Poisson's distribution , Frustration solitaire

Question

I m not sure how to solve this question from Introduction to Probability & statistics for scientists and engineers,Sheldon M. Ross. Need help.

The game of frustration solitaire is played by turning the cards of a randomly shuffled deck of 52 playing cards over one at a time. Before you turn over the first card, say ace; before you turn over the second card, say two, before you turn over the third card, say three. Continue in this manner (saying ace again before turning over the fourteenth card, and so on). You lose if you ever turn over a card that matches what you have just said. Use the Poisson paradigm to approximate the probability of winning. (The actual probability is .01623.)

Thanks :)

Answer 1

Comment.

It is difficult to simulate small probabilities with good accuracy. Nevertheless, a simulation of a million plays of this game gives just good enough results to distinguish between the approximate answer $e^{-4} \approx 0.018$ and the exact answer $\approx 0.016.$ [The simulation also shows the number of 'hits' (matches between what is called and what is revealed), assuming the game is continued until the deck is exhausted. Four is the average number of hits; similar to the Poisson as suggested by @AndreNicholas; also, the most likely number.]

 m = 10^6;  call = rep(1:13, times=4);  hit = numeric(m)
 for (i in 1:m) {
   shuf = sample(call, 52)
   hit[i] = sum(call == shuf) }
 mean(hit == 0)
 ## 0.016293    # approx of actual answer P(no hits)
 mean(hit)
 ## 4.000411    # expected number of hits.
 exp(-4)   
 ##  0.01831564 # Poisson approximation

 round(table(hit)/m, 4)  # approx dist'n of hits
 ## hit
 ##      0      1      2      3      4      5      6      7      8 
 ## 0.0163 0.0691 0.1443 0.1977 0.2011 0.1617 0.1051 0.0586 0.0280 
 ##      9     10     11     12     13     14     15     16     17 
 ## 0.0118 0.0043 0.0014 0.0004 0.0001 0.0000 0.0000 0.0000 0.0000 

The figure below shows the simulated distribution of hits (histogram bars) along with the PDF (open red circles) of the approximating $Pois(4)$ and the PDF (solid blue dots) of $Binom(52, 1/13).$ This problem is fine as an exercise in using the Poisson approximation, but the binomial is more accurate and it is easy enough to compute $(12/13)^{52} = 0.01557$, more accurate than $e^{-4} = 0.01832.$ Only the values at 0 are directly relevant to the Question.