Math question, extra credit (semi-lo content)

[SonOfAkira]

So if you play poker long enough you will see every possible scenario, though this one is limited to onine play:

On two tables, at the same time, I am dealt:



They were blue, however. This is incidental.
This is a hand I rank high in terms of the suited 1-gappers (close behind J9s, the uber nuts) and wouldnt mind seeing a cheap multiway flop with, preferably with position. I manage this on both tables (button, CO) and the flop comes down on the first table:



and on the second table:



Weird. I win a small pot on one and a mid-large pot on the other. This is also incidental.
The question is, obviously, what are the odds of this happening, being dealt the exact same hand on two tables at the same time and having the flop come down nearly identical, numbered identically with only one pair being unsuited?
Also, how much more improbable is the the flop being identical, or in the same order (Q, 10, 2)?
Or both hand being dealt in the same position?
Or all of the above?

This isnt an official contest; I won't be sending anybody $5 in their poker stars account (I don't have one), though if I can't figure it when I tackle it tomorrow, maybe i'll think of something; besides, of course, the knowledge being our reward.

[WildBobAA]

28.9%

[swiggidy]

28.9%

I'll take the under

[johnnyBuz]

.289%
.0289%

maybe one of those.

[pokerroomace]

The odds of being dealt the exact same hand on both tables in any order is 1 in 2,300,000. The same as being dealt a royal flush of spades (or a particular suit) with 5 cards.

[Pyroxene]

The question is...

The odds of a specific table being dealt the exact same two card hand that is in play at another specific table is:

1 in 52!/((50!)*(2!)) or 1 in 1326 (.0754%)

the odds of the same hand happening and then matching the 3 card flop are:

1 in 1326 * 50!/((47!)*(3!)) or 1 in 25,989,600.

the general odds of the same hand happening and then numerically matching the 3 card flop with 2 matching suits and 1 unmatched (and no pairs) are:

1 in 1326 * 50!/((47!)*(54)) or 1 in 2,887,733.3 (the key here is that there are 54 combinations of the same ranks with 2 matching suits and 1 unmatched).

the more specific odds of all that happening and flopping two pair matching your hole cards

1 in 2,887,733.3 * 50!/((47!)*3*3*44*6) or 1 in 142,928,213

The odds of being dealt the exact same hand on both tables in any order is 1 in 2,300,000. The same as being dealt a royal flush of spades (or a particular suit) with 5 cards.

I want to point out that this is not quite the questsion that was asked. While I agree that the odds of being dealt the same hand on both tables are 1 in 2,598,960 (52!/(47!*5!)), that would match situations where the hole cards on table one are :As: :Ks: and the flop is :Qs: :Js: :Ts: and the hole cards on table 2 are :Ts: :Js: and the flop is :Qs: :Ks: :As:. His question was a bit more specific that the hole cards had to match and then the flop had to match; which, of course, is less likely by a fair amount.

[SonOfAkira]

28.9%

Detention, smart-aleck.

[Monty9]

Not that often??

I had a situation happen to me the other night where I was delt AK on two tables and was beaten and busted out by JJ on both tables.

This reminds me of the question:

If a chicken and a half laid an egg and a half in a day in a half...How long does it take a grasshopper with a wooden leg to kick the seeds out of a dill pickle??

[swiggidy]

If a chicken and a half laid an egg and a half in a day in a half...How long does it take a grasshopper with a wooden leg to kick the seeds out of a dill pickle??

28.9%

[Monty9]

Guess it was that obvious.... gg swiggidy

[pokerroomace]

The question is...

The odds of a specific table being dealt the exact same two card hand that is in play at another specific table is:

1 in 52!/((50!)*(2!)) or 1 in 1326 (.0754%)

the odds of the same hand happening and then matching the 3 card flop are:

1 in 1326 * 50!/((47!)*(3!)) or 1 in 25,989,600.

the general odds of the same hand happening and then numerically matching the 3 card flop with 2 matching suits and 1 unmatched (and no pairs) are:

1 in 1326 * 50!/((47!)*(54)) or 1 in 2,887,733.3 (the key here is that there are 54 combinations of the same ranks with 2 matching suits and 1 unmatched).

the more specific odds of all that happening and flopping two pair matching your hole cards

1 in 2,887,733.3 * 50!/((47!)*3*3*44*6) or 1 in 142,928,213

The odds of being dealt the exact same hand on both tables in any order is 1 in 2,300,000. The same as being dealt a royal flush of spades (or a particular suit) with 5 cards.

I want to point out that this is not quite the questsion that was asked. While I agree that the odds of being dealt the same hand on both tables are 1 in 2,598,960 (52!/(47!*5!)), that would match situations where the hole cards on table one are :As: :Ks: and the flop is :Qs: :Js: :Ts: and the hole cards on table 2 are :Ts: :Js: and the flop is :Qs: :Ks: :As:. His question was a bit more specific that the hole cards had to match and then the flop had to match; which, of course, is less likely by a fair amount.

Ye. I didn't think of that. I meant to write 1 in 2.6mil. But it's higher than that as you said, because the two cards in your hands on both tables have to match as well.

[siknd]

the chances might be better than they appear because these calculations are assuming that he is playing ONLY two tables. if hes playing 8, say, then this is hardly remakable. in fact, im suprised it never happened sooner.



i have no superstitions, really, but often when im dealt the same hand simultaneously i think it might be lucky so ill play both. this was fun for awhile, but then it just cost money.

last night i got 98o three times in a row at the same table. boy was i getting bored with that.

[swiggidy]

If he is playing 8 tables then there are 8!/(6!*2!) = 8*7/2 = 28

so take 1 in 142,928,213, divide by 28

1 in 5 104 579, I would not call that common