## Math question probabilities

### #1

Posted 2019-February-22, 13:17

I ask this because I carefully played the hand to make 10 tricks, but when it turned out both suits broke 3/3, I had 11 available. I could provide the hands if they are relevant, but I'm really more interested in the probabilities.

### #2

Posted 2019-February-22, 13:33

HardVector, on 2019-February-22, 13:17, said:

I ask this because I carefully played the hand to make 10 tricks, but when it turned out both suits broke 3/3, I had 11 available. I could provide the hands if they are relevant, but I'm really more interested in the probabilities.

This is the type of problem that simulations are very well suited for...

(Much easier than trying to figure out all the conditional probabilities)

### #3

Posted 2019-February-22, 13:51

HardVector, on 2019-February-22, 13:17, said:

I ask this because I carefully played the hand to make 10 tricks, but when it turned out both suits broke 3/3, I had 11 available. I could provide the hands if they are relevant, but I'm really more interested in the probabilities.

Obviously if one has AK10x opposite xxx or H10xx opposite Hxx, that is the suit to attack, since we gain not only on 3-3 but also when the suit is either QJ opposite xxxx or, more commonly, Hx behind the 10, so that we reduce to leading x towards 10x with Hx onside. I assume that your holding with the 10 was with the 10 in the 3 card holding. Of course that is still the suit to tackle, all other things being equal, since one may find the QJ tight (but one would know if this were possible after just one round of the suit..assuming that the overall hand suggested that one could lead a top card initially, which it sounds as if one could.

As for the odds, the 'book' odds may not help much, because one almost always knows or can infer quite a bit about the opponents' shape outside of these suits. Not always, of course, but even if one can't test other suits first, they had to lead something and follow suit, etc, and often one has inferences.

Plus, on some hands, one would be catering to an opp being 4=4 in our two suits, and playing for a squeeze, and so on. I tend not to worry to much about a priori odds on most hands....at the table one usually has sufficient information to render a priori odds of little assistance

*Johann Hari*

### #4

Posted 2019-February-22, 14:21

Odds *at least one* of two suits breaking 3-3 = ~57.86%

Pavlicek dual suit break calculator:

http://rpbridge.net/cgi-bin/xds1.pl

### #5

Posted 2019-February-22, 15:41

Opponents silent in bidding? That probably rules out any 7 card or longer suits, and good 6 card suits, and 5 card suits with an opening hand. That will shift the percentages a bit.

### #6

Posted 2019-February-22, 18:57

Stephen Tu, on 2019-February-22, 14:21, said:

Odds *at least one* of two suits breaking 3-3 = ~57.86%

Pavlicek dual suit break calculator:

http://rpbridge.net/cgi-bin/xds1.pl

Thanks, that's what I was looking for.

### #7

Posted 2019-February-23, 03:44

You may be interested in getting Glauert's excellent book on combining chances and calculating the probabilities- I cannot remember the exact title as I loaned it to a partner long ago.

Jeff Ruben's book would be in that category also I expect/

### #8

Posted 2019-February-23, 12:17

### #9

Posted 2019-February-23, 12:25

So, here is the hand for anyone interested. The bidding isn't what happened, just where we ended up. The lead was the 5♦, low, 9, J. I had forgotten when I made the first post here that I actually had 3 7 card 4/3 fits. On this one, I assumed the diamonds were breaking 4/2 or 5/1 based on the lead.

### #10

Posted 2019-February-23, 13:42

### #11

Posted 2019-February-24, 00:28

The real hand has some problems. The 9 tricks are based on LHO having DQ - 4 in D with finesse, + 5 top cards in the other 3 suits. SJ make proper to test H split but a club return next release a trick (or more) for opponents. Would be nice that small to small trick in heart to be given to LHO (if lead was not in clubs will have a reason to refrain to do it now). Spades can give 3 tricks not only for 3-3. If opponents have Qx vs T9xx proper play is A,K,2 to J and standard play fails. It is not just math for two 4-3 suits to be in opponents hands 3-3 or 4-2.

### #12

Posted 2019-February-24, 16:58

In the actual play, I played low setting up a potential double squeeze if neither suit broke. When it turned out both suits broke 3/3, I regretted the club duck and wondered if it was correct...it was not.

### #13

Posted 2019-February-24, 20:36

You start with the probability that the suit will break 3-3, which is 0.355. The probability that it WON'T break 3-3 is 1.000 - 0.355, or 0.645. This probability includes it breaking 4-2, 5-1, or 6-0.

The probability that BOTH suits do not break 3-3 is 0.645 x 0.645, or 0.416.

So the probability that AT LEAST ONE SUIT will break 3-3 is 1.00 - 0.416, or 0.584.

In your case, with three 7-card suits, the probability of at least one of them breaking 3-3 is 1.00 - (0.645 x 0.645 x 0.645) or 0.732.

### #14

Posted 2019-February-25, 02:05

HardVector, on 2019-February-24, 16:58, said:

In the actual play, I played low setting up a potential double squeeze if neither suit broke. When it turned out both suits broke 3/3, I regretted the club duck and wondered if it was correct...it was not.

Your play missed the chance to find Qx in spades in any of opponents hand. At trick 4 small spade to dummy's A and duck a spade from dummy could reveal spade Q at trick 5. If so 10 tricks are secured and club return can be taken by dummy's A. 4th winners in diamonds and spades are a potential squeeze for 11th trick. If not, there is one more chance that south win trick 5 an chance to play for 3-3 in both majors is still there.