Question 1138212: Suppose it costs $15 to roll a pair of dice. You get paid 2 dollars times the sum of the numbers that appear on the dice. If the cost of the game is not taken into consideration, what is the expected value of the game? Is it a fair game?
Answer by ikleyn(52846) (Show Source):
You can put this solution on YOUR website! .
The space of events is the (6 x 6)-matrix of pairs (i,j), where i and j are integer numbers from 1 to 6.
All events have the same probability of .
The expected value of the game, as it is formulated in the problem, is the doubled sum of all pairs of the numbers (i,j) in the matrix.
To calculate the sum, notice that each number 1, 2, 3, 4, 5, 6 goes to the sum 6 times as " i " and 6 times as " j ",
therefore, the sum of all components (i+j) from 36 pairs (i,j) is
SUM of (i+j) over all pairs (i,j) = 6*(1+2+3+4+5+6)*2 = 12*21 = 252.
Thus the expected value of the game is = 14 dollars.
You pay $15 to roll a pair of dice.
Having all these data, you can decide on your own if this game is fair.
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I got your comment, thanks for it !
The way, which "they" propose to solve the problem is correct but "stupid" : it assumes to do a lot of excessive calculations.
My way is much more straightforward and ELIMINATES these excessive (and not needed repeating calculations).
Actually, the entire sense and meaning of solving this problem is to find (and to understand) this simple, "economic" and straightforward way.
If you want to learn BETTER this class of problems, look into and read my lesson
- Rolling a pair of fair dice
in this site.
Also, you have this free of charge online textbook in ALGEBRA-II in this site
- ALGEBRA-II - YOUR ONLINE TEXTBOOK.
The referred lesson is the part of this online textbook under the topic "Solved problems on Probability".
Save the link to this textbook together with its description
Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson
into your archive and use when it is needed.
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