LetYouDown or Aaron does this look correct to you?

VivaLaViking · #1 08-07-2005, 07:38 PM

On a flop if there are N solutions for a straight, the probability is 2% x N. eg assume you have Q J for down cards so the solutions are A K T, K T 9 and T 9 8 and the probalility for the three solutions is 6%. The actual numbers per solution are ~1.95% ~3.91% ~5.87% and ~7.83% but when you're sitting at a table, 2%, 4%, 6% and 8% are close enough.

LetYouDown · #2 08-07-2005, 09:06 PM

Doesn't look right to me. You hold Q-J. There are 4 * 4 * 4 flops of A-K-10 or 64 out of 19600 flops. 64/19600 isn't 2%. You're mixing combinations and permutations. If you're considering order in the top, you need to consider it in the bottom as well.

VivaLaViking · #3 08-07-2005, 09:48 PM

I think you're correct. I just ran this on 20K+ trials and the numbers were way off, about 5x. Thanks again and I'll make the correction.

VivaLaViking · #4 08-07-2005, 10:23 PM

LetYouDown, thanks for the correction. To those interested substitute .33% for 2%, the actual numbers on the flop per solution are ~.32%, ~.65%, ~.98% and ~1.30% but the .33% should be suitable for an on-the-fly approximation. Also for coders change 384 to 64.

BruceZ · #5 08-08-2005, 08:06 PM

This was a great example of how NOT to use computer technology. Always do a hand calculation first, at least to get a reasonable idea of where the answer should be, and just use the computer to nail down the last tenth of a percent or so. Any other use of the computer, and you might as well get your answer from a random number generator, as that is invariably what you will be producing. In this case, there was no reason to use the computer at all as a hand calculation could easily compute the exact answer.

VivaLaViking · #6 08-09-2005, 02:14 PM

Bruce, I read your comment yesterday and thought just to ignore it but decided you needed a reply. Had you bothered to read the associated posts you would have discovered that the math was done first and I erred in dividing the number of permeations by the number of combinations and used the computer to test my result with 20K+ hands, a valid use of a computer. As three digits may be permeated 6 ways my initial result was exactly six too great and do not feel comfortable enough with combinatorial math yet. I am an EE and combinatorial math is far removed from the type of math I do.

Further, there are many questions that may be unknown or too difficult to calculate such as my recently answered question, what is the correct order of preferred openers per the number of players at the table. And yes, I used a computer.

Lastly, I will guess that you are under 25 years old. If you are, before you were born, in a spare room in my house there was my computer (hint: it had over 40 lights). It was given to me because of my involvement in it's design. Two points I am making is I have been involved in computers for a long time at a high level and have been writing software (and firmware) longer. So one last question, why do you think you can comment on what a computer should or shouldn't be used for?

AaronBrown · #7 08-09-2005, 06:42 PM

I think BruceZ just read your second post, which by itself gives the impression you ran the simulation first. I think he would agree that it makes sense to use the computer to check a hand calculation. The hand calculation gives a precise answer, but if you make an error you can be off by a huge amount. The computer has random noise, but usually gets pretty close (unless there's a bug in the code, but then it's unlikely to agree with your incorrect hand calculation).

I have most faith in a calculation when I do it by hand, then check it by simulation. It's easy to make a mistake by hand, and often the problem has to be simplified. But I agree with BruceZ that a simulation by itself hurts more often than it helps.

gol4pro · #8 08-09-2005, 08:00 PM

n = number of possible 3 card straights

Flopping straight with QJ(n=3) = n(4c1)^3/50c3
= 3(4c1)/50c3
= 3(64)/19600
= 192/19600 = about 1/100

Flopping straight with JT(n=4)
= n(4c1)^3/50c3
= 4(4c1)^3/50c3
= 4(64)^3/50c3
= 256/19600
= slightly better than 1/80

Chance of flopping straight with KQ(n=2)
= n(4c1)^3/50c3
= 2(4c1)^3/50c3
= 2(64)/19600
= 128/19600 = about 1/150

Chance of flopping a straight with AK(n=1)
1(64)/19600
= about 1/300

VivaLaViking · #9 08-09-2005, 08:24 PM

Aaron, thank you. BTW: You have stepped up a notch in my book by helping the 14 year-old kid yesterday with his standard deviation problem. If I felt confident I would have replied. Also, I too feel uncomfortable when I have to "Monte Carlo" for an answer mainly because it's correctness cann't be verified.