Help on Expected Value?

Question

I have this problem on expected value:

A carnival game consists of three dice in a cage. A player can bet a dollar on any of the numbers 1 through 6. The cage is shaken, and the payoff is as follows. If the player's number doesn't appear on any of the dice, he loses his dollar. Otherwise, if his number appears on exactly k of the three dice, for k = 1, 2, 3, he keeps his dollar and wins k more dollars. What is his expected gain from playing the carnival game once?

My Answer:

If I understand it right, since there are 3 dice, the probability that any of the numbers 1 to 6 appears after the throwing the dice is 36/216 which gets simplified to 1/6. so the probability of winning back your dollar is 1/6 and losing it would be 5/6. so intially, i have this expected value >> 1/6- 5/6 = -4/6. Bu then it said there that if you win, you will have additional k dollars depending... so i tried to compute the different winning combinations... if i am correct, then there 56 different possible outcomes.. of those 56, 20 of them composed of numbers without replacement i.e. (a number only appear once in each combination), 30 of them contains those combination which some numbers appear twice i.e {112, 334, 551, ...} and 6 combinations of the numbers appearing three times i.e. {111, 222, ...} so the probability that you get an additional dollar to your winning is 20/56, 2 dollars is 30/56 and 3 dollars is 6/56.

Is it correct to say that the expected gain from playing the game once is equal to E[X] = 1(1/6) - 1 (5/6) + 1(20/56) + 2(30/56) + 3 (6/56)?

Or is it equal to,

E[X] = (1(20/56) + 2(30/56) + 3 (6/56) + 1)(1/6) - 5/6?


Please, I just want to be clarified... Thanks a lot!

Answer 1

I think there are some errors in the probabilities calculated above. Please do look at the following method, since I think it's correct.



Let k denote the number of matches, where k = 0, 1, 2, 3.



Then k = 0 when all 3 dice "miss", each with a probability of 5/6. So



P(k = 0) = 5³/6³.



k = 3, when all dice "hit", each with a probability of 1/6. So



P(k = 3) = 1³/6³ = 1/6³.



k = 2, when 2 dice "hit", each with a probability of 1/6, and one die misses, with a probability of 5/6. Since there are 3 candidates for the missing die,



P(k = 2) = 3·1²·5/6³ = 3·5/6³.



Similarly,



P(k = 1) = 3·1·5²/6³ = 3·5²/6³.



The gain associated with each k is given by g(0) = -1, g(k) = k for k = 1, 2, 3.



The expected gain is then



∑_(i = 0 → 3) P(k = i) g(i)

= (5³/6³)(-1) + (3·5²/6³)(1) + (3·5/6³)(2) + (1/6³)(3)

= (-125 + 75 + 30 + 3) / 216

= - 17 / 216

≈ - 0.0787,



i.e. the player stands to lose about 8 cents per dollar per turn.

Answer 2

David's answer is correct.



It is useful to add that the question "What is the probability of n successes in m trials, given independent trials with individual probability of success p?" is answered by the binomial distribution.



The probability mass function of the binomial distribution is m-choose-n × p^n × (1-p)^(m-n).



To calculate the expected payoff, we must multiply each outcome's probability times the payoff for that outcome, then add the products.



-1 × 3-choose-0 × (1/6)^0 × (5/6)^3 = -1 × 1 × 1 × 125/216 = -125/216

0 × 3-choose-1 × (1/6)^1 × (5/6)^2 = 0 × 3 × 1/6 × 25/36 = 0

1 × 3-choose-2 × (1/6)^2 × (5/6)^1 = 1 × 3 × 1/36 × 5/6 = 15/216

2 × 3-choose-3 × (1/6)^3 × (5/6)^0 = 2 × 1 × 1/216 × 1 = 2/216



Sum: -108/216 = -1/2

Answer 3

There are 216 possible outcomes



There are 3 ways that you can get one number correct

3(1/6)(5/6)(5/6)=75/216 ways you can break even

win $1 - pay $1= 0 return



There are 3 ways that you can get two numbers correct

3(1/6)(1/6)(5/6)=15/216 ways you can win 1 dollar

win $2 - pay $1 = $1



There is one way to get all 3 correct

1/216 ways to win 2 dollars

win $3 - pay $1 = $2



There is one way to get all three incorrect

(5/6)(5/6)(5/6)=125/216 ways to lose 1 dollar

win 0 - pay $1 = -$1 loss



((75/216)*0)+((15/216)*1)+((1/216)*2)+

((125/216)*-1)= -.50

you could expect to lose 50 cents for every game

Answer 4

Value of the game is (1/38 * $245 + 37/38 * $0) - $7 = -$0.55 I expect to lose $550 if I played 1000 times.