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    Greetings, so just working on some stuff from Probability: An Introduction by Lord Grimmett himself, so i got stuck on this question (early on ) and require assitance from one of you greater beings:


    Show that the probability the each of the four players in a game of bridge receives one ace is

    \displaystyle\frac{24\cdot48!13^  4}{52!}

    Now I can clearly see 13^4 has something to do with 4 hands of 13 cards, and the rest corresponds to the recipricol of \displaystyle \binom{52}{4} but i am unsure as to why these things relate to the probability asked for.

    Help pl0x.
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    The probability that each player gets an ace is equal to the number of ways that each of the four aces land in exactly one of the hands divided by the number of ways the cards can be dealt.

    Can you count out those two numbers?
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    Dealt in anyway, the number of ways the cards can be dealt is clearly 52!.
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    Now, let's suppose we partition the set of the cards into 4 sets of size 13, and let's order each set in the order of that they are dealt to the person, and so each set will represent the cards a person is dealt. Now, take on of these sets. There are 13 ways in which one of them can be an ace in any given set: the first is an ace, or the second is an ace, or ...., or the thirteenth is an ace.

    But surely we can rearrange the aces amongst themselves, and the 'non-aces' amongst themselves?
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    Which ace goes into each set does not matter, so we can exchange the four aces how we want between the four sets - there are 4! ways of arranging them. The remaining 48 'non-ace' can also be rearranged and exchanged, and there are 48! ways of doing this.

    So how many ways are there so that there is exactly one ace in each hand? Hence what is the probability?
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    ah thankyou
 
 
 
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Updated: January 16, 2010
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