[Trinidad]

I was checking Mike Lawrence's Bridgeclues website today. It had the following play problem:

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You are in 6♠ on the lead of the ♥J. You try the Q, but it is covered and you take your ace. You play the ♠K, get the 6 from West while East discards a heart. The question is: what is the best line to still make your contract?

Since you are obviously going to need the diamond finesse, let's take a winning finesse straight away in our minds and see what cards are going to be left then:

---

Now what are you going to do now?

Mike plays the ♦A and discards a club, plays ♣KA, ruffs a club, entering dummy in trump and discarding the heart on the last club. He is playing for clubs to be 3-3.

An alternative line would be to take the ♠A, play the ♦A, discard a heart and ruff a heart. Take the ♣K, play a club to the ace and as long as nothing has happened exit with a trump. West will be endplayed if clubs were 0-6, 1-5 or 2-4 (with west having the shorter clubs). (If west ruffs the small club to the ace, you can even be flashy and play the ace anyway )

Everything else being equal, odds for clubs splitting 3-3 are 35.5 %, leaving 64.5 % for the other distributions, half of which will have west with the short clubs. That is 32.25%, making the play for 3-3 the better line.

But everything else isn't equal. Spades were 3-0 and West is already placed with at least three diamonds to the queen. On top of that, West is known to have the ♥J from the opening lead while East is known to have the ♥K3. Presumably, East also followed to three rounds of diamonds. How does this change the calculation?

Greetings,

Rik

[Fluffy]

don't know the odds, but you might try ♣K before diamonds, east has nothing left id spades, hearts or diamonds, he is favourite to give real count on the suit.

[dicklont]

My answer reading bridgeclues was excactly the alternative line: eliminate hearts and diamonds, partialy eliminate clubs exit with trumps.

In practice I would still do that, placing west with the short clubs. I am very curious what the statistics have to say about it.

I found the quiz very instructive anyway because the main line given by Mike Lawrence never even occured to me.

[MFA]

It takes some counting to determine the best line.

A: Mike's line. Works with 3433, 3343, 3253 in west. *.
B: OP's line. Works with 3532, 3442, 3352, 3631, 3541, 3451, 3730, 3640, 3550.
*) If west has 6 diamonds, we will notice and switch to plan B, which will now clearly be best.

So it's a matter of counting. West has "shown" ♥JT (or Jx) & ♦Q so far. The ♥3 in east is irrelevant, it's just a small heart in our eyes, and we know that he does have small hearts.

3433. H:15 (ways to have JTxx), D:21, C: 20. Total: 6300
3343. H:6, D:35, C:20. Total: 4200
3253: H:7, D:35 C:20. Total: 4900.

Plan A: 15400.

3532. H:20, D:21, C:15. Total: 6300.
3442. H:15, D:35, C:15. Total: 7875.
3352. H:6, D:35, C:15. Total: 3150.
3631. H:15, D:21, C:6. Total: 1890.
3541. H:20, D:35, C:6. Total: 4200.
3451. H:15, D:35, C:6. Total: 3150.
3730. H:6, D:21, C:1. Total: 126.
3640. H:15, D:35, C:1. Total: 525.
3550. H:20, D:35, C:1. Total: 700.

Plan B: 27916.

If I haven't made a mistake underway, it's not close. The partial elimination is much better.

[han]

www.rpbridge.net

[Fluffy]

If instead of counting West's distributions, you counted Easts, would it give you same results?

[Trinidad]

Jlall, on Nov 12 2007, 11:37 AM, said:

nice spot on this hand, my instinctive line was Lawrence's but I wouldn't be surprised if the OP's line was a bit better, would like to see some good math analysis of this (HANNIE??)

I tried to do the math, assuming the following:
West has shown ♠QJx ♥J ♦Qxx and has six other places that could be clubs. East has shown ♠- ♥Kx ♦xxx and has eight other places that could be clubs.

Then, if you start distributing the clubs over East and west, the odds that the first club goes to west are 6/(8+6). Now one place is occupied and the odds for the next club to go to west are 5/(8+5), etc.

