SOLUTION: Two marbles are drawn in succession from a bag containing 3 red marbles and 5 blue marbles. The first marble is returned to the bag before the second is drawn. You win $5 if both

Question 1025098: Two marbles are drawn in succession from a bag containing 3 red marbles and 5 blue marbles. The first marble is returned to the bag before the second is drawn. You win $5 if both of the marbles are red, and $3 is both of the marbles are blue. You lose $1.50 if the two marbles are different colors. What price would make this a fair game?
Found 2 solutions by Fombitz, mathmate:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!

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Answer by mathmate(429) (Show Source): You can put this solution on YOUR website!

Question:
Two marbles are drawn in succession from a bag containing 3 red marbles and 5 blue marbles. The first marble is returned to the bag before the second is drawn. You win $5 if both of the marbles are red, and $3 is both of the marbles are blue. You lose $1.50 if the two marbles are different colors. What price would make this a fair game?

Solution:
A fair game is a game in which the expected earning equals the price to play the game.
So here we are looking for the expected receipt, E[X].
By definition,
E[X]=Σp(x)G(x)
where G(x) is the gain for outcome x, and p(x) is the corresponding probability.
The probabilities p(x) and payout G(x) of the outcomes x are:
P(RR)=(3/8)(3/8)= 9/64 $5
P(RB)=(3/8)(5/8)=15/64 -$1.5
P(BR)=(5/8)(3/8)=15/64 -$1.5
P(BB)=(5/8)(5/8)=25/64 $3
Total probabilities = (9+15+15+25)/64=1 (checks!)
Here we have

x......p(x).....G(x)....p(x)G(x)
RR 9/64 $5 45/64
RB 15/64 -$1.5 -22.5/64
BR 15/64 -$1.5 -22.5/64
BB 25/64 $3 75/64
Σp(x)G(x)=(45-22.5-22.5+75)/64=75/64

Since there is an expected net gain E[x]=75/64=$1.171875, this is the price to pay for a fair game.
(Note: Usually in a game, it is rare to have to pay to play, PLUS you could lose additional amounts for unfavourable outcomes).

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