ML Interview Q Series: Binomial Probability Analysis: Comparing Winning 3/4 vs 5/8 Games Likelihood

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You play draughts against an opponent who is your equal. Which of the following is more likely: (a) winning three games out of four or winning five out of eight; (b) winning at least three out of four or at least five out of eight?

Short Compact solution

Let X and Y be the numbers of wins in 4 and 8 games respectively. When you play 4 games, there are 2^4 = 16 equally likely outcomes (for example, WLWW indicates 3 wins). From basic counting arguments, the number of ways to get j wins in 4 games is (4 choose j). Since each outcome has probability (0.5)^4, we get P(X=3) = 4 × (0.5)^4 = 0.25.

For 8 games, there are 2^8 = 256 equally likely outcomes, and the number of ways to get 5 wins is (8 choose 5). Hence P(Y=5) = 56 × (0.5)^8 = 0.2188. So winning exactly 3 out of 4 (which has probability 0.25) is more likely than winning exactly 5 out of 8 (which has probability 0.2188).

For part (b), consider P(X >= 3) in 4 games, which covers 3 or 4 wins. We have P(X >= 3) = P(X=3) + P(X=4) = 0.25 + 0.0625 = 0.3125. For 8 games, P(Y >= 5) is P(Y=5) + P(Y=6) + P(Y=7) + P(Y=8) = 0.2188 + 0.1094 + 0.0313 + 0.0039 = 0.3633. Thus it is more likely to win at least five out of eight (0.3633) than to win at least three out of four (0.3125). This illustrates that the chance of a drawn series diminishes as the number of games increases.

Comprehensive Explanation

Overview of the Problem

You are comparing probabilities in a fair, binary-outcome setting (like a coin flip, or an equally skilled draughts game where each player has a 50% chance of winning each game). There are two main comparisons:

1. (a) Which is more likely?

   • Winning 3 out of 4 games

   • Winning 5 out of 8 games

2. (b) Which is more likely?

   • Winning at least 3 out of 4 games

   • Winning at least 5 out of 8 games

Binomial Distribution as the Underlying Model

Because each individual game is assumed to have a 50% chance of winning (and presumably 50% chance of losing), the total number of wins in n games follows a Binomial distribution with parameters n = number of games and p = 0.5.

In general, if X is a binomial random variable that counts the number of wins (successes) out of n trials, each with success probability p, then:

where:

• n is the total number of independent games

• k is the number of wins (successes)

• p is the probability of winning any single game

• (n choose k) = n! / (k! (n-k)!) is the binomial coefficient

Part (a): Comparing Winning Exactly 3 out of 4 vs. Exactly 5 out of 8

Winning 3 out of 4

• The total number of possible outcomes for 4 games is 2^4 = 16.

• The number of ways to get exactly 3 wins out of 4 is (4 choose 3) = 4.

• Each outcome (sequence of wins/losses) has probability (0.5)^4 = 1/16, because p=0.5 per game and we assume independence.

So: P(X=3) = (4 choose 3) × (0.5)^4 = 4 × (1/16) = 4/16 = 0.25.

Winning 5 out of 8

• The total number of possible outcomes for 8 games is 2^8 = 256.

• The number of ways to get exactly 5 wins is (8 choose 5) = 56.

• Each of the 256 equally likely sequences has probability (0.5)^8 = 1/256.

So: P(Y=5) = (8 choose 5) × (0.5)^8 = 56 × (1/256) = 56/256 = 0.2188 (approximately).

By comparing 0.25 vs. 0.2188, we see that it is more likely to win exactly 3 out of 4 (0.25) than to win exactly 5 out of 8 (0.2188).

Part (b): Comparing at Least 3 out of 4 vs. at Least 5 out of 8

Winning at Least 3 out of 4

We compute: P(X >= 3) = P(X=3) + P(X=4).

We already know P(X=3) = 0.25. For P(X=4):

• (4 choose 4) = 1

• Probability = 1 × (0.5)^4 = 1/16 = 0.0625

Hence: P(X >= 3) = 0.25 + 0.0625 = 0.3125.

Winning at Least 5 out of 8

We sum P(Y=5) + P(Y=6) + P(Y=7) + P(Y=8).

