0
$\begingroup$

A game involves you choosing one number (between 1 to 6 inclusive) and then throwing three fair dice simultaneously. If none of the dice shows up the number that you have chosen, you lose Rs. 1. If exactly one, two or three dice show up the number that you have chosen, you win Rs. 1, Rs. 3 or Rs. 5 respectively. What is your expected gain?

Please suggest some other questions of this kind, if you have any.

$\endgroup$
  • 1
    $\begingroup$ By symmetry, it doesn't matter what number you choose. Suppose you choose $6$. What is the chance you get no $6$s? Each other number of $6$s? It is a binomial distribution. Where are you stuck? $\endgroup$ – Ross Millikan Aug 8 '18 at 5:03
  • 1
    $\begingroup$ The number of dice show up the chosen number, $X \sim \text{Binomial}(3, 1/6)$. The outcome is precisely $2X - 1$. $\endgroup$ – BGM Aug 8 '18 at 5:04
  • $\begingroup$ @RossMillikan could you please tell me the final answer and explain the procedure? $\endgroup$ – Pandey Aug 8 '18 at 6:01
  • $\begingroup$ This is a version of en.wikipedia.org/wiki/Chuck-a-luck - if you want to practise more, it might be an idea to show what you have tried here $\endgroup$ – Henry Aug 8 '18 at 6:32
0
$\begingroup$

Say you choose $6$. The chance you get exactly two sixes is ${3 \choose 2}$ ways to choose the dice that come up six times $(\frac 16)^2$ that they come up six times $\frac 56$ that the other comes up something else. Do the same calculation for each number of sixes. Multiply each probability by the return of that result and add.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.