[inquiry]

Matchpoints.

West leads a diamond and it goes AK and another diamond.. West follows to all three rounds.

They play Standard american, but will open better minor with 3 in each, but always open their longer minor if they have one.

Lets assume that EAST is good enough to drop the club Queen from Qxx if you cash the AK (since you assume he has the queen for his opening bid, one that would not be that good, playing AK would in theory give you a fairly accurate count.

There are three lines of play....

Play one, play two rounnds of turmps, make if 3=2 or East has stiff honor.

Play two, one round of trumps, cash two clubs, ruff fourth heart in dummy and lead low spade.

Play three, ruff a club in dummy, cash three roufnds of hearts, one top spade, then lead a low spade from dummy.

Each line deals with one of the distributions shown in the poll. Which distribution has the highest probability assuming opener does nto have four clubs or a five card major?

---

It would be most helpful if you state what you think the relative odds of each holding was as a percentage or a fraction or something -- perhaps showing your math as well. I think I have the right answer. There are some shortcuts to getting to the correct solution, I can think of a couple. If you used a short cut or some "trick" to get teh right answer what was it. Paraprhasing the words of Rubens, watch for the evilness of The Great Dealer.

[gnasher]

I'm confident that 3433 is the most likely of the three. Look at the suit breaks in each case:

4333: 4-1, 4-3, 3-3, 5-3
4432: 4-1, 4-3, 3-3, 6-2
3433: 3-2, 4-3, 3-3, 5-3

A specific even distribution is more likely than a specific uneven distribution. There are also more 3-2 breaks than 4-1 breaks, and more 5-3 breaks than 6-2 breaks. Therefore

3433 > 4333 > 4432

I haven't looked at the lines of play that you suggest.

This post has been edited by gnasher: 2011-February-14, 12:23

[gnasher]

There's a line that works against both 4333 and 3433: ♠A, ♣AK, club ruff, four rounds of hearts. If LHO follows, ruff with the 10, endplaying East.

[nigel_k]

Line 1 appears to need East to be 3433. But line 2 works if East is 3433 or 4432 and line 3 works if East is 3433 or 4333. So in fact we can disregard the probability of 3433 as we are always making on that layout. It is just 4432 vs 4333. I would be inclined to favour 4333 as it has more permutations of spot cards, but only if I place a lot of faith in their always opening the better minor.

[ceeb]

Since it costs nothing to cash the ♠A early we always make when West has a singleton trump honor (as well as any doubleton). After ♣,♣,♠,♥,♥,♥, both ♥ lies are probably (if the ♥ spots are meaningless) equally likely, and we are advised that RHO plays the ♣Q and a ♣x. Hence the dealing probabilities are in the ratio of RHO being dealt ♣x vs. ♣xx -- (7 choose 2):(7 choose 1) = 3:1 favoring ♣ play next if we accept the stipulations given.

However, notwithstanding that it "should" be easy for RHO to find the play of the ♣Q from Qxx, real life is not perfect. This adds to Nigel's wonder about how rigid are the opponents' opening bid rules. These considerations tend toward justifying playing a 4th ♥.

Some 3♠ bid!

[dellache]

I'm not sure that counting card combinations like gnasher did is enough, because East has got to have 12-13 HCP, and that has an influence on the whole distribution probabilities.
BTW, you didnot say who has the ♦J, important card.

CASE 1: West had ♦J

As East is balanced, he has got all the missing honor cards (12 HCP), and the combinations have to be made on the 'x' slots.
WHOLE EAST DIST --- SLOTS E ------ SLOTS W
QJxx xxx AKx Qxx -- xx xxx . xx -- x xxxx xxx xxxxx
QJxx xxxx AKx Qx -- xx xxxx . x -- x xxx xxx xxxxxx
QJx xxxx AKx Qxx -- x xxxx . xx -- xx xxx xxx xxxxx

If you count the combinations on the major suits only regarding the 'x' cards, you get the same for the 3 distributions.
Only the club suit leads to a difference, and we have 4333=3433 > 4432.
The ratio 4432/3433 = B(7,1)/B(7,2) = 1/3.
Hence 4333=3433=3/7, and 4432=1/7.

CASE 2 : East had ♦J.

Now all the honors are known except the place of the ♠J who acts "as a small card".
The table has become :
WHOLE EAST DIST --- SLOTS E ------- SLOTS W
Qxxx xxx AKJ Qxx -- xxx xxx x xx -- x xxxx xx xxxxx
Qxxx xxxx AKJ Qx -- xxx xxxx x x -- x xxx xx xxxxxx
Qxx xxxx AKJ Qxx -- xx xxxx x xx -- xx xxx xx xxxxx

We now clearly have 3433 > 4333 > 4432.
4333/3433 = B(4,1)/B(4,2)= 2/3
4432/4333 = 1/3.
Looks like (3433,4333,4432) = (9/17,6/17,2/17) now [edited]

All these figures don't tell us what to do in reality if the ♣Queen falls on the King.
[edited in red on feb the 15th]

[AlexJonson]

Lot of comments and few votes, as I look at it.

