I've managed to still play a lot of bridge in the past week (210 hands and counting since Monday) despite working 9-630 with rehearsal 730-10 all week. So, I should have some hands to write about. And I've done practically nothing today except mow the lawn. I needed the day of rest. I don't get another one for weeks. Anyway, there are several interesting hands I will write about later, but for now a suit combination.
You're in 6S and your trump suit is Q97632 opposite AT85. You have all the side suit tricks so you just need to play spades for 1 loser. How do you do this? Do you make the same play at imps and matchpoints?
Playing low to the Q first is a safety play that guarantees the contract on any lie of the cards, but it also gives up the chance of an overtrick. If KJx is in front of the Q, the Q will win a trick. If KJx is behind the Q, you can finesse for the jack later.
Playing the A first goes down when KJx is behind the Q, an 11% chance. But it gains an overtrick when the K in singleton in either hand, a 26% chance.
At imps, the overtrick is worth 1 imp but going down costs 13 if non-vulnerable and 17 if vulnerable, so the odds overwhelmingly favor making the safety play. In matchpoints, it depends on how many pairs you think will be in the slam and how many people in the field you think will be in the slam. Most of the time, though, the odds still probably favor the safety play, especially if it's not an easy slam to reach.
How about leading toward the Ace? Play the ace if lho shows out, cover his card if he follows suit. You will never lose two tricks this way and you will lose no tricks if
ReplyDeletelho has the stiff king.
I suppose that is correct :(. An overtrick 13% of the time and making all the other times. As long as you don't lay down the ace first or try leading the Q.
ReplyDeleteit would probably be a good idea for me to not write in such a hurry, while doing other things. eh.
ReplyDeleteWell, I wrote mine while watching for my luggage at the carousel at Hartsfield. BTW, only a mathematician would care, but isn't the probability of a specific opponent having a stiff king 12.5% so the probability of either opponent having it is 25%?
ReplyDeletethe odds tables say a 2-1 split is 78% and the probability of the 1 being the K 1/3 so 26%. low to to the 10 still loses a trick to half of those.
ReplyDeleteAh, that was the danger in calculating the probabilities at the carousel. I was just taking the number of possible holdings LHO could have in the trump suit (8) and dividing by the number of these holdings that were the stiff king (1). When I saw what the tables said, I then had to figure out how they came up with those percentages. The probability that LHO has a singleton trump is (the number of different singleton trump holding he can have)X(the number of ways he can hold 12 of the other 23 opponents' cards)divided by the number of ways he can hold 13 of the 26 opponents' cards) or 3C1 X 23C12/26C13. This isn't quite as easy to calculate in my head as 1/8 was, but my calculator did tell me the result was .39. Thus the probability of a 2-1 split is twice that, or .78.
ReplyDeleteDid someone say mathematician? I am still that, for another few weeks.
ReplyDeleteAlternate way to do things: think of dealing cards one-by-one; the probability of the next card landing in a hand is equal to (# of spaces in that hand)/(total # of spaces). Probability that all three cards are in the same hand: 12/25 x 11/24 = exactly 22%. I like this way because it's just what you'd do if each hand had infinitely many cards (it'd be 1/2 x 1/2, as in Emory's second post), with each fraction just tweaked a little to account for the finiteness of the vacant spaces.
With a 2-1 split, the formula is 13/26 x 12/25 x 13/24 x 3C1 x 2 = exactly 78%. The 3C1 is because the product of the fractions is simply the probability that lefty gets cards 1 and 2 and righty gets card 3, but there are 3C1 choices of labellings that lead to different holdings. And the 2 is because we can switch lefty and righty. (This comes in when the splits are uneven.)
This generalizes completely: say to compute a 3-3 split, we get
13/26 x 12/25 x 11/24 x 13/23 x 12/22 x 11/21 x 6C3 = 286/805 = 35.53%. (Note no factor of 2.) Or a 4-2: 13/26 x 12/25 x 11/24 x 10/23 x 13/22 x 12/21 x 6C2 x 2 = 48.45%.
I recommend a little old book called "Bridge Odds for Practical Players" by Kelsey and Glauert. More restricted choice and vacant spaces than you can shake a stick at.
Also the following article that I just found online: Some Remarks about Bridge Probabilities, P. F. Zweifel, Mathematics Magazine, Vol. 59, No. 3 (Jun., 1986), pp. 153-157
ReplyDeletehttp://www.jstor.org/stable/2690205?seq=1
Patrick,
ReplyDeleteWhy will you no longer be a mathematician in a few weeks?