Starting Hands

[CredoFear]

Does anyone know the chances of you being dealt the same starting hand in a hold'em game 3 times in a row? Or even 4 or more times in a row. If no one knows exactly could some of you venture a guess as to what the chances are of it happeneing 3 or more times in a row.

[Ponks]

Depends what the hand is.
Suited?
Pocket Pair?
Unsuited?
I dont know off the top of my head, but it doesnt seem like it would be that hard to calculate.

If I wasnt playing right now I'd offer to help more, but maybe someone else can.

Ponks

[Songwind]

[ QUOTE ]
Does anyone know the chances of you being dealt the same starting hand in a hold'em game 3 times in a row? Or even 4 or more times in a row. If no one knows exactly could some of you venture a guess as to what the chances are of it happeneing 3 or more times in a row.

[/ QUOTE ]

The exact same hand, or just the same face values?

Exact same hand would be:
((1/52) * (1/51)) ^ 3 = 18651791807 to 1

Same face values, assuming it's not a pair would be:
((4/52) * (4/51)) ^ 3 = 4553659 to 1

Pair, suits unimportant would be:

((4/52) * (3/51)) ^ 3 = 10793861 to 1

[Jdanz]

those numbers are way off, since what the first hand is doesn't really matter it should be ^2 not ^3, and the chance are way way better then those listed.

[Songwind]

[ QUOTE ]
those numbers are way off, since what the first hand is doesn't really matter it should be ^2 not ^3, and the chance are way way better then those listed.

[/ QUOTE ]

One particular hand three times in a row means the first hand does matter.

If you don't care what you start with then you are of course correct.

[ScottyAA]

Those numbers are way off...
however the probability is different for pocket pairs and other cards. so i show the probability to be dealt pocket aces 3 times in a row.

assuming 3 trials,
the probability that you will be dealt pocket aces 3 times is
6^3
----
(52!/2!*50!)^3

for four times in a row replace 3 witha 4, the probabilities for other hands is different
-scottyAA

[kem]

[ QUOTE ]
Exact same hand would be:
((1/52) * (1/51)) ^ 3 = 18651791807 to 1

[/ QUOTE ]

I think you're confusing the probability of being dealt the same hand, and the same hand in the same way. For instance, you're dealt A/images/graemlins/heart.gif K/images/graemlins/heart.gif, don't you want to count the case where the next hand you are dealt K/images/graemlins/heart.gif first, and then the A/images/graemlins/heart.gif second? In this case, it's (2/52)*(1/51) or 2/2652.. First you need one of two cards (A/images/graemlins/heart.gif or K/images/graemlins/heart.gif), then you need one exact card (whichever remains from A/images/graemlins/heart.gif K/images/graemlins/heart.gif)..

And as someone pointed out, you don't count the first time it happened -- similar to the problem: what's the probability that the guy standing next to you has the same birthday as you? It isn't (1/365)^2.... it's 1/365. Your birthday is already fixed, just like the first standing hand is fixed.