[Bbradley62]

By using a simple "choose 10 random cards from a pool of 20 specific cards" and including only the three distributions on your list, I get:
P(4333)=3/11
P(4432)=2/11
P(3433)=6/11
Breaking down these numbers to distinguish between honors and not:
QJxx, xxx, AKx, Qxx = 6.1%
Qxxx, xxx, AKx, Qxx = 2.0%
Jxxx, xxx, AKx, Qxx = 2.0%
QJxx, xxx, AKx, xxx = 10.2%
Qxxx, xxx, AKx, xxx = 3.4%
Jxxx, xxx, AKx, xxx = 3.4%
QJxx, xxxx, AKx, Qx = 2.7%
Qxxx, xxxx, AKx, Qx = 0.9%
Jxxx, xxxx, AKx, Qx = 0.9%
QJxx, xxxx, AKx, xx = 8.2%
Qxxx, xxxx, AKx, xx = 2.7%
Jxxx, xxxx, AKx, xx = 2.7%
QJx, xxxx, AKx, Qxx = 6.1%
Qxx, xxxx, AKx, Qxx = 6.1%
Jxx, xxxx, AKx, Qxx = 6.1%
xxx, xxxx, AKx, Qxx = 2.0%
QJx, xxxx, AKx, xxx = 3.4%
Qxx, xxxx, AKx, xxx = 13.6%
Jxx, xxxx, AKx, xxx = 13.6%
xxx, xxxx, AKx, xxx = 3.4%

You didn't mention the ♦J; I have put it in West's hand. From the above chart, eliminate whatever hands East would not have opened. So, if you think East wouldn't have opened without at least 11HCP, the remainder would be:
QJxx, xxx, AKx, Qxx = 6.1 shares = 25.5%
Qxxx, xxx, AKx, Qxx = 2.0 shares = 8.5%
QJxx, xxxx, AKx, Qx = 2.7 shares = 11.3%
Qxxx, xxxx, AKx, Qx = 0.9 shares = 3.8%
QJx, xxxx, AKx, Qxx = 6.1 shares = 25.5%
Qxx, xxxx, AKx, Qxx = 6.1 shares = 25.5%
It's not clear to me that you want to eliminate all of those other possibilities, but you probably get the idea. Not surprisingly, East shouldn't be shortest in the suit that E/W have the most of.

This post has been edited by Bbradley62: 2011-February-14, 16:32

[dellache]

Bbradley62, on 2011-February-14, 15:14, said:

By using a simple "choose 10 random cards from a pool of 20 specific cards" and including only the three distributions on your list, I get:
P(4333)=3/11
P(4432)=2/11
P(3433)=6/11
Breaking down these numbers to distinguish between honors and not:
(...snip)
QJx, xxxx, AKx, Qxx = 2.0%
(...snip)
xxx, xxxx, AKx, Qxx = 2.0%

It's hard to believe that those 2 distributions can have the same probabilities (3 cases for ♠QJx/xx, and only one case for ♠xxx/QJ, other suits being the same).
It looks like there's something wrong doesn't it ?
Or do I miss something obvious ?

[Bbradley62]

dellache, on 2011-February-14, 15:33, said:

It's hard to believe that those 2 distributions can have the same probabilities (3 cases for ♠QJx/xx, and only one case for ♠xxx/QJ, other suits being the same).
It looks like there's something wrong doesn't it ?
Or do I miss something obvious ?

Good catch! I will revise...

[Bbradley62]

AlexJonson, on 2011-February-14, 14:55, said:

Lot of comments and few votes, as I look at it.

I think this is because people are pointing out that the proposed lines of play don't necessarily correspond to the three possible distributions, so people are hedging.

[gnasher]

nigel_k, on 2011-February-14, 13:01, said:

line 3 works if East is 3433 or 4333

I don't think that's right. When East is 3433, West may win the second spade and play a club to promote a trump.

[gnasher]

AlexJonson, on 2011-February-14, 14:55, said:

Lot of comments and few votes, as I look at it.

I don't see any merit in using democracy to analyse a bridge hand, especially when it's a play problem that has an exact solution.

[ceeb]

gnasher, on 2011-February-14, 17:48, said:

Nigel said:

line 3 works if East is 3433 or 4333

I don't think that's right. When East is 3433, West may win the second spade and play a club to promote a trump.

