[abbadagg]

In a recent teams match I was the declarer in a 3NT contract. I bought the contract on the idea of running a seven-card club suit missing the queen. It ended down 5 because West held Qxx. I did a calculation of the probability the suit would run, which resulted in a surprisingly high 0.77293. But is the calculation correct? Your feedback is welcome.

------------------------ probability calculation-------------------

Problem: calculate the probability that in a NT contract with a 7-card club suit missing the queen and no entry to dummy you'll be able to run the suit for seven tricks.

Solution

Use notation P[e] for the probability that event e occurs, P[e,u] for the probability that both e and u occur, and P[e|u] for the probability that e occurs given that u occurs; all these probabilities under the assumption that South has a 7-card club suit missing the queen.

P[can run suit] = P[Q in dummy] + P[Q not in dummy, Q is singleton or doubleton]

P[Q in dummy] = 13/39

P[Q not in dummy, Q singleton or doubleton] =
2 x P[Q with West, Q singleton or doubleton] =
2 x P[Q with West] x {P[Q singleton | Q with West] + P[Q doubleton | Q with West]}

P[Q with West] = 13/39

P[Q singleton | Q with West] = (26/38) x (25/37) x (24/36) x (23/35) x (22/34) = 0.13105

P[Q doubleton | Q with West] = 5 x (12/38) x (26/37) x (25/36) x (24/35) = 0.52835

P[can run suit] = 0.77293

[PhantomSac]

Dunno about your math, but 77 % feels about right. 1/3rd partner has the queen, ~52 % to run if partner has 2, ~80 % to run if partner has 3, significant chances if partner has a stiff (dunno the percentages).

[vuroth]

Your math seems wrong to me somehow. The probability of Queen-sixth is 1, and Qx + Qxx is greater than 1.

I thought it was:
nCr = n!/r!(n-r)!

Q = (13C0)(26C5)/(39C5) = 65780/575757 = 0.1142
Qx = (13C1)(26C4)(39C5) = 13 * 14950/575757 = 0.3376
Qxx = (13C2)(26C3)/(39C5) = 0.3522
Qxxx = (13C3)(26C2)/(39C5) = 0.1614
Qxxxx = (13C4)(26C1)/(39C5) = 0.0323
Qxxxxx = (13C5)(26C0)/(39C5) = 0.0022

(Check: these values add up to 0.9999)

So Q or Qx is 0.4518. So 1/3 + 2/3 x 0.4518 = 0.6345, or around 64%. Still way over 50%

[655321]

Edit, sorry, I think I misread Vuroth's answer

This post has been edited by 655321: 2010-February-09, 18:59

[nigel_k]

Below are the possible distributions (regardless of who has length), the probability of each, and how often we succeed on each one:

006 0.00158 succeed 1/3 of the time
015 0.03077 succeed 1/3 * (1 + 1/6)
024 0.10257 succeed 1/3 * (1 + 1/3)
033 0.07521 succeed 1/3 * (1/2 + 1/2)
114 0.11111 succeed 1/3 * (1 + 1/3 + 1/3)
123 0.53333 succeed 1/3 * (1 + 1/2 + 1/2)
222 0.14545 succeed always

Therefore probability of success is 0.00053 + 0.01197 + 0.04559 + 0.02507 + 0.06172 + 0.35555 + 0.14545 = 0.64588

This may be subject to minor rounding errors. I don't understand either of the above methods.

[cherdanno]

abbadagg, on Feb 9 2010, 08:23 AM, said:

P[Q doubleton | Q with West] = 5 x (12/38) x (26/37) x (25/36) x (24/35) = 0.52835

This should be
P[Q doubleton | Q with West] = 5 x (12/38) x (26/37) x (25/36) x (24/35) x (23/34)

Vuroth, you need to use 12 instead of 13, and 38 instead of 39 (one spot is taken by the queen already).

[Cascade]

nigel_k, on Feb 10 2010, 03:00 PM, said:

Below are the possible distributions (regardless of who has length), the probability of each, and how often we succeed on each one:

006 0.00158 succeed 1/3 of the time
015 0.03077 succeed 1/3 * (1 + 1/6)
024 0.10257 succeed 1/3 * (1 + 1/3)
033 0.07521 succeed 1/3 * (1/2 + 1/2)
114 0.11111 succeed 1/3 * (1 + 1/3 + 1/3)
123 0.53333 succeed 1/3 * (1 + 1/2 + 1/2)
222 0.14545 succeed always

Therefore probability of success is 0.00053 + 0.01197 + 0.04559 + 0.02507 + 0.06172 + 0.35555 + 0.14545 = 0.64588

This may be subject to minor rounding errors. I don't understand either of the above methods.

