Probability of guessing red/black correctly (14 times) when flipping over the top card of a shuffled deck

Question

Shuffle a deck of cards. The deck is placed face down. There are trillions of card combination. One is asked to guess before a card is flipped over if it is red or black. What is the probability that a person could guess 14 correct in row.

Take into account that you learn something from the previously exposed card.

This is not the same as a coin flip (.5 * .5 * .5 ....)

Answer 1

There is some ambiguity in the statement of the problem. I’m going to assume that guessing "14 in a row" should be interpreted as correctly guessing the suits of the first 14 cards drawn, since I think that is what was meant.

There is also ambiguity in the statement "Take into account that you learned something from the previously exposed card". While this does imply that the cards are drawn from the deck without replacement, it doesn't say how the guesser should "take this into account". The most obvious interpretation is that if more red cards have been drawn, then we guess black next, while if more blacks have been drawn, then we guess red next. If the same number of reds and blacks has been drawn, then our next guess may be red or black randomly.

Intuitively, this strategy should improve our chances of correctly guessing the suits of the first 14 draws compared to the more naive strategy of simply guessing each suit randomly (not taking the previous draws into account). I'm going to solve the problem using both strategies and compare the results. I'll start with a simpler problem, then extend the solution to the more complicated problem.

First suppose we are only guessing at the suits of the first two draws. Suppose also that we aren't taking the result of the first card draw into account, i.e., we guess randomly for both cards (strategy 2). There are 4 possible guesses we might make {RR,RB,BR,BB}, each equally-likely since we are guessing randomly at both draws. Likewise, there are four possible draws {RR,RB,BR,BB}. The draws are not equally-likely because the first card drawn is not returned to the deck. For instance, the draw RR has probability (26/52)(25/51) while the draw RB has probability (26/52)(26/51).

Let C be the event that we correctly guess the suit of all cards drawn. This is the same as the probability that the pair we guess matches the pair drawn. The law of total probability gives

P(C) = P(RR drawn) * P(C | RR drawn) + P(RB drawn) * P(C | RB drawn) +

P(BR drawn) * P(C | BR drawn) + P(BB drawn) * P(C | BB drawn).

Now, each of the conditional probabilities above is 1/4 because the guesses are made randomly. So

P(C) = P(RR drawn) * (1/4) + P(RB drawn) * (1/4) + P(BR drawn) * (1/4) + P(BB drawn) * (1/4)

= [ P(RR drawn) + P(RB drawn) + P(BR drawn) + P(BB drawn) ] * (1/4)

= [ 1 ] * (1/4)

= 1/4.

Now suppose we guess at the suits of the first 3 draws. The possible guesses and draws are {RRR,RRB,RBR,RBB,BRR,BRB,BBR,BBB}. We have

P(C) = P(RRR drawn) * P(C | RRR drawn) + P(RRB drawn) * P(C | RRB drawn) + ... +

P(BBB drawn) * P(C | BBB drawn)

= [ P(RRR drawn) + P(RRB drawn) + ... + P(BBB drawn) ] * (1/8)

= 1/8.

Similarly, if we guess randomly at the suit of each of the first 14 draws, the probability that we guess all 14 correctly is (1/2)¹⁴ which is about 0.0000610. Very poor odds. Perhaps using strategy 1 instead will improve our chances?

Return to the scenario where we only guess at the suits of the first two draws, but suppose we base our guess at the second draw on the result of the first draw. If a red is drawn first, we guess black for the second draw. If a black is drawn first, we guess red for the second draw. As before, we have

P(C) = P(RR drawn) * P(C | RR drawn) + P(RB drawn) * P(C | RB drawn) +

P(BR drawn) * P(C | BR drawn) + P(BB drawn) * P(C | BB drawn),

but now some of the conditional probabilities above are zero. For instance, if an R is drawn first, then our strategy dictates that we must guess B for the second draw. So P(C | RR drawn) = 0. Similarly, P(C | BB drawn) = 0. On the other hand, P(C | RB drawn) and P(C | BR drawn) both simplify to P(first suit is guessed correctly). For instance, if we assume the draw is RB, then by correctly guessing R for the first draw, our strategy will force us to guess B for the second draw, which will also be a correct guess. The probability that the first suit is guessed correctly is clearly 1/2. Thus,

P(C) = P(RR drawn) * 0 + P(RB drawn) * (1/2) + P(BR drawn) * (1/2) + P(BB drawn) * 0

= [ P(RB drawn) + P(BR drawn) ] * (1/2)

= [ (26/52)(26/51) + (26/52)(26/51) ] * (1/2)

= 13/51.

Compared the strategy 1, we have have slightly improved our odds (13/51 versus 13/52).

Now suppose we guess at the suits of the first 3 draws using the strategy 1. Recall that the possible guesses and draws are {RRR,RRB,RBR,RBB,BRR,BRB,BBR,BBB}. The law of total probability gives

P(C) = P(RRR drawn) * P(C | RRR drawn) + P(RRB drawn) * P(C | RRB drawn) + ... +

P(BBB drawn) * P(C | BBB drawn).

Each of P(C | RRR drawn), P(C | RRB drawn), P(C | BBR drawn), and P(C | BBB drawn) is zero, for the same reason that P(C | RR drawn) and P(C | BB drawn) are both zero when considering just two draws.

