Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have this small homework I don't quite understand.

The problem is formulated as a game. (Who wants to be millionaire)

You start with 1£. money = 1.0£

You can choose to quit at anytime So you can quit immediately without answering and still have 1£.

You will be given a question with a probability p of answering correctly.

If you answer wisely you double up and get 2£. If you answer poorly you lose everything and get 0£.

The probability p is not exactly known but is a uniform distributed variable between 0.3 and 1.0

What is the expected value of the money you'll earn? E(money). The correct answer is 1.357 if there is only one question.

My best guess is p = (1.0 + 0.3)/2

E(money) = 1.0 + 2.0*p - 1.0*(1-p) = 1.95 which is the wrong answer...

My homework is about finding an algorithm for more than 1 question, however i fail to understand for just 1.

Thanks for any help.

share|cite|improve this question
"What is the expected value of the money you'll earn?" Assuming you follow which strategy? – Did Apr 27 '14 at 9:22
No strategy needed. The probability that you answer correctly is somewhere between 0.3 and 1.0 – ColacX Apr 28 '14 at 6:41
Sure but at any time you have to decide whether to quit or to continue. – Did Apr 28 '14 at 7:02
I suppose the strategy then is to quit right after when you've won the expected value of money. But the strategy is not really the main concern. Oh when i mean the correct answer is 1.357 i meant if there is only 1 single question. – ColacX Apr 28 '14 at 8:19
Then the expected value of the money one will earn is 1.3, not 1.357. Sorry but your question is unclear. – Did Apr 28 '14 at 8:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.