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Using Mathematical Expectation to Scientifically Tackle Problems

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The text discusses the mathematical expectations of scoring on single-choice and multiple-choice questions based on different strategies of guessing answers. For a single-choice question worth 5 points with four options (A, B, C, D) and two correct answers, the expected scores for three individuals (Zhang San, Li Si, and Wang Wu) are calculated based on their guessing probabilities. Zhang San, who has no knowledge, has an expected score of 1.25, while Li Si, who has a 90% chance of guessing correctly, has an expected score of 4.5. Wang Wu, who guesses three options, has the lowest expected score of 0.5. In the case of a multiple-choice question, the expected scores vary more significantly depending on how many options are guessed. Zhang San, who guesses one option (A), has an expected score of 1.2, while Li Si, who guesses two options (AB), has a lower expected score of 0.9, and Wang Wu, who guesses three options (ABC), has an even lower expected score of 0.5. The analysis suggests that when unsure of the answers, selecting fewer options can lead to a higher expected score. However, when some answers are known to be correct or incorrect, the strategy changes, and the expected scores can vary based on the number of options chosen. The text concludes that in scenarios where one option is confirmed as incorrect, choosing two options yields the highest expected score, but it raises the question of whether one would take the risk of potentially scoring zero by guessing two options when only one is confirmed correct.

First, let's look at single-choice questions#

Let's assume a single-choice question is worth 5 points, with four options: A, B, C, and D. There are exactly two correct answers, meaning there are four reasonable ways to answer the single-choice question: A, B, C, and D.
Here we assume there is a single-choice question:

Zhang San completely cannot answer this question. For Zhang San:
The probability of choosing incorrectly is $\frac{3}{4}$
The probability of choosing correctly is $\frac{1}{4}$
The mathematical expectation is: $5 \times \frac{1}{4} + 0 \times \frac{3}{4}= \frac{5}{4} = 1.25$

Li Si has a $\frac{9}{10}$ probability of answering this question correctly. For Li Si:
The probability of choosing incorrectly is $\frac{1}{10}$
The probability of choosing correctly is $\frac{9}{10}$
The mathematical expectation is: $5 \times \frac{9}{10} + 0 \times \frac{1}{10}= \frac{45}{10} = 4.5$

Single-choice questions seem meaningless, next, let's look at multiple-choice questions#

Let's assume a multiple-choice question is also worth 5 points, with four options: A, B, C, and D. The correct answers are two or three of them; getting all correct earns 5 points, missing answers earns 2 points, and not answering or answering incorrectly earns no points.
The possible answers are AB, AC, AD, BC, BD, CD, ABC, ABD, ACD, BCD, totaling 10 types.
The reasonable ways to answer the multiple-choice question are A, B, C, D, AB, AC, AD, BC, BD, CD, ABC, ABD, ACD, BCD, totaling 14 types.
Here we assume there is a multiple-choice question:

Zhang San cannot answer this question at all and randomly chooses A. For Zhang San:
The probability of scoring 0 points is: $\frac{2}{5}$
The probability of scoring 2 points is: $\frac{3}{5}$
The probability of scoring 5 points is: 0
The mathematical expectation is: $0 \times \frac{2}{5} + 2 \times \frac{3}{5} + 5 \times 0= \frac{6}{5} = 1.2$

Li Si also cannot answer this question at all, but he is very brave and randomly chooses AB. For Li Si:
The probability of scoring 0 points is: $\frac{7}{10}$
The probability of scoring 2 points is: $\frac{1}{5}$
The probability of scoring 5 points is: $\frac{1}{10}$
The mathematical expectation is: $0 \times \frac{7}{10} + 2 \times \frac{1}{5} + 5 \times \frac{1}{10}= \frac{9}{10} = 0.9$

Wang Wu also cannot answer this question at all, but he is even braver and randomly chooses ABC. For Wang Wu:
The probability of scoring 0 points is: $\frac{9}{10}$
The probability of scoring 2 points is: 0
The probability of scoring 5 points is: $\frac{1}{10}$
The mathematical expectation is: $0 \times \frac{9}{10} + 2 \times 0 + 5 \times \frac{1}{10}= \frac{5}{10} = 0.5$

From the above, we can see that for a question that one cannot answer at all, the fewer options chosen, the higher the mathematical expectation.

However, in reality, there are often certain options that are definite. For example, at this time, Xiao Ming, Xiao Jin, and Xiao Tao are sure that option A is correct.
At this time, the possible answers are AB, AC, AD, ABC, ABD, ACD, totaling 6 types.

Xiao Ming decides to only choose A. For Xiao Ming:
The probability of scoring 0 points is: 0
The probability of scoring 2 points is: 1
The probability of scoring 5 points is: 0
The mathematical expectation is: $0 \times 0 + 2 \times 1 + 5 \times 0= 2$

Xiao Jin decides to choose AB. For Xiao Jin:
The probability of scoring 0 points is: $\frac{1}{2}$
The probability of scoring 2 points is: $\frac{1}{3}$
The probability of scoring 5 points is: $\frac{1}{6}$
The mathematical expectation is: $0 \times \frac{1}{2} + 2 \times \frac{1}{3} + 5 \times \frac{1}{6}= \frac{3}{2} = 1.5$

Xiao Tao decides to choose ABC. For Xiao Tao:
The probability of scoring 0 points is: $\frac{5}{6}$
The probability of scoring 2 points is: 0
The probability of scoring 5 points is: $\frac{1}{6}$
The mathematical expectation is: $0 \times \frac{5}{6} + 2 \times 0 + 5 \times \frac{1}{6}= \frac{5}{6} \approx 0.83$

For example, at this time, Xiao Ming and Xiao Jin are sure that options AB are correct.
At this time, the possible answers are AB, ABC, ABD, totaling 3 types.