Literally true but it is easy to change the order of tricks a little and lead the second spade from hand.

(Does this new interface omit nested quotes (I had to do the above manually), or is there a procedure I'm not seeing?)

[inquiry]

WORD OF CAUTION: I am mathematically challenged.

Thanks for dellache for adding some math to the problem, and the few others who joined in. I am sorry i got the discussion distracted by throwing in a red herring on why you might care which distribution existed based on lines of play. What I was interested in was how people calculate the particular possible holdings. To answer an earlier question, the opener had the diamond jack.

I am not fond of the 3♠ bid, but this was real world, and that is what my partner bid. The auction, other than the opening bid is probably not relative, so I probably should have "fixed" it to be more standard.

Bottom line, is can you calculate the odds, and if so, how did you go about it? I think the answers above are right on target, taking not only possible distributions, but the effect of certain honors in one hand or the other. Before I post my final thoughts on the actual math of the situation (in a later post), here is an example of what I might have thought about. In the past, I probably would have approached this by brute force (with math shortcuts) or through oversimplification. Let me give an example of each. First, I think both are technically wrong (although the math one might be "right" without taking some additional facts specific for the bidding into account).

To simplify the hand, I might have plugged away at it using open spaces. It turns out, if you start with the assumptions as stated, Opener has 2 open spaces and his partner has one (Opener has to have 3S+3H+3D+2C+2 more non-diamonds but not in the same suit, his partner 1S+3H+3D+5C+1 more non-diamond card). So the logic might go that opener is twice as likely to have 4♠ as east to have 2♠. So however I played it, I would have started with one round of spades, and seeing no honor, tried one of the endplay lines.

I will tell you right now, this doesn't work in this situation, because not all the splits are equally likely. The 5=3 ♣ split is much more likely than the 6=2 split ((expressed mathematically, C(8,2) = 28 possible 6-2 splits, but C(8,5) = 56 splits, and if you cash AK of clubs and don't see the club Queen, you can reduce the possible 28 splits by seven). So it is 56/28 = 2 to 1 that clubs will be divided 5=3. The open space calculation will get you , however, to chose the winning line in the real world.

One brute force way is to calculate the total number of combinations for the possible distributions. Without tricks, this will surely take a calculator, but here goes, openers partner hands could be:
5C - 3H - 2S - 3D = C(8,5)*C(7,3)*C(5,2)*C(6,3) = some large number (it is 392,000)
5C - 4H- 1S - 3D = C(8,5)*C(7,4)*C(5,1)*C(6,3) = some large number = (196,000)
6C - 3H - 1S - 3D = C(8,6)*C(7,3)*C(5,1)*C(6,3) = some large number = (98,000)

At the table, we could never come up with numbers like 392,000 etc. However, using the brute force approach, here is a shortcut you can throw in. The C(6,3) terms for diamond combinations cancel out, and surprize, C(7,3) = C(7,4), so the heart terms cancel out. Simplifying to

5C-1S which is C(8,5) * C(5,1) = 56 * 5
5C-2S which is C(8,5) * C(5,2) = 56 * 10
6C-1S which is C(8,6) * C(5,1) = 28 * 5

The first term is divisible by 28 (56/28 = 2, 28/28 = 1), the second term by 5, simplifying to 2:4:1 for the three chances. If you take the large numbers from the unsimplied equation, and divide each by the smallest (98000) you get 392000/98000 = 4, 196000/98000 = 2, and of course 98,000/98000 = 1, the same ratios).

Even the calculations to get the 56 and 28 might be too tough, but those could have been simpler as well.

C(8,5) is really 8*7*6*5*4*3*2*1/(5!*(8-5!)
C(8,6) is close to the same 8*7*6*5*4*3*2*1/(6!*(8-6!)

These lead to the huge numbers (56 and 28), but notice, 5! can be taken out of the top and bottom to give 8*7*6/(3*2*1) and for the second one, 6! can be taken out to give 8*7/(2*1). Eight times seven is common to both, as is 2 in the denominator so they can cancel out, greatly simplifying the calculations
C(8,5) becomes 6/3 = 2
C(8,6) becomes 1

Thus your three groups become
5C-1S which is C(8,5) * C(5,1) = 2 * 5 = 10
5C-2S which is C(8,5) * C(5,2) = 2 * 10 = 20
6C-1S which is C(8,6) * C(5,1) = 1 * 5 = 5

Divide all three by five, gets us back to 2:4:1 ratio, no calculator needed..

So this seems to suggest a 4 spades with opener is less than even money (out of the 2:4:1 ratio, the 2 and 1 have 4♠, so it is 3:4 that opener has 4♠.

