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High School
Statistics and Probability
High school probability

hannahkatesumile

Answered question

2022-04-01

Answer & Explanation

alenahelenash

Expert2023-04-26Added 556 answers

Let

X

and

Y

be the outcomes of rolling two dice. Each die can show a number from 1 to 6. The probability mass function

p_{X}

X

is given by

p_{X} (k) = \frac{1}{6}, k = 1, 2, \dots, 6

since each outcome is equally likely. Similarly, the probability mass function

p_{Y}

Y

is also given by

p_{Y} (k) = \frac{1}{6}, k = 1, 2, \dots, 6

Let

Z = X + Y

be the sum of the two dice. Then, the probability mass function

p_{Z}

Z

is the convolution of

p_{X}

and

p_{Y}

, given by

p_{Z} (k) = \sum_{i + j = k} p_{X} (i) p_{Y} (j)

for

k = 2, 3, \dots, 12

. This means that

p_{Z} (2) = \frac{1}{36}, p_{Z} (3) = \frac{2}{36}, p_{Z} (4) = \frac{3}{36}, p_{Z} (5) = \frac{4}{36}, p_{Z} (6) = \frac{5}{36}, p_{Z} (7) = \frac{6}{36},

p_{Z} (8) = \frac{5}{36}, p_{Z} (9) = \frac{4}{36}, p_{Z} (10) = \frac{3}{36}, p_{Z} (11) = \frac{2}{36}, p_{Z} (12) = \frac{1}{36} .

Now, let

E

be the event that the sum of the two dice is even, and let

O

be the event that the sum of the two dice is odd. Then,

P (E) = \sum_{k = 2, 4, \dots, 12} p_{Z} (k) = p_{Z} (2) + p_{Z} (4) + \dots + p_{Z} (12) = \frac{1}{2}

since the sum of two dice is even if and only if both dice show the same parity (i.e., both even or both odd). Similarly,

P (O) = \sum_{k = 3, 5, \dots, 11} p_{Z} (k) = p_{Z} (3) + p_{Z} (5) + \dots + p_{Z} (11) = \frac{1}{2}

since the sum of two dice is odd if and only if the two dice show different parities (i.e., one is even and the other is odd).
Therefore, the chances of the player getting an odd sum is as good as the chances that the students will get an even sum. In other words, this is a fair game. However, the expected value of the player's payoff is

E (payoff) = 10 \times P (O) - 10 \times P (E) = 10 \times \frac{1}{2} - 10 \times \frac{1}{2} = 0

which means that the player can expect to neither gain nor lose money in the long run. Therefore, whether the player should play this game depends on their personal preference for risk and reward.

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