View Full Version : The odds on big pairs

10-04-2004, 11:02 AM
Anyone know the percentage or odds on, if I have a big pocket pair (jacks or better) the odds that one of my opponents in a full ring game will also have a big pocket pair (again jacks or better.

Really need this info, thanks ahead CASHIZ

10-04-2004, 11:29 AM
Here is a quick way to get a rough approximation: The average number of other big pairs you are up against is 9*19/(50 choose 2) = 171/1225 = .13959. That is greater than the probability you are up against a big pair because there may be more than one. To estimate the probability, the probability that a Poisson distribution with mean x is greater than 0 is 1-exp(-x). 1-exp(-171/1225) = .130287.

That isn't exact because the Poisson distribution is a rough approximation here. I'm not sure it is even better than 1-(1-19/1225)^9 = .131238, which assumes your opponents' hands are independent. If I get bored, I may do the exact calculation later, using this method (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=849658&page=&view=&sb =5&o=&vc=1). However, I think these estimates are off by less than 0.2%.

10-04-2004, 11:54 AM
I'm really bad at math, does that mean 2 percent. Can you please elaberate a little more. It seems to me just out of experiance that it happens quite often that you have a big pair and so does one of your opponents. Of course you don't see every hand but just for example i've seen it four times in the last 6 hours.

10-04-2004, 04:38 PM
does that mean 2 percent

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.131 = 13.1%, or between 6:1 and 7:1. The 0.2% I mentioned was an error estimate for the technique I used. Rather than saying the probability is exactly 13.1%, I'm claiming that the probability should be between 12.9% and 13.3%.

10-04-2004, 08:04 PM
thanks bro, your [censored] smart.