Steem Riddle of the Week #4: Black Cards (Prize for first correct answer)

in #life6 years ago (edited)

Riddle #4: Black Cards

This weeks riddle goes as follows:

There is a dealer with a standard, shuffled deck of 52 cards with 26 red cards and 26 black cards. One by one, he flips a card over. At any point you can tell him to pause and bet $1000 that the next card is a black card. After that, the dealer flips over the next card and if it red you lose your $1000, if it is black you double your money. The game ends at that point and the dealer reshuffles the deck.

You must bet at some point during the game, so if the dealer flips the first 51 cards and you haven't yet bet, then you are forced to bet on the final card.

What is the optimal strategy for the game and how much money on average will you win per round?

Prize

First of, congratulations to @mtness for giving a close, but not complete, solution last week and earning half of the prize.

As usual, for this week, the first person to upvote and comment with a valid solution (as decided by me and only me) prior to 12-31-2017 midnight UTC will get a 1 SBD prize!

Don't forget to follow and to check out my past riddles

First off, thanks to everyone for all the support! If you'd like to browse my older riddles you can do so at the links below:

https://steemit.com/life/@droopy/steem-riddle-of-the-week-3-alternating-truth-prize-for-first-correct-answer
https://steemit.com/life/@droopy/steem-riddle-of-the-week-2-flipping-cards-prize-doubled

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The chance of getting a black is 50%. So technically you will just break even at the end of the game. To optimize winning, let the dealer flip first and observe how many raw cards have been flipped. If more red cards have been flipped, your chances of winning get higher. If let's say 30 cards have been flipped and from the 30, 20 are red so the remaining deck has 22 cards with 6 red and 16 black. So, the chance of winning is now 16/22 and the average winning amount can be 16 x 2 = 32 (with 16 black remaining) and 6 x 1 = 6 (with 6 red remaining) so 32 - 6 = 26,000. Hence, 26000/22 remaining cards is the average amount you will win.

Good general idea, but how often does that happen and more importantly how often does it not happen?

Generally, your chance of winning in the 1st to 51st card is 50% and you have 100% chance of winning in the last card. The best time to bet which will give you a higher chance of winning is on the 2nd card when the 1st card flipped is red (it will give you 50.9% chance of winning $2000) or on the 28th card provided 14 red and 13 black have been flipped (it will give you 52% chance of winning $2000). Hence, your chances of winning are higher on betting at the moment when cards flipped have more than red than black...

that's true but to be sure the strategy would profit you need to figure out how often you end up in a state where more red cards have been flipped than black cards. It is possible for the dealer to go through the entire deck without every being in a situation where more red cards flipped than black cards.

I would win $1000 and bet on the last card, as if I bet on black and he comes to the second last card, and it is red, so I have counted 26 red cards, therefore black is the last card.

What is the way the deck was shuffled, the bottom card is red? You'd never get to 26 red while counting.

Hmm, interesting, so it's multiple bets right? I think I'd bet on every card that gets flipped from the start, as that way even if he flips all red cards first, I'd at least break even and if there was a run of black cards I'd be in profit. I'd keep doing that until I was about 10 k up (or as much as is still possible) then stop betting...

Ah, good question, I edited to clarify. Once you decide to bet, the dealer flips over the next card, and you win or lose and the game ends and the dealer takes back all 52 cards and reshuffles the deck. You only get to make one bet, but you can choose when to bet based on what cards have come.

Ah ok, tough. I'd probably bet on the first card to give me a 50/50 chance, as after that if he pulls more black than red before I bet my chances get worse. Obviously in the reverse case chances get better but it's a risk.

I would allow him to keep on flipping the cards while I count how many of each has appeared and how many is left. This gives me a better shot at getting the next card to be what I stake my money on.

That's a good idea, but you'll need to be more specific for it to actually count as a solution :)

This post has received a 7.33 % upvote from @booster thanks to: @droopy.

I would suggest using the Martingale Strategy which states that should a loss occur, the wager is doubled the next time. Should a win occur, the wager is reset to the original $1000 base amount. The strategy could result in a small net profit assuming the player has an unlimited amount of resources to continue doubling the bet every time a loss occurs.

Good, idea, but unfortunately a martingale strategy still has expected value 0. Furthermore, in this game the question is what is the optimal strategy within a single round where there is only a single bet and how much does it win.

https://math.stackexchange.com/questions/83904/on-martingale-betting-system

Wow, great resource. Thanks.
I gave your riddle more thought and the following strategy is offered as a solution..... I will record every single outcome from the game from the first card flip and continue counting how many red cards versus black cards appear each time. There are only 26 red and 26 black so after having counted the cards that are revealed, as soon as 26 red cards are revealed then the next card is guaranteed to be a black (or vice-versa)....I would place a bet after either 26 red or 26 black cards have been counted.....if 26 red have been revealed then I would bet on black, if 26 black cards were previously revealed then I would bet red...... By counting the cards revealed, you would be guaranteed to win by the final card flip.....all 51 cards were revealed so you should know exactly what colour card remains on the final 52nd card flip.

Good idea, but you are only allowed to bet that the next card is black. Otherwise that would work :)

Ok, well if one can only bet on black then I would wait for 26 red cards to appear. As soon as the 26 red cards appear then there will be a guaranteed black card next. Alternatively, one will have to keep track of the number of black cards revealed too.... if there is a trend of red cards being revealed then the odds would be weighted towards a black card appearing. For example, let's assume 20 red cards have been revealed and only 15 black cards, then the odds would be weighted towards black.....
Waiting until more cards are revealed provides a greater level of predictability. Ideally, you would want 26 red cards to appear before the last card is revealed.... any time before the 52nd-card is shown, if 26 red appears, the bet on black would be a guaranteed win.

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