[ajfonty]

Let's imagine you're the declarer and dummy comes down. You have 6 ♣ cards and dummy has 3 ♣ cards.

4 ♣ cards are thus missing. From basic probabilities I've seen that there is a 50% chance that the remaining four cards break 3-1 between opponents, and 41% chance that they break 2-2.

But, I was taking a look at overall hand distribution probabilities, and saw that 6-3-2-2 distribution is much more likely than a 6-3-3-1 distribution (5.64% vs 3.65%). If I'm playing the hand, what probability do I rely on? I get the feeling that the missing card probabilities are more "accurate" and that I should play for the 3-1 break, but I'm interested as to why one methodology is more accurate than another.

Thanks in advance!

[Cyberyeti]

You use the first basic probability calculation.

If you know one of the opponents opened 2♥ so he has 6 of them, or you know the distribution of another suit from the play, you can adjust it for that, but not otherwise.

[PhilKing]

There are two ways that the suit can break 3-1. Anyway, by the time you have to decided what to do with this amazing stat, other things will change. For instance, say you are missing the queen and the first player follows small twice, you can eliminate all holdings except Qxx onside and Qx offside.

[gszes]

Apples to apples the generic concept that a 31 split is more likely than a 22 split (and by how much) is a generic approximation based on little information about the opps cards. The "generic" approximation becomes less and less "useful" (though it may by coincidence remain highly accurate) as we gather more and more information.
Your problem indicates you have a huge amount of information about 1 of your oppA (you also have a lot of specific information about oppB based on the information you gathered from oppA). Your decision should be based on the probabilities of only the remaining distributional chances and the generic approximation (at least in this case) is completely useless.

There are only ten card combinations left (lets assume the cards are ABCD) for every card combination oppA can have oppB has the alternate.
oppA / oppB can have the following holdings:

A BCD
B ACD
C ABD
D ABC
AB CD
AC BD
AD BC
BC AD
BD AC
CD AB

Note that the odds are 6 to 4 in favor of a 22 break (very far from the original estimation of 31 outnumbering 22 but sort of close to the 5.6 vs 3.6 of overall hand pattern chances. NOW look closely and you will see that both original suit calculations are completely useless and misleading at this point IF you are missing 1 key card that is finesseable (is that even a word?). Let us assume A is a missing Q in the key suit. It makes perfect sense to try and drop the singleton A from oppA since it is impossible for oppB to have a singleton but when you play the next card of the suit and oppB follows low it appears you are on a 5050 guess. It is probably best to play for a more even suit distribution if the opps bidding has been relatively tame and take the finesse if the opps bidding has been spirited other than that its a coin flip. Note how 50% is a far cry from either original suit break or hand pattern probability. Nothing takes the place of good detective work:))))))))))))))))))))))))))))) I hope this is what you were looking for because my fingers are tired.

[apollo1201]

I think you guys are slightly overdoing. The OP was wondering why, having a 6-3 trump suit, the 3-1 break was more likely, while for a hand, 6331 was less likely than 6322.

I guess it is some Baeysian thing, 6331 and 6322 are absolutes, while in the case you already know a hand has 3 cards (dummy). Or maybe the probabilities of the other 4 slightly differ between 54 and 63?

I suspect 1st but my mind started its holidays this evening while my body follows tomorrow only!

[Stephen Tu]

ajfonty, on 2015-August-03, 08:57, said:

4 ♣ cards are thus missing. From basic probabilities I've seen that there is a 50% chance that the remaining four cards break 3-1 between opponents, and 41% chance that they break 2-2.

But, I was taking a look at overall hand distribution probabilities, and saw that 6-3-2-2 distribution is much more likely than a 6-3-3-1 distribution (5.64% vs 3.65%). If I'm playing the hand, what probability do I rely on? I get the feeling that the missing card probabilities are more "accurate" and that I should play for the 3-1 break, but I'm interested as to why one methodology is more accurate than another.

Neither methodology is more accurate than the other. They give the same result, you just have to know what you are counting properly. Also it helps to use more accurate percentages.
Method 1, 2-2 break is 40.7%, 3-1 break is 49.74%. Ratio is 49.74:40.7 = 1.22:1

Method 2, from the tables I can find, 6322 distribution is 5.6425%. 6331 distribution is 3.4482% (not 3.65% as you claimed).
But giving you 6 cards, in only 1/3 of 6322 breaks is the 3 card fragment your partner, while for 6331 your partner has 3 cds 2/3 of the time. And only 1/4 of the time do you have 6 cards.
5.6425 * (1/3) * (1/4) = 0.4702
3.4482 * (2/3) * (1/4) = 0.5747
And thus you see the ratio of 3-1 vs. 2-2 is 0.5747:0.4702 = 1.22:1 same as method 1.

In math if two different methods are giving you different answers, usually you are doing something wrong.

Now as for how to use this in the play, that's another matter, like in dropping or finessing queens often by the critical juncture you have eliminated certain possibilities and/or the bidding/previous play have tilted the odds in some manner.

[ajfonty]

Thank you for the responses! The math makes a lot of sense. I guess in hindsight it's reasonable to deduce!

[Trick13]

Actually the more balanced distribution is always the most likely: a 2=2 split is 40.7% and a 3=1 is only 24.9% (but when we talk of 3-1 breaks we normally combine the probability of 3=1 and 1=3).

The 4-4-3-2 distribution is more likely than a 4-3-3-3 only because there are twelve six possible 4-4-3-2 distributions but only four 4-3-3-3.

Since there are the same number of 6-3-3-2 distributions as 6-3-3-1, the more balanced will be more frequent.