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juantamad
07-01-2010, 03:48 AM
Thanks everyone! Now I understand it, the keyword is independent! :)


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!

Link
07-01-2010, 10:45 AM
Your answer seems a little off. Think of the situation. You bet on a number, X. So the 3 dice get rolled and you want to know your probability of hitting on at least 1 of the 3 dice. If it were just 1 die, that would be easy: 1/6.

However, you have three of them. In this situation, it'd be easiest to work backwards. The probability of not hitting your number on die #1 is 5/6. The probability of not hitting on die #2 is also 5/6. So is die #3. Now since the three dices are independent of each other, you can calculate the total probability of not hitting any of the three dice by just multiplying. So, (5/6)^3 = 125/216, which is the probability of not hitting any of the three dice. That means the probability of hitting at least 1 die is 1 - 125/216.

See if you can figure it out from there.

BGM
07-01-2010, 01:28 PM
The number of times that the number you bet appear in the 3 dice
\sim Binomial(3, \frac {1} {6})

And then working the expectation with your simple payoff function should be easy.