MHB Expected Value of Rolling a Pair of Dice - Fair Price to Play

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In the dice game, if the sum is 7, the player pays $28, while if it is not 7, the player receives an amount equal to the sum. The expected value of the opponent's winnings is approximately $5.83, while the player's expected winnings are about $4.67. To make the game fair, the player should pay $35 to participate. This calculation ensures that both parties have equal expected outcomes in the game.
rymatson406
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We roll a pair of dice. If the sum of the dice is 7, you pay me $28. If the sum is not 7, I pay you the number of dollars indicated by the sum of the dice. What is the price that you should pay to play the game that would make the game fair?
 
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I deleted the duplicate of this thread in the Advanced Probability subforum. We ask that you post a question only once, as doing otherwise can lead to duplication of effort on the part of our helpers. I'm sure you can understand that the time and effort of our helpers is valuable, and we don't want to see it wasted. :D

edit: I did the same with your other question.
 
rymatson406 said:
We roll a pair of dice. If the sum of the dice is 7, you pay me $28. If the sum is not 7, I pay you the number of dollars indicated by the sum of the dice. What is the price that you should pay to play the game that would make the game fair?

Among the 36 possible results of the roll of two dice, those whose sum is k are k-1 for k ranging from 2 to 7 and 12 - (k-1) for k ranging from 8 to 12. The expected win your opponent is ...

$\displaystyle E\ \{ O \} = \frac{(1 + 2 + 3 + 4 + 5)\ 14}{36} = 5.833... \text{dollars}$

Your expected win is...

$\displaystyle E\ \{ Y \} = \frac{28}{6} = 4.666... \text{dollars}$

... and it is lower. For a fair game You should ask 35 dollars...

Kind regards

$\chi$ $\sigma$
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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