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Say I'm playing a game and I get to roll $5$ dice once. If I roll $3$ or more sixes I lose. If I roll at least $1$ one I lose. I can lose if both losing conditions happen at the same time too. If I don't lose, I win. What is the probability that I win?

I attempted to figure this out by calculating the possible individual losing conditions and summing them together. But I think I had some overlap since I got $\sim43\%$ but when simulated with $20,000$ rolls in excel the win rate was $\sim38\%$.

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    $\begingroup$ Do you win if you don't lose? $\endgroup$ – Bram28 Dec 6 '17 at 21:32
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    $\begingroup$ The problem statement is a little unclear. Do you lose if you roll four or five $6$'s, or does it have to be exactly three? Similarly, by "a single one" do you mean exactly one $1$ or at least one $1$? $\endgroup$ – kccu Dec 6 '17 at 21:33
  • $\begingroup$ I edited it to reflect your questions. $\endgroup$ – James Dec 6 '17 at 21:36
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HINT

You're right: if you just sum the probabilities of the individual events then you're not taking into account that they might both happen at the same time, and indeed that can easily happen by e.g. throwing one $1$, three $6$'s and a $4$.

But, it should be easier to calculate the probability of both events not to happen. That is, what is the probability of not getting any $1$'s at all, and getting at most two $6$'s? And that can be further broken down to:

What is the probability of not getting any $1$'s at all, and getting no $6$'s?

What is the probability of not getting any $1$'s at all, and getting one $6$?

What is the probability of not getting any $1$'s at all, and getting two $6$'s?

Add up those three probabilities and you'll have the probability that you win.

(and yes, my calculations show that's just a little under $38$%, confirming your Excel findings)

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  • $\begingroup$ Heh, yeah I don't know why I didn't think to go about it this way, thanks! $\endgroup$ – James Dec 6 '17 at 21:46
  • $\begingroup$ @James You're welcome! :) $\endgroup$ – Bram28 Dec 6 '17 at 21:47

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