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astarck
08-07-2005, 11:30 PM
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
08-07-2005, 11:40 PM
[ 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)!

[ 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 ]

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

BruceZ
08-07-2005, 11:48 PM
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
08-07-2005, 11:55 PM
Sorry for the stupid question, but what is the difference between P(52,20) and C(52,20).

thanks

BruceZ
08-08-2005, 12:01 AM
[ 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
08-08-2005, 12:02 AM
Ok thanks.

BruceZ
08-08-2005, 12:08 AM
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
08-08-2005, 07:47 AM

Thanks!

MickeyHoldem
08-08-2005, 08:10 PM
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
08-08-2005, 08:21 PM
[ 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.

MickeyHoldem
08-08-2005, 08:27 PM
Yes... thank-you... I calculated something else.... maybe you could check it....

The over/under for seeing a particular hand.

BruceZ
08-08-2005, 10:10 PM
[ QUOTE ]
Yes... thank-you... I calculated something else.... maybe you could check it....

The over/under for seeing a particular hand.

[/ QUOTE ]

To have a 50% chance of seeing a particular hand, you need 8.36 * 10^36 hands or 2.65 * 10^29 years.

If N is the number of hands, then

[(N-1)/N]^n = (1 - 1/N)^n = (1 - 1/N)^(N*n/N) =~ exp(-n/N) = 0.5

n = -N*ln(0.5) =~ -(1.21 * 10^37)*(-0.693)

= 8.36 * 10^36 hands or 2.65 * 10^29 years.

astarck
08-08-2005, 11:36 PM
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Now getting any hand twice wouldn't take nearly as long.

[/ QUOTE ]

Does this mean that you will be far more likely to get duplicate hands in your search for a specific hand, but not THE specific hand in question?

BruceZ
08-09-2005, 01:57 AM
[ QUOTE ]
[ QUOTE ]
Now getting any hand twice wouldn't take nearly as long.

[/ QUOTE ]

Does this mean that you will be far more likely to get duplicate hands in your search for a specific hand, but not THE specific hand in question?

[/ QUOTE ]

Yes. It's like the birthday problem. You need an average of 365 people before you find one that shares your birthday, but you only need 23 people to have a 50% chance that any two people share a birthday.

It should still take an astronomical amount of time to get a match. I'll work that out later.

MagicMan08
08-09-2005, 02:46 AM
I wish i took more math, maybe i should.