anonymous
  • anonymous
We roll a pair of dice. If the sum of the dice is 7, you pay me $25. If the sum is not 7, I pay you the number of dollars indicated by the sum of the dice. What is your expected value for the game?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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dan815
  • dan815
|dw:1377901582726:dw|
dan815
  • dan815
|dw:1377901701332:dw|
dan815
  • dan815
every 1/6th you will get 25 dollars

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More answers

dan815
  • dan815
you need to know whats the expected average value is when its not 7
dan815
  • dan815
every 5/6th of the time you will be paying the guy that much
anonymous
  • anonymous
Choices: 1.67, -1.67,8.33,-8.33
anonymous
  • anonymous
So 1.67?
dan815
  • dan815
|dw:1377902218400:dw|
anonymous
  • anonymous
1.67
dan815
  • dan815
@abb0t
abb0t
  • abb0t
?
dan815
  • dan815
how do you do this question
abb0t
  • abb0t
Expected value = 2(\(\frac{1}{36}\)) + 3(\(\frac{2}{36}\)) + 4(\(\frac{3}{36}\)) + 5(\(\frac{4}{36}\)) + 6(\(\frac{5}{36}\)) - 25(\(\frac{6}{36}\)) + 8(\(\frac{5}{36}\)) + 9(\(\frac{4}{36}\)) + 10(\(\frac{3}{36}\)) + 11(\(\frac{2}{36}\)) + 12(\(\frac{1}{36}\))
dan815
  • dan815
so 14(1+2+3+4+5) 25(6) ------------- + ---------- 36 36
dan815
  • dan815
1.67 is right
abb0t
  • abb0t
Yes.
dan815
  • dan815
@abb0t so where was i going wrong in my method, 7 is the average so 5/6th of the time you end up getting 7 and 1/6th of the time you lose 25 so for avg of 6 turns you gain 35 and lose 25 every 6 turns we get 10 dollars right
dan815
  • dan815
ohh i see so for 1 turn we get on avg 10/6 = 1.667

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