Xiao Ming decides to choose AB. For Xiao Ming:
The probability of scoring 0 points is: 0
The probability of scoring 2 points is: $\frac{2}{3}$
The probability of scoring 5 points is: $\frac{1}{3}$
The mathematical expectation is: $0 \times 0 + 2 \times \frac{2}{3} + 5 \times \frac{1}{3}= 3$

Xiao Jin decides to choose ABC. For Xiao Jin:
The probability of scoring 0 points is: $\frac{2}{3}$
The probability of scoring 2 points is: 0
The probability of scoring 5 points is: $\frac{1}{3}$
The mathematical expectation is: $0 \times \frac{2}{3} + 2 \times 0 + 5 \times \frac{1}{3}= \frac{5}{3} \approx 1.67$

At this point, we seem to have drawn a conclusion: in cases of incomplete certainty, the fewer options chosen, the higher the mathematical expectation. But is it really so?

For example, at this time, Xiao Ming, Xiao Jin, and Xiao Tao are sure that option A is incorrect.
At this time, the possible answers are BC, BD, CD, BCD, totaling 4 types.

Xiao Ming decides to only choose B. For Xiao Ming:
The probability of scoring 0 points is: $\frac{1}{4}$
The probability of scoring 2 points is: $\frac{3}{4}$
The probability of scoring 5 points is: 0
The mathematical expectation is: $0 \times \frac{1}{4} + 2 \times \frac{3}{4} + 5 \times 0= \frac{3}{2} = 1.5$

Xiao Jin decides to choose BC. For Xiao Jin:
The probability of scoring 0 points is: $\frac{2}{4}$
The probability of scoring 2 points is: $\frac{1}{4}$
The probability of scoring 5 points is: $\frac{1}{4}$
The mathematical expectation is: $0 \times \frac{2}{4} + 2 \times \frac{1}{4} + 5 \times \frac{1}{4}= \frac{7}{4} = 1.75$

Xiao Tao decides to choose BCD. For Xiao Tao:
The probability of scoring 0 points is: $\frac{3}{4}$
The probability of scoring 2 points is: 0
The probability of scoring 5 points is: $\frac{1}{4}$
The mathematical expectation is: $0 \times \frac{3}{4} + 2 \times 0 + 5 \times \frac{1}{4}= \frac{5}{4} = 1.25$

For example, at this time, Xiao Ming, Xiao Jin, and Xiao Tao are sure that option A is incorrect and option B is correct.
At this time, the possible answers are BC, BD, BCD, totaling 3 types.
Xiao Ming decides to only choose B. For Xiao Ming:
The probability of scoring 0 points is: 0
The probability of scoring 2 points is: 1
The probability of scoring 5 points is: 0
The mathematical expectation is: $0 \times 0 + 2 \times 1 + 5 \times 0= 2$

Xiao Jin decides to choose BC. For Xiao Jin:
The probability of scoring 0 points is: $\frac{1}{3}$
The probability of scoring 2 points is: $\frac{1}{3}$
The probability of scoring 5 points is: $\frac{1}{3}$
The mathematical expectation is: $0 \times \frac{1}{3} + 2 \times \frac{1}{3} + 5 \times \frac{1}{3}= \frac{7}{3} \approx 2.33$

Xiao Tao decides to choose BCD. For Xiao Tao:
The probability of scoring 0 points is: $\frac{2}{3}$
The probability of scoring 2 points is: 0
The probability of scoring 5 points is: $\frac{1}{3}$
The mathematical expectation is: $0 \times \frac{2}{3} + 2 \times 0 + 5 \times \frac{1}{3}= \frac{5}{3} \approx 1.67$

For example, at this time, Xiao Ming and Xiao Jin are sure that option A is incorrect and options BC are correct.
At this time, the possible answers are BC, BCD, totaling 2 types.
Xiao Ming decides to only choose BC. For Xiao Ming:
The probability of scoring 0 points is: 0
The probability of scoring 2 points is: $\frac{1}{2}$
The probability of scoring 5 points is: $\frac{1}{2}$
The mathematical expectation is: $0 \times 0 + 2 \times \frac{1}{2} + 5 \times \frac{1}{2}= \frac{7}{2}= 3.5$

Xiao Jin decides to choose BCD. For Xiao Jin:
The probability of scoring 0 points is: $\frac{1}{2}$
The probability of scoring 2 points is: 0
The probability of scoring 5 points is: $\frac{1}{2}$
The mathematical expectation is: $0 \times \frac{1}{2} + 2 \times 0 + 5 \times \frac{1}{2}= \frac{5}{2} = 2.5$

From the above, we can see that when one option is confirmed to be incorrect, the mathematical expectation is maximized by choosing two options.
But when only one option is confirmed to be correct, would you dare to take the risk of scoring 0 points by choosing two?

This article is synchronized and updated to xLog by Mix Space
The original link is https://blog.nepuko.cn/posts/default/Multiple-choice-probability-analysis

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