ML Interview Q Series: Expected Value Analysis of a Conditional Payoff Dice Game
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You may simultaneously roll one red die and three blue dice. The stake is $1. If none of the blue dice matches the red die, you lose your stake; otherwise, you get paid k+1 dollars if exactly k of the blue dice match the red die. In the case that exactly one blue die matches the red die, you get paid an additional $0.50 if the other two blue dice match each other. What is the expected payoff of the game?
Short Compact solution
Define the random variable X as the total payoff from the game. One finds:
P(X=0) = 125/216
P(X=2) = 60/216
P(X=2.5) = 15/216
P(X=3) = 15/216
P(X=4) = 1/216
Then the expected payoff is:
Since the game costs $1 to play, the net expectation to the player is 0.956 - 1 = -0.044. Therefore, the game is unfavorable to the player, with a house edge of about 4.4%.
Comprehensive Explanation
Overview of the Dice Game Setup
You roll a single red die (six possible outcomes) and three blue dice (each also six possible outcomes). We analyze the payoffs as follows:
If none of the blue dice matches the red die: You receive 0 dollars. Your stake of $1 is completely lost.
If exactly k of the blue dice match the red die (k=1,2,3): You receive k+1 dollars.
For k=1: You receive 2 dollars, plus an additional $0.50 if the other two blue dice match each other. This yields a final payoff of either 2 or 2.5.
For k=2: You receive 3 dollars.
For k=3: You receive 4 dollars.
In every scenario, the cost of playing is $1, so the net profit is X (the payoff) minus 1. However, when we speak of the expectation of X itself, we do not subtract 1 in the formula. Only afterward do we interpret E(X) - 1 to see how profitable the game is for the player.
Computing the Probability Distribution
Total sample space: 6^4 = 1296 equally likely outcomes (6 for the red die times 6 for each of the three blue dice).
P(X=0): No blue die matches the red die. For a given red-die outcome, each blue die has 5 ways (out of 6) to avoid matching. The probability each blue die avoids matching is 5/6, so for three dice (5/6)^3 = 125/216.
P(X=2) and P(X=2.5): Exactly one of the three blue dice matches the red die. We must split this scenario further into:
Probability that exactly one blue die matches AND the other two do NOT match each other: This yields a payoff of 2 dollars.
Probability that exactly one blue die matches AND the other two match each other: This yields a payoff of 2.5 dollars.
P(X=3): Exactly two of the blue dice match the red die (k=2), so the payoff is 3 dollars.
P(X=4): All three blue dice match the red die (k=3), so the payoff is 4 dollars.
From the short solution, these combined probabilities are:
P(X=0) = 125/216
P(X=2) = 60/216
P(X=2.5) = 15/216
P(X=3) = 15/216
P(X=4) = 1/216
Expected Payoff Calculation
Using the probability mass function for X:
Here:
X=0 with probability 125/216
X=2 with probability 60/216
X=2.5 with probability 15/216
X=3 with probability 15/216
X=4 with probability 1/216
So:
Since each play costs $1, the expected net profit to you is 0.956 - 1 = -0.044, meaning on average you lose 4.4 cents per game. Therefore, the house edge is approximately 4.4%.
Practical Verification Through Simulation
One could write a quick Python simulation to confirm this analytical result:
import random
import statistics
def simulate_dice_game(num_trials=10_000_00):
payoffs = []
for _ in range(num_trials):
red = random.randint(1, 6)
blue = [random.randint(1, 6) for _ in range(3)]
matches = sum(1 for b in blue if b == red)
if matches == 0:
payoff = 0
elif matches == 1:
# exactly one matches
payoff = 2
other_two = [b for b in blue if b != red]
if other_two[0] == other_two[1]:
payoff += 0.5
elif matches == 2:
payoff = 3
else: # matches == 3
payoff = 4
payoffs.append(payoff)
return statistics.mean(payoffs)
avg_payoff = simulate_dice_game()
print("Approximate average payoff:", avg_payoff)
print("Approximate net (avg_payoff - 1):", avg_payoff - 1)
Repeated runs of this simulation will yield results close to 0.956 for the expected payoff, giving a net of around -0.044 per game.
Potential Follow-Up Questions
How would the result change if the dice were biased?
