# The Top-to-Random Shuffle II

In the last post, I introduced the top-to-random shuffle. In particular, we considered why this sort of procedure was important as an alternative to choosing an ordering afresh, and how to we would go about measuring how close we had got to randomness.

In this post, I want to develop the second of these points. The intuition might be that we can get very close to uniform randomness if we repeat the shuffle often enough. Recall this means that even if we choose our bet in a really complicated and careful way, we still couldn’t make much profit by knowing the actual distribution of the ordering. But we might also suspect that the pack will never be exactly random, in the same way that the distribution of the proportion of heads seen on a repeatedly-flipped coin will eventually get very close to 1/2, but will not be exactly 1/2.

This intuition is extremely sensible, and in general is true. It is a nice fact, however, that it fails for the top-to-random shuffle, where we do in fact get to a uniformly random deck. Recall that we approximated how long it would take to get to a state that was roughly random by calculating the time taken for the original bottom card to rise to the top of the deck. This time was:

$n+\frac{n}{2}+\frac{n}{3}+\ldots+\frac{n}{n-1},$

where n is the total number of cards. A further shuffle is required to send the original card back into the pack somewhere. The claim is that the pack is now uniformly random. Note that if we ever actually use this, we have to be careful because although we can calculate the average time at which this event happens, the time itself is random. Rather than worry about that, let’s see why it is true.

As a motivating example, let’s assume that the pack was originally in the order 1,2,3,…,n, and consider the final relative order of the cards numbered 1 and 2. There is some small probability that on the first go (and then again on the second go possibly also) that the card number 1 stays put. Let’s ignore that possibility and progress to the first time that the 1 has been moved into the interior of the pack, so the 2 is now on top. When we do this, we are choosing a number between 2 and n uniformly at random, and moving the card numbered 1 to this position. Let’s call this number X.

Now, when we try to shuffle the 2, we are choosing a number between 1 and n uniformly at random, and moving the card numbered 2 to this position. Let’s call this number Y. Under exactly what circumstances does 2 end up above 1? The clearest example is if Y=X. Then card 2 has been moved to the position previously occupied by card 1. So card 1 moves up a position (since the card 2 is no longer on the top). The final configuration therefore includes card 1 directly above card 2. So we can say

$\mathbb{P}(\text{2 above 1})=\mathbb{P}(Y

where the fact that the inequality in the second probability is strict is important. To calculate this second probability, we want to exploit the symmetry of the situation. The only problems are that the case X=Y is not symmetric as then 1 ends up above 2 as described above, and also that X cannot be 1. So we account for these separately. Note

$\mathbb{P}(Y=1)=\frac{1}{n},\quad \mathbb{P}(Y=X)=\frac{1}{n}.$

The second result holds overall since it holds whenever we condition on a particular value of X. These events are also disjoint. Then

$\mathbb{P}(Y1, X\neq Y)\mathbb{P}(Y>1,X\neq Y)$

$= \frac{1}{n}+\frac{1}{2}(1-\mathbb{P}(Y>1,Y\neq X))$

$=\frac{1}{n}+\frac{1}{2}(1-\frac{1}{n}-\frac{1}{n}) = \frac{1}{2}.$

In summary, the cards 1 and 2 will be in uniformly random order. We might like to extend this idea, but it will get complicated when we add card 3 to the mix, as it is possible (if unlikely) that 1 and 2 will be further mixed before this. This shouldn’t affect the result much, but it will get complicated to define the notation required to carry this sort of argument all the way up to the nth card.

Even using induction is not going to make life substantially easier. Knowing that once we have inserted cards 1,2,…,k into the pack they are in uniformly random order is not enough to make inference about what happens once we put k+1 into the pack. We have to know something about the current positions of 1,2,…,k. For example, if one of these cards is definitely on the bottom of the pack, then the probability that k+1 ends up last among 1,2,…,k+1 is 1/n rather than 1/k+1 as it should be. So in fact we would have to control an annoying amount of information jointly.

In the argument we attempted above, we were looking at the first times some card k got folded back into the pack. Note that this division of time is different to the one we were using for the coupon collector approach to the mixing time in the previous post. Let’s try to use that instead here.

Now we consider the times at which a card is moved below card n. We deliberately decline to say what these cards are. But rather, we want to prove that, conditional on the cards below n being $A_k=\{a_1,\ldots,a_k\}$, the ordering of these is uniform on $S_{A_k}$, that is, every possibility is equally likely. Now this is easy to prove by induction. For, by conditioning on $A_k$ and $a_{k+1}$ being the new card to be moved below n, we are conditioning on the set of cards below n now being $A_{k+1}=A_k\cup\{a_{k+1}\}$. The position of the new card is uniformly random within this, by construction of the top-to-random shuffle, and so the new arrangement is uniformly random on the (k+1)! possibilities.

To see why we have proved the original result we wanted, note that this argument works at the time when the original bottom card is now at the top. So the remaining cards are uniformly randomly ordered. Inserting card n at random gives an arrangement that is uniformly random overall. So as we suggested before, working out how long it takes to get close to randomness in this case reduces to working out how long it is before the original bottom card hits the top and is re-inserted, as at that point, the pack genuinely is uniformly random.