In the card game bridge, each of 4 players is dealt a hand of 13 of the 52 cards. What is the probability that each player receives exactly one Ace? (You may use a calculator to compute the probability, but answer as an exact number. Entering a few decimal places of a nonterminating decimal is not considered exact; if you reach such an answer, enter it as a fraction.)

Guest Apr 9, 2015

#1**+5 **

For the first player:

The total number of possible hands of 13 cards from a deck of 52 is _{52}C_{13}.

The total number of possible hands for this player that contains one ace and twelve other cards:

_{4}C_{1}·_{48}C_{12} (which is choosing 1 ace from 4 aces and 12 other cards from the 48 remaining cards)

So, the probability for the first person getting one ace is: ( _{4}C_{1}·_{48}C_{12 }) / _{52}C_{13 }

For the second player:

Probability = ( _{3}C_{1}·_{36}C_{12 }) / _{39}C_{13 }

because: there are only 3 aces left, only 36 other cards left, and only a total of 39 cards left.

For the third player:

Probability = ( _{2}C_{1}·_{24}C_{12 }) / _{26}C_{13 }

because: there are only 2 aces left, only 24 other cards left, and only a total of 26 cards left.

For the fourth player:

Probability = ( _{1}C_{1}·_{12}C_{12 }) / _{13}C_{13 }

because: there is only 1 ace left, only 12 other cards left, and only a total of 13 cards left.

Multiply these four probabilities together to get the total probability. ...

geno3141 Apr 10, 2015

#1**+5 **

Best Answer

For the first player:

The total number of possible hands of 13 cards from a deck of 52 is _{52}C_{13}.

The total number of possible hands for this player that contains one ace and twelve other cards:

_{4}C_{1}·_{48}C_{12} (which is choosing 1 ace from 4 aces and 12 other cards from the 48 remaining cards)

So, the probability for the first person getting one ace is: ( _{4}C_{1}·_{48}C_{12 }) / _{52}C_{13 }

For the second player:

Probability = ( _{3}C_{1}·_{36}C_{12 }) / _{39}C_{13 }

because: there are only 3 aces left, only 36 other cards left, and only a total of 39 cards left.

For the third player:

Probability = ( _{2}C_{1}·_{24}C_{12 }) / _{26}C_{13 }

because: there are only 2 aces left, only 24 other cards left, and only a total of 26 cards left.

For the fourth player:

Probability = ( _{1}C_{1}·_{12}C_{12 }) / _{13}C_{13 }

because: there is only 1 ace left, only 12 other cards left, and only a total of 13 cards left.

Multiply these four probabilities together to get the total probability. ...

geno3141 Apr 10, 2015