Probability to win a solo card game

card-gamesprobability

I discovered a simple game with cards:

You have a deck of 52 French Cards, this deck has four sets of cards with values from 1 to 10 and J Q K (which values are 11, 12, 13 respectively)

The game is very simple: you reveal the first card and say ONE, reveal the second card and say TWO and so on… After saying THIRTHEEN you restart the voice counting

If the revealed card value is the same as the said number, you have lost. If you finish the deck you have win.

Now comes my question: how can I compute manually the probability of winning to this game? (I already computed with programming so I simply want to know how to compute this probability by hand 🙂 )

Here I attach some example of games:

Drawed cards:
2 4 6 8 5… -> lost because of 5 is the fifth card

4 2… -> lost because of 2 is the second card

13 12 11 10 9 8 7 6 5 4 3 2 2 1 -> lost because 1 is the 14th card (but after 13 we have to restart the count, so 1 correspond to saying ONE)

I hope that some of you can answer to my question 🙂

Best Answer

This same problem occurred to me when I was an undergraduate, $50$-odd years ago. I wasn't able to solve it it, though I thought about it off and on for about $15$ years. Then I ran across rook polynomials, and was surprised how easy it was.

There's a good article on rook polynomials here.

In actually doing the problem, it's easier if you think about calling the cards in the order $1,1,1,1,2,2,2,2,...$ which obviously doesn't change the probability of success. Then the dark chessboard of forbidden positions consists of $13$ $4\times4$ chessboards, arranged in block diagonal fashion. Since the rook polynomial of a $4\times4$ board is $$24x^4+96x^3+72x^2+16x+1,$$ (see the wiki article,) the rook polynomial of the entire dark chessboard is $$\left(24x^4+96x^3+72x^2+16x+1\right)^{13}$$

Then just apply the formula.

If we have $s$ suits and $r$ ranks (so in the question $s=4, r=13$) then as $s$ remains fixed and $r\to\infty$ one can show that the probability of success approaches $e^{-s}$. This is the basis of my comment that the probability is approximately $e^{-4},$ though frankly, I can't remember how good that approximation is when $r=13$. I seem to remember that it came out to be about $1/60$, but I wouldn't bet on it.

Note that when $s=1$, this is just the problem of derangements, and it's well-known that the probability of success approaches $e^{-1}$.

EDIT

One other thing I'd like to mention is that friends I showed the problem to often answered $$\left(\frac{12}{13}\right)^{52},$$ incorrectly treating the events as independent, but this doesn't give a bad first approximation, since $$\left(\frac{12}{13}\right)^{52}=\left(\left(1-\frac1{13}\right)^{13}\right)^4\approx e^{-4}$$

EDIT

Couldn't resist it. I wrote this python script

from sympy import Symbol, factorial

x = Symbol('x')
p = 24*x**4+96*x**3+72*x**2+16*x+1
c = (p**13).as_poly().coeffs()[::-1]
n = sum((-1)**k*c[k]*factorial(52-k) for k in range(53))

prob = n/factorial(52)

The probability of success is

$$\frac{4610507544750288132457667562311567997623087869}{284025438982318025793544200005777916187500000000}\\\approx0.016232727467194636$$

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