You draw three cards uniformly at random (without replacement) from a standard 52-card deck. You reveal only the color of each card in order -- red (R) or black (B).
Let $A$ be the event that all three cards have distinct ranks, and let $B$ be the event that the color pattern is RRB.
Compute $P(B)$ and $P(A)$.
Compute $P(A \mid B)$.
A bookmaker offers even odds on whether the three ranks are all distinct. You get to see the color pattern before deciding whether to bet. For which color patterns (if any) is this bet positive expected value, and why?
Hints
Think about how drawing without replacement affects the count of available cards of each color and rank -- the color pattern constrains which cards are available for later draws.
For $P(A \cap B)$, count card-by-card: after fixing a red card, how many remaining red cards share its rank? Each rank has exactly 2 red and 2 black cards.
For part (c), compute $P(A)$ unconditionally first. It is around 0.83 -- ask whether any color pattern could shift it below 0.5.
Worked Solution
How to Think About It: This is a conditional probability problem where you need to figure out whether observing the color pattern tells you anything about rank distinctness. The key tension: colors and ranks feel like they should be independent, but the sampling-without-replacement mechanics create a subtle coupling. Before diving into computation, ask yourself -- does knowing the color pattern change the number of "available" cards of each color for later draws, and does that affect rank collisions?
Quick Estimate: There are $\binom{52}{3} \times 3! = 52 \times 51 \times 50$ ordered draws. For $P(A)$: first card is anything, second must avoid 3 cards of the same rank out of 51, third must avoid 6 cards (two ranks seen) out of 50. So $P(A) = 1 \times \frac{48}{51} \times \frac{44}{50} \approx 0.941 \times 0.88 \approx 0.828$. For $P(B)$: about $(26/52)(25/51)(26/50) \approx 0.5 \times 0.490 \times 0.520 \approx 0.1275$. These should be close to the exact values.
Approach: Count ordered triples directly, conditioning on color and rank constraints.
Formal Solution:
Part (a): Work with ordered draws of size 3 from 52 cards. The sample space has $52 \times 51 \times 50 = 132{,}600$ equally likely sequences.
$P(B)$: The pattern RRB means the first two cards are red and the third is black.
Numerically, $P(A) \approx 0.8282$ and $P(B) \approx 0.1275$.
Part (b): We need $P(A \mid B) = P(A \cap B) / P(B)$.
Count sequences with pattern RRB and all distinct ranks. Card 1: any of 26 red cards. Card 2: must be red and a different rank from card 1. There are 25 remaining red cards, but 1 of them shares rank with card 1 (if card 1's rank has another red card) or 0 do (if card 1's rank has no other red card). Since each rank has exactly 2 red cards and 2 black cards, exactly 1 of the remaining 25 red cards shares card 1's rank. So card 2 has
5 - 1 = 24$ valid choices.
Card 3: must be black and a different rank from both cards 1 and 2. There are 26 black cards total, and 2 of those ranks are "used" (cards 1 and 2 each contribute a rank). Each used rank has exactly 2 black cards, so
\times 2 = 4$ black cards share a rank with cards 1 or 2. Card 3 has
6 - 4 = 22$ valid choices.
Part (c): At even odds, the bet on $A$ has positive EV when $P(A \mid \text{pattern}) > 0.5$. Since the unconditional $P(A) \approx 0.828$, and the color pattern creates only a small perturbation, $P(A \mid \text{pattern})$ stays well above 0.5 for every pattern.
By symmetry there are only three distinct pattern types: all same color (RRR or BBB), two-and-one (RRB, RBR, BRR, BBR, BRB, RBB), and alternating (RBR, BRB -- already counted). The key insight is that same-color patterns make rank collisions slightly more likely (you are drawing from a pool of only 26 cards sharing a color, so ranks are more constrained), but even in the worst case $P(A \mid \text{pattern})$ remains far above 0.5.
For the all-same-color pattern (say RRR): $P(A \mid RRR) = \frac{24}{25} \cdot \frac{22}{24} = \frac{22}{25} = 0.88$ -- actually higher because you are forced into distinct suits within a color, but with only 2 red suits per rank the collision rate is similar.
Therefore, the bet is positive EV for every color pattern. The color pattern shifts $P(A)$ only slightly, and $P(A)$ is so far above 0.5 that no pattern can push it below the breakeven threshold.
Answer: $P(B) = \frac{13}{102}$, $P(A) = \frac{352}{425}$, $P(A \mid B) = \frac{264}{325} \approx 0.812$. The even-odds bet on distinct ranks is positive EV for all color patterns, because $P(A)$ is high enough that the modest dependence between color and rank distinctness cannot bring it below 0.5.
Intuition
This problem tests whether you understand the subtle dependence between color and rank when sampling without replacement. Intuitively, colors and ranks feel independent -- and in an infinite deck they would be. But in a finite deck, observing that all three cards are red means you are drawing from only 26 cards spanning 13 ranks (2 cards per rank), which slightly changes collision probabilities compared to drawing from the full 52. The practical lesson: in finite-population sampling, attributes that seem independent can become coupled through the sampling mechanism. This shows up constantly in trading -- for example, conditioning on a stock being in a particular sector (analogous to color) changes the distribution of other characteristics (analogous to rank) because sectors have finite membership.
The part (c) twist is a nice test of quantitative intuition. The probability of distinct ranks is so high (~83%) that you should immediately recognize no color pattern can push it below 50%. A good interviewer wants to see you make that argument quickly rather than computing all cases.