Answer to Question #313799 in Statistics and Probability for hane

Question

You play a game with two six-sided dice. If you roll a sum of 3 or 8, you win ₱600. If you roll a sum of 10, you win ₱400. However, you lose ₱500 for anything else. If you continue to play the game, how much do you expect to win or lose in the game?

Answer 1

Let X be a random variable representing the financial result of the game, then

pairs that satisfies X = 3 - (1,2); (2,1)

$P(X=3)={\frac 2 {36}}$

pairs that satisfies X = 8 - (2,6); (6,2); (3,5); (5,3); (4,4)

$P(X=8)={\frac 5 {36}}$

pairs that satisfies X = 10 - (4,6); (6,4); (5,5)

$P(X=10)={\frac 3 {36}}$

$P(lose)=1-P(X=3)-P(X=8)-P(X=10)={\frac {26} {36}}$

$E(X)={\frac {2+5} {36}}*600+{\frac 3 {36}}*400-{\frac {26} {36}}*500=-{\frac {7600} {36}}=-{\frac {1900} 9}\approx-211$

You expect to lose 211 per game, so for n games you can expect to lose -211*n