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?

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 $8.664\%$.