Odds of getting one Ace (math help)

[EdSchurr]

Could someone explain to me how you find the odds for exactly one Ace to be dealt, and for exactly two Aces to be dealt preflop to any of 18 cards. Assume two of the kings are missing.

I can't get my simple AND and OR probability math to work here. It looks like one needs the Choose function, but I can't really figure it out.

[PokerSlut]

Do you mean two aces to the same player, or two aces among all 18 cards?

[EdSchurr]

Two situations: the one ace dealt to any one of the players, exactly (not one or more Aces); and, the two aces to any of the players (one player could have both, or one player could have one if another player has one).

This is to get at the KK versus Axx flop question w/ Bayes's Theorem. But I need to find these initial odds first.

[MKR]

[ QUOTE ]
Two situations: the one ace dealt to any one of the players, exactly (not one or more Aces); and, the two aces to any of the players (one player could have both, or one player could have one if another player has one).



[/ QUOTE ]

1. Situation one, the probability of exactly one ace in 18 cards: You want one ace out of four aces and 17 non-aces out of 48 cards. You can choose the ace in C(4,1) ways and the non-aces in C(48,17) ways, so you can choose one ace and then choose 17 non-aces in C(4,1)*C(48,17). Since you can choose 18 cards out of 52 in C(52,18) ways the probability of exactly one ace is C(4,1)*C(48,17)/C(52,18) or .398 (about 3:2 against)

2. Situation 2, the probability of exactly two aces in 18 cards: You can choose 2 aces in C(4,2) ways and you can choose 16 non aces in C(48,16) ways and the probabibiliy of two aces in 18 cards is C(4,2)*C(48,16)/C(52,18) = .317 or about 2.15:1 against.

C(n,r) stands for n!/((n-r)!*r!) so C(4,2) = 4!/((4-2)!2!) = 6.

MKR

[EdSchurr]

Thanks! Makes perfect sense.