Suppose the probability for A to win a game against B is 0.4. If A has an option of playing either a

Q: Wait, why does the probability of winning decrease when we play more games? Shouldn't more games give me more chances to recover?

This is the most common trap! Think of it this way: if you are an underdog ($p < 0.5$), you are fighting against the 'Law of Large Numbers'. As the number of games $n$ increases, the actual win percentage tends to converge toward the expected value, which is your win probability $p$. Since $0.4 < 0.5$, the more games you play, the more the 'average' result dominates, and the harder it becomes to pull off an upset. You want fewer games to rely on the variance of luck!

Q: Can I use the complement rule to solve this faster?

Absolutely! For the 'Best of 5', instead of calculating $P(X=3) + P(X=4) + P(X=5)$, you could calculate $1 - [P(X=0) + P(X=1) + P(X=2)]$. It is mathematically equivalent. However, in this specific problem, the number of terms is small enough that direct calculation is just as fast and less prone to subtraction errors.

Q: Is there a general formula to determine if I should choose 'Best of 3' or 'Best of 5' without doing all this math?

Yes, there is a beautiful intuition here. If your win probability $p$ is less than $0.5$, you should always choose the shortest possible match (Best of 1, if allowed). If your win probability $p$ is greater than $0.5$, you should choose the longest possible match to let your skill dominate the variance. Since $0.4 < 0.5$, the 'Best of 3' is mathematically superior to the 'Best of 5'.

Question

Suppose the probability for $A$ to win a game against $B$ is 0.4. If $A$ has an option of playing either a "best of 3 games" or a "best of 5 games" match against $B$, which option should be choose so that the probability of his winning the match is higher? (No game ends in a draw).

Sigma Percentile · Accepted Answer

### The Underdog's Strategy: A Lesson in Probability

Imagine you are standing on the court, racket in hand, facing an opponent who is statistically better than you. Your probability of winning a single game is $p = 0.4$.

You are the underdog. The tournament organizers offer you a choice: play a 'Best of 3' match or a 'Best of 5' match. Which one do you choose?

Most students instinctively think, "More games give me more chances to win!" But in the world of probability, intuition can be a dangerous trap. Let us dissect this using the power of the **Binomial Distribution**.

### Phase 1: The Best of 3

In a 'Best of 3' match, you win the match if you win at least 2 games. This means you can win exactly 2 games or exactly 3 games.

We use the binomial formula:
$$P(X=k) = \binom{n}{k} p^k q^{n-k}$$
where $n=3$, $p=0.4$, and $q=0.6$.

For exactly 2 wins, we calculate:
$$\binom{3}{2} (0.4)^2 (0.6)^1 = 3 	imes 0.16 	imes 0.6 = 0.288$$

For exactly 3 wins, we calculate:
$$\binom{3}{3} (0.4)^3 (0.6)^0 = 1 	imes 0.064 	imes 1 = 0.064$$

Adding these together, your total probability of winning the 'Best of 3' is $0.288 + 0.064 = 0.352$. You have a **35.2% chance of victory**.

### Phase 2: The Best of 5

Now, let us look at the 'Best of 5'. Here, you need at least 3 wins to take the match. This means you could win 3, 4, or 5 games.

Using the same binomial logic with $n=5$:

For 3 wins:
$$\binom{5}{3} (0.4)^3 (0.6)^2 = 10 	imes 0.064 	imes 0.36 = 0.2304$$

For 4 wins:
$$\binom{5}{4} (0.4)^4 (0.6)^1 = 5 	imes 0.0256 	imes 0.6 = 0.0768$$

For 5 wins:
$$\binom{5}{5} (0.4)^5 (0.6)^0 = 1 	imes 0.01024 	imes 1 = 0.01024$$

Summing these up, your probability of winning the 'Best of 5' is $0.2304 + 0.0768 + 0.01024 = 0.31744$. Your chances have dropped to approximately **31.7%**.

### The Philosophical Insight

Why did your chances drop? This is the core of the lesson.

When you are the underdog ($p < 0.5$), you rely on 'variance'—the lucky streaks that allow you to win a short series. As the number of games $n$ increases, the **Law of Large Numbers** takes over.

The actual outcome of the match begins to mirror the expected probability more closely. Since your expected win rate is below 50%, the longer the match, the more certain it becomes that you will lose.

Therefore, to maximize your chances as an underdog, you must minimize the number of games. Choose the 'Best of 3' and hope for a lucky streak! The math confirms it: **$0.352 > 0.31744$**. Always trust the numbers over your gut feeling.

Visualized Solution

Defining Probabilities $p$ and $q$

Winning Condition for Best of 3

Binomial Formula for $n = 3$

Calculating $P (Best of 3)$

Winning Condition for Best of 5

Binomial Formula for $n = 5$

Calculating $P (Best of 5)$

Final Comparison and Conclusion

The Sigma Insight: Binomial Distribution

The Underdog's Strategy

Phase 1

Phase 2

The Philosophical Insight

Similar Questions

A multiple choice examination has 5 questions. Each question has three alternative answers of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is

The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least 90% is :

A student appeared in an examination consisting of 8 true-false type questions. The student guesses the answers with equal probability. The smallest value of $n$ , so that the probability of guessing at least ' $n$ ' correct answers is less than $1/2$ , is

In a bag there are 20 cards 10 names $A$ and another 10 names $B$ . Cards are drawn randomly one by one with replacement then find probability that second $A$ comes before third $B$ .

The probability that an event $A$ happens in one trial of an experiment is $0.4$ . Three independent trials of the experiment are performed. The probability that the event $A$ happens at least once is

Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than 99% is :

For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is 4/5, then the probability that he is unable to solve less than two problems is :

Let in a Binomial distribution, consisting of 5 independent trials, probabilities of exactly 1 and 2 successes be 0.4096 and 0.2048 respectively. Then the probability of getting exactly 3 successes is equal to: