consider this problem:
Draw cards repeatedly, without replacement, from a standard 52-card deck until we find the ace of spades. What is the probability that we draw between 20 and 30 cards?
The solution I came across:
$$P(A)=\frac{11}{52}$$
where the numerator denotes the sum from the 20th to the 30th draw, and the denominator the sum of all possible draws from the 1st until the 52nd.
I've thought about this problem a lot, and I just do not understand the thinking put behind it, this problem clearly requires more attention, order does matter and the fact that the cards drawn can not be replaced made me doubt the above solution.
The solution that I believe is correct:
$$P(A)=\frac{1}{33}+\frac{1}{32}+\frac{1}{31}+…+\frac{1}{23}$$
Why?
Well the first 19 drawn cards must have resulted in other values than the ones required and considering the fact that cards drawn can not be replaced, the denominator or sample space always decreases by 1 until exactly the criterion of having to draw between 20 and 30 cards is met.
I would like to be assured if first of all the solution I gave is correct, or corrected if I am wrong, I would also appreciate if you could provide other helpful ideas, methods which I could to tackle these kind of problems.
Best Answer
When you shuffle the deck, the ace of spades is as likely to be in any spot as any other. You win if it is in $11$ of the $52$ spots. You can do the calculation your way, but you have to start with the chance that the ace is not drawn in the first $19$ cards. The chance of that is $\frac {51}{52} \cdot \frac {50}{51} \cdot \dots \cdot \frac {33}{34}=\frac {33}{52}$ Then the chance that it is in the next $11$ cards is $\frac {11}{33}$. The product of these is $\frac {11}{52}$ as it should be. If you want to keep going your way, the chance it is the $20$th card given it has not been found is $\frac 1{33}$. The chance it is the $21$st card, given that it is not found by card $19$, is $\frac {32}{33} \cdot \frac 1{32}=\frac 1{33}$ You can keep going this way, but it is a much harder approach and will still come out $\frac {11}{52}$ in the end.