On the other hand, each of P(C | RBR drawn), P(C | RBB drawn), P(C | BRR drawn), and P(C | BRB drawn) is 1/4. Why? Consider P(C | RBR drawn). We have a 26/52 chance of guessing the first suit correctly. Since we are assuming the first suit is R, our second guess will correctly be B. On the third guess, having already drawn one R and one B, the remaining cards are in equal proportions again. So we guess randomly at the third draw, having a 25/50 chance of being correct. So P(C | RBR drawn) = (26/52) * (1) * (25/50) = 1/4.

Thus, after removing all the zero probabilities, we are left with

P(C) = P(RBR drawn) * P(C | RBR drawn) + P(RBB drawn) * P(C | RBB drawn) +

P(BRR drawn) * P(C | BRR drawn) + P(BRB drawn) * P(C | BRB drawn)

= P(RBR drawn) * (1/4) + P(RBB drawn) * (1/4) + P(BRR drawn) * (1/4) + P(BRB drawn) * (1/4)

= [ P(RBR drawn) + P(RBB drawn) + P(BRR drawn) + P(BRB drawn) ] * (1/4)

= [ (26/52)(26/51)(25/50) + (26/52)(26/51)(25/50) + (26/52)(26/51)(25/50) + (26/52)(26/51)(25/50) ] * (1/4)

= 4 * [ (26/51)(1/2)^2 ] * (1/4)

= 13/102.

Let's increase the number of draws to 4. This gives 16 possible sequences of guesses and draws:

{RRRR,RRRB,RRBR,RRBB,RBRR,RBRB,RBBR,RBBB,BRRR,BRRB,BRBR,BRBB,BBRR,BBRB,BBBR,BBBB}

With strategy 1, the only way we will guess all four suits correctly is if the draw sequence is one of RBRB,RBBR,BRRB,BRBR. For instance, assume the sequence of draws is RBBR. Since the first draw is R, our strategy will force us to guess B for the second draw. Since the first three draws are RBB, our strategy will force us to guess R for the second draw. So we will guess the entire sequence RBBR correctly if and only if we guess the first and third draw correctly. This has probability 1/4. On the other hand, if the sequence is, say, RBRR, then we will certainly guess wrong on the fourth draw (and possibly sooner), because our strategy forces us to guess B on the fourth draw. Therefore, we have

P(C) = P(RBRB drawn) * P(C | RBRB drawn) + P(RBBR drawn) * P(C | RBBR drawn) +

P(BRRB drawn) * P(C | BRRB drawn) + P(BRBR drawn) * P(C | BRBR drawn)

= P(RBRB drawn) * (1/4) + P(RBBR drawn) * (1/4) + P(BRRB drawn) * (1/4) + P(BRBR drawn) * (1/4)

= [ P(RBRB drawn) + P(RBBR drawn) + P(BRRB drawn) + P(BRBR drawn) ] * (1/4)

= 4 * [ (26/52)(26/51)(25/50)(25/49) ] * (1/4)

= 325/4998.

At this point, we would like to proceed to the case of 14 card draws. There are a total of 2¹⁴ possible sequences of Rs and Bs, too many to list. The case of 4 card draws revealed a critical piece of information. Of the 2¹⁴ possible sequences, the sequences that have any chance of being guessed correctly are precisely the sequences that have the form a₁a₂a₃a₄a₅a₆a₇ where each a_i comes from the set {RB,BR}. There are 2⁷ such sequences. Each of these sequences has the same probability of being drawn:

(26/52)(26/51)(25/50)(25/49)(24/48)(24/47)(23/46)(23/45)(22/44)(22/43)(21/42)(21/41)(20/40)(20/39)

= (26/51)(25/49)(24/47)(23/45)(22/43)(21/41)(20/39) * (1/2)⁷

And given that one of these sequences is drawn, the probability that we succeed in guessing the entire sequence correctly is (1/2)⁷. Therefore, the probability of guessing all 14 suits correctly is

P(C) = 2⁷ * [ (26/51)(25/49)(24/47)(23/45)(22/43)(21/41)(20/39) * (1/2)⁷ ] * (1/2)⁷

= (26/51)(25/49)(24/47)(23/45)(22/43)(21/41)(20/39) * (1/2)⁷

= 6325/88744131

which is about 0.0000713. Recall, using the purely random guess strategy, the probability of guessing all 14 suits correctly is about 0.0000610. So strategy 1 does indeed improve the odds, ever so slightly. But your chances are highly unlikely with either guessing strategy.

Answer 2

Hi Ben,

In this situation, the prompt suggests that each flipped over card is not placed back into the pile and reshuffled by the line that says "Take into account that you learn something from the previously exposed card." This means that each card flip after the first one is a dependent event, which complicates the calculations from the coin flip significantly, as you noted. After understanding this fact, you need to know the basic information that a standard deck has 52 cards, 26 being red, and 26 being black.

It should then be pretty straightforward that the first guess has a 26/52 chance or 1/2 chance of being correct. Let's then examine the next flipped card after that, following the assumption that the first flipped over card was black. That means that there are 26 red cards and 25 black cards left in the deck. If one were to guess that the next flipped card was also black, the probability of that being correct would be 25/51, while the probability of the card being red would be 26/51.

Following this same process and logic, you could determine that further guesses will have even more varying probabilities depending on both the previous cards pulled as well as the guess being made. Thus, it is nearly impossible to answer generically what the probability of 14 correct guesses is. Based on the given situation, it is possible to determine the probability of a specific patterns of reds and blacks being revealed by multiplying the probabilities as determined above together for all 14 cards flipped, but unless you run a computer simulation to go through all 2^14 (or 16,384) possibilities of series of guesses as well as flipped results, it is unrealistic to calculate an answer to the probability of 14 correct guesses in a row by hand.