Need some help from binomial sharps on a question.
June 11th, 2012 at 5:55:35 PM
permalink
Parameters of the question. You are playing Hold Em (being dealt 2 cards) on 3 internet poker tables. For sake of simplicity, let's say that each table's hands are dealt at identically the same time, and every hand lasts the same exact amount of time before the next hand is simultaneously dealt again to the 3 tables. Let's say we are dealing 500 hands per day to each of the 3 tables.
Now, I know that the chances of having the identical 2 cards dealt on all 3 tables at the same time is 1,758,275-1. Here is the question :
Using the parameters explained above, what are the odds/% that we will have the same hand dealt at all 3 tables at least once in : A. 1 day (500)
B. 1 week (3500)
C. 1 month (15,000)
D. 1 year (182,500)
Thank you so much to the sharp cat who will take on this question.
Now, I know that the chances of having the identical 2 cards dealt on all 3 tables at the same time is 1,758,275-1. Here is the question :
Using the parameters explained above, what are the odds/% that we will have the same hand dealt at all 3 tables at least once in : A. 1 day (500)
B. 1 week (3500)
C. 1 month (15,000)
D. 1 year (182,500)
Thank you so much to the sharp cat who will take on this question.
June 11th, 2012 at 7:30:34 PM
permalink
Are the correct answers?
A. .000286
B .00199
C .00849
D .0986
A. .000286
B .00199
C .00849
D .0986
June 11th, 2012 at 7:55:30 PM
permalink
Quote: shakhtarAre the correct answers?
Yes.
June 11th, 2012 at 8:12:10 PM
permalink
You can use the Poisson approximation when n is large and p is small.
I get answers close to those listed. I also say:
yes.
I get answers close to those listed. I also say:
yes.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
June 12th, 2012 at 6:38:37 AM
permalink
I also agree, but you don't need Poisson distribution - since the OP is asking for the probability that it happens at least once in N deals, and each deal is independent, it's 1 minus the probability that none of the deals have all three hands with the same two cards, which is:
1 - (1 - 1/1758275)N
If you want to know how many deals it will take before the probability that at least one will have all three hands with the same cards is P, it is:
N = log (1 - P) / log (1 - 1/1758275)
For example, if P = 0.1 (10%), N = 185,253
1 - (1 - 1/1758275)N
If you want to know how many deals it will take before the probability that at least one will have all three hands with the same cards is P, it is:
N = log (1 - P) / log (1 - 1/1758275)
For example, if P = 0.1 (10%), N = 185,253