When you work this out you will get: P(n clubs with East)=Combin (6,n)*fact(8)*fact(6)*fact(8)/fact(8+6)/fact(n)/fact(8-n). There fact(x) stands for the factorial of x [x*(x-1)*(x-2)*...*2*1] and Combin (x,y) stands for the combinatorial of x and y, i.e. fact(x)/fact(y)/fact(x-y).

The calculation gave:

probability distribution of clubs
0.00033 6-0
0.01598 5-1
0.13986 4-2
0.37296 3-3
0.34965 2-4
0.11189 1-5
0.00932 0-6

I was happy to see that the sum of these probabilities was 1 .

Thus, the probability for a 3-3 distribution was 37.3%. The sum of the probabilities for the 2-4, 1-5 and 0-6 distributions was 47.1%.

Making the daring assumption that I did all this math correctly, the alternative line (playing for 0-6/1-5 or 2-4) would then be 1.26 times more likely to succeed than the given line (playing for 3-3).

Rik

[cherdano]

Trinidad, your math analysis isn't quite correct. Maybe I will write a longer explanation later, the way you take the diamond spots into account isn't correct - see http://www.rpbridge.net/7z75.htm for an explanation.

Anyway, just using that spades are 3-0 the odds already favor your suggested line by a significant amount (roughly 44% instead of 34%, http://www.rpbridge.net/xsb2.htm).
Additionally we know that LHO has ♥JT, not ♥K, and that he has ♦Q, and that he has at least 3 diamonds -- all this will make the case for your line even better. I am sure you are right and it is not close.

I didn't check the details but MFA's numbers look good.

[Halo]

Very interesting.

I would definitely play oppo for 3532 or similar and play the partial elimination in clubs rather than 33.

[kenrexford]

Interesting problem.

As an aside, it seems that LHO has nothing to gain by any falsecard play of the diamond Queen into the AJ, for a complicated reason.

I'm not sure what this does to the odds, but there also seems to be some nuance tweaks to the odds when one considers restricted choice and the option to lead a heart rather than a club, unless the bidding accounted for that choice. I would not expect :HO, for instance, to have QJ10 in clubs and opt to lead a heart.

[han]

Jlall, on Nov 12 2007, 11:37 AM, said:

nice spot on this hand, my instinctive line was Lawrence's but I wouldn't be surprised if the OP's line was a bit better, would like to see some good math analysis of this (HANNIE??)

But is this really a math problem JLALL?? The fact that west lead the heart jack affects the likelyhood of different club holdings. For example, if west had had a singleton club then she might have lead it. West might also have lead QJ10 or J109 of clubs (depending on the auction and west's heart holding).

As a simple model I will assume that west has J10 of hearts and was forced to lead the heart jack.

I think that Trinidad's analysis was close to correct:

Trinidad, on Nov 12 2007, 02:28 PM, said:

I tried to do the math, assuming the following:
West has shown ♠QJx ♥J ♦Qxx and has six other places that could be clubs. East has shown ♠- ♥Kx ♦xxx and has eight other places that could be clubs.

This is dangerous stuff. Given that west has at least 3 diamonds and 3 spades east must indeed have at least one small heart. But by taking away one of the nine free spots for east you are actually placing a known heart card in east's hand, that is not correct. Similarly you have to be careful with the diamond cards. The proper assumption that we should make is the following:

West has shown ♠QJx, ♥J10 ♦Q and has at least 2 more diamonds (one could argue about the heart 10). East has shown ♥K and has at least 3 diamonds (if east shows out on the second or third round of diamonds then that changes then that changes the odds).

We can now use the dual suit calculator (7 missing diamonds and 6 missing clubs, west having 7 free spaces and east 12) in Pavlicek's website. We should only look at the cases where both hands have enough diamonds and add the odds. We get that the clubs split with the following probabilities:

0-6: 1.3892 + 1.0419 + 0.2501 = 3.0810
1-5: 6.2515 + 8.3353 + 3.7509 = 18.3377
2-4: 6.2515 + 15.6287 + 12.5030 = 34.3832
3-3: 1.3892 + 8.3353 + 12.5030 = 22.2275

Now note that these probabilities don't add up to 1 (as they shouldn't) but we only care about the relative probabilities so that's ok.

So the total odds for 0-6, 1-5 and 2-4 are 55.8019 which is about 2.5 times as much as 22.2275.

