Potential noobish question about Ax vs. Ax AIPF probability
Join Date: Apr 2008
Posts: 6,402
Just wanting to get a little insight as to the specific probability of a hand like AK losing to A8, A9, etc all-in pre-flop over multiple consecutive trials. I was under the impression that Bernoulli's formula would answer it, but have been told otherwise (due to the binomial therom)... Say for 9 straight out of 10 trials AK losing. I had thought...
(10/9)(.25)^9(.75)^1=0.000003=0.0003%
Can someone be kind enough to explain why this is wrong?
Thanks in advance.
Join Date: Mar 2009
Posts: 6,730
Are you specifically making a distinction between 9 consecutive out of 10 trials vs. 9 out of 10 trials? That is, is the "consecutive" essential to your question?
If so, then the standard binomial formula would have to be modified since, of course, N consecutive events out of T trials are less likely than N events (consecutive or not) out of T trials.
Is that what the (10/9) in your formula above is attempting to capture? I don't think that is correct, but I'll await your reply before going further.
Join Date: Apr 2008
Posts: 6,402
Correct, 9 consecutive events... To be honest, the T variable doesn't really matter now that I think about it. I'm more wondering about that many consecutive losses.
Thanks.
Join Date: Mar 2009
Posts: 6,730
There is always more to the story, but if you are trying to determine the probability of getting 9 failures in 9 trials, where the trials are independent and where the prob of success of each trial is P, then it is:
(1-P)^9.
What we typically add to this is that the "real" probability is undoubtedly higher than this for several reasons:
(1) The streak is noticed only after it happened
(2) The streak is part of a much longer series of trials
(3) Many people are experiencing series of trials, and maybe only one person got such an unlucky streak
(4) The trials may be correlated in some way
(5) The true probability of success may be higher for some reason.
Join Date: Apr 2008
Posts: 6,402
