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Ace In The Hole


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During the pre-cards bull session at this weeks friendly game it was obvious Coop was a little anxious. No doubt still reeling from loss that was tucked neatly in my wallet just in case. Shortly after Vern finally arrived, it was Coop who pushed for the game to begin and Coop who cussed under his breath as Rob, seated to his left, drew high card for first deal. Prop bets are not really used to juice the action at our game but are more a matter of pride. Even so they are usually not proffered in the first round of deals. But when it came to Coop's turn he couldn't wait any longer. "Ok, here's how it works. The buy in's $10 each." He was dealing cards face up by suit to Rob, Huck, Vern and I. "Each of you now has 13 cards from 2 to Ace, aces are high. Shuffle and place them face down and everyone flip over your top card. You'll all turn over cards in succession until you're shot down by the bullet. For every card (not counting the first one) that is flipped and higher than all previous cards, you score a point. No points loses, one point gets you five of your ten back, two points is a push, three points wins $1, four points wins $5, five points - $10, six points - $20, seven points - $50, and eight or more points wins you $100. We played several rounds. Not a bad game all in all. But I'm still wondering; what is the expected (average) number of points for this game? And secondly, what is the expected value of the $10 bet?

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yes indeed superprismatic, thanks for asking and allowing me to clarify; when a player turns over an ace, that is essentially his last card for that game. also want to say each player is using only their individual stack. realized when I said "in succession" that was a little ambiguous. Each player turns over the top card in their respective stack and keeps their own individual score.

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It's easier if we start with smaller decks.

Let E(k) be the expected number of points with a deck of k cards.

E(1)=0, because the first card does not score

For larger decks, imagine adding the lowest card so far. In other words, if we already know the value for a deck of size k, think of them as the cards from (n-k+1) to n. For example, if k=4, then they are J, Q, K, A

Some of the permutations begin with the new (and lowest) card, the remainder do not.

The permutations that begin with the new card score 1+E(k), because the new card is the lowest, making the first card of the deck of k score a point.

the permutations that do not begin with the new card score E(k), because the new card is hidden inside the deck, so the first card of the deck of k still does not score a point.

The fraction of perms beginning with the new card are k! out of (k+1)! = 1/(k+1)

The fraction of perms with new card hidden are ( (k+1)! - k! )/(k+1!) = (k+1-1)*k! / (k+1)! = k/(k+1)

So E(k+1)= (1+ E(k))*1/(k+1) + E(k)*k/(k+1) = 1/(k+1) + E(k)

This means the total expectation E(n) = sum (1/i) from i = 2 to N

In particular, for n = 13, this sum is 2.18

PS: no additional calculation needed to deal with "being shot down by the bullet". That simply reflects the fact that there are no additional points after the Ace appears. We have to count all permutations, and I believe the calculation above does so.

Thanks for a good puzzle, plainglazed, I thought this was going to be impossibly complicated. (Notice I'm not touching the question about the value of the game...)

Edited by CaptainEd
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Points | Number of Hands

0 | 479,001,600

1 | 1,486,442,880

2 | 1,931,559,552

3 | 1,414,014,888

4 | 657,206,836

5 | 206,070,150

6 | 44,990,231

7 | 6,926,634

8 | 749,463

9 | 55,770

10 | 2,717

11 | 78

12 | 1

Total | 6,227,020,800

Total points = 13,575,738,240 Expected Number of points ≈ 2.1801338

Total return = -4,134,540,612 Expected return ≈ -0.66396767

Expected value of $10 bet ≈ 9.33603233

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yes indeed, nicely done CaptainEd and superprismatic. like the Cap'n, I too thought the first question was most interesting in its simplification. needed a spreadsheet for the second part and don't know of a simple way to express the formulations as they involve subsequent cycles of the permutations. am guessing superprismatic did similarly with a computer program?

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yes indeed, nicely done CaptainEd and superprismatic. like the Cap'n, I too thought the first question was most interesting in its simplification. needed a spreadsheet for the second part and don't know of a simple way to express the formulations as they involve subsequent cycles of the permutations. am guessing superprismatic did similarly with a computer program?

Yes, I went through all 13! permutations.

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