Discard $2$ cards from a standard deck (shuffled) and $10$ cards are dealt to each player. Find the probability that one in five players has $2$ Aces.

Question

Discard $2$ cards from a standard deck (shuffled) and $10$ cards are dealt to each player. Find the probability that at least one in all of five players in has exactly $2$ Aces.
My plan of solving:
Divide the problem into three case:
If we discard $0$ Aces, the number of ways that a player has $2$ Aces is $_{4}\textrm{C}_{2}\cdot _{48}\textrm{C}_{8}.$
If we discard an Ace, the number of ways that a player has $2$ Aces is $_{3}\textrm{C}_{2}\cdot _{48}\textrm{C}_{8}.$
If we discard $0$ Aces, the number of ways that a player has $2$ Aces is $_{2}\textrm{C}_{2}\cdot _{48}\textrm{C}_{8}.$
So the probability that a player has $2$ Aces is $$\frac{_{4}\textrm{C}_{2}\cdot _{48}\textrm{C}_{8}+ _{3}\textrm{C}_{2}\cdot _{48}\textrm{C}_{8}+ _{2}\textrm{C}_{2}\cdot _{48}\textrm{C}_{8}}{_{50}\textrm{C}_{10}}= 0.367,$$ is this right ? I need to the help. However, other solutions are appreciated and welcome. Thank you.

Answer 1

So...

First, make the simplification that we do not discard any cards. This simpler scenario will have exactly the same probabilities as the original scenario. The discarding of the two cards is a red herring that only serves to confuse people into thinking it is relevant. The only reason why it would become relevant is if we were to have looked at the cards that were discarded and used the information of what we saw being discarded to update our expectation of the probabilities involved. For a smaller problem related to this, see If you draw two cards what is the probability that the second card is a queen.

Now... let us look at the probability that specifically the first player draws two aces. The first player draws ten cards from the deck. Any collection of ten cards are equally likely to have been drawn. Your work attempt above was close, but you were thrown off by the whole "discard two cards" bit and so modified your calculation in an incorrect way.

The probability that player1 has exactly two aces can be seen by the hypergeometric distribution and is:

$$\frac{\binom{4}{2}\binom{48}{8}}{\binom{52}{10}}\approx 0.143116$$

If you insist on keeping track of the cards discarded, the correct calculation would have been to look at each case separately and condition each case on the probability of arriving in that case before adding together per the law of total probability. That would be $$\frac{\binom{4}{0}\binom{48}{2}}{\binom{52}{2}}\cdot\frac{\binom{4}{2}\binom{46}{8}}{\binom{50}{10}} + \frac{\binom{4}{1}\binom{48}{1}}{\binom{52}{2}}\cdot\frac{\binom{3}{2}\binom{47}{8}}{\binom{50}{10}}+\frac{\binom{4}{2}\binom{48}{0}}{\binom{52}{2}}\cdot\frac{\binom{2}{2}\binom{48}{8}}{\binom{50}{10}}\approx 0.143116$$ which of course equals the same as before.

Now, this is the same probability as the probability that player 2 has two aces, or player 3, etc... We can just add these probabilities together. However, in doing so we will have overcounted the cases where two players simultaneously have two aces.

The probability player 1 has exactly two aces while player 2 also has two aces will be:

$$\frac{\binom{4}{2}\binom{48}{8}}{\binom{52}{10}}\cdot \frac{\binom{2}{2}\binom{40}{8}}{\binom{42}{10}}\approx 0.00748$$

Similarly, that is the probability the first and third, or first and fourth, or third and fifth, etc... players simultaneously have two aces each.

Correcting our count per inclusion-exclusion principle, we subtract the probability of two players having two aces each $\binom{5}{2}$ times, corresponding to the number of pairings of players possible that could have had two aces each.

In a more complicated problem, we might have then subtracted too much and would need to look at three players simultaneously achieving the condition, or then further subtracting the probability that four players do, etc... however here it is impossible for more than two players to achieve the desired condition as there are only four aces in the deck. To have three or more players having two aces each would have required a minimum of six aces in the deck.

Our final answer then is:

$$5\cdot \frac{\binom{4}{2}\binom{48}{8}}{\binom{52}{10}} - \binom{5}{2}\cdot\frac{\binom{4}{2}\binom{48}{8}}{\binom{52}{10}}\cdot \frac{\binom{2}{2}\binom{40}{8}}{\binom{42}{10}} = \frac{6939}{10829}\approx 0.640779$$