Question 1176904
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When rolling two dice, in all, 36 equally likely outcomes are possible.


Of them, there are 2 outcomes that produce the sum of 11 :  (5,6) and (6,5).


Also, there are 6 outcomes that produce the sum of 7: (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1).


So, you can win $20 with the probability  {{{2/36}}}  and

    you can win  $5 with the probability  {{{6/36}}}.


So, your winning expectation is  {{{20*(2/36) + 5*(6/36)}}} = {{{(40+30)/36}}} = {{{70/36}}} = 1.94  dollars.


From it, you should subtract the cost of the play of $3,  so the expectation of the game is  $1.94 - $3.00 = - $1.06.


In other words, you will lose 1.06 dollars statistically at each game, in average.
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Solved.


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See the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Rolling-a-pair-of-fair-dice.lesson>Rolling a pair of fair dice</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Probability-and-statistics/Math-expectation-of-winning-in-games-with-rolling-pair-of-dice.lesson>Math expectation of winning in games with rolling pair of dice</A> 

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