[bonanova]

In a we parsed a deck of 52 cards into sequentially higher

and lower piles, and pondered the expected high card value in the low pile.

The answer was found to be Let's say this answer is correct.

Observe that the ratio r₅₂ = 43.55 / 52 = 0.8375.

Let's also assume that as the number of cards n in the deck increases without

limit, the ratio r_n converges to some value r_o.

I'm guessing there is a closed form expression for r_o that involves the number e.

Anyone care to derive it?

[superprismatic]

Well, I've been working on this problem on and off since you posed it.

My gut says that the limit is 1, but mathematics has a habit of making

a fool of my gut. So, I decided to do some more simulations. Doing

one million plays using a deck of 20,000 I get .9911. Here are values

I've used to "sneak" up on whatever the limit is:

One million plays in each trial:
deck size r
1,000 .9610
2,000 .9723
3,000 .9774
4,000 .9803
5,000 .9824
10,000 .9875
20,000 .9911

Since my analytic efforts led to dead ends, my official guess is 1,

(or e⁰, for the way you posed the problem). I am very

curious to see a proper derivation of the limit.

[bonanova]

Well, I've been working on this problem on and off since you posed it.

My gut says that the limit is 1, but mathematics has a habit of making

a fool of my gut. So, I decided to do some more simulations. Doing

one million plays using a deck of 20,000 I get .9911. Here are values

I've used to "sneak" up on whatever the limit is:

One million plays in each trial:
deck size r
1,000 .9610
2,000 .9723
3,000 .9774
4,000 .9803
5,000 .9824
10,000 .9875
20,000 .9911

Since my analytic efforts led to dead ends, my official guess is 1, (or e⁰, for the way you posed the problem). I am very curious to see a proper derivation of the limit.

I continued the simulation for some larger decks and got further evidence the limit is 1.

Deck size r
50,000 0.9949
100,000 0.9972
1,000,000 0.99926

This is, of course, obvious -- once it's demonstrated to be true. Take the cards that are in the top 10% of the deck. As the deck size increases, so does the likelihood two of them will be adjacent. In half of those cases, the smaller will follow the larger and end up in the left pile. Thus, at some deck size, it becomes almost certain that r will be .9 or higher. Make the same argument for the cards in the top 1% of the deck. At some deck size, it becomes almost certain that r will be .99 or higher. And so on ... So the rigorous proof is simply the derivation of the probability of a pair of cards in a subset of the deck being adjacent after a randomizing shuffle. The results for smaller decks, are also interesting, beginning with r₂ = r₃ = 0.500...

Deck size r
2 0.500.. 2 permutations: 12 and 21 each give 1. 1/2 = .5
3 0.500.. 6 permutations: 123, 213, 312 give 1, 132, 231, 321 give 2. (Avg=1.5)/3 = .5
4 0.541..
5 0.576..
10 0.671..
20 0.753..
30 0.794..
40 0.819..