I recently saw some friends play a game where the goal is to get a series of 4 educated and/or random guesses on 4 playing card flips. Thinking about the probability of winning seemed as simple as multiplying the chances of each event by each other, however, these were the 4 questions you had to get correct:
Flip 1: Color of the card (red/black). 0.5 (obviously)
Flip 2: If this card will be higher/lower. More permutations, AA,A2,23...2A,22,23...
Flip 3: If this card will be in the range between or outside the values of Flip 1 and Flip 2. (I've seen it played both on a closed and open interval. Let's just say open
(Flip 1 < Flip 3 < Flip 2) == True
Flip 4: The suit of the card (diamond, club, heart, spade). 0.25, but actually 13/49 due to previous flips without replacement.
Note I: Let value of Ace be 1 (even though it doesn't really matter)
Note II: After each flip you make the guess after knowing the previous flip(s). True random would probably be different than an optimal odds table.
Note III: Cards are flipped from a full 52 card deck WITHOUT replacement.
Not sure how to approach it. Maybe we could write a program to just count all the possible permutations in the sample space? Or is it as simple as an nCr or nPr kind of problem.
OR ... Is it as simple as: $$ \frac{1}{2} \cdot \frac{1}{13} \cdot \frac{1}{12} \cdot \frac{13}{49} = \frac{1}{15,288} $$ $$ \hspace{3.72cm}\approx 6.5445\%? $$
To remove ambiguity I will assume that the player looses the game in the flips 2 and 3 if the flipped card value coincides with that of the previously flipped card(s).
Obviously the probability to win the first flip is $\frac12$. The best strategy to win the second flip is apparently to guess "higher" if the first card has the value less than $7$ and "lower" if it exceeds it. If the value of the first card is $7$ both guesses are equally good. Thus the probability to win this flip is: $$ \frac1{13}\left[\sum_{i=1}^6\frac{52-4i}{51}+\frac{24}{51}+\sum_{i=8}^{13}\frac{4i-4}{51}\right]=\frac{160}{221}. $$
If we pass the second flip we have two different values of two previous cards separated by some gap (the gap is the number of values between the two given). There are altogether $\binom{13}2$ possible combinations. The best strategy to win the third flip is to guess "outside" if the gap is $5$ or less, and "inside" if it is $6$ or larger. Obviously there are 12 combinations with gap $0$, 11 combinations with gap $1$ and so on till 1 combination with gap $11$. Thus the probability to win the flip is: $$ \frac1{\binom{13}2}\left[\sum_{g=0}^{5}(12-g)\frac{44-4g}{50} +\sum_{g=6}^{11}(12-g)\frac{4g}{50}\right]=\frac{17}{25}. $$ Finally in the last flip there is at least one suit which did not yet show. Therefore if one guesses this suit (which is the best strategy) one wins the flip with probability $\frac{13}{49}$.
The probability to win the game is therefore: $$ \frac12\cdot\frac{160}{221}\cdot\frac{17}{25}\cdot\frac{13}{49}=\frac{16}{245}\approx6.53\%. $$
It is interesting to compare this result with the best possible blind guess (the guess for all four flips before the trial).
The probability to guess the correct color in the first flip is still $\frac12$. For the second flip both guesses give the same probability $\frac{24}{51}$ The third flip is more interesting; it appears that the probability to win with the guess "outside" is 2 times higher than with "inside": $\frac{44}{75}$ vs. $\frac{22}{75}$. By the forth flip it is advantageous to guess a suit with color opposite to that one predicted for the first flip; this has the probability $\frac{13}{51}$.
The probability to win for the optimal blind guess is therefore: $$ \frac12\cdot\frac{24}{51}\cdot\frac{44}{75}\cdot\frac{13}{51}=\frac{2288}{65025}\approx3.52\%, $$ almost two times less than with "educated" guess.