[HardVector]

I just had a hand come up that I have some questions on. The scenario is you are in a 3n contract with 9 top tricks, but it's MP so you are looking for all the overtricks you can get. There are no real finesse options, so it all comes down to setting up suits. You have 2 suits that are 7 card fits (both 4/3 splits). In both suits you have the A and K and pretty much nothing else (one suit has the T but not the 9). The odds favor 4/2 breaks in each suit individually, but what are the odds that one of the two suits will break 3/3? Is this also a situation in which if one suit breaks 3/3 it increases the likelihood of the other breaking 3/3 as well?

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.

[hrothgar]

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

I just had a hand come up that I have some questions on. The scenario is you are in a 3n contract with 9 top tricks, but it's MP so you are looking for all the overtricks you can get. There are no real finesse options, so it all comes down to setting up suits. You have 2 suits that are 7 card fits (both 4/3 splits). In both suits you have the A and K and pretty much nothing else (one suit has the T but not the 9). The odds favor 4/2 breaks in each suit individually, but what are the odds that one of the two suits will break 3/3? Is this also a situation in which if one suit breaks 3/3 it increases the likelihood of the other breaking 3/3 as well?

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)

[mikeh]

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

I just had a hand come up that I have some questions on. The scenario is you are in a 3n contract with 9 top tricks, but it's MP so you are looking for all the overtricks you can get. There are no real finesse options, so it all comes down to setting up suits. You have 2 suits that are 7 card fits (both 4/3 splits). In both suits you have the A and K and pretty much nothing else (one suit has the T but not the 9). The odds favor 4/2 breaks in each suit individually, but what are the odds that one of the two suits will break 3/3? Is this also a situation in which if one suit breaks 3/3 it increases the likelihood of the other breaking 3/3 as well?

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

[Stephen Tu]

If one suit known to split 3-3, odds of other suit also breaking 3-3 is 37.15% sans additional suit break info.

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

Pavlicek dual suit break calculator:
http://rpbridge.net/cgi-bin/xds1.pl

[TylerE]

Gotta consider the whole hand.

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.

[HardVector]

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

If one suit known to split 3-3, odds of other suit also known breaking 3-3 is 37.15% sans additional suit break info.

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.

[dsLawsd]

All excellent responses.
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/

[ncohen]

The probabilities might have presented an interesting choice. Suppose you duck once in each of those suits, and the defense plays another suit, where you can also duck. Do you do so? It wins if neither suit splits but there's a squeeze (a defender holds both suits), but loses in the actual situation, when both suits split 3-3. Then, Pavlicek's site says there's a 13.2% chance that both suits split 3-3 (if nothing is known), but roughly a 13% chance (I was too lazy to add up all the chances exactly) that a defender guards both suits -- at least 4 of both -- ie. one defender is 4-4, 5-4, 4-5 in the relevant suits. In this case, knowing something about the defenders' distributions would be very helpful.

[HardVector]

---

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.

[apollo1201]

Doesn’t the presence of the SJ make it more appealing to play on S (either SQ with S, or SQ stiff or 2nd in N, or 3-3 split)? Because seeing the C situation, it is unlikely we will be able to test both majors.

[Buty2008]

In Axx vs Kxxx (declarer vs dummy) the standard safe play to check opponents split (3-3 or 4-2) is to use blank trick (x from both hands) before using A, K and see if last x is the last in the suit. No risk to give a trick if opponents have 4-2 split in the suit. So just play x-x from in both suits than use AK to check the split in opponents hands. I suppose the hand has no problem to win the lead, play x to x in one suit, win next trick, play x to x in 2nd 4-3 suit and win next trick - you have 2 losers and time to check both 4-3 suits. The only risk is that opponents may develop a winner to be played at trick 5. I am not expert at table but I read a lot and solved problems, so I hope my post here answer pretty well to the question.
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.

[HardVector]

So, the way I played the hand, was after winning the first trick, I ducked a heart, south winning with the ten. Trick 3 was the J♥, K, low, low. I thought at the time that hearts were not breaking because it didn't make sense to continue hearts with 3 of them (as I noted, the defense was not great). After winning the K♥, I thought about playing 3 rounds of spades, but I didn't want to lose control of the spade suit if that wasn't splitting, so I ducked a spade, north winning. North finally found the correct defense and came back with the T♣. This is the point that generated the original question I had about the hand. How likely, at this point, am I going to get a 3/3 spit in either hearts or spades? The answer is about 58%, so I should play for that and go up with the A♣.

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.

[carmelbobc]

Here's a simple way in which to calculate odds like this. It is called the "indirect" method.

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.

[Buty2008]

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

So, the way I played the hand, was after winning the first trick, I ducked a heart, south winning with the ten. Trick 3 was the J♥, K, low, low. I thought at the time that hearts were not breaking because it didn't make sense to continue hearts with 3 of them (as I noted, the defense was not great). After winning the K♥, I thought about playing 3 rounds of spades, but I didn't want to lose control of the spade suit if that wasn't splitting, so I ducked a spade, north winning. North finally found the correct defense and came back with the T♣. This is the point that generated the original question I had about the hand. How likely, at this point, am I going to get a 3/3 spit in either hearts or spades? The answer is about 58%, so I should play for that and go up with the A♣.

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.