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Here's the assignment: " In a dice-game for two players, two dice (A & B) are used. Each player gets to choose a die. The one who throws the highest number wins".

(Die A has six sides; two of them have two dots and four are five dots. Die B has six sides; four of them have two dots, two of them have six dots).

a) Which of the dice has the largest probability to win? Explain why.

b) Give suggestions on how you can change the worse die in order for it to have a larger probability than the other die.

c) Give a suggestion on how you can double the probability of winning on the dice. "

My thoughts:

a) Die A: It's hard because you have a 2/3 chance of getting 5, which is a large number, but also 1/3 of getting a one, that's bad. Die B: You have a 2/3 chance of getting two, which beats the sides with one dot on dice A, and 1/3 chance of getting six dot sides, which instantly makes you win.

I really don't know how to choose one die ?, Die B seems good though.

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I'm kind of confused about your question because you state that the die A has 2 sides with 2 dots but in your thoughts you tell us that the die A has 2 sides with one dot.

Anyway you can either calculate the probability of die A winning, which is only the case if A rolls a 5 and B rolls a two. Then the probability is (because each throw is independent). $$ p = \frac{2}{3} \cdot \frac{2}{3} = \frac{4}{9} $$, which means in the case of your description of your thoughts means that the probability of B winning is the complementary event (there is no draw) and calculates as $$ q = 1-p = \frac{5}{9} $$.

An alternative approach would be to compare the expected values of each die and compare them to each other.

To change the probability of the worse die to be better than the better one I would recommend you to experiment on changing one side of the worse die and recalculate the probability. This would give you the exercise to understand the calculations above better.

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