# Math Probability question please help?

Mr. A and 3 other players are dealt 5 cards each from an ordinary deck. Given that Mr.

A has gotten exactly 1 ace, what is the conditional probability that none of the other 3

players has more aces than Mr. A?

### 1 Answer

- MathMan TGLv 710 years agoFavorite Answer
Assume Mr A's hand has been dealt,

leaving 3 aces and 44 other cards to deal to the others.

Now choose one of the other players to receive more aces than 1:

3 choices for that player.

He can receive 3 aces or 2 aces.

He gets 3 aces in 1 way * 44C2 ways to choose his other cards

or he gets 2 aces and 3 other cards in

3 (ways to choose the aces) * 44C3 (ways to choose the other cards)

So out of the total number of deals,

the number where one of other players has 2 or 3 aces =

3 * (1*44C2 + 3*44C3)

The total number of ways this designated player can be dealt any

hand is 47C5 (since Mr A's hand has already been dealt).

(We can also figure how many ways to deal Mr A's and the other two players hands:

4 * 44C4 and (42C5 * 37C5) respectively, but then those factors cancel out

of the probability because they are included in both the "someone has 2 or 3 aces"

calculation AND the "how many deals in total" calculation.)

The probability that one of the players has 2 or 3 aces is then

3 * (44C2 + 3 * 44C3) / 47C5 =

= 0.07956 that someone does have more aces than Mr A,

and so it is 92.044% that none of them do. ◄ Answer

I also did a simulation, and it confirms this result.

(Three trials of 100000 deals yielded 7831, 7954, and 7844 times

when there was a hand with 2 or 3 aces among the other three.)

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