Hi everyone. I play cards with friends, and I ran into this problem:

Say I have a deck of 52 cards. Now I add one joker to the deck. Now I pull the cards one at a time, checking the card each time. What are the odds that I draw the joker afternumber of draws?n

Logic tells me that it has to add up to 1 after 53 draws, but I don't know the formula. I tried a few things, but none of them made sense (like adding to one after 20 or so draws). My next intuition says something like, for the first 3 draws:

$\displaystyle \frac{1}{51}${$\displaystyle 1-[\frac{1}{52}(1-\frac{1}{53})]$}

But I could be way off my rocker here. I could test this formula, but I don't mean to be petulant when I say that there seems to be no easy way to run through this formula...53 iterations is really complicated. If this is the solution, or even if it is not, is there an easier way to do the math?

Just to be clear, I am not looking necessarily for the 53rd iteration, in case anyone was thinking of doing something tedious. I'm just interested to know the general formula, and perhaps a simpler way of calculating it if the one I gave is correct (hopefully it's not!)

Incidentally, what if there were 2 jokers in the deck? How would the probability for the 2nd joker change after drawing the first one? I am totally lost on that one!

Thanks for any tips/advice! Links of course are also appreciated. I tried searching for this particular question, but everything I put it sends me to simpler probability problems...or at least I don't think they apply to mine. Thanks for reading!

EditWhen I tried a different approach, imagining a deck of 2 cards with 1 joker, I realized that my formula wouldn't work. In light of that, the only thing that I can think of is that your probability to draw the card remains constant and equal to the initial probability, despite having removed cards from the deck. In other words, after 10 draws in a 53 card deck, my probability would be 10/53 to have drawn the card, even though on that 10th draw my probability is only 1/43. Is this correct?

But that would still leave a bit of a dillema in that if I were to add up the individual probability for each of the first 10 draws, I would not have 10/53 by the end. So I guess back to square one.

For example, if I had 3 cards in a deck, one of which is a joker, then I have a 2/3 chance to draw the joker after 2 draws; however, on the second draw, I must have a 1/3 chance to draw it, even though only two cards are remaining. This is because I could not add 1/3 from the first draw, then 1/2 from the second draw to reproduce the 2/3 total. This seems like an extended example of that game show with the 3 curtains, but my logic is faltering at that point where I have 2 cards in my hand and only a 1/3 chance to pick the joker instead of 1/2. That doesn't make sense.

Anyone with some advice for the correct direction? Thanks!