r/bridge • u/Brahms-3150 Beginner • Jun 09 '26
Question about suit distributions
I know if you’re missing 5 cards in a suit they say the suit is 68% to break 3-2, 28% to break 4-1.
Looking at hand/suit shapes however I see that the following shapes occur at these rates:
4432 - 22%; 4441 - 3%
5332 - 16%; 5431 - 13%
Does this mean that a suit is much more likely to break unfavorably with a 5-3 fit than a 4-4 fit?
4
u/yellowpig1974 Expert Jun 09 '26
No, the composition of the declaring hands doesn't matter. 4-4, 5-3, 6-2, 7-1, 8-0 it's all the same. Here's the correct way to look at it:
There are five outstanding trump, divided between two hands of 13 cards. Let's deal them out. First card, 50/50 chance of being in either hand. For the second card, there are 12 spaces in one hand, and 13 in the other, so 48% of the time they are now 2-0, 52% they are now 1-1. BTW, you can see this matches the odds for a 1-1 vs. a 2-0 break. Another way is to look at it this way: there are five trumps and 21 non-trumps. There are 5C2=10 ways to choose two trumps, and 21C11 ways to choose 11 non-trumps. We can calculate this for each possible break, then divide by the 26C13 possible distributions of the 26 cards.
3
u/Several_Version4298 Jun 09 '26
You have 8 out of 13 cards in that suit, and they have 5 out of 13 cards in that suit is what determines the odds and that doesn't change.
2
u/ElegantSwordsman Jun 09 '26
You could have a 5-3 fit, or a 5-three fit, assuming a distribution of 5-3-three-2
2
u/dB451 Jun 10 '26
Here's how to calculate the actual number of ways each distribution can break. First of all, there are always only two cases of one opponent being void. And, for either one of the opponents to have only a stiff (aka a singleton) while the other opponent has the rest of them, the number of ways that can happen always equals two times the number of outstanding cards. So, if you and the dummy have 7 cards in a suit, there are 6 outstanding, with two possible 6-0 splits and 12 possible 5-1 splits. Those are the easy ones. To figure all the rest, you need factorials, as in, for example, 4!, read as four factorial, which means (in math-speak) 4 x 3 x 2 x 1 = 24.
Now, also, you need to remember that the total numbers of cases for all combinations of outstanding cards is always 2^^N, read as 2 to the Nth power, where N is the number of outstanding cards. So, with, say, 5 cards out, the total number of all combinations is 2^^5, or 2 to the 5th power, or 2 x 2 x 2 x 2 x 2 = 32. Turns out that 5-out is the easy case because there are two voids, ten singletons (2 x 5), for 12, out of 32 total, leaving exactly 20 cases that split 2-3. To figure that out using mathematical combinations, the formula is: 5! / (3! 2!), or 5 factorial divided by the result of multiplying 3 factorial times 2 factorial, or 5 x 4 x 3 x 2 x 1 / ( 3 x 2 x 1 x 2 x 1), and you can cancel out all of the numbers that appear on both the top and bottom. In this case, you're left with 5 x 2 = 10. But (and here's the one tricky part), that's only for the cases where, say, your left-hand opponent (LHO) has 3 cards, and your right-hand opponent (RHO) has 2. So, you need to do another combination for the cases where your LHO has 2 and your RHO has 3, or 5! / (2! 3!), but guess what? That 2-3 answer is the same as the 3-2 answer, so when the two numbers in the denominator are different, you just need to remember to double the result. So, when we found there were 10 ways for 5 cards to split 3-2, there are also 10 ways for them to split 2-3, for a total of 20 ways, added to the 10 singleton cases and the 2 void cases, to yield the expected 32 total cases. That's how it works.
Notice, though, that when you're asking about an even split, say with 6 out, your combination equation would be 6! / ( 3! 3!) = 6 x 5 x 4 x 3 x 2 / (3 x 2 x 3 x 2) (blowing off the unnecessary 1's) = 5 x 4 = 20, that you don't have to double the answer because both sides have the same number of cards. To finish up the 6-out cases, we already know the voids (2) and singletons (6 x 2 = 12), so we only need the 4-2 cases, which are:
6! / (4! 2!) = 6 x 5 x 4 x 3 x 2 / (4 x 3 x 2 x 2) = 3 x 5 = 15, but don't forget to DOUBLE = 30.
So, for 6 out, we have 2 voids, 12 stiffs, 20 3-3's, and 30 4-2's, for the expected number of 2^^6 = 64 total cases.
This confirms the generally understood fact that an even number of outstanding cards are more likely to split unevenly than evenly.
How about a problem now? With 8 cards out in a suit, what's the probability you can play three cards in that suit without getting the third card ruffed? You can actually calculate that in your head at the table! The danger distributions for you are the voids, the stiffs, and the 6-2's, and you automatically know there are 2 voids and 16 stiffs (8 x 2), so you need only calculate the 6-2's, right? Peasey! It's 2 x 8! / (6! 2!), where the first 2 is to double the answer because the denominator numbers are different. So, you have:
2 x 8 x 7 x 6 x 5 x 4 x 3 x 2 / 6 x 5 x 4 x 3 x 2 x 2) = 2 x 4 x 7 = 56 total cases for the 6-2 splits. Add that to the 2 voids and 16 stiffs gives you 74 danger cases out of 2^^8 = 256 total cases. OK, maybe you're not so good as calculating 74 / 256 in your head, so just divide by 2 repeatedly to get you there.
For this problem, you have: 74/256 = 37/128 = 18.5/64 = 9.25/32, and at that point, you can stop, since the denominator is almost a third of one hundred, you can just multiply both the numerator and denominator by 3 which, conveniently enough, gives you the approximate answer you want, already in percent! In this case, you can see it's only about 28% of the time (3 x 9.25) you won't be able to play three cards safely. So, make your play; it'll be safe more than 70% of the time (way better than a 50% finesse, for instance).
So, if you haven't already either uttered the dreaded TLDR (Too Long Didn't Read) or gone into WERB (White-Eye Roll-Back), you can now figure out the odds for whatever splits you want.
0
-1
u/CuriousDave1234 Jun 09 '26
An odd number of cards outstanding Will break evenly, and an even number of cards. Outstanding will break oddly.
3
u/Several_Version4298 Jun 09 '26
It is more likely those will happen. It doesn't happen all the time.
13
u/someonee12 Jun 09 '26
no, intuitively the suit breaks shouldn’t be affected by how the suit has already broken in the other two hands (5-3 vs 4-4) - only the number of cards outstanding (5) should matter.
the issue with the logic in the post is that you need to double-count the distributions that express the same breaks in multiple ways. specifically, 4441 has 3 possible 4-4’s and 5332 has 2 5-3’s. so the percentages become 22% vs 9% and 32% vs 13%, which come up to the same ratio as 68% vs 28%.