If I toss a six-sided dice and record its outcome. Then toss it repeatedly until I get an outcome which is greater than or equal to that on the first toss. What would be the expected number of tosses needed?
My attempt: If we don't include the first toss, then we can use the formula $E[X]=\sum\limits_{i=1}^niP(n=i)$ for $i=1,\dots,n$ and $n$ be the toss when we get an outcome greater than or equal to that on the first toss. But I'm not sure if the formula is correct, and how to go on.
If the first roll is a $1$ it takes one more roll to be equal or greater. If it is a $2$ it takes $\frac 65$ of a roll on average because the chance of a lower roll is $\frac 16$. If you toss a $3$ it takes $\frac 64$ and so on. Counting the first roll the expected number of rolls is $$1+\frac 16\sum_{i=1}^6\frac 6i=\frac {69}{20}=3.45$$