Source
r/mathriddles•
possiblywrong

Deck roulette

Medium

Shuffle a deck of r=26 red cards and b=26 black cards. Starting with one dollar, at each turn wager any fraction 0<=f<=1 of your current bankroll, betting on the color of the next card dealt. Continue until the deck is depleted.

What is the maximum expected return from playing this game?

Bonus: is there a betting strategy that guarantees this return?

(This is not a new problem, but I've now seen it three times in recent weeks, so thought this group might find it interesting.)

Posted Jul 14, 2022, 12:25 PM UTC

187
lukewarmtoasteroven•

!Let E(r,b) denote the best possible expected value with r red cards and b black cards when starting with 1 dollar. Letting f vary between -1 and 1, the expected value for betting f on red or -f on black is E(r,b)=r/(r+b) * E(r-1,b) * (1+f) + b/(r+b) * E(r,b-1) * (1-f)=1/(r+b) * (rE(r-1,b) + bE(r,b-1)) + f * 1/(r+b) * (rE(r-1,b) - bE(r,b-1)). Since the first term does not depend on f, we maximize the expected value by setting f=1(all in on red) when rE(r-1,b) - bE(r,b-1) is positive and f=-1(all in on black) when it's negative. When it's 0 it doesn't matter, so we might as well go all in on one of the colors in that case. From this we know there is an optimal strategy which involves going all-in every time. A reasonable all-in strategy ends up with no money every time except for 1 particular arrangement of red and black, in which case it wins 2^52 dollars. Therefore the best expected value is (2^(52))/(52 choose 26), which is approximately 9.08 dollars.!<

6
lordnorthiii•

I have seen this problem before so these aren't exactly original thoughts but a few notes:

a)>!If you wanted to make a $5 bet on red, another $2 bet on red, and a $4 bet on black in the same turn, you can equivalently combine all three bets into one $3 bet on red. Thus, you can effectively bet $1/(52 chooes 26) dollars on every possible sequence of cards simultaneously, and be guaranteed that one of your bets will win you that $9.08 in the end. This solves the bonus.!<

b) >!Notice that strategy above only goes "all in" on one of the colors if that is literally the only color left. But you showed that you should always go all in as long as rE(r-1,b) - bE(r,b-1) is non-zero. This means it is zero as long as there are cards left of both colors. This leads to the remarkably counter-intuitive conclusion that, in terms of expected value, it doesn't matter what you do most of the time!!<

3
possiblywrongOP•

Correct! >!That's what I really like about this problem, that as long as you go all in when you know the color of the next card, you can do anything you want the rest of the time.!<

!Another way of describing strategy that guarantees the expected return: when r red cards and b black cards remain, bet f=(r-b)/(r+b) on red.!<

2
Mate_Bingo•

One clarification, if you wager x for red, you get 2x if the color is red and 0 if black?

2
possiblywrongOP•

Good question-- not exactly, to clarify, if you currently have w dollars, and bet a fraction f (with 0<=f<=1) on red, you're effectively "risking" wf dollars on the table; if the card is red, you end up with w(1+f) dollars, if the card is black, you end up with w(1-f) dollars.

0
Mate_Bingo•

Yeah, that's what I meant.

I am getting maximum expected payoff of>! (2^26)*(26!)/(49*47*45*.....*1)!<

The strategy is - >!You choose the color when you are exactly sure what the color will be. When you are unsure, you can bet any fraction of your amount to the higher probability color.[This means if you are left up with BBR in your deck, you can wager any fraction w for B]!<

1
Tony_343•

!>!Hi, so what I did is in the first round, if one guesses right he gets f*1 dollars. In the 2nd turn if he guesses right he gets 1+2*f + f^2 dollars, which is (f+1)^2. In the 3rd turn if guesses right he gets 1+3f+3f^2+f^3 dollars, which is (f+1)^3. In the 4th time he would get 1+4f+6f^2+4f^3+f^4, which is (f+1)˘^4, and so on until the 52nd turn, which gets him (1+f)^n dollars, and if we want maximum, we set f = 1 (assuming f needs to be fixed throughout) and we get 2^52 dollars. An easier way of looking at it is that each turn you get fraction of the previous amount more, which means multiplying it each time with (1+f) n times!<. This is obviously the maximum amount of money you can get.

-1