How to calculate odds Tim hortons has a roll-up the rim to win contents.  your chances of winning in a single play is 1 in 6.  so you have a 5/6 (83% chance) of winning nothing.
How do I calculate the odds of losing multiple times in a row?  In 2 plays, what are my chances of losing both times?  and what are my chances of losing 41 times in a row?  because that's what I'm at now?
 A: The answer Ross Millikan gave applies if you have played $41$ times and lost every time. However, the chance that you will have a streak of at least $41$ losses at some point is $1$ if you keep playing. Did you play a total of $41$ times, or did you have some successes before the streak of $41$ losses?
The average wait between streaks of length at least $L$ is $p^{-L}/(1-p)$, where $p$ is the chance of "success" or $5/6$ here. For $L=41$, you have to wait about $10,582$ trials between streaks of length $41$.
The probability that there is a streak of length at least $L$ in a session of length $S$ where $S$ is much bigger than $L$ is about $1-\exp(-(S-L)/\text{average wait})$. By exact calculations with transfer matrices, the probability that you have a streak of length at least $41$ in $100$ trials is $0.614\%$. In $1000$ trials it is $8.748\%$. These are close to the estimates of $0.556\%$ and $8.664\%$.
A: Assuming the chances are independent, the probabilities are multiplied.  So you would say the chance of losing $n$ in a row is $\left(\frac{5}{6}\right)^n$  For $n=2$, this is $\frac{25}{36}$ or just over $\frac{2}{3}$.  As Zach says, for $n=41$ this is $0.05\%$, so maybe you want to revise your estimate of the chance of winning.
