Odds of Being Dealt Same Hand on 2 Tables At Same Time

[GrunchCan]

I was dealt 8[img]/images/graemlins/club.gif[/img]8[img]/images/graemlins/diamond.gif[/img] in that order at two tables at the same exact moment.

What're the odds?

I fugured them to be:

52 * 51 * 2 = 5304:1

Seems like much better odds than I would expect, since in 100k hands this is the first time I remember this happening.

[Tom1975]

That's correct, since you stated 'in that order'. If the order didn't matter you would divide by two.

[GrunchCan]

Actually, thinking a little more, I think the odds were even better than I first thought.

Your'e dealt 8[img]/images/graemlins/club.gif[/img]8[img]/images/graemlins/diamond.gif[/img] on table A.

What are the chances of being dealt 8[img]/images/graemlins/club.gif[/img] on table B? 1/52
Having been dealt one card, what are the chances that the second card on table B is 8[img]/images/graemlins/diamond.gif[/img]? 1/51

Applying the mutiplication rule, the chance of being dealt the exact same hand in order at 2 tables while 2-tabling are: (1/52)(1/51) = 2652:1

Funny I don't remember this ever happening. Then again, maybe not, since most deals are very unremarkable like 7[img]/images/graemlins/spade.gif[/img]2[img]/images/graemlins/diamond.gif[/img]

[ QUOTE ]
Actually, thinking a little more, I think the odds were even better than I first thought.

[/ QUOTE ]

If you consider 8 [img]/images/graemlins/diamond.gif[/img] 8 [img]/images/graemlins/club.gif[/img] to be the same hand as 8 [img]/images/graemlins/club.gif[/img] 8 [img]/images/graemlins/diamond.gif[/img] then the odds are twice that (1 in 1326)

[LetYouDown]

If you're saying that you already have a specific hand at one table and you're looking for the odds of getting the same hand at another in that order:

1/52 * 1/51 = 1/2652 = 2651 to 1. Not sure why you're multiplying by 2...but my brain hurts right now.