Probability of straight flush at full table

[Dex]

Assume 10 players at a hold 'em table. All players are dealt in, the board is dealt, and all hands go to a showdown.

What is the probability that at least one player has a straight flush?

[cwes]

Let me give it a try.

There are ten different straights, namely A-5 to 10-A. There are 4 different colors. Hence there are 40 5-card combinations that make a straight flush.
In Hold'Em you get 7 cards, but you do not care about those two other cards, so they may be any of the remaining 47 cards.

So there are '40 * (47 choose 2) = 43,240' seven card combinations that make a straight flush. Divide them by the total amount of card combinations '52 choose 7' and you get the probability for a hand becoming a straight flush ex ante. Call it 'Pr(SF)'.

The probability of a hand NOT becoming a straight flush obviously is 'Pr(No SF)=1-Pr(SF)'. The probability of none of the ten players getting a straight flush then is 'Pr(No SF)^10'. This makes the probability of at least one straight flush '1-Pr(No SF)^10 = 310^-1'.