Problem :
An experiment consists of removing the top card from a well shuffled pack of 52 playing cards. Calculate the probability that (a) the first card is an ace, (b) the next card is an ace.
Attempt:
The (a) is quite simple, there are only 4 aces, so the probability that I get an ace at first attempt is $4/52=1/13$.
The (b) part, we can see that if the first one is not an ace, then there are 48 choices, for the 1st take, and 4 ace choices for the 2nd take. So the probability is $$ \frac{48 \times 4}{52 \times 51} $$
But the answer key for (b) says the probability is $1/13$. If we look in different point of view, say a stack of 52 cards after a shuffle, then the probability that the card at the 2nd place on top is an ace is $1/13$.
Is my analysis accuratr or not? Thanks.
Best Answer
Your reasoning is okay, but the first card may be an ace, or it may not.
The probability that the first is an ace and so is the second is $\tfrac {4}{52}\tfrac{3}{51}$
The probability that the first is something else and the second is and ace is $\tfrac {48}{52}\tfrac{4}{51}$
You need to add them together (re: Law of Total Probability) to get the probability that the second card is an ace whatever the first may be. And since $3+48$ conveniently equals $51$... the result is :$$=\dfrac 1{13}$$
As anticipated.
Hey, now, what is the probability that the fifteenth card down the deck is an ace?