# Betting Problem

Here is an interesting problem I came across.

Let us play a game of cards, shall we? I will be the dealer, and you will do the betting. The rules are simple. I take a standard deck of 52 playing cards and shuffle it well. I begin to flip over the cards in the deck, in sequence, one at a time. Before each flip, you have the opportunity to bet any amount of money that you have, from $0 to everything you have, on the color of the next card. That is, you can bet any amount of money up to what you have that the next card will be red, or bet that it will be black. A correct bet of$x wins you $x; an incorrect bet you lose your$x.

You begin the game with $100. At any point in the game, you can recall perfectly the sequence of cards that has been flipped. When you choose to bet, you can bet any positive amount of money, and are not restricted to betting just multiples of cents. Now the question is: What is the maximum amount of money you can be guaranteed to have once the deck is through, and what betting strategy should you use to achieve this outcome? As a place to start, notice that there is a simple way to guarantee$200. Just wait until I am about to flip the 52nd card, and at that point, bet your entire $100 on the color you know that card to be. How much better can you do? Note by Alex Porush 7 years, 6 months ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. 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(a = remaining money b = number of cards left in the deck of the color that is in most abundance c = number of cards left in the deck of the color that is in least abundance d = number of cards left in the deck of both colors.) Examples: There are 5 cards left. 4 are black and 1 is red. I have 100 dollars. I bet 60 dollars on black. 100*((4-1)/5) or 100(3/5) 1 card left. It is black. I have 100 dollars. I bet all of my money. 2 cards are left, 1 black, 1 red. I bet no money as neither bet guarantees a larger amount of money than I already have. 100((1-1)/2) or 100(0/2) or 0 (no bet) 3 cards are left, 1 is black, 2 are red. I bet 33.33 dollars on red. The idea is that if my luck goes bad on that particular bet, my luck increases and I have more of a chance on the next bet to double my betted money. I think in the worst situation I earn 9.081329549 times my original sum of money. So with 100 dollars I would earn somewhere around 908.13 dollars (This is when I get the first 26 guesses wrong). It is also important to note that to do this I need to be able to bet amounts much smaller than a penny. This is clarified when the creator explains that you can bet amounts that are not multiples of 0.01 or one penny meaning you can bet .001 or 33.3333333(three repeating) Actually the more I think about it, maybe I am guaranteed that amount of money 100% of the time. - 7 years, 6 months ago Log in to reply Adam, my solution is the same as yours. Can you provide your reasoning for why the optimal amount to bet is a((b-c)/d) on the color that is most abundant? - 7 years, 6 months ago Log in to reply If you have $n$ cards left with $n - 1$ of color A and $1$ of color B, essentially to maximize the profit you want to make it such that the amount of money you would get if you bet wrong is equal to the amount of money you would get if you bet right. For example, if you have 3 cards (1R 2B) then you would bet one-third of your money. If you are wrong, then you still have $\frac{2}{3}$ of your money left, then you bet it all on the next two for $\frac{8}{3}$. If you are right that time, now you have $\frac{4}{3}$. You don't bet any on the next round, and then on the round after that you bet it twice, and you end up with $\frac{8}{3}$. Extending this example, it is clear that for larger $n$, you get even more money. For example, if you have 4 cards (1R 3B), then you would had three cases: R B B B, B R B B, B B RB/BR. The money gained in these equations is modeled by: $(1 - b_1)(8), (1 + b_1)(1 - b_2)(4), (1 + b_1)(1 + b_2)(2)$. With equality at $b_1 = \frac{1}{2}, b_2 = \frac{1}{3}$, and the max being $4$. If someone really wants me to show it I could come up with a generalize it, but I think you get the point -- from this, we can conclude that the "guaranteed" case will be a different case than this -- i.e., 2R 2B, or something like that. (Request to problem creator: Can we work with non-discrete numbers, and assume that the "100$" thing was just for clarity of the problem? I don't want to have to worry about betting "$33.33" vs.$33.34" when betting $\frac{1}{3}$)

- 7 years, 6 months ago

Michael, the problem creator clarified the 33.33 vs 33.34 predicament with his statement involving "multiples of cents" otherwise i would have to round and my solution could fail, thus it would not guarantee about 9 times the original amount of money (im sorry if I'm not really clear, I'm trying to type this out on an iphone) edit: also michael you will realize that your bets are a product of my poorly drawn out equation.

- 7 years, 6 months ago

When you choose to bet, you can bet any positive amount of money, and are not restricted to betting just multiples of cents.

You mean this? I think replacing $100 with any other positive amount does not matter. - 7 years, 6 months ago Log in to reply Man, I really didn't read the problem carefully. Haha. - 7 years, 6 months ago Log in to reply If nobody posts a solution by Monday I will put mine up. - 7 years, 6 months ago Log in to reply If you want a guaranteed outcome I think$200 is the best you can do since you may be extremely unlucky and lose all of the bets. If you want the maximum expected value though...

- 7 years, 6 months ago

You can certainly guarantee more than $200. At a certain point there will be only one card of a particular color left. The worst case scenario is that there are only two cards of the other color left (if there are more than two cards of the other color left, you can guarantee even more than what I am about to demonstrate below). Bet 1/3 of your$100 on the two-card color.

If you are correct, you now have $\100 \times 4/3$ and there is one red card and one black card remaining. Do not bet this round, and then wager your entire stack on the last card which you are guaranteed to get right. This results in $\100 \times 8/3$ at the end.

Alternatively, if you are wrong, then you are left with $\100 \times 2/3$, but the remaining two cards are both of the same color, which you will get right, so you bet your entire stack both times and end up with $\100 \times 8/3$, again.

So we can guarantee at least $\100 \times 8/3 > \200.$ Who can do better?

- 7 years, 6 months ago

- 7 years, 6 months ago

So are we looking for the strategy with the highest guaranteed outcome (i.e., if strategy $X$ yields any outcome of dollars from the set $\{Y_1, Y_2, Y_3, ... Y_n\}$ then we take $\displaystyle \min_{i = 1 \to n} Y_i$) or are we looking for the strategy with the highest expected value?

Now that I read your post more carefully, I assume it's the former.

- 7 years, 6 months ago

Guaranteed one, I'd say.

- 7 years, 6 months ago

Yes, I am looking for the greatest guaranteed amount.

- 7 years, 6 months ago

Progress so far:

The strategy consists of betting a certain amount under every certain circumstance, thus the "guaranteed" amount of money occurs when all of these circumstances work against you to allow you to win the least amount of money. I am pretty sure this occurs when card colors alternate (this is because if there are $m$ red cards and $n$ black cards left with $m < n$, then you gain less money in this situation than in the situation when there are $m$ red cards and $j$ red cards, where $j > n$ -- haven't formally shown this to be true, but I'm pretty sure it is). Assuming what I have said previously is correct, now I just have to come up with a strategy that would have the greatest minimum profit in the scenario that the card colors alternate like this.

- 7 years, 6 months ago

How about writing a Dynamic Program ? This should give you an optimal solution. At any point of time the state is simply a 4-tuple $(x_1,x_2,x_3,x_4)$ corresponding to the number of cards remaining in the deck for each color. Although the resulting program will be pretty complex.

- 7 years, 6 months ago

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- 1 year, 1 month ago

taugh

- 7 years, 6 months ago