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?
Answer
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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