Odds of being deal same hand twice

[ericslagle]

Ok, so I know the odds of being dealt any specific 5 card hand is 1 in 2,598,960. And I know that if you get about 30 people in a room it's likely that 2 share a birthday. So it got me thinking - how many hands do you need to be dealt to get the same hand twice in that session. So, it doesn't matter which hand gets repeated, and it doesn't matter if it's hand 4 and 546 that match. A bet a friend that it's probably likely that you get the same 5 card hand at least 2 times a year and he didn't believe me.

[Isura]

If you want complete certainty, then you need 2,598,960 hands on average to get the same hand twice. But similar to the birthday problem, you can get a fairly high probability of having two of the same hand with a much smaller sample of hands. This problem is equivalent to the birthday problem, with the birthdays replaced by hands. In other words, it's like saying for each person there are ~2,600,000 possible birthdays (ie possible hands) and if we have only say 1000 people there is a fair probability of two having the same hands. Offcourse in the birthday problem we NEED atleast 366 people in the room to guarantee that two have the same birthday (ie with probability 100%), but with 30 people that probability number is still fairly high. I need to think a bit to do the calculation but it is fairly straightforward.

[Gebre]

[ QUOTE ]
Ok, so I know the odds of being dealt any specific 5 card hand is 1 in 2,598,960. And I know that if you get about 30 people in a room it's likely that 2 share a birthday. So it got me thinking - how many hands do you need to be dealt to get the same hand twice in that session. So, it doesn't matter which hand gets repeated, and it doesn't matter if it's hand 4 and 546 that match. A bet a friend that it's probably likely that you get the same 5 card hand at least 2 times a year and he didn't believe me.

[/ QUOTE ]

Like the birthday problem, it takes a surprisingly few number of hands before it is likely that a hand will repeat. The probability that at least one hand repeats in n trials is

1 - (2598959/2598960 * 2598958/2598960 * ... * (2598960-n-1)/2598960)

This probability becomes greater than 50% for n = 1899.

This probability becomes greater than 75% for n = 2685.

This probability becomes greater than 90% for n = 3460.

This probability becomes greater than 99% for n = 4892.

[Isura]

Nice numbers. Ah, this mathematics thing is a strange beast.