If each die were not fair, probabilities like 5/6 or 1/6 would need to be adjusted to reflect the bias. The methodology remains the same, but you must plug in the correct probabilities for each face outcome. You would still compute the expected payoff via a sum over all possible outcomes, but each outcome’s probability would now depend on the dice’s bias distribution.
What is the variance of the payoff?
You can calculate variance using Var(X) = E(X^2) - [E(X)]^2. You would compute E(X^2) by summing x^2 * P(X=x). This can help determine the riskiness of the game. Even though the expected value might be slightly negative, a high variance could lead to occasional large wins.
Could you design a strategy to improve your odds?
No, there is no skill-based strategy in a fair dice game of this sort. The outcomes are purely random. You cannot alter the probabilities of rolling certain faces, so you have no way to improve your expected payoff beyond not playing.
How does the law of large numbers apply here?
Over a large number of plays, the average payoff will converge to the expected value of 0.956 per game. This means that in the long run, you can expect to lose about 4.4 cents per play on average.
Could the house adjust the payouts to attract players but still retain a positive edge?
Yes, the house can tweak payouts (for example, increase the payoff on certain outcomes) to make it appear more attractive while still maintaining a positive expected advantage. By carefully rebalancing payouts, the house ensures the overall expected value to the player remains below 1 dollar, preserving profitability.
These follow-up points highlight how probability theory underlies both the fairness and the profitability structure of such dice games.
Below are additional follow-up questions
If we changed the bonus condition (for instance, awarding the extra $0.50 in a different scenario), how would it affect the expected value?
If you alter the current bonus rule—where you receive an additional $0.50 only when exactly one die matches the red die and the other two match each other—you would need to re-derive the probabilities for the newly defined bonus condition and recalculate the payoff. For example, suppose you modify the bonus to apply whenever exactly two dice match each other (regardless of matching the red die). This would affect the overall distribution of payoffs:
You need to break out new distinct events (for example, “one die matches the red die but two do not, yet they match each other” versus “exactly two dice match the red die” and so on).
Each event now has different probability and payoff values.
The calculation method remains the same: identify all relevant outcomes, multiply by respective payoffs, and sum to find E(X).
A key pitfall is failing to correctly partition events. Overlapping cases can lead to double-counting or missing some outcomes entirely. For instance, the event “two dice match each other” and the event “one die matches the red die” might partially overlap, so you must carefully define disjoint conditions.
Would the order in which the dice fall matter for any reason?
In principle, each of the blue dice is indistinguishable except for their final values. The game’s payoff is determined by the count of matches and whether the remaining dice match each other. Therefore, the order in which the dice fall does not matter as far as the final payout calculation is concerned. All permutations of the same combination (e.g., red=3, blue={3,1,1} versus red=3, blue={1,3,1}, etc.) lead to the same payoff.
However, a subtle point is that if the game rules were expanded to track the sequence or position of the dice in some special way (for example, awarding a bonus if the matching die occurs in a particular position), then the order would become relevant. In that modified game, you would need to account for permutations, which expands the sample space and changes the probability of each distinct outcome.
What happens if we play only a few rounds versus many rounds?
When playing only a few rounds (say 5 or 10), the outcome could deviate greatly from the expected net loss of 4.4%. You might get lucky and see multiple high-paying results in just a handful of rolls. Over many rounds, however, the law of large numbers says the average payoff per game will converge closer to the theoretical expected value of 0.956 (net -0.044).
A common pitfall is to confuse a short-term winning streak with a positive expected value. Variance can cause large fluctuations in small sample sizes, potentially leading players to believe they have a winning system even though the underlying mathematics remains unfavorable.
How does this game illustrate the concept of expected value versus utility in real-life decision-making?
Real-world gamblers might assign a subjective utility to each outcome. For instance, winning $3 when you’re low on funds might feel disproportionately more valuable than losing $1. Even though the pure expected monetary value is negative, some players might still find psychological “value” in the excitement or potential for a big payoff.
A pitfall here is to conflate the theoretical monetary expectation with a personal utility function. While mathematically the game’s EV is negative, an individual might still choose to play if the entertainment value or psychological payoff is worth the cost. In corporate or high-stakes environments, however, decision-making typically aligns more with strict monetary expectations rather than intangible utility, unless explicitly stated otherwise.