Following this logic the "odds" of 4=3=3=3 is 2/7 ~ 29%
3=4=3=3 is 4/7 ~52%
4=4=3=2 is 1/7 ~14%

Sadly, I don't think even this is the correct answer to this problem -- well it might be the mathematically correct one before you apply bridge world conditions to the problem. Not clearly spelled out is the following; If opener has 4♠'s, the only case that matters is when he has both the Queen and the Jack. This also means East can not have ♠Qx or ♠QJ. You need to take that into account for your calculations... that is not all the 4-1 splits in spades are possible, nor are all 3-2 ♠ splits

So the question is, how to go about calculating that. For those who want the “correct line”, once you realize 5=3 clubs is twice as likely as 6=2, you will ruff a club in dummy and make on the endplay. This is about ways to think about and calculate odds.

[Mbodell]

gnasher, on 2011-February-14, 12:19, said:

I'm confident that 3433 is the most likely of the three. Look at the suit breaks in each case:

4333: 4-1, 4-3, 3-3, 5-3
4432: 4-1, 4-3, 3-3, 6-2
3433: 3-2, 4-3, 3-3, 5-3

A specific even distribution is more likely than a specific uneven distribution. There are also more 3-2 breaks than 4-1 breaks, and more 5-3 breaks than 6-2 breaks. Therefore

3433 > 4333 > 4432

I haven't looked at the lines of play that you suggest.

Right, and the rough quantification, if I remember my "Bridge, probability and information" reading would have us say:

C1(3433) = 3-2, 4-3, 3-3, 5-3 = C
C2(4333) = 4-1 (1/2), 4-3, 3-3, 3-3 = 1/2 C
C3(4432) = 4-1 (1/2), 4-3, 3-3, 6-2 (1/2) = 1/4 C

So 3433 is more than twice as likely as the other two distributions combined based just on the dealing (not allocating high cards or figuring out if AKx in diamonds and Qxx in clubs is always opened 1♦).

Edit to add: I posted this before Inquiry added his long post which got to the same point through longer showing of work which you never need to do. 3-2 => 4-1 is 2 (second number) / 4 (first number) as likely = 2/4 = 1/2. 4-4 => 5-3 is 4/5 as likely but 5-3 => 6-2 is 3/6 = 1/2. If you need to knock out conditions like opener has to have the Q then when you were distributing the spades ss-sss and now you want to distribute it ss-Qss then you know that you've gone from C(5,3) to C(4,2) which is 10 to 6 so only 3/5 as many cases. If you need it to be ss-QJs then it is C(3,1) so 10 to 3. Similarly if one line has spades s-ssss and you need it to be s-Qsss you are going from C(5,4) = 5 to C(4, 3) = 4 so only 80% as often. This type of calculation shouldn't really be that hard to do mentally at the table with only a bit of practice.

This post has been edited by Mbodell: 2011-February-15, 00:45

[Bbradley62]

Mbodell, on 2011-February-15, 00:33, said:

Right, and the rough quantification, if I remember my "Bridge, probability and information" reading would have us say:

C1(3433) = 3-2, 4-3, 3-3, 5-3 = C
C2(4333) = 4-1 (1/2), 4-3, 3-3, 3-3 = 1/2 C
C3(4432) = 4-1 (1/2), 4-3, 3-3, 6-2 (1/2) = 1/4 C

So 3433 is more than twice as likely as the other two distributions combined based just on the dealing (not allocating high cards or figuring out if AKx in diamonds and Qxx in clubs is always opened 1♦).

With your numbers, C1 is not more than twice as likely as the other two combined: C1=4/7=57%, C2=2/7=29% and C3=1/7=14%, so C1 is more likely than the other two combined, but not twice as likely.

[Mbodell]

Bbradley62, on 2011-February-15, 00:30, said:

With your numbers, C1 is not more than twice as likely as the other two combined: C1=4/7=57%, C2=2/7=29% and C3=1/7=14%, so C1 is more likely than the other two combined, but not twice as likely.

Yes, you are right, that was a mistake. I meant merely more likely then the other two combined, an odds-on favorite.

[gnasher]

Dellache, are you sure you've dealt with the jacks correctly? Unless I've misunderstood, your calculations assume that:
- If East has ♦J, he must also have ♠J
- If East does not have ♦J, he need not have ♠J either

Presumably that's not what you intended.

[dellache]

gnasher, on 2011-February-15, 03:01, said:

Dellache, are you sure you've dealt with the jacks correctly? Unless I've misunderstood, your calculations assume that:
- If East has ♦J, he must also have ♠J
- If East does not have ♦J, he need not have ♠J either

Presumably that's not what you intended.

Yes you are of course right, I didnot reread last minute modifications of my post.
The maths and results remain the same.
I'll edit my post.