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A similar question to mine was answered here on stackexchange:

Probability of winning the game "1-2-3"

However, I am unable to follow the formulas so perhaps someone could show the calculation and the way they arrived at it to answer this question.

My card game is very similar, except the counting of cards goes all the way from 1 (Ace) through to King and then restarts. If you can make it through an entire deck of cards without hitting the same ranked card you win. So, for example, if you call out Ace - 2 -3 - 4 - 5 - 6, etc. and hit "7" when you've just called out "7", you lose.

What is the chance of winning this card game?

Community
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Bill
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4 Answers4

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It turns out there is an easier solution to this, following Byron's approach in this answer. The rook polynomial for each $4\times4$ block of excluded matches is $R(x)=24x^4+96x^3+72x^2+16x+1$, and the desired probability is

$$ \frac1{52!}\int_0^\infty x^{52}R(-1/x)^{13}\mathrm e^{-x}\mathrm dx\;. $$

Wolfram|Alpha refuses to evaluate this, but in Sage

y=(-1/x)
a=(x^52 * (24*y^4 + 96*y^3 + 72*y^2 + 16*y + 1)^13).full_simplify ()
integral (a * exp (-x),0,infinity)/factorial (52)

yields

$$ \frac{4610507544750288132457667562311567997623087869}{284025438982318025793544200005777916187500000000} $$

in agreement with the result quoted by Barry.

Community
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joriki
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  • Am I correct that if we generalized this to $n$ cards and $k$ suits ($k|n$), we'd simply replace 52 by $n$ and 13 by $\frac{n}{k}$? – MichaelChirico Jul 12 '15 at 00:54
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    @MichaelChirico: No, you'd also need to use the rook polynomial

    $$R_k(x)=\sum_{i=0}^k\binom ki^2i!x^i;.$$

    The one I used was $R_4$ for $k=4$.

     – joriki Jul 12 '15 at 00:57
  • i.e., $$P(n,k)=\frac{1}{n!}\intop\limits_0^{\infty}x^n R_k(-\frac{1}{x})^{\frac{n}{k}}e^{-x}dx$$? – MichaelChirico Jul 12 '15 at 02:10
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    @MichaelChirico: Sorry, I missed that comment -- yes, that's right. – joriki Jul 21 '15 at 03:20
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I think you'll find what you want in the paper Frustration solitaire by Doyle, Grinstead, and Snell at http://arxiv.org/pdf/math/0703900.pdf -- it looks like they get the answer

$$\begin{align} {R_{13} \over 52!} &= {4610507544750288132457667562311567997623087869 \over 284025438982318025793544200005777916187500000000}\cr \cr &= 0.01623272746719463674 \ldots \end{align}$$

Barry Cipra
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  • The referenced paper answers the question thoroughly. Thank you! Just for play I re-ran my computer simulation of the game with an iteration of one billion and the winning percentage was 1.623995. Close enough :) – Bill Sep 18 '13 at 03:11
  • It turns out there's actually a simpler way to do this (see my answer). – joriki Jul 12 '15 at 01:02
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The probability that a card causes you to fail is $\frac 1{13}$ An approximation, which shouldn't be too far off, is that each card is independent. This is not true, as the fact that the first Ace didn't make you fail increases (slightly) the chance that the next two makes you fail. If we make that approximation, the chance of winning is the chance that no card makes you fail, which is $\left(\frac {12}{13}\right)^{52}\approx 0.0157$

Ross Millikan
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    I ran one billion computer simulations of this card game using two different random number generators to shuffle the deck. As the simulations ran the win percentage quickly converged even after a few million simulations at around 1.62% so this approximation is close, but I was wondering if there was an exact mathematical answer. – Bill Sep 17 '13 at 03:15
  • There is an exact mathematical answer. One way to find it is to list all $\frac {52!}{4!^{13}}\approx 9.2\cdot 10^{49}$ decks and count the winning ones, but it is unrealistic. I don't know another way, but maybe somebody else does. – Ross Millikan Sep 17 '13 at 03:33
  • I just added one (based on Byron's answer to a similar problem). – joriki Jul 12 '15 at 01:00
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After our first draw the probability of winning will be $\frac {12}{13}$. For the second card we have to split the problem into two cases. The first one is we draw two in the first draw and the second is we didn't draw two in the first draw. So the probability will be: $\frac{12}{13} \cdot \frac{47}{51} + \frac{1}{13} \cdot \frac{48}{51}$. So the probability of not losing after two draws will be:

$$\frac {12}{13} \cdot \left(\frac{12}{13} \cdot \frac{47}{51} + \frac{1}{13} \cdot \frac{48}{51}\right) \approx 85,2 \% $$

For the third draw the calulcation are even more complicated. $\frac{12}{13} \cdot \frac{47}{51} \cdot \frac{46}{50} + \frac{1}{13} \cdot \frac{48}{51} \cdot \frac{47}{50} + \frac{12}{13} \cdot \frac{4}{51} \cdot \frac{47}{50} + \frac{1}{13} \cdot \frac{3}{51} \cdot \frac{48}{50}$

So after the third draw our chance of not losing will be $\approx 78.65 \%$, but note that:

$$78.65 \% \approx \left(\frac{12}{13}\right)^3$$

I don't know whether this will continue, but maybe it's good try:

I tried to make a program that will compute the chance of winning after 52 draws(although I must admit I didn't have very good skills in programing), and it computed $\approx 0.0162$ which is very close to the result Ross Millikan's formula gave$

Stefan4024
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