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QUESTION (TL;DR)

Let's assume you are handed a deck of 52, standard cards which are already shuffled. You take the the deck and mindlessly shuffle it for a few minutes. After the shuffle, you turn over the deck to find that all of the cards are split into two piles-- red on top, black on bottom (or black on top, red on bottom.) However, each color-pile is randomly sorted.

What are the chances of such an event occurring?

BONUS POINTS

Follow-up question, which doesn't have to be answered. How would the probability change if the deck was pre-color sorted?

In other words, if one started with two randomly sorted color piles and then shuffled the deck a random number of times, how would the probability change that the deck were to return to a state similar to how is started? Assume a traditional shuffle, splitting the deck into two piles and then fuzzy merging every 1 or 2 cards on each side, what would be the chances that the deck would resort itself into two color piles after an small but optimized number of shuffles?

WHY I ASK

This is important to me for my sanity. After playing a game with some friends in late-high school, I remember taking the deck of cards we were using (with no jokers) and mindlessly handling it. There is chance that at one point I sorted the deck by color or suits. I was mindlessly fiddling with the deck, so even though I never recall sorting it in any way, I know that if given a deck to hold, that is something I would very likely, mindlessly do.

After handling the deck for a while, I distinctly recall shuffling it at least a few times, turning it over and observing a color sorted deck. My heart pounded as I thumbed through the cards because I couldn't believe what I was seeing. It was... impossible.

To be fair, I've never felt that this was realistically possible. I'm too rational to accept that I started with a well shuffled deck. Before I shuffled the cards, I must have done something to the deck. For the scope of this exercise, what I did doesn't matter. I've resolved that I must have done something that my short term memory had forgotten before shuffling the deck 3-10 times which must have been a factor in the outcome.

Regardless of what really happened, this has been a math problem that has stuck in my head for over 16 years! What are the chances of this happening?

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  • $\begingroup$ Is it necessary for the black and red piles to be (separately) unsorted? If so, how would you define unsorted? $\endgroup$ – Arthur Oct 7 '14 at 12:31
  • $\begingroup$ Imagine if you took one side of the deck (says the red) and you saw { KH, 2D, 5D, 7H, 2H, AH, 3D, ... } The same would be true for the black half { 4C, QC, JC, 6S, 6C, 2S, ...}. In other words, within each color pile, there is no defined order. The cards of one color appear to be in a random order. $\endgroup$ – RLH Oct 7 '14 at 12:37
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The answer is $$ \frac{26!\cdot 26!\cdot 2}{52!} $$ If we know that the red cards are all first, and the black cards are last, there are $26!$ ways for each of those piles to be shuffled. So there are $26! \cdot 26!$ ways of organizing a deck of cards with reds first, blacks last, and there are $52!$ total ways of organizing the deck. The last factor $2$ is there because blacks first, reds last is also a solution with the same probability.

If you want to specify that the reds and blacks are not to be ordered, you have to swap $26!$ with $(26! - n)$ where $n$ is the number of combinations you'd consider "sorted".

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  • $\begingroup$ Ha! It is just as I had suspected. I must be crazy. This is practically impossible. ;) $\endgroup$ – RLH Oct 7 '14 at 12:50

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