Suppose I keep drawing cards from a deck of cards(with replacement, meaning after I draw a card, I switch a new deck of cards but keep the card I draw) until I have drawn both a 1 and a 2. Then What is the expected number of times I would have to draw cards until I can stop?
I have no clue about how to compute such expectation numbers.
On
Assuming a 52-card deck with 4 1s (aces) and 4 2s, this is relatively simple to calculate.
Each draw there is a 2/13 chance you will draw either a 1 or a 2. After you have drawn either a 1 or 2, you have a 1/13 chance of drawing whatever the other number you need to draw is.
IDK but according to WNG (the other answerer) it takes 1/p times on average before succeding. So 13/2 + 13/1 = 21.5
Apparently it will take on average 21.5 draws to get both a 1 and a 2. Seems kindof high to me so IDK if this is right :(
Edit: 13/2 + 13/1 is actually 19.5
If you try something with probability $p>0$ of succeeding, you'll have to try on average $1/p$ times before succeeding (cf https://en.wikipedia.org/wiki/Geometric_distribution )
Thus, with $n\ge2$ cards in the deck, you'll draw the first 1 or 2 in $\frac{n}{2}$ tries and the second one in $n$ tries, so $\frac{3n}{2}$ is the answer