Is it the same as a coin flip? What are the odds of correctly guessing every red or black in a 52 card deck? How about a half deck, assuming is was evenly 13 reds 13 blacks?
The chances of guessing the color of one card drawn at random from a full deck are 0.5, same as a coin flip.
The chances of guessing the color of each card in the entire deck correctly depends on the order of the cards. Let’s say you draw the first card. The chance it is black is exactly 0.5. Suppose it turns out to be red. Now the chances that the next card is black is 0.5098. This continues through the entire deck. Just to give you the idea of the order of magnitude, if you were to make 52 guesses in a row with replacement (that is, use a full deck each time), the chances of 52 correct guesses in a row is 1 out of 4,503,599,627,370,496.
Yes, odds are 50:50.
Of course those odds would change after you make a guess at the first card and then remove it from the deck.
If for example you guessed the first card was black, pulled it to find it was the 3 of clubs, then set it aside, your next guess would be 49:51 in favor of red.
The chances of 26 straight correct guesses in a deck with replacement would be much better, 1 out of 67,108,864.
It’s actually 1 in 495,918,532,948,104 and there’s no strategy you can use to change it. As long as you guess “black” 26 times and “red” 26 there’s no way to improve or worsen this figure.
Is it the same if you do remove a card from the deck with each correct guess?
I was about to post “1 in 500 trillion”. The reason that’s the best you can do is, every possible sequence of guesses is equally likely; you may as well guess that the first 26 cards will be red and the last 26 cards black.
What do you mean? If you guess wrong then… You failed to guess all the colors
Assuming best strategy you can get the first 26 one time in 47,681,723 with no replacement.
If the deck is shuffled, isn’t the best strategy to assume half of the 26 cards in the half-deck will be red and half black? Other distributions are less probable.
If you have to guess all 52 there’s no set of guesses better than the other. Only guessing 26 you will have 13 pure coin flips, and 13 guesses that are slightly better than a coin filp assuming good strategy.
Yes, right , that’s simpler. Any set of guesses containing 13 reds and 13 blacks will have the same optimum probability.
My figure is 2^52. Each correct guess has a chance of 1/2, repeated 52 times.
How did you derive your figure?
I didn’t notice you were talking about “with replacement”. I solved the question in the OP and misread your post.
It’s basically (26/52)times(25/51) etc until times(1/27)
Bloody markdown turns an asterisk into italics.and removes my ellipse
The chances when you remove each card are different than if you replace the card each time, as I explained in post #2.
And what you just said here makes things a lot more complicated if you remove cards only for correct guesses.
What I mean is, you remove a card, place it in front of them, they guess, you reveal the card either way and discard it on a discard pile.
Shouldn’t it be 2^51, since you’ll know the final card’s color?
No, he’s doing it “with replacement”, not as in the OP. There won’t be a final card.
The odds only differ significantly from 50:50 if an unusually large number of cards of one color have been guessed correctly and removed. And this could only come about if the guesser were guessing against the odds - because whenever one excess card of one color has been removed, the best strategy is to guess the other color for the next card.
That’s why removing correctly guessed cards doesn’t improve your odds for a run of correct guesses significantly.
The reason I didn’t solve the problem exactly as asked is that it I think it depends on what remains in the deck, and also whether you are told what color is drawn each time before you make your next guess. For example, the first card drawn is red, and you are told it was a red card. Now you know there are 26 black cards remaining out of 51, and if you are smart you are going to guess black because the chance of the next card being black is >0.5. Your chances of being correct on the second draw are 26/51, not 25/51 as in your sequence. The chances of guessing right on the third card are different depending on whether the second card was black or red. The chances of making the right guess on any given draw depend on the proportion of black/red cards, which depends on all of the preceding draws. It seems to me that solving this analytically is a very complex problem.
For this type of problem, sometimes these variations come out in the wash, so that it doesn’t matter what the order is; you get the same answer. But I am not sure how to do that analytically for this problem. I will run some simulations as time permits.
Nope, each sequence of red and black has exactly the same chance. As long as you know there are exactly 26 of each, you cannot improve on your chances with new info. You may as well guess blind from the start. Always playing the odds, (betting R on straight 50/50) wins if the sequence goes RBRBRB… throughout the deck. Guessing 26 blacks to start wins if it goes BBBB…RRR. Neither is more likely.