Proof by Induction:
Let P(n) denote the statement: "No strategy exists which gives a winning probability strictly greater than 1/2 for a deck of n cards, where n is even and there are n/2 black cards and n/2 red cards."
We will prove that P(n) holds for all even n ≥ 2, using mathematical induction.
Base Case (n=2):
When n=2, the deck consists of 1 black card and 1 red card. Since the cards are randomly shuffled, and there is no way to distinguish between the two cards, the probability of correctly guessing when the black card will appear is exactly 1/2. No strategy can increase this probability because both cards are equally likely to appear next. Therefore, P(2) holds.
Inductive Step:
Assume that P(n) holds for some even n≥2, i.e., for a deck of n cards (with n/2 black cards and n/2 red cards), no strategy can give a winning probability strictly greater than 1/2. We need to prove that P(n+2) holds.
By the inductive hypothesis, for the deck of n cards (with n/2 black and n/2n red cards), any strategy has a winning probability of at most 1/2. Adding one black and one red card does not change the relative proportion of black to red cards in the deck. Since the ratio remains balanced and unaffected by the addition of one black and one red card, no additional information or advantage is introduced. Therefore the probability remains same for n + 2.
Thus, by the principle of mathematical induction, P(n) holds for all even n≥2 and therefore we can conclude no strategy exists which gives probability of wining higher than 1/2 for deck of 52 cards.