The question is to show that the probability that each of the four players in a game of bridge receives one ace is $$ \frac{24 \cdot 48! \cdot13^4}{52!}$$ My explanation so far is that there are $4!$ ways to arrange the 4 aces, $48!$ ways to arrange the other cards, and since each arrangement is equally likely we divide by $52!$. I believe the $13^4$ represents the number of arrangements to distribute 4 aces among 13 cards, but I don't see why we must multiply by this value as well?
[Math] The probability that in a game of bridge each of the four players is dealt one ace
combinatoricsprobability
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Best Answer
Imagine $4$ groups of $13$ spots for the $4$ hands
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An ace can be placed in each group in any of $13$ places, hence $13^4$
The aces' suits can be distributed between groups in $4!$ ways, hence $24$
The remaining cards can be placed in 48! ways,
and 52! is the unrestricted ways of placing the cards
thus (putting the terms in the order you have), $Pr = \dfrac{24\cdot48!\cdot13^4}{52!}$
SIMPLER WAY:
There is a much simpler way to get the same result.
We need an ace in each group of 13, how the rest of the cards go doesn't matter !
The first ace has to be in some group, each of the other aces have to fall in a different group, so the $2nd$ ace has $39$ permissible spots out of $51,$ and so on
thus $Pr = \dfrac{39}{51}\cdot\dfrac{26}{50}\cdot\dfrac{13}{49}$