You and I each get one six-sided die and each roll it once. If you roll a 1, you keep rolling until you get something higher. If we tie, I win. What is the probability of you winning?
The probability of you winning should be calculated as:
- when you get a 2 I should get 1,
- when you get a 3 I should get 1 or 2
- when you get a I should get 1 or 2 or 3 and so on
so , (1/6*1/6+1/6*2/6+1/6*3/6+1/6*4/5+1/6*5/6) = 15/ 36.
Please tell me what am I doing wrong?
You're using the wrong probabilities. Since I can't roll a $1$, the probability for each of my numbers is $\frac15$, not $\frac16$, so you need to multiply your result by $\frac65$, yielding
$$ \frac{15}{36}\cdot\frac65=\frac12\;, $$
so, interestingly, the advantage of winning ties and the advantage of rerolling $1$s cancel. This is true for all $n$-sided dice: There are $n-1$ results where I win because you roll the lowest number that I can't get, and of the remaining $(n-1)(n-1)$ results, you win $n-1$ ties; and by symmetry we both win in half of the remaining results.