Question 1166511
<pre>
The probability that the 1st roll is a 4 is 
{{{1/6}}}, the 1st term
The probability that the 1st roll is not a 4 and the 2nd one is a 4 is
{{{(5/6)*(1/6)}}}, the 2nd term
The probability that the 1st 2 rolls are not 4's and the 3rd one is a 4 is
{{{(5/6)(5/6)*(1/6)=(5/6)^2*(1/6)}}}, the 3rd term
The probability that the 1st 3 rolls are not 4's and the 4th one is a 4 is
{{{(5/6)(5/6)(5/6)*(1/6)=(5/6)^3*(1/6)}}}, the 4th term
...
The probability that the 1st x-1 rolls are not 4's and the xth one is a 4 is
{{{(5/6)^(x-1)*(1/6)}}}, the xth term

So

{{{f(x)=sum(((5/6)^(n-1)*(1/6)),n=1,x)}}}

Notice that theoretically (mathematically) it could take you a billion
rolls, or a million years, [lol] to roll your first 4. [if you had an
extremely stubborn die!!!]

The sum of all values of f(x) is an infinite series and the sum of the
infinite series is:

{{{s[infinity]=a[1]/(1-r^"")}}}

where a<sub>1</sub> = first term = 1/6, and r = 5/6

{{{s[infinity]=expr(1/6)/(1-expr(5/6)^"")=(1/6)/(1/6) = 1}}}

Edwin</pre>