Re: ODDs math question
Thanks for you answer. Here are my comments.
I'm really asking for the answer to the general probability question:
If any one of 3 events a,b, or c, will cause event y to occur, and the probabilities of a, b, and c occurring are:
a=d
b=e
c=f
Is the probability that y will occur equal to?:
d + e + f = Probability of y occuring
And if that is the correct formula for the odds of y occuring, where did I go wrong in my calculations.
With a pocket pair we have the following probabilites for a,b, and c.
a = (2 outs)/(52-2 or 50 cards) = .04
b = (2 outs)/(52-3 or 49 cards) = .041
c = (2 outs)/(52-4 or 48 cards) = .042
So why aren't the odds of y (getting trips) equal to:
.04 + .041 + .042 = 12.3%
The web site says it's 10.8%
Remember, we're ONLY concerned with the case of having any pocket pair, and getting 3 of a kind on the flop. NOT any other case. Let's keep it simple.
If you have any further insight, please comment.
Thank you for your help.
Eric
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