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I came up with this question after watching a Numberphile video on the number of riffle shuffles required to randomize a deck. One question posed in the video was "How many one-card riffles are needed to randomize a deck?" The answer to this question was roughly 236 shuffles: \begin{equation} \sum_{k = 1}^{52} \frac{52}{i} = \frac{52}{1} + \frac{52}{2} + \frac{52}{3} + . . . + \frac{52}{52} \approx 236 \end{equation} By taking the top card off the deck and poking it somewhere in the deck, it has a $\frac 1 {52}$ probability of going under the last card, say the king of diamonds. By repeating this procedure, we will slowly move the king of diamonds to the top of the deck, and when the king of diamonds is shuffled, the deck is now random. For each card moved under the king of diamonds, the probability is increased by $\frac 1 {52}$, resulting in the summation we have.
Now, for the question of a two-card riffle, I am having trouble understanding the casework behind it. There are now several ways for the last card to move up: either one of the cards being riffled goes under the last card, both go under, or neither go under. For both to go under, the probability is $\frac 1 {52} * \frac 2 {52}$. For one of the cards to go under, is it $2 * \frac 1 {52} * \frac {50} {52}$? What would the summation look like?

Note: A two-card riffle would involve taking the top two cards of the deck and reinserting each of the two cards somewhere in the deck at random (independently of each other).

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  • $\begingroup$ You described how a one-card riffle works, but you didn't describe the two-card riffle that you actually want to use, so we don't know what you're trying to do. $\endgroup$
    – joriki
    Commented Nov 30, 2019 at 20:28
  • $\begingroup$ @joriki I have clarified what a two-card riffle is. $\endgroup$
    – Gerard L.
    Commented Nov 30, 2019 at 20:39
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    $\begingroup$ Unfortunately you haven't :-) Do you mean both cards together? Or separately? If separately, how are the possible insertions distributed? The only natural interpretation that comes to mind is that you mean you uniformly randomly insert first one and then the other -- but that would just be two one-card riffles (except you take the second card before you do the first riffle), so I suspect that's not what you mean. $\endgroup$
    – joriki
    Commented Nov 30, 2019 at 20:44
  • $\begingroup$ @joriki: your last is how I interpreted it. Take the top two cards off the pack, put the top one into the pack somewhere, then put the second into the pack somewhere. As you say, the only difference is avoiding the first card going back on top and being the second card riffled. $\endgroup$
    – Ross Millikan
    Commented Nov 30, 2019 at 20:55

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The only way that the two card riffle differs from two steps of the one card riffle is that it ensures that the first card is not put back on the top and used for the second card. This happens only once in $52$ times. We would therefore expect it to take about $118-2=116$ two card riffles, where we divided $236$ by $2$ and subtracted the two wasted riffles we would expect.

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