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It is known that 7 random riffle shuffles are enough to make almost every configuration equally likely in a deck of 52 cards.

Perfect Shuffle is when you cut the cards exactly in half and perfectly interleave the cards from the two halves.

In-Shuffle is Perfect Shuffle in which top card from the top half remains top after the shuffle, while Out-Shuffle is one in which bottom card from the first half becomes bottom (in other words top card becomes second) after the shuffle.

Question:

If randomly chosen In and Out Shuffles are performed on a deck of 52 cards, how many shuffles are needed to make almost every configuration equally likely?

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  • $\begingroup$ In/out-shuffle are still mixed up. Out-shuffle is the one where top stays at top and bottom at bottom. $\endgroup$
    – Sleafar
    Commented Oct 11, 2015 at 18:13
  • $\begingroup$ What do you mean by "almost every configuration"? $\endgroup$
    – xnor
    Commented Oct 12, 2015 at 0:18

2 Answers 2

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No amount of in-shuffles and out-shuffles will approach a random shuffle.

Let's say that two cards mirror each other if they are the same distance from the middle of the deck. This means that the $i$th and $j$th cards from the top mirror each other if and only if $i+j=53$.

If two cards mirror each other, then after the deck is either in-shuffled or out-shuffled, they will still mirror each other. This means that the result of several in/out-shuffles is uniquely determined by the top 26 cards, so that at most $2^{26}\times 26!$ of the $52!$ deck permutations can be achieved by in/out shuffling.

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I think it is impossible. Using $52$ cards there are $52!=8.06*10^{67}$ possible combinations of cards. Using in/out-shuffles randomly we have potentially $2^n$ possibilities after $n$ shuffles. To reach at least $52!$ combinations this way we would need at least $226$ shuffles.

Let's look at the situation after $104$ shuffles. The interesting feature of in/out-shuffles is that $8$ out-shuffles in a row return the deck to the starting position. The same applies to $52$ in-shuffles. So after $104$ shuffles we have at least $2$ cases with the starting position again, and not enough other cases to cover all other combinations. After a total of $208$ shuffles we have more than $4$ cases with starting position and not enough cases to cover the rest, and so on.

Therefore we will never reach a situation where all shuffles are equally likely.

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  • $\begingroup$ I think we need some numerical constraints, like where "at least six sigmas of shuffles have probabilities within a relative variation of $\pm 5\%$", to be able to give a numerical answer. $\endgroup$
    – user88
    Commented Oct 11, 2015 at 18:57
  • $\begingroup$ @JoeZ. Sorry, can you repeat that for a non-math-student? $\endgroup$
    – Sleafar
    Commented Oct 11, 2015 at 19:14
  • $\begingroup$ Basically, the likelihood of any given permutation being more than 5% away from the average probability (1 in 52!) should be less than about 1 in 500 million. en.wikipedia.org/wiki/68–95–99.7_rule $\endgroup$
    – user88
    Commented Oct 12, 2015 at 5:25

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