Game of Trieze as described in "The People of the Book," by Geraldine Brooks Start with a perfectly shuffled deck of 52 cards. Deal the cards one at a time. What is the probability that the first card is an Ace? What is the probability that the second card is a deuce? etc. The game stops if the nth card is an n, where n = 1,2 ... 13 and Jack is 11, Queen is 12, and King is 13.
 A: Here's my approach
If the game ends on card 13, then 12 cards have already been dealt, none of which is a King. So the probability of ending on card 13 = P(13) = 4/40, because 4 Kings remain in the deck and 40 cards remain in the deck.
Similarly P(12) = 4/41, P(11) = 4/42, …, P(1) = 4/52
The probability that the game does not end (i.e., goes all the way to King without a match) is
 (1-P(13)(1-P(12) … *(1-P1) = 39*38*37*36/(52*51*50*49) = 0.304 approximately
Comments? 
A: Second attempt is messier (but hopefully correct).
The probability that the game ends when the first card is dealt is P(1) = 4/52 because there are 4 Aces in a deck of 52 cards.
Now P(2) needs to take into account whether a deuce was dealt on the first card. If there was no deuce, then there are 4 deuces in a pack of 51 cards. This occurs with probability 48/52. But if the first card was a deuce, then there are 3 deuces left in a pack of 51. This occurs with a probability 4/52. Overall we have
P(2) = (4/51)(48/52) + (3/51)(4/52) = 4/52
Similarly, P(3) needs to take into account whether a trey was dealt on the first or second card (or both). If no trey has appeared, then there are 4 treys in a pack of 50 cards. This occurs with a probability of (48/52)(47/51). If a trey appeared on the first card, but not the second card, then there are 3 treys in a pack of 50 cards. This occurs with a probability of (4/52)(48/51). If a trey appeared on the second card, but not the first, then there are again 3 treys in a pack of 50 cards. This occurs with a probability of (48/52)(4/51). Finally, if a trey appeared on both the first card and the second card, there would be 2 treys in a pack of 50 cards. This occurs with a probability of (4/52)(3/51).
Hence P(3) = (4/50)(48/52)(47/51) + (3/50)(4/52)(48/51) + (3/50)(48/52)(4/51) + (2/50)(4/52)(3/51)
Now I don't intend to keep going with this because it is just tedious. But can someone comment as to whether this is a correct approach and whether there are any simplifications to be had?
A: Third try. There's nothing wrong with the second attempt. If you work it through, you find P(3) = 4/52. Well, that was not what I expected.
But then the solution became clear. And it's very simple.
The probability that the third card dealt is a trey is exactly the same as the probability that it is a six or a King or any other rank, i.e. 1/13. Same for any other card dealt.
An easy way to look at it is just to ask about the nth card dealt, and assume we know nothing about any previously dealt cards, because we don't. It has an equal chance of being any rank.
P(n) = 1/13 for n = 1,2,..., 13
