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I recalled this "trick" I read in some magazine years ago and thought I'd try to explore it a bit more.

You have standard deck of 52 cards. The cards are on the table, facing up. Your friend secredly selects a number, $n_0$, from 1 to 13. Then your friend lifts $n_0$ cards one by one. Then he selects the value of the $n_0$th card, say $n_1$, as his new number and lifts again $n_1$ cards and takes the value of the $n_1$th as the new value. This continues until all the cards have been lifted. Let the value he has after the final card be $n_v$. All this is done silently.

The trick is, the magazine stated, that you can, with somewhat high probability, predict his final number. The idea is to choose your own number, $n_0^*$, and do the above process yourself as your friend does his counting. There is, apparently, high probability that your final value is the same as his. Not a very good trick, in my opinion.

I did some simulations. It seems that the probability that the values are the same is a bit lower than $0.7$.

How should one analyze this kind of a situation and what is the correct probability?

I remember the article mentioned Markov chains, but I cannot see how I could use them in this case. The process is finite and I don't think this has the Markov's property. I tried constructing some large transition matrix, but nothing came of it.

I tried some simplifications, like using only three cards with numbers 1,2 and 3, but the calculations become very messy rather quickly.

Edit: I did some more simulations, 200,000 runs for each pair $\{n_0,n_0^*\}$ and got the following matrix for probability of same result, rounded to save space.

$\left( \begin{array}{ccccccccccccc} 1. & 0.7 & 0.7 & 0.7 & 0.69 & 0.69 & 0.69 & 0.68 & 0.68 & 0.68 & 0.67 & 0.67 & 0.67 \\ 0.7 & 1. & 0.7 & 0.69 & 0.69 & 0.69 & 0.68 & 0.68 & 0.67 & 0.67 & 0.67 & 0.66 & 0.66 \\ 0.7 & 0.7 & 1. & 0.69 & 0.69 & 0.68 & 0.68 & 0.67 & 0.67 & 0.67 & 0.66 & 0.66 & 0.66 \\ 0.7 & 0.69 & 0.69 & 1. & 0.68 & 0.68 & 0.67 & 0.67 & 0.67 & 0.66 & 0.66 & 0.65 & 0.65 \\ 0.69 & 0.69 & 0.69 & 0.68 & 1. & 0.68 & 0.67 & 0.67 & 0.66 & 0.66 & 0.65 & 0.65 & 0.65 \\ 0.69 & 0.69 & 0.68 & 0.68 & 0.68 & 1. & 0.67 & 0.66 & 0.66 & 0.66 & 0.65 & 0.65 & 0.64 \\ 0.69 & 0.68 & 0.68 & 0.67 & 0.67 & 0.67 & 1. & 0.66 & 0.66 & 0.65 & 0.65 & 0.64 & 0.64 \\ 0.68 & 0.68 & 0.67 & 0.67 & 0.67 & 0.66 & 0.66 & 1. & 0.65 & 0.65 & 0.64 & 0.64 & 0.63 \\ 0.68 & 0.67 & 0.67 & 0.67 & 0.66 & 0.66 & 0.66 & 0.65 & 1. & 0.64 & 0.64 & 0.64 & 0.63 \\ 0.68 & 0.67 & 0.67 & 0.66 & 0.66 & 0.66 & 0.65 & 0.65 & 0.64 & 1. & 0.64 & 0.63 & 0.63 \\ 0.67 & 0.67 & 0.66 & 0.66 & 0.65 & 0.65 & 0.65 & 0.64 & 0.64 & 0.64 & 1. & 0.63 & 0.62 \\ 0.67 & 0.66 & 0.66 & 0.65 & 0.65 & 0.65 & 0.64 & 0.64 & 0.64 & 0.63 & 0.63 & 1. & 0.62 \\ 0.67 & 0.66 & 0.66 & 0.65 & 0.65 & 0.64 & 0.64 & 0.63 & 0.63 & 0.63 & 0.62 & 0.62 & 1. \\ \end{array} \right)$

Actually, as the order doesn't matter, I simulated only cases with $n_0^* > n_0$ and copied the results to the lower triangle. The diagonal elements are all one, for if both players select the same starting number, the results are identical.

If I calculated it correctly, for the elements outside the diagonal, a 95% two-sided confidence interval is $\text{value}\pm 0.00219131.$ In any case, the best strategy seems to be to select $n_0^* = 1$.

Valtteri
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  • If you could manage to keep track of what happens for all $13$ possible initial states and then guess the number that most of these states lead to, the probability would be even slightly higher. – joriki Jun 04 '16 at 19:20
  • Presumably the choices of $n_0$ and $n_0^$ make a difference. Are you assuming that these are uniformly randomly drawn? That wouldn't be realistic -- certainly not for $n_0^$, since you should choose it such as to maximise your chances, and perhaps also not for $n_0$. It seems a better question to ask is, for each pair $(n_0,n_0^*)$, what is the probability for the two processes to end with the same number when applied to a uniformly randomly permuted deck? – joriki Jun 04 '16 at 19:34
  • You could improve the odds (and make calculations easier for the general public) by making all face cards equal to $10$, say. Or make them all equal to $1$... – Théophile Jun 10 '16 at 15:18
  • Is the deck of cards shuffled? If so, it is not true that if both players select the same starting number, the results are identical. – Intelligenti pauca Jun 10 '16 at 15:19
  • @Aretino The deck is suffled, yes, but if you have a suffled deck in front of you and you select, say 2, and your friend selects 2 as well, then you both change your value at same cards and have same values at every state. You do the same process. – Valtteri Jun 10 '16 at 15:22
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    This trick is called the "Kruskal count". – Michael Lugo Jun 10 '16 at 15:26
  • @MichaelLugo Ah, thank you, I didn't know that. Might be that the article mentioned the name or that you are supposed to use only numbers from 1 to 10 and such restrictions and I just forgot. But it is nice to know the proper name for this. – Valtteri Jun 10 '16 at 15:36

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