I'm working on the following probability problem and would like verification of my proof:
Problem Statement
Consider a deck of 52 regular playing cards, 26 red and 26 black, randomly shuffled. The cards are face down. Cards are drawn one at a time from the top of the deck and turned face up. At any point before the deck is exhausted, you must declare that you want the next card. If that next card is black, you win; otherwise, you lose. The question is: Is there a strategy that gives a probability of winning strictly greater than 1/2?
My Proof Attempt
The decisions any strategy will make will fall into one of two cases:
- When both red and black cards remain.
- When either black or red cards remain.
We can calculate the probability of wining for each of them:
Probability of winning for decisions that are made when both red and black cards remain:
There are 52!/26!*26! Arrangements of black and red cards. In all possible arrangements of 26 black and 26 red cards, for any given position, due to symmetry exactly half of the arrangements will have a black card in that position, and half will have a red card. This means that if after a number of cards if a strategy decides on the next card, the number of combinations where the next card is black equals the number of combinations where the next card is red. This means the overall probability of wining is 1/2 for such decisions.
Probability of winning for decisions that are made when either black or red cards remains:
We can only win this scenario only if black cards are remaining cards. The number of combinations whereby the last remaining cards are only black equals the number of combinations whereby the last remaining cards are only red. Therefore the probability of getting into a scenario where only black card remains equal 1/2 .
[Note that I'm referring to overall probability here not the probability in individual games which can increase or decrease. The result of making such decisions is that they do not produce wining probability greater than 1/2 for the reasons provided.]
Thus, both types of decisions yields a probability no greater than 1/2. Therefore the overall wining probability for any strategy cannot be grater than 1/2.
My Question
My specific doubt is whether my reasoning in the second case is correct i.e that the probability of wining for decisions when only one color remains is indeed 1/2. Also does the conclusion I’ve derived from both cases (that no strategy can yield a probability greater than 1/2) logically follows or not.
Thank you in advance for your insights and corrections!
