WHAT ARE THE ODDS????????
 

A question for all you probability and statistic gurus.... I just got off a 1/2 online game that still leaves me dumbfounded. In a little over 2 hours of play and some 120 hands later I saw the board pair approximately 54 times! For those in the know, what statistical formula could be used to figure the odds of such an event happening? Of the aforementioned 120 hands, 30 never went to a showdown so the entire board was never dealt out...leaving 90 hands played to the river and 54 of these being dealt pairs on the board. [img]/forums/images/icons/confused.gif[/img] I'm sure the odds against something like this happening are astronomical. Comments appreciated! [img]/forums/images/icons/tongue.gif[/img]

Re: WHAT ARE THE ODDS????????
 

The probability of the board pairing in 5 cards is 49.3%. This is 1 - 52*48*44*40*36/P(52,5). In 90 hands played to the river, you would expect the board to pair 44 times, and you saw it pair 54 times. The standard deviation of this binomial distribution is sqrt(44) = 6.6, so you are 1.5 standard deviations above average. If we use the normal approximation, this has probability of about 7%. No big deal.

Re: WHAT ARE THE ODDS????????
 

You think your stat is something.......try this one.....

Playing online over the summer, I saw well over 20 hands where opponnents had 4 of a kind or better in a matter of 3 days.

This is wrong
 

An exact calculation with a spreadsheet shows the odds to actually be 2.7% for the board pairing 54 or more times out of 90. My earlier approximation using the normal distribution isn't accurate, and this may be instructive.

While it is true that the standard deviation of the binomial distribution is sqrt(90*.493), you cannot simply refer to the normal distribution with this same standard deviation in order to compute the probability of being greater than 54 and expect to get a very accurate answer. There is, however, a correct way to make this approximation.

The standard deviation we must use turns out to be sqrt[90(p-p^2)] where p is the probability of a given board being paired (p=.493). That is, the overall variance is 90 times the variance of a given hand. The central limit theorem tells us that the number of paired boards is distributed as a normal distribution with this standard deviation (4.74) and mean of 90*.493=44.37. The probability of getting 54 or more is then (54-44.37)/4.74=2.03 standard deviations above average, or 2.1%

The exact answer can be computed as:

sum[k=54 to 90]C(90,k)*(p)^k*(p)^(90-k)

where p = 1 - 52*48*44*40*36/P(52,5) = .493

This comes out to 2.7%. So it's rare, but still not a cause for concern unless it's repeatable.

Re: WHAT ARE THE ODDS????????
 

that is amazing.
my top one is in 8 single table HE satellites on True Poker I saw three straight flushes, and that was using both cards in the players hand.

I think i've seen one in the eight months since.
needless to say, that was the only 8 tourneys i played there.

Re: WHAT ARE THE ODDS????????
 

Yawn.

In my last 40 hours of play, I've made a 5-high straight flush, quad Kings, and quad 7's all by myself.

Re: WHAT ARE THE ODDS????????
 

I've been playing at my B&M for over a year now, probably around 300-400 hours total. I have never had a straight flush in hold'em even a 1 card straight flush.

Re: WHAT ARE THE ODDS????????
 

And I thought that I was going to be the last person in Minnesota to get a Canterbury hat. [img]/forums/images/icons/grin.gif[/img] Maybe we can both get one on tuesday on the same hand. [img]/forums/images/icons/laugh.gif[/img]

later,
Bob T.

Re: WHAT ARE THE ODDS????????
 

Same here. I've logged close to 1,000 hours between B&M and online play, all in the last year, and no straight flushes. A guy at my table Friday night had 2 in 4 or 5 hours.

You'd think with the way the online card rooms are rigged I'd have a few dozen by now. [img]/forums/images/icons/smile.gif[/img]