[inquiry]

When I posted this, I said assume West was good enough to play the Queen from Qxx on the second round of clubs. The reason being clear: it seems he has to have the club queen and if you play ♣AK and the Queen does not show up, the distribution will be 4=4=3=2, nuff said. The "good news" is that the majority of players will not be that good.

But let's start with clubs. In calculating the potential club combinations, we said opener would have two or three clubs, and we showed a calculation that indicated that the odds of Opener having 3 clubs was twice as likely as two clubs.There is a problem, adequately addressed by dellache. Here we will show, hopefully, why he was right.

If opener has to have the club queen, then that leaves only 7, no 8 clubs to be distributed between the two hands. So, we place the club queen in openers hand, and then us 7 as the number of unspecificed clubs. So instead of C(8,2) and C(8,3) to express the possible club holdings for opener, you have C(7,1) and C(7,2) plus the "known" club queen.....

C(8,2) is 28, C(8,3) is 56.. for a 2:1 ratio
C(7,1) is 7, C(7,2) is 21 for a 3:1 ratio

So by placing the known club QUEEN in openers hand you have greatly increased the likelyhood that he has a third club. This is good news if you are planning to ruff a club.

All the diamonds are known, and all the hearts are immaterial, but a similar calculation can be applied to spades, and here is where I have two question/considerations.

We know West has to have at least one spade honor (missing Queen might be possible if he opens 11 hcp junk, so the club queen above discussion assumes he has to have 12). With one known spade in his hand, the spade calculation becomes C(4,3) and C(4,2) instead of C(5,4) and C(5,3).

C(4,2) is 6, C(4,3) is 4
C(5,4) is 5, C(5,3) is 10

So without taking into account the "known' location of the spade queen, there are twice as many cases where opener has 3♠ than where he has 4♠. If you force opener to have the spade Queen, the 3 card spade suit is only a 3-to-2 favorite.

So working with these numbers, we find,
4=4=3=2 = C(4,3)*C(7,1) = 4*7 = 28
4=3=3=3 = C(4,3)*C(7,2) = 4*21 = 84
3=4=3=3 = C(4,2)*C(7,2) = 6*21 = 126

We can correct this by dividing all the numbers by the lowest common denominator (14) an the ratio is 2:6:9 with is exactly the 2/17, 6/17, and 9/17 posted very early by dellache. This is the actual real number as to which one is most likely.

Now for where I have a question, and one that might need a tiny bit of explainations. The question wasn't which was most likely, it was which one would you play opener for. This is a related question, but not exaclty the same question. Your contract is in basically no trouble if opener has only one top spade. So if you are considering the probability of the hands, you might want to take into account only possibilities where opener has both spade honors. If you do this, the term C(4,3) and C(4,2) become C(3,2) and C(3,1). It turns out that C(3,2) = C(3,1). This dramatically affects the odds. Let's look...

4=4=3=2 = C(3,2)*C(7,1) = 3*7 = 21
4=3=3=3 = C(3,2)*C(7,2) = 3*21 = 63
3=4=3=3 = C(3,1)*C(7,2) = 3 *21 = 63

This drops to 1:3:3, and suggest that 4 spades is equally likely to 4 hearts,,, well, not in "reality equal", but in conditions where it might make a difference (opener with both missing spade honors), the odds of 4=3=3=3 is equal to 3=4=3=3. That is why I place other in the poll... in that it really doesn't matter, you play for a 4333 (any reasonable 4 card major) and then I was going to show the line where you go ahead and ruff a club that was already shown.

How valid do you think it is to consider only the condition where opener has BOTH major honors in calcuating odds like this since if he has only one, there is no problem. I think the correct answer to the poll should be either 4=3=3=3 or "other" with the explaination the relavent 4333 are functionally the same chance. Probably the 2:6:9 ratio is actually correct, however, and it hopefully it was clear how those numbers were calculated.

Useful shortcuts here, was to eliminate like terms (diamonds same, so they are gone, and reciprical combinations are the same .. like C(3,1)=C(3,2) so they can cancel, or C(7,4)=C(7,3) for hearts so they cancel).

Other useful shortcut is how to cancel out like terms. In a simple one off calculation the combination equation is:
C(8,4) = 8!/(4!*(8!-4!)) = 8*7*6*5*4*3*2*1/(4*3*2*1)*(4*3*2*1).....

Since 4*3*2*1 is in the numerator and denominator, you can cancel them out, reducing too....
= 8*7*6*5/(4*3*2*1)

that is somewhat hard to deal with, you can cancel other terms, like six in top and 3 in bottom becomes 2 in top, and 4 in top and 8 in bottom a 2 in top, and the 2 in bottom with one of the new twos in top both drop out. leaving 2*7*5 = 70

However, if you compare a couple of like distributions, and you have C(8,4) in one and C(8,5) in another, you can use the 8,7 in one to cancel the 8,7 in another....so

C(8,4) = 8!/4!*4! = 8*7*6*5/4! = 8*7*6*5/4 = 5/4 = 1.25
C(8,5) = 8!/5!*3! = 8*7*6/3! = 8*7*6/1 = 1 = 1

The 3! cancels out in the numerator of both, leaving C(8,4) divided by 4, then the 8*7*6 cancel out, leaving 5/4 compared to 1

If you are curious, C(8,4) = 70, C(8,5) = 56, it should not surprize you that 70/56 = 1.25. I can't do 8 times 7 times 6 etc at the table, but I can visualize how to cancel out the terms and get a relative relationship.

[gszes]

a tremendous exercise in the use of mathematics. Now let us work on a different tack (a shortcut if you will).

we begin with the probability that the opps spades will break 32 just shy of 70% (after both follow to 3 rounds of diamonds.

Since there is nothing in the bidding or play thus far that prevents W from having the spade J and
even assuming it is 100% that the only viable distributions for east are 4432 4333 or 3433
the two combinations with e holding 4 spades are nowhere near the 70% probability.

If you are playing MP and looking for a top (or imps when trailing and needing to make up ground quick)
THEN it might be reasonable to look for the second best LOP. At IMPS and MP you will be much better served
taking the known 70% vs a much more speculative guess. You will not get in the papers this way but you will
spend much less mental energy looking for ways to avoid the obvious for no good reason.

[gnasher]

gszes, on 2011-February-17, 08:00, said:

a tremendous exercise in the use of mathematics. Now let us work on a different tack (a shortcut if you will).

So that we can arrive at the wrong answer?