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I'm having a rough time finding 4 different probabilities. Say you have a deck of only 4 different cards, you draw one card, put it back in the deck & shuffle it. You continue drawing one card, put it back & shuffle the deck until you draw one card you've seen before.

I want to find the probabilities of the cycle ending at the 2nd, 3rd, 4th & 5th draw.

I realize that you will draw at least 2 cards out of the deck before ending the cycle, in the case of drawing the same card both times. And if you get to the 5th card the cycle will end because that means you've drawn all 4 different cards once.

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The game ends on the second draw if your card happens to be the one of four that you've already seen: $\frac14$.

The game ends on the third draw if it does not end on the second draw ($\frac34$), and if the third card is one of the two you're already seen ($\frac24$). That gives us: $\frac34\frac24=\frac38$.

The game ends on the fourth draw if it does not end on the second draw ($\frac34$), and it does not end on the third draw ($\frac24$), and if the fourth card is one of the three we've already seen ($\frac34$). That gives us: $\frac34\frac24\frac34=\frac9{32}$

The remaining probability $(1-\frac14-\frac38-\frac9{32}=\frac3{32})$ should be the probability of lasting until the fifth draw, which is the probability of drawing four different cards on the first $4$ draws. To double-check, we calculate this. The second card is not a repeat with probability $\frac34$, the third card with probability $\frac24$, and the fourth with probability $\frac14$. The product: $\frac34\frac24\frac14=\frac3{32}$.

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The probability that the first $(n-1)$ cards are all different:

$$\frac {4!}{\Big(4-(n-1)\Big)!\,4^{n-1}}.$$

Times the probability that the $n^{th}$ card is one that has already been seen: $\frac {n-1}{4}.$

Therefore, the probability that the $n^{th}$ card ends the game $(n$ in $[2,5])$ is:

$$\frac {(n-1)\cdot4!}{\Big(4-(n-1)\Big)!\,4^n} = \frac {(n-1)\cdot3!}{(5-n)!\,4^{n-1}}$$

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  • $\begingroup$ Great answer! I'm not sure how familiar the OP is with combinatorics, but perhaps the first bit could be better explained $($it is simply $\binom4n \cdot (n-1)!$ to account for permutations$)$. $${}$$ I especially like that this answer easily generalizes to decks with $m$ cards. $\endgroup$ – Fimpellizieri Nov 6 '17 at 22:28

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