What is f(x) for Loaded Die?

[Jason76]

What would be f(x) - given the below information?

A die has its six faces loaded so that P(roll is i)=K*x for x=1,2,3,4,5,6. It is rolled until an even number appears. Let X be the number of rolls needed.

$$K + 2K + 3K + 4K + 5K + 6K = 1$$

is the correct format so

$$21K = 1$$

so

$$K = \dfrac{1}{21}$$

$$f(x) = \dfrac{1}{21}x$$ ??

 

[I like Serena]

What would be f(x) - given the below information?

A die has its six faces loaded so that P(roll is i)=K*x for x=1,2,3,4,5,6. It is rolled until an even number appears. Let X be the number of rolls needed.

$$K + 2K + 3K + 4K + 5K + 6K = 1$$

is the correct format so

$$21K = 1$$

so

$$K = \dfrac{1}{21}$$

$$f(x) = \dfrac{1}{21}x$$ ??

Hi Jason76,

Let me take the liberty to rephrase your problem statement a bit, since I believe $i$ and $x$ are supposed to have different meanings.

A die has its six faces loaded so that P(roll is i)=K*i for i=1,2,3,4,5,6. It is rolled until an even number appears. Let X be the number of rolls needed.
Let $f(x)$ be the probability that $x$ rolls are needed until an even number appears.​

Does that look right to you?

It would mean that f(1) is the probability that the very first roll is even, after which we stop.

So:
f(1) = P(1st roll is even) = P(roll is 2 or 4 or 6) = P(roll is 2) + P(roll is 4) + P(roll is 6) = 2/21 + 4/21 + 6/21 = 12/21

and:
f(2) = P(1st roll is odd and 2nd roll is even) = P(1st roll is odd) P(2nd roll is even)

What would f(2) be?