# 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?

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There is a 0.05% chance of losing 41 times in a row. :( – Zach Langley Mar 19 '11 at 1:35

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 $8.664\%$.

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A group of us (cube mates) at work, keep track of total losses since the last win. Since the last time anyone won, there have been 41 losses with no wins. I figured the odds of of 41 losses seemed very low considering Tim Hortins publishes that the chance of winning is 1 in 6. However, there is a finite number of prizes, so each time a prize is one, it is removed from the possible pool of prizes. – Jeremy Mar 19 '11 at 3:58
Good point about the specification of the problem. On the other hand, the losing streak is one-sided (still continuing) instead of any time in a long string. Many of these probability things depend very sensitively on the assumptions. – Ross Millikan Mar 19 '11 at 4:02

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.

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