[BruceZ]

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Can someone help me with this, I've noticed something on a poker site and would like to test it.

4 tabling what is the chance of being dealt the same hand on 2 tables, and what is the chance of this happening regardless of suit?

The site im playing on, uncannily you get the same hands every now and then but very frequently, it's weird and I'd like to know the actual % chance so I can test this.

Thanks!

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Regardless of suit:

1 - P(91,4)/91^4 =~ 6.5%

That is 1 minus the probability that all 4 hands are different. There are C(13,2) = 13*12/2 hands with 2 different ranks, plus 13 pairs makes 91 different hands. There are 91^4 ways to be dealt the 4 hands, and P(91,4) = 91*90*89*88 ways for them all to be different.

So you should have at least 2 hands the same on average about every 15.5 hands.

If you care about suit, then there are C(52,2) = 52*51/2 = 1326 different hands, so the probability that at least 2 are the same is

1 - P(1326,4)/1326^4 =~ 0.45%

So this should happen on average about every 221 hands, the same rate at which you would get AA on any given hand.