# Question regarding Conditional Probability with a deck of card

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.

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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?

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Thanks a lot for your hint. I understand the idea, however I thought the number of aces "after" putting into the second-half deck should be fixed. The part that I don't understand is why do we have to take the rest of the cards (in 1st half and 2nd half) into account? –  Chan Oct 2 '11 at 0:30
@Chan: Because the original distribution of the aces determines how many aces are in the $27$-card pack. If all $4$ aces were originally in the first half-pack, you’ll have only one chance in $27$ of drawing an ace, but that situation is pretty unlikely. It’s much more likely, for instance, that the aces were originally split evenly between the half-packs, in which case you’ve $3$ chances in $27$ of drawing an ace. Since these two cases aren’t equally likely to occur, you can’t simply average $1/27$ and $3/27$ (and the same goes for the other two cases). –  Brian M. Scott Oct 2 '11 at 0:37
Thanks a lot. Now it makes sense, I keep forgetting the "equally likely" property. –  Chan Oct 2 '11 at 0:47

The probability of having $i=0,1,2$ or $3$ ace in the second half-deck when dividing the whole deck at the beginning, is multinomial: $$\frac{{4 \choose i}{48 \choose 26-i}}{{52 \choose 26}}$$ then...

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The second deck before adding the ace contains 26 cards all of which have a chance of 3/51 to be an ace. The split is irrelevant for the probabiltiy of the other cards being an ace, there are 3 unseen aces remaining out of 51 possible cards. Now you add the ace to a deck of 26 cards creating a 27 card deck and draw from that. You then have a probability of 1/27 to draw that ace but 26/27 to draw one of the other cards which still have a 3/51 chance of being an ace, all that is known about them is that they are not the one specific ace.

Thus the result is: 1/27 * 1 + 26/27 * 3/51 = 17/459 + 26/459 = 43/459

It's not neccesary to think about the distribution of aces between the decks happening by the split. Look at it this way, if you draw a card from a deck it is 1/52 chance to be an ace regardless of burning cards first. Splitting the deck can be seen as just burning half the deck.

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