Could correlation among dice affect the outcome?
Ordinarily, dice rolls are independent events. But if, in an unusual situation, the rolls of different dice were correlated (for example, a defective or physically linked set of dice that tend to land in certain combinations), the standard probabilities (1/6 for matching, 5/6 for non-matching) might not hold.
The pitfall is assuming independence when it does not exist. If there is correlation (e.g., a particular face on one die is strongly associated with a particular face on another die), your probability distribution changes significantly. You would then need a joint probability table for (red, blue1, blue2, blue3) that reflects the correlation structure, and recalculate E(X) based on that new distribution.
What if the stake is not $1 but some other fixed amount?
If the entry cost to play is C dollars, then the net outcome per game is X - C, where X is the random payoff. The expected net would be E(X) - C. If you scale payouts proportionally while changing the stake, the sign of the expectation could remain the same. However, if you only change C but not the payouts, then:
The expected payoff E(X) remains the same numerically.
The expected net becomes E(X) - C, which may be more or less negative (or positive) depending on C.
A practical pitfall is to assume changing the stake alone could flip the game from being negative EV to positive EV—this rarely works unless the rules themselves are modified to alter probabilities or payouts in a way that increases E(X).
How would adding a fourth blue die change the analysis?
If a fourth blue die is introduced, you must account for new ways to match the red die:
Probability that none of the four blue dice match.
Probability exactly one matches, plus potentially more complicated bonus conditions (e.g., whether the other three match among themselves, or any pair within those three).
Probability exactly two match, exactly three match, or all four match.
Each of these events must be meticulously enumerated (or determined via combinatorial methods), multiplied by the new payoff amounts, and summed. The sample space grows to 6^5 = 7776 total outcomes. As a result, the overall expected payoff might shift significantly depending on how the new payouts are structured.
A subtle pitfall is incorrectly counting the number of ways certain patterns can appear. With four dice, scenarios like “exactly one matches the red die, while exactly two of the remaining three match each other” become more numerous and more complex.
What if we impose a limit on how many times you can win the bonus?
Consider a rule where the extra $0.50 bonus can only be awarded once every five games, or some limit is placed on consecutive bonus payouts. This artificially alters the distribution of payoffs across a sequence of games, creating dependence from one roll to the next (the current game’s bonus payout might be disallowed based on previous game results).
From a mathematical standpoint, you can no longer treat each game as an independent trial with the same probability distribution. Instead, you have to track states reflecting whether the bonus is “locked” or “unlocked.” This becomes a Markov chain problem, where the expected value in each state must be calculated, and then you combine them to get the long-run expectation.
A common pitfall in this scenario is to assume you can still multiply a single-round expectation by the number of rounds, ignoring the new dependency constraints. The state-based approach is necessary to capture how the bonus constraint modifies the probabilities over multiple plays.
Is there a closed-form generating function for this game?
Yes, you can construct a probability generating function (PGF) or moment generating function (MGF) for the random variable X. In a simpler matching-dice scenario (without the extra $0.50 bonus), the distribution is small enough to handle easily. However, with the bonus rule for exactly one match and the other two matching each other, you can still theoretically build a piecewise generating function:
Identify each outcome’s probability.
Represent each outcome as a term in the generating function with coefficient equal to that probability, and exponent or factor reflecting the payoff.
The pitfall is that constructing a generating function for a piecewise random variable can be cumbersome. You must carefully isolate the event “one match plus the other two match each other.” Nonetheless, once established, the MGF or PGF can give you direct formulas for E(X) or other moments. However, it can be more complex than direct enumeration since we have multiple discrete payoffs and a small custom bonus.
In a commercial setting, how might a casino or game provider justify offering a negative-EV game?
Casinos rely on the principle that the game is entertaining or has an exciting “hook” that draws players in. Despite the negative expected value, players often focus on the possibility of big payouts or on short-term winning streaks. In addition, casinos can claim transparency: if the expected return is publicly posted, the player can make an informed choice.
A subtlety is that some jurisdictions require a minimum payback percentage or require that the house edge be below a certain threshold. Game providers must ensure their game design complies with these regulations. They do so by tweaking payouts or adjusting bonus rules to keep the expected value for the player high enough to meet legal requirements while still retaining a margin for the house.