Number of hand combinations

[astarck]

What are the odds that at a 10-handed table everyone will be dealt a certain hand, and all the community cards will be the same?

Also, if each hand can be played out in 1 second, how long on average until you reach the same combination of hands?

[BruceZ]

[ QUOTE ]
What are the odds that at a 10-handed table everyone will be dealt a certain hand, and all the community cards will be the same?

[/ QUOTE ]

I assume you mean 5 particular board cards in a certain order.

P(52,20)/2^10 * P(32,5) =~ 7.23 x 10^36 to 1.

where P(n,k) = n!/(n-k)!




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Also, if each hand can be played out in 1 second, how long on average until you reach the same combination of hands?

[/ QUOTE ]

With the same board, about 2.29 x 10^29 years.

[BruceZ]

If you don't care about the order of the 3 flop cards, then divide those results by 6 to get 1.2 x 10^36 and 3.8 x 10^28 years.

[KJL]

Sorry for the stupid question, but what is the difference between P(52,20) and C(52,20).

thanks

[BruceZ]

[ QUOTE ]
Sorry for the stupid question, but what is the difference between P(52,20) and C(52,20).

thanks

[/ QUOTE ]

P is permutations, so it counts all possible orderings of the 20 cards. C is combinations, which ignores order. C(52,20) = P(52,20)/20! = 52*51*50*49*...*33/20!

[KJL]

Ok thanks.

[BruceZ]

I also typed the definition of P wrong in the original post, but it's fixed now.

P(n,k) = n!/(n-k)!

C(n,k) = n!/(n-k)!/k!

P(52,20) = 52*51*50*49*...*33 = 52!/(52-20)!

C(52,20) = 52*51*50*49*...*33/20! = 52!/(52-20)!/20!

[astarck]

As usual Bruce, great reply.

Thanks!

[MickeyHoldem]

Bruce.... I like your numbers but I'll take on another *10 for the placement of the button in relation to the hands .... and

[ QUOTE ]
Also, if each hand can be played out in 1 second, how long on average until you reach the same combination of hands?

[/ QUOTE ]
Hmmmm... does the OP want the average time until a hand repeats itself given a random shuffle... if so I get 1.55 x 10^29 years (with unique button placement)

[BruceZ]

[ QUOTE ]
Bruce.... I like your numbers but I'll take on another *10 for the placement of the button in relation to the hands .... and

[ QUOTE ]
Also, if each hand can be played out in 1 second, how long on average until you reach the same combination of hands?

[/ QUOTE ]
Hmmmm... does the OP want the average time until a hand repeats itself given a random shuffle... if so I get 1.55 x 10^29 years (with unique button placement)

[/ QUOTE ]

It should just be my number of years times 10, or 3.8 x 10^29 years. If there are N possible hands, the probability of getting a particular hand again is 1/N, so on average that will take N hands.

Now getting any hand twice wouldn't take nearly as long.