Odds of hitting Broadway...

[Ixnert]

After the fourth time I was a victim to this happening in the space of about nine SNGs, I tried to figure out what the odds were, and I'm not sure how.

If I have two non-paired cards 10 or higher in my hand, what are the odds of my making the ace-high straight if I see all five cards?

[Ixnert]

Ok, I think I figured out how to do this, can someone tell me if this looks right?

Assume, without loss of generality (since any two non-paired 10+ cards should be as good as any other), I have JT, and my opponent has no 10+ cards (say, a lower pair).

I need an ace, a king, and a queen.

There are 4 ways to get the ace, out of 50 cards. After that, there are 4 ways to get the king, out of 49, and 4 ways to get the queen, out of 48. And then two cards that can be anything. Then multiply by 5! (= 120) since I don't care about the order.

So (4/50) * (4/49) * (4/48) * 120 = 64/117600 * 120 = 0.065.

If instead of a lower pair, my opponent had, say, AK, it would be:

(3/50) * (3/49) * (4/48) * 120 = 36/117600 * 120 = 0.037.

These seem like the right magnitude to me (unlike the numbers I was getting before, that were in the 35% range), am I making any major mistakes here?

[LetYouDown]

I believe you're double counting:

[C(4,1) * C(4,1) * C(4,1) * C(47,2)]/C(50,5) = 3.265%

If your opponent held A-K, I think it should be:

[C(4,1) * C(3,1) * C(3,1) * C(45,2)]/C(48,5) = 2.08%

Although that would include the times you lose to a boat/quads.

[KJL]

Can someone explain why it is C(4,1) rather than C(16,1)on the first card.

[LetYouDown]

I'm disregarding order. You need the board to contain an A, K and Q. There's 4 Aces, 4 King and 4 Queens which, by themselves can come a total of 64 ways (different suit combinations).

[KJL]

Ohh yea i forgot about that.
Thanks