• P(Y=5) = 56 × (0.5)^8 = 56/256 = 0.2188

• P(Y=6) = (8 choose 6) × (0.5)^8 = 28 × (1/256) = 0.1094

• P(Y=7) = (8 choose 7) × (0.5)^8 = 8 × (1/256) = 0.0313

• P(Y=8) = (8 choose 8) × (0.5)^8 = 1 × (1/256) = 0.0039

Summing these: 0.2188 + 0.1094 + 0.0313 + 0.0039 = 0.3633.

Comparing 0.3125 vs. 0.3633, it is more likely to win at least 5 out of 8 games than to win at least 3 out of 4. This is consistent with the intuition that the series has more total games, so having a “drawn series” (where neither player dominates) is slightly less probable as the number of games grows, thus the probability of one player reaching a certain threshold of wins can become larger.

Additional Insight

The results show that when the total number of games is small (like 4), you can achieve 3 wins fairly easily (0.25 probability to get exactly 3). But in 8 games, the probability of getting exactly 5 wins is smaller (around 0.2188). However, once you broaden the condition to "at least 3 out of 4" versus "at least 5 out of 8," the sum of those higher counts (5, 6, 7, 8) eventually outweighs the 3-or-4 possibility in 4 games.

Below is a short Python snippet illustrating how to compute these probabilities:

import math

def binomial_prob(n, k, p=0.5):
    # (n choose k) * p^k * (1-p)^(n-k)
    return (math.comb(n, k)) * (p**k) * ((1-p)**(n-k))

# Probability of winning at least 3 out of 4
p_at_least_3_of_4 = binomial_prob(4, 3) + binomial_prob(4, 4)

# Probability of winning at least 5 out of 8
p_at_least_5_of_8 = sum(binomial_prob(8, k) for k in range(5, 9))

print("P(X >= 3) when X ~ Binomial(4, 0.5):", p_at_least_3_of_4)
print("P(Y >= 5) when Y ~ Binomial(8, 0.5):", p_at_least_5_of_8)

Potential Follow-up Questions

What if the probability of winning each game is not 0.5?

In a real-world scenario, you might not have a 50-50 chance of winning each game. If the probability of winning each game is p (not necessarily 0.5), the binomial distribution becomes:

The key differences are:

• You would replace 0.5 with p in all formulas.

• If p > 0.5, then your chances of hitting a threshold of wins (like at least 5 out of 8) generally increase, while if p < 0.5, they decrease.

• The relative comparisons (like X >= 3 in 4 games vs. Y >= 5 in 8 games) can change based on p.

How does the probability of a drawn series evolve with more games?

A drawn series can happen if each player wins the same number of games (e.g., for an even number of games, you get an equal split). If the probability of each game is 0.5, the binomial distribution’s mean is n * 0.5. As n grows large, the distribution becomes more spread out (standard deviation is sqrt(n * p * (1-p))), but crucially, the relative probability of exactly half the wins (n/2) tends to decrease in the sense of proportion to the entire distribution. This is because there are more possible ways to deviate from half as n increases. Hence, the chance of a perfectly drawn series (like 4-4 in 8 games) typically diminishes as n grows (in a relative sense).

How can we intuitively understand that “the chance of a drawn series falls as the series gets longer”?

When the number of games is small, it’s relatively easy for both players to end up in a tie. For example, with 2 games, the chance to finish 1-1 is (2 choose 1)*(0.5^2) = 0.5. But as the number of games grows, the distribution of total wins becomes “wider,” so there are more ways to end up with counts other than an exact tie. While the absolute number of tied outcomes does increase with n, the probability of that exact tie out of all possible outcomes is smaller.

What might happen in real tournaments?

• Fatigue, skill gaps, or psychological factors might invalidate the assumption that p=0.5 is constant over all games.

• Some draws might occur if the game results in an actual “draw” outcome, which changes the distribution beyond just a win/loss model.

• The strategy might change if you are behind or ahead in a series, making the independence assumption questionable.

Regardless, the binomial framework is a good idealized baseline for analyzing the probability of achieving a certain number of wins under the simplest fair assumptions.