Let's say we have a deck of cards excluding face cards, so cards from Ace to 10.
Which of these is the correct way of computing the probability that the sum of 4 randomly chosen cards is equal to 5?
Method #1:
$P(X=5)=\frac{8}{40}\cdot\frac{7}{39}\cdot\frac{6}{38}\cdot\frac{4}{37}$
My logic is that for the first card, there are 8 possibilities. 4 Aces and 4 Two's. For the second card, if the first card was an Ace, there are, 7 possibilities. 3 Aces and 4 Two's. If the second card was an Ace, the third card can be chosen from 2 Aces and 4 Two's. Finally, if the third card was an Ace, the fourth card has to be a Two, so its sample space is just 4 Two's.
Method #2:
$P(X=5) = \frac{8}{40}\cdot\frac{4}{39}\cdot\frac{3}{38}\cdot\frac{2}{37}$
Like Method #1, the first card can be chosen from 4 Aces and 4 Two's. This time however, we will assume that the first card turned out to be a Two. By definition, all the remaining cards have to be Aces.
Method #3:
$P(X=5)=\frac{4}{40}\cdot\frac{4}{39}\cdot\frac{3}{38}\cdot\frac{2}{37}$
We know that in order to have a sum of 5, we specifically need 1 Two card and 3 Ace cards. The first fraction represents the probability of choosing a Two, and the rest pertains to Aces.
I think Method #3 is the most plausible of them all, but I still feel that they're all wrong and that I should just use the hypergeometric formula to find out the probability.
Best Answer
As other responses have indicated, the first two choices are wrong.
The third choice, which (like the other two choices) is attempting to use a probability-oriented approach, is also wrong, specifically because it wrongly assumes that the deuce must be the first card chosen. In fact the deuce can be chosen 1st, 2nd, 3rd, or 4th.
Staying with the probability-oriented approach, and relying purely on intuition to remedy the third choice, you could reason that by considerations of symmetry, the third choice is off by a factor of $~4.~$
The idea is that by considerations of symmetry, if you are given that the first four cards chosen are a deuce and three aces, in some order, the deuce is just as likely to have been the 1st card chosen as the 2nd card chosen, the 3rd card chosen, or the 4th card chosen.
In fact, I will formally demonstrate that the third approach is off by a factor of $~4.~$ That is, I will first show how you would use a probability-oriented approach to the problem, and then I will show how you would use a combinatorics approach to the problem.
$\underline{\text{Probability Oriented Approach}}$
There are $~4~$ mutually exclusive cases to deal with, depending on whether the deuce is the
1st card drawn:
The probability of this happening is
$\displaystyle \frac{4}{40} \times \frac{4}{39} \times \frac{3}{38} \times \frac{2}{37}.$
2nd card drawn:
The probability of this happening is
$\displaystyle \frac{4}{40} \times \frac{4}{39} \times \frac{3}{38} \times \frac{2}{37}.$
3rd card drawn:
The probability of this happening is
$\displaystyle \frac{4}{40} \times \frac{3}{39} \times \frac{4}{38} \times \frac{2}{37}.$
4th card drawn:
The probability of this happening is
$\displaystyle \frac{4}{40} \times \frac{3}{39} \times \frac{2}{38} \times \frac{4}{37}.$
So, you have $~4~$ separate terms that must be added together, where each term is equal to
$$\frac{4 \times 3 \times 2 \times 4}{40 \times 39 \times 38 \times 37}.$$
Therefore, the correct answer is
$$\frac{4 \times 3 \times 2 \times 4}{40 \times 39 \times 38 \times 37} \times 4 = \frac{16 \times 4!}{40 \times 39 \times 38 \times 37}.$$
$~\underline{\text{Combinatorics Oriented Approach}}$
Determining the probability combinatorically, you have that the number of different ways of selecting one deuce and three Aces is
$$\displaystyle N = \binom{4}{1} \times \binom{4}{3} = 16.$$
Then, the number of different ways of selecting any four cards from the deck is
$$\displaystyle D = \binom{40}{4} = \frac{40 \times 39 \times 38 \times 37}{4!}.$$
Therefore, the overall probability is
$$\frac{N}{D} = \frac{16 \times 4!}{40 \times 39 \times 38 \times 37}.$$