Problem
A deck of card is shuffled then divided into two halves of 26 cards each. A card is drawn from one of the halves, it turns out to be an ace. The ace is then placed in the second half-deck. The half is then shuffled, and a card is drawn from it. Compute the probability that this drawn card is an ace.
My attempt was,
The second half which contains 27 cards could have:
- 1 ace
- 2 aces
- 3 aces
- 4 aces
Hence the total probability of drawing an ace from this half is:
$$\dfrac{ 1 + 2 + 3 + 4 }{27}$$
However the solution from the book is $$\dfrac{43}{459},$$ and I have no idea how they came up with this number. I personally think this problem is straightforward, but I might have missed something. Could anyone give me a hint? Thank you.
Best Answer
Your answer is more than $1/3$, which is clearly not reasonable. If the four cases that you enumerate were equally likely, the desired probability would be $$\frac14 \cdot \frac{1+2+3+4}{27} = \frac{5}{54},$$ which is considerably more plausible. However, the four cases aren’t equally likely.
If there is just one ace in the larger pack, the first half-deck must have contained all four aces. There are $\binom{48}{22}$ ways to pick $22$ cards to go with the four aces, compared with $\binom{52}{26}$ ways to choose the first half-pack without any restrictions, and $\binom{48}{22}/\binom{52}{26}$ is considerably less than $1/4$.
Can you take it from here?