Poker 10 Decks, 10 cards, independent 4 Aces

[mattmns]

Silly question of very little importance

My dad just asked me the following question: You have 10 standard independent decks (52 cards), you draw 1 card from each deck 1 at a time, what is the probability (count) of getting 4 aces (not necessarily distinct suits) out of the 10 cards you draw.
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I am thinking that the number of ways to get such a "hand" of 10 cards is the following:
$$\binom{10}{4}\left(\frac{4}{52}\right)^4\left(\frac{48}{52}\right)^6$$

Reasoning being that you want 4 aces and there are 4 per deck, hence the (4/52)^4 and for the other 6 cards you want you have (48/52)^6, then 10 choose 4 for all the different ways you can get the 4 aces out of the 10 cards.

And if we wanted the aces to have distinct suits we would have 4! instead of 4^4

Does this look correct? Thanks.

 

[mighty2000]

Yes, your reasoning and calculation look correct. You have correctly accounted for the number of ways to choose 4 aces out of 10 cards and the probability of getting each ace in the chosen cards. The only thing to note is that the probability should be multiplied by 10 in order to account for the different ways the 4 aces can be chosen out of the 10 cards. So the final probability would be:

10 * \binom{10}{4}\left(\frac{4}{52}\right)^4\left(\frac{48}{52}\right)^6

For the case of distinct suits, you would indeed multiply by 4! instead of 4^4. Great job on your calculation!