Rolling a 15 sided dice

Question

A game of chance involves rolling a 15-sided die once. If a number from 1 to 3 comes up, you win 2 dollars. If the number 4 or 5 comes up, you win 10 dollars. If any other number comes up, you lose. If it costs 4 dollars to play, what is your expected net winnings?

I am not looking for the answer, but someone to help me figure out work the formula with me, so I can actually learn this. I am online all night!

score 0 · Answer 1 · answered Sep 28 '14 at 00:11

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Hint: The expected net winning (in dollars) is $$ \sum_z (z-4) \Pr[\text{you win $z$ dollars}] = \sum_z z \Pr[\text{you win $z$ dollars}] - 4. $$ Here the summation goes over all possible values you could win, in your case $0,2,10$ ($z=0$ corresponds to the case in which you lose).

answered Sep 28 '14 at 00:11

Yuval Filmus

57,157

Wonderful, thanks for that help buddy! – Amanda Sep 28 '14 at 00:17
No, you'll have to solve it yourself. It's plugging numbers in a formula. – Yuval Filmus Sep 28 '14 at 00:18

score 0 · Accepted Answer · answered Sep 28 '14 at 00:13

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Imagine instead "rolling" 15 times, and you get each value once. How much money would you win doing this? How much would you spend? Divide your net winnings for this scenario by 15 to discover your expectation.

answered Sep 28 '14 at 00:13

Dan Uznanski

11,025

Rolling a 15 sided dice

2 Answers2