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 March 5th, 2014, 07:21 PM #1 Senior Member   Joined: Jan 2014 Posts: 196 Thanks: 3 Expected value and p.m.f. for pair of dice. Please correct any errors, and let me know if there are more efficient ways of approaching a problem like this. Thank you! In the casino game called high-low, there are three possible bets. Assume that $1 is the size of the bet. A pair of fair six sided dice is rolled and their sum is calculated. If you bet low, you win$1 if the sum of the dice is $\{2,3,4,5,6,\}$. If you bet high, you win $1 if the sum of the dice is $\{8,9,10,11,12}\$ If you bet on $\{7\}$, you win$4 if a sum of $7$ is rolled. Otherwise, you lose on each of the three bets. In all three cases, your original dollar is returned if you win. Find the expected value of the game to the bettor for each of these 3 bets. To start, I assume I need the p.m.f. for $p(x)$ for each sum of two dice. * 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 $**$Is there a more general, less labor intensive way of getting the probabilities? For $E[u(x)]=\sum u(x)f(x)$ If the bet is for 2-6, then $u(x)= \begin{cases} 1, \:\: \: \=2,3,4,5,6 \\ -1 , \: \: x=7,8,9,10,11,12\end{cases}" /> then $\sum u(x)f(x)= \frac{1}{36}(1+2+3+4+5) -1/6 -\frac{1}{36}(1+2+3+4+5)$ $=$\$$-1/6$ And the same technique for the other 2 bets? Thanks for any help!
 March 6th, 2014, 03:31 PM #2 Global Moderator   Joined: May 2007 Posts: 6,821 Thanks: 723 Re: Expected value and p.m.f. for pair of dice. Your approach is correct.

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# in the casino game called high-low there are three possible bets

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