Question.

I came up with [(2/52)(1/51)]^3

When are you starting you 3 hands in a row? From when you first receive Ah7h? Or from the next deal?

I guess I would be interested to know the results for both, thanks.

Presumably you witnessed this sequence of hands? So as a7 suited is not a particularly remarkable hand, I'd guess what would have caught your attention and produced a similar query would be, a suited hand, followed by the same suited hand of a different suit and finally the same hand again of a different suit from the first two?

A common fallacy is "I saw a number come up twice in a row at roulette - that's well over 1000-1". When what they witnessed was only 36-1*, because the first number that "came up twice" could have been any number.

I can't remember off hand how many suited hands there are, I think it might be 312, so if that's right, the prob of "a suited hand, followed by the same suited hand of a different suit and finally the same hand again of a different suit from the first two" would be 312/1326 x 3/1326 x 2/1326.

It's also worth considering that the longer you play the more "unlikely" sequences of hands you will notice, but it's just meaningless background noise.

*single zero

Hi, thanks for the responses.

Yeah that’s what happened,dealt the first A7, then the next two in the next two hands.

No tinfoil hat here, just genuinely interested in the odds. I guess if it’s something like 1:1000000 then maybe I’m a chance to win the lottery after all? ��

The thing is, if you play enough, it would be extremely unlikely that it never happens. It is extremely unlikely that if you picked three random hands you would get them in the next three hands. It is almost certain that you would get them in order at some point.