

I also know that missing 4 cards in the suit that they are more likely to be 3-1 rather than 2-2.

Can someone explain the math on playing for the drop.

Thank you

a specific 3-1 is less likely than a specific 2-2. For instance, West having 3 and East having 1 might be 25 %, 2-2 might be 40 %, and thus east having 3 and west having 1 is also 25 %.

Here, you would be playing for 1 person specifically to have 3 which would be less likely than that player having 2 given those numbers.

[chasetb]

awm nailed it, but his explanation is a tad overdone. The key is the fact that when the suit divides Qxx - x , the Queen will be tripleton 37.3% Just because you find out that one opponent has 2 Diamonds, the 2-2 break is STILL 40.7% (this is all 'a priori'). So the Drop is the slight favorite, but other factors of the hand can override that, like knowing that RHO has 3+ more cards accounted for.

[TWO4BRIDGE]

dickiegera, on 2012-July-01, 13:54, said:

I know that it is said that with 9 in a suit you play to drop a missing Queen holding the Ace and King.

I also know that missing 4 cards in the suit that they are more likely to be 3-1 rather than 2-2.

Can someone explain the math on playing for the drop.

Thank you

With no inferences, the Bridge Encyclopedia states that " the finesse is a 50% probability of success holding 8 cards, while the drop has a 53% holding 9 cards ".

[Statto]

The odds still favour the drop if you start with the high card on the 2nd round. The crucial factor is that the Queen is a significant card, so the 3-1 splits with a stiff Q have been eliminated (as well as, obviously, the 4-0 splits). On the other hand, if you had all the top cards, after 1 round of the suit it would still be more likely to split 3-1, as only the 4-0 splits have been eliminated - nothing can be eliminated from the specific cards played by opps, as they were effectively played at random.

[Yu18772]

You can also look at it from free space perspective:
Once you played 2 round and the opponent follows low, the Q is the only missing card. The opponent that already played low card has 1 space less than the opponent that did not play to hold that card. The difference in % is so small though that if you have any indication about the Q from bidding or play you should finesse.
Yu

[Statto]

But beware of occupying spaces in opponents' hands when you shouldn't. For example, you have AQxx in a suit opposite Kxx. When you have played 2 rounds in the suit, free spaces suggest it is now more likely to split 3-3 than 4-2. But that is not the case. You have eliminated 6-0 and 5-1 splits, and that's it. It's still more likely to be breaking 4-2, absent other information. You can read nothing more into the fact that opps have followed to 2 rounds. On every 4-2 split they would have done that anyway, and you have eliminated none of them.