Wow, my calculation gives that the OP is 2.5 times as good as the line suggested by Mike Lawrence. That's a huge ratio. I did the additions by hand so perhaps non-mathematicians should check them.

Assuming that my calculation is correct (I will give it more thought later) it really suggests that the OP is much better. The assumption that west has the heart 10 seems reasonable and although the choice of leads should affect the odds, not by such a wide margin.

So I pitch the heart. I also think it is a more beatiful line.

[han]

cherdano, on Nov 12 2007, 03:10 PM, said:

Trinidad, your math analysis isn't quite correct. Maybe I will write a longer explanation later, the way you take the diamond spots into account isn't correct - see http://www.rpbridge.net/7z75.htm for an explanation.

Anyway, just using that spades are 3-0 the odds already favor your suggested line by a significant amount (roughly 44% instead of 34%, http://www.rpbridge.net/xsb2.htm).
Additionally we know that LHO has ♥JT, not ♥K, and that he has ♦Q, and that he has at least 3 diamonds -- all this will make the case for your line even better. I am sure you are right and it is not close.

I didn't check the details but MFA's numbers look good.

He, now I noticed that Arend made the same comment, I'm in good company!

[han]

MFA, on Nov 12 2007, 09:37 AM, said:

3253: H:7, D:35 C:20. Total: 4900.

Your method looks correct. I believe that you need H:1 here (only 1 way to have J10 doubleton ), which would give only 700 instead of 4900. We both may have made some more mistakes which would explain the difference in our results.

[MFA]

Hannie, on Nov 12 2007, 06:03 PM, said:

MFA, on Nov 12 2007, 09:37 AM, said:

3253: H:7, D:35 C:20. Total: 4900.

Your method looks correct. I believe that you need H:1 here (only 1 way to have J10 doubleton ), which would give only 700 instead of 4900. We both may have made some more mistakes which would explain the difference in our results.

The H:7 was deliberate, allowing for any ♥Jx lead. Of course questionable, but since we don't know about bidding etc., I felt it was better to include it.

[han]

Ah ok, that's fine then. That explains why we got different answers.

[dburn]

Put more simply, no line will work if West has four or more clubs. Lawrence's line will work if West has three clubs; Trinidad's line will work if West has 0, 1 or 2 clubs.

The a priori chance that West has exactly three clubs is about 36%. The a priori chance that he has more than three clubs is therefore about 32% (half of the 64% chance that he doesn't have exactly three).

But when West has three spades and East none, the chance that East has more clubs than West increases by... well, by an amount that would seem significant even if you didn't know what it was. The vast array of figures in the posts above are no doubt designed to contradict my assertion that mathematicians cannot count. They still can't, but ever since the proof of the four-colour theorem, they have had to pretend that they believe computers can.

Now, suppose that West had played the queen of diamonds [a] under the king or [b] when you led the second round. How would, or should, you play then?

[gnasher]

dburn, on Nov 13 2007, 12:44 AM, said:

Now, suppose that West had played the queen of diamonds [a] under the king or [b] when you led the second round. How would, or should, you play then?

The same way as I was going to play before he played DQ.

[kenrexford]

The play of the diamond Queen does nothing substantial. It must be a falsecard to matter. The falsecard could be because it is about to drop anyway (Qxx) or from Q10xx(x)(x). The falsecard would be done to suggest something to Declarer, but these two plausible holdings each suggest something different about clubs. Plus, the Queen as a falsecard is too obvious. Thus, playing it is relatively meaningless from LHO's perspective and will not be done when it matters (or will, randomly).

That being said, it seems that LHO has a likelihood of holding some honor in clubs, because he did not lead them. He also does not seem to be QJ10. So, it seems that one of his cards is an honor. This eliminates out any x, xx, and xxx holdings, somewhat. If the "x" is not the 10, then three "x" patterns are eliminated, three "xx" patterns, and one "xxx" pattern. If you include the 10 as an "x", then you can eliminate four "x" patterns, six "xx" patterns, and four "xxx" patterns.

So, the club-honor problem seems to somewhat increase the odds above pure mathematical expectation for the 3-3 split.

[dburn]

I believe you, young Andy. It's the thousands who wouldn't that you have to convince.