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My question is more for my own knowledge, and is based off of a board game I'm playing.

I'm rolling for an event and I'm rolling 2d10 (two 10 sided dice). The chances of succeeding on a roll if I were only using one die is 70% (a 4 or higher). What is the percentage of success if I'm rolling 2 dice and I have to take the lower result? And how (in simple terms) did you come to that conclusion?

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  • $\begingroup$ If a 4 or higher is a success then you have a 70% chance of succeeding with one roll, not 60% (you have 10 possible outcomes, only a 1,2 or 3 fails, i.e. 30%). Thus, for two successful rolls the probability would be $0.7^2=0.49$ or 49%. $\endgroup$ – ekkilop Nov 12 '15 at 17:25
  • $\begingroup$ @ekkilop My guess is Nicole is using standard gaming d10 dice, which have numbers from 0 to 9. $\endgroup$ – Empiromancer Nov 12 '15 at 17:27
  • $\begingroup$ @user164385 Sweet, I've never seen one of those =) I stand corrected. $\endgroup$ – ekkilop Nov 12 '15 at 17:31
  • $\begingroup$ Haha, nope. I was totally wrong. It is a 70% chance. For some reason I had in my head that a 4 was a fail as well, even though I wrote it correctly in the question. 4+ was success. Thanks ekkilop. $\endgroup$ – Nicole Nov 12 '15 at 17:34
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First of all, if you're interested in dice probabilities I'd recommend playing around on anydice.com - it's a great tool for calculating these sorts of probabilities. If you run output [lowest 1 of 2d10], you can play around with various probabilities associated with taking the lowest of 2d10.

For a rigorous mathematical answer, though: If you're rolling twice and taking the lower result, you want the probability of both dice being a success. Hence you want to multiply the probability of the first die being successful (70%) with the probability that the second die is successful (also 70%, since the dice are identical and each roll is independent of the others).

Thus, we have $(0.7)^2 = 0.49$, so you have a 49% chance of success.

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  • $\begingroup$ Sorry, I wrote my question wrong, it should be 70% chance success. That would then be (.7)squared = 49%? $\endgroup$ – Nicole Nov 12 '15 at 17:43
  • $\begingroup$ @Nicole Yes indeed. I've edited my answer to reflect this. :) $\endgroup$ – Empiromancer Nov 12 '15 at 17:44
  • $\begingroup$ Thanks a lot! The link and the statement to find the answer are really helpful. :) $\endgroup$ – Nicole Nov 12 '15 at 17:47

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