Let us play a game of cards, shall we? I will be the dealer, you will be the bet

Question

Let us play a game of cards, shall we? I will be the dealer, you will be the bettor. 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. Immediately 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 card that I am about to flip. So, for instance, if you have $5 and I am about to flip a card, you may either do nothing, bet any amount of money up to $5 that the card will be red, or bet any amount of money up to $5 that the card will be black. A correct bet of $x wins you $x; an incorrect bet costs you $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. To simplify things, assume that dollars are continuously divisible. That is, whenever you chose to bet, you can bet any positive number of dollars; you need not stick to multiples of a cent.

Now the questions 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 yourself $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?

Explanation / Answer

I would approach this situation using the following strategy: Bet an amount of money equal to a((b-c)/d) on the color that is most abundant. (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).

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.