0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.