## Betting on the next card.

Have an interesting puzzle? Let's hear it!

### Betting on the next card.

As mentioned, the value for Victor, using a 52 card deck, should be $57.70+, if my calculations are correct. I'd really like to check with a 7-digit log table, but . . . [My math calculation skills are iffy enough at the best of times. I see no errors, but won't commit myself to being correct.] Spoiler! : Having had time to consider it, and making a triangle of what Victor should bet, and a second triangle of the value of outcome given a stake of$1, here is my solution. As noted by Fingolfinm the optimal bet is (W-L)/(W+L), where W is the number of cards left in the deck you are betting on (the color with higher cards left) and L is the number of losers in the deck.

This leads to a minimal value, at the end of the game, of 2^(w+L)/[Bin (W, L)], where [Bin (W, L)] is the binomial coefficient of W and L = (W+L)!/ (W! x L!). That is, it forms Pascal's triangle, but in inverse form: each element is 2^n divided by the corresponding element of Pascal's triangle.

If we do not have to settle up at the end of each bet, and can bet fractions of a penny, the final expected outcome with a deck of 52 cards is then (2^52) x 26! x 26! / 52!

As calculated by logs to five places, hoping I did not make any errors, this value is about $57.7014. Which is rather higher than the$2.00 suggested above for Victor, if he bets only on the last card.

Although the problem says that he can bet "any fraction of his current worth", I'm not sure what limiting his bets to whole cents would do. I'm thinking that he can settle up at the end, rounding to the nearest penny if needed. I suspect he will need to round, given that the denominator of the final value fraction includes the following factors: 51, 49, 47, 43, 41, 37, 31, 29, and 9.

I hate not to finish by considering what the value of the game will be for Victor if he must settle up after each bet, but rounding does make the math much messier, and takes more calculations. I have considered the first few bets using this assumption:
First bet=0¢
Second bet = 100/51 = 2¢
Third bet = 0¢ or 100x2/50=4¢ depending on whether the first two cards are (both the same, or one of each color)
Fourth bet = 2¢ or 6¢ (depending on whether the first three cards are all three the same, or of both colors--2 and 1)
and so on.
bmonk
University

Posts: 133
Joined: Thu Feb 10, 2011 4:03 pm

### Re: Betting on the next card.

I have the same formula, but I think $\frac{2^{52}}{52 \choose 26} \simeq 9.08$
de-mil
Kindergarten

Posts: 2
Joined: Sun Mar 04, 2012 5:58 am

### Re: Betting on the next card.

de-mil wrote:I have the same formula, but I think $\frac{2^{52}}{52 \choose 26} \simeq 9.08$

Well, I did warn you about my math calculation skills.

There are three kinds of mathematicians: Those who can count, and those who can't. I'm the latter.

9.08 does sound more like it, but it's still far better than $2.00. bmonk University Posts: 133 Joined: Thu Feb 10, 2011 4:03 pm ### Re: Betting on the next card. Yup that's what I got. For each turn, you would bet the following fraction $\frac{\left|\mbox{number red cards remaining - number black cards remaining}\right|}{\mbox{total number of cards remaining}}$ of your stack. In particular, whenever the number of red cards and black cards remaining are equal, you pass (bet$0).

Then one can calculate that the money you will have left at the very end will be:
$(1+1)(1+1/3)(1+\frac{1}{5})\cdots(1+\frac{1}{51})\approx 9.08$

Another way to write this formula when starting with n black and n red cards is:
$\frac{2^{2n}}{{2n} \choose n}$
and substitute $n=26$.

For asymptotics, you can use Stirling's formula to get that the amount of money you will have at the end asymptotic to:
$\sqrt{\pi n}$
Math - It's in you to give.

SpikedMath