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i have a problem and i can't figure out any solution.

Suppose i have this game: i throw a die untill i get a 6. Every time i throw the dice i pay -1 and when i get the 6 i win 5. (Nb: when i obtain the first 6 the game end).

I call $X$ a radom variable that is the number of attempts to get a 6. $X$ has a geometric distribution with parameter $\theta = \frac{1}{6}$.

Now define a variable that tells me how much i win/loss: $Y = 5 - X$; this is my transformation.

The problem is that i don't know how to compute the Cumulative distribution function for the $Y$.

I tried: $F_Y(y) = P(Y \le y) = P(5 - X \le y) = P(X \ge y - 5) = 1 - P(X < y-5) = 1 - F_X(y -5 -1)$ sustituting: $F_Y(y) = 1 - (1 - (\frac{5}{6})^{y - 6}) = (\frac{5}{6})^{y - 6}$

That is wrong! Can you help me finding the mistake?

Thank you, bye.

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    Yes, maybe in this specific case it is easier...but i'm searching for a general procedure to solve this kind of problems..2012-03-21

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$ \mathbb{P}(Y \leqslant y) = \mathbb{P}(5-X \leqslant y) \stackrel{\color\red{\text{!!!}}}{=} \mathbb{P}(X \geqslant 5-y) = 1-\mathbb{P}(X < 5-y) = 1 - F_X(4-y) $ Compare this with your answer $1-F_X(y-6)$.

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    @Aslan986 Than$k$s, I have corrected the typo2012-03-22