Probability of winning/losing a lonely card game when you shout an integer progressively each time you draw Take a standard face-down 52-deck of cards, including 13 ranks of four suits (for simplicity, assign value 1 to 13 from ace to King). Draw the cards one by one, shouting the integers from 1 to 13 (and then starting again) in the natural order (i.e. {1,2,3,4...}) at each draw. For example, you shout 'one' and draw a 7, then you shout 'two' and draw an ace, then you shout 'three' and draw a King, and so on and so forth. You lose instantly if the number you've shouted coincide with the rank of the card you've just drawn. What is the probability of winning/losing the game (whichever is easier to compute)? 
**Additional information:
The problem is related to the first-encounter probability of two walks (one deterministic and one random): a deterministic walk jumping to the right by one lattice site at each time-step, and resetting after 13 steps, and another walk randomly jumping among the same thirteen sites (with the constraint that no site can be visited more than four times).

This is the so-called "Frustration solitaire". This is a good reference: article by Doyle,  Grinstead and Snell. The probability of winning is about $1/e^4\approx 0.0183$ where $4$ is related to the number of suits. That is about $2$%, this game is really frustrating!!
P.S. As pointed out by Michael Lugo below, at page 8 of the linked article we can find the exact winning probability for a 52-card deck which is about $0.0162$. Even for $n=13$, the asymptotic probability $1/e^4\approx 0.0183$ gives a reasonable estimation of the exact probability.
