Originally Posted by diabloblanco
That story from the OP really actually is kind of freaky if you figure the probability of hitting the same random hand on 2 seperate tables twice in a row. That's math I can't even do.
Originally Posted by robwhufc
Chance of being dealt 2 lots of AA at same time is therefore 48,400/1 (220 x 220).
The odds robwhufc calculated are frighteningly huge, but that is largely because the odds of catching specifically AA + AA + 82 + 82 are that huge. Diablo's question hinted at the much more general same hand, two tables, twice in a row - not specifically this sequence. The possibility of that is much larger.
Hands number 1 and 3 can be anything. It's 2 and 4 we need to have specifics on. 1326 unique starting hands (not counting order of cards - 52 choose 2), just means we get 1326^2 to 1. Or 1,758,276 to 1.
But that's if you need it to be exactly the same cards. If it just has to be two cards with the same value (but not suit), chances go up. Of course, that gets trickier to calculate depending on what the requirements are. Does the first hand need to be a pair?
The probability of getting the same hand (if we don't require the suits to be the same) twice in a row is 1.16%. The probability of THAT happening twice in a row is therefore 1.16% ^2 or .013%. Which is 7403 to 1, and therefore well within the boundaries of likelyhood.