I am not sure how you get the success rates - 1/3 * (1 + 1/6) etc

I calculated them differently and got different rates in two cases:

006 1/3 of the time partner has the long suit

015 1/3 of the time partner has the long suit and the remaining 2/3 of the time the queen is short 1/6 of the time which is 1/3 + 2/3 * 1/6 = 1/3 * (1 + 1/3) in your format

024 1/3 of the time partner has the long suit and the remaining 2/3 of the time the queen is short 1/3 of the time which is 1/3 + 2/3 * 1/3 = 1/3 * (1 + 2/3)

033 2/3 of the time partner has a long suit and half of that time it includes the queen. That is 2/3 * 1/2 = 1/3

114 1/3 of the time partner has the long suit and the remaining 2/3 of the time the queen is short 1/3 of the time which is 1/3 + 2/3 * 1/3 = 1/3 * (1 + 2/3)

123 1/3 of the time partner has the long suit and the remaining 2/3 of the time the queen is short 1/2 of the time which is 1/3 + 2/3 * 1/2 = 1/3 * (1 + 1)

222 always.

[vuroth]

cherdanno, on Feb 9 2010, 09:38 PM, said:

Vuroth, you need to use 12 instead of 13, and 38 instead of 39 (one spot is taken by the queen already).


Q = (12C0)(26C5)/(38C5) = 65780/501942 = 0.1131
Qx = (12C1)(26C4)(38C5) = 12 * 14950/501942 = 0.3574

Overall: 1/3 + 2/3 x (0.1131 + 0.3574) = 0.6470

Whoops. You are, of course, completely correct.

Q =
Qx =

[vuroth]

cherdanno, on Feb 9 2010, 09:38 PM, said:

Vuroth, you need to use 12 instead of 13, and 38 instead of 39 (one spot is taken by the queen already).

Whoops. You are, of course, completely correct.

Q = (12C0)(26C5)/(38C5) = 65780/501942 = 0.1131
Qx = (12C1)(26C4)(38C5) = 12 * 14950/501942 = 0.3574

Overall: 1/3 + 2/3 x (0.1131 + 0.3574) = 0.6470

[Cascade]

Cascade, on Feb 10 2010, 04:41 PM, said:

nigel_k, on Feb 10 2010, 03:00 PM, said:

Below are the possible distributions (regardless of who has length), the probability of each, and how often we succeed on each one:

006  0.00158  succeed 1/3 of the time
015  0.03077  succeed 1/3 * (1 + 1/6)
024  0.10257  succeed 1/3 * (1 + 1/3)
033  0.07521  succeed 1/3 * (1/2 + 1/2)
114  0.11111  succeed 1/3 * (1 + 1/3 + 1/3)
123  0.53333  succeed 1/3 * (1 + 1/2 + 1/2)
222  0.14545  succeed always

Therefore probability of success is 0.00053 + 0.01197 + 0.04559 + 0.02507 + 0.06172 + 0.35555 + 0.14545 = 0.64588

This may be subject to minor rounding errors. I don't understand either of the above methods.

I am not sure how you get the success rates - 1/3 * (1 + 1/6) etc

I calculated them differently and got different rates in two cases:

006 1/3 of the time partner has the long suit

015 1/3 of the time partner has the long suit and the remaining 2/3 of the time the queen is short 1/6 of the time which is 1/3 + 2/3 * 1/6 = 1/3 * (1 + 1/3) in your format

024 1/3 of the time partner has the long suit and the remaining 2/3 of the time the queen is short 1/3 of the time which is 1/3 + 2/3 * 1/3 = 1/3 * (1 + 2/3)

033 2/3 of the time partner has a long suit and half of that time it includes the queen. That is 2/3 * 1/2 = 1/3

114 1/3 of the time partner has the long suit and the remaining 2/3 of the time the queen is short 1/3 of the time which is 1/3 + 2/3 * 1/3 = 1/3 * (1 + 2/3)

123 1/3 of the time partner has the long suit and the remaining 2/3 of the time the queen is short 1/2 of the time which is 1/3 + 2/3 * 1/2 = 1/3 * (1 + 1)

222 always.

This gives a probability of 0.658975207

vuroth said:

Q = (12C0)(26C5)/(38C5) = 65780/501942 = 0.1131

I think you transposed the digits in your answer here it should be 0.1311 and then your answer concurs with the number above.

[nigel_k]

Looks like I carelessly missed out a couple of chances. It should be:

006 0.00158 succeed 1/3 of the time
015 0.03077 succeed 1/3 * (1 + 1/6 + 1/6)
024 0.10257 succeed 1/3 * (1 + 1/3 + 1/3)
033 0.07521 succeed 1/3 * (1/2 + 1/2)
114 0.11111 succeed 1/3 * (1 + 1/3 + 1/3)
123 0.53333 succeed 1/3 * (1 + 1/2 + 1/2)
222 0.14545 succeed always

Therefore probability of success is 0.00053 + 0.01368 + 0.05698 + 0.02507 + 0.06172 + 0.35555 + 0.14545 = 0.65898

So vuroth and I did get the same answer (after Wayne corrected both of our mistakes).

Where I have said, for example, 015 has a success rate of 1/3 * (1 + 1/6 + 1/6) that means we succeed always when partner has 5, 1/6 when partner has 1 and 1/6 when partner has zero.