Sequence of red and black cards in one single deck

Hi, I would like to know what the odds are on guessing the color( black of red) of 3 consecutive cards on a single shuflled deck? What about 4 or 5 consecutive cards?
Regards

Interesting question.


For the first guess, you have a 26/52 (50%) chance of being right.

For the second guess, if you select the color that already appeared, you have a 25/51 (49.02%) chance of being right. If you select the other color, you have a 26/51 (50.98%) chance.

For the third guess, if the first two cards were the same color, then you have a 24/50 (48%) chace if you pick that color or 26/50 (52%) chance if you pick the other color. If the first two cards were different, then you have a 25/50 (50%) chance.


The odds get more and more gruesome as you go along.

Is there a simple way to figure it out? I don't know, but now you got me thinking and doing research...

Quote: burebista

...the odds are on guessing the color( black of red) of 3 consecutive cards on a single shuflled deck...

I'll reword the question as "Guess the color of each of the first 3 cards of this single shuffled deck in the right order. What's the probability that you will get all three right."

I wouldn't guess red-red-red or black-black-black because my probability of getting all three right would be only (26*25*24)/(52*51*50)= 0.118. So, I would guess any other permutation, and my probability would be (26*26*25)/(52*51*50) = 0.127. (And expressed as "odds in favor of," the answer is 26*26*25 to 52*51*50-26*26*25.)

I still don't understand if you get to see the first card before guessing the second, and the first two before guessing the third.

Quote: Wizard

I still don't understand if you get to see the first card before guessing the second, and the first two before guessing the third.

Good point. Yes, it's vital to make the problem clear. So, the problem could be one of at least two different questions:
1. "Before I reveal any cards, guess the color of each of the first 3 cards of this single shuffled deck in the right order. What's the probability that you will get all three right."

2. "Before I reveal the first card of this single shuffled deck, guess its color. After I reveal the first card, guess the color of the second card, and after I reveal the second card, guess the color of the third card. What's the probability that you will get all three right."

My answer to question 1: If I guess red-red-red or black-black-black, the probability that I'm right is 26*25*24/(52*51*50). If I guess any other permutation, it's 26*26*25/(52*51*50). So, I wouldn't guess red-red-red or black-black-black.

My answer to question 2: The probability that I'm right on the first card is 26/52 whether I guess red or black. I'd guess red. If the first card is revealed as black, I'd lose. But if it's red, I'd be wise to guess black for the second card, and the probability that I'd be right would be 26/51. If the second card is revealed as red, I'd lose. But if it's black, I'd guess red or black for the third card, and the probability of my being right would be 25/50. So my answer for my probability getting all three right is 26*26*25/(52*51*50).

I'm surprised that my answers to question 1 and question 2 are the same!

Quote: ChesterDog

I'm surprised that my answers to question 1 and question 2 are the same!

Without working all the way through this, I suspect this is because it doesn't matter whether you lose on the first, second, or third card, so long as you lose. In cases where you win, you are going to follow the same strategy either way. Did I get that right?

Quote: ChesterDog

I'm surprised that my answers to question 1 and question 2 are the same!

You see to be assuming that the first two cards are different colors. If they were the same, the probability of guessing the third correctly, by going with the opposite color, would be 26/50.

I suppose I was assuming that his strategy for question #1, instead of being "any other permutation" would actually be "red-black-either" or "black-red-either" and for question #2 would be the same thing only with the announcement of his decisions delayed. If the first two cards are different colors, then ChesterDog's probabilities are correct, I think. If the first two cards are the same color, then he had the wrong strategy and loses. I think that outcome just falls into the 1-P(win) category.

If your strategy for either question is "red-red-black" or "black-black-red", I think you get the probability as suggested by the Wizard for the first two cards being the same color -- higher probability of getting the third card correct but lower probability of surviving until the third card. Doesn't this give

P=(26/52)(25/51)(26/50), which is the same probability that ChesterDog suggested?

Am I missing something here?

Quote: ChesterDog

I'm surprised that my answers to question 1 and question 2 are the same!

Well, as you pointed out, the probability does not depend on your strategy as long as you are not naming the same color three times. But if the strategy is irrelevant, it should not be surprising that learning any additional info along the way does not change your odds.

Quote: Wizard

You see to be assuming that the first two cards are different colors. If they were the same, the probability of guessing the third correctly, by going with the opposite color, would be 26/50.

But then the second one is 25/51, so the final product is still 26*26*25/52*51*50

In case an apology is due to weaselman, I did edit my earlier post without saying so, probably at the exact time while he was typing. I did not originally post the same conclusion as the second half of his post.

Quote: weaselman

But then the second one is 25/51, so the final product is still 26*26*25/52*51*50

Okay, the light bulb just came on.

Quote: Wizard

You see to be assuming that the first two cards are different colors. If they were the same, the probability of guessing the third correctly, by going with the opposite color, would be 26/50.

But the correct play would be to guess the opposite colour for the second card, so you only get to the third card if the first two are different colours.