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This question already has an answer here:

When you roll with disadvantage is the chance to get a 1 5+5 for 10% or 5X5 for 25%? Just sat down and rolled 100 times with disadvantage got 18 natural 1s and 0 natural 20s. Still don't know.

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marked as duplicate by Oblivious Sage, okeefe, Thomas Jacobs, nitsua60 dnd-5e Mar 31 '17 at 2:26

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • \$\begingroup\$ "rolled 100 times with disadvantage got 18 natural 1s and 0 natural 20s" Do you mean you got no 20's on either die at all, or that no result of rolling with disadvantage came up double 20's? \$\endgroup\$ – LegendaryDude Mar 30 '17 at 13:44
  • \$\begingroup\$ @LegendaryDude The second one sounds more probable, since the chance for a nat 20 with disadvantage is 1 in 400 (which this simulation hasn't gotten close to). Not rolling a single 20 in 200 dice rolls is 1 in 28,000. Though, that is almost twice the expected number of nat 1's. Quite unlucky. \$\endgroup\$ – user27327 Mar 30 '17 at 15:49
  • \$\begingroup\$ @markovchain That's exactly what I was thinking, and I was wondering if that difference is what lead to the question being asked. \$\endgroup\$ – LegendaryDude Mar 30 '17 at 16:55
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The odds of a natural 1 are 9.75% and the odds of a natural 20 are 0.25%.

When you roll 2d20 you have 400 possible combinations of numbers. For a natural 20 there is exactly 1 possible way to get a 20 when rolling with disadvantage (20 on both dice) so the probability of getting that value is 1/400 (or .0025 or .25%)

For a natural 1 there are 3 different possibilities that can result in a 1 when rolling with disadvantage, 1-1, 1-X, and X-1 (where X is not 1). There is 1 way to get 1-1, 19 ways to get 1-X and 19 ways to get X-1. Adding all of those up we get 39 different ways to get 1, so the probability is 39/400 (or .0975 or 9.75%).

As a note your 5% + 5% comes close, but you are double counting the 1-1 combination.

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There is a \$9.75\%\$ chance that you will roll a 1 with disadvantage.

Math

The chance to roll a 1 on a fair d20 is \$1\div20\$, or \$5\%\$. From this we can use statistics to determine what the real odds are.

We care about a couple of scenarios here:

  1. Odds we roll a 1 on the first die and the second die.

  2. Odds we roll a 1 on the first die and not the second die.

  3. Odds we roll a 1 on the second die and not the first die.

The above scenarios are all the ways we can end up with a 1 when rolling with disadvantage. The fourth scenario will be all the other results.

  1. Odds we roll no 1s.

For the first scenario we have a \$5\%\$ chance, twice. That means we can multiple the percentages:

\$5\% \times 5\% = .05 \times .05 = .0025 = .25\% \$

Fun note, this is the same as the odds of rolling a natural 20 with disadvantage!

Now, let's check the next two. For scenarios 2 and 3, the math will actually be the same. We'll take the odds of rolling a 1, and multiply them by the odds of rolling not-1. The odds of not doing something are simply 100% minus the odds of doing that thing. That means the odds of rolling a 1 are still 5%, and the odds of rolling not-1 are \$100\% - 5\% = 95\%\$.

Scenario 2:

\$5\% \times 95\% = .05 \times .95 = .0475 = 4.75\%\$

Scenario 3:

\$95\% \times 5\% = .05 \times .95 = .0475 = 4.75\%\$

Finally, let's check the odds that we don't actually roll a 1 at all! This will be the odds of rolling not-1 twice in a row!

Scenario 4:

\$95\% \times 95\% = .95 \times .95 = .9025 = 90.25\%\$

Now with a little bit of math we can get to our final answer!

To get the odds than any of the first three scenarios will occur, we can add them together:

\$.0025 + .0475 + .0475 = .0975 = 9.75\%\$

This means that, according to our math, there is a 9.75% chance that you will roll a 1 with disadvantage. We can check that number, though, to be sure our math is correct. That's where our fourth scenario comes in! In statistics, the sum of all possible scenarios must add up to 100%, otherwise you're missing a scenario. To be sure we didn't leave anything out, let's add up our numbers:

\$.0975 + .9025 = 1 = 100\%\$

So you can see that, unless we made some larger errors in our theory, we did in fact cover all of the possible scenarios. You might also notice that there's a short-cut to getting out answer here. If we simple take 100% and subtract the odds that we roll no 1s, we can get the final answer much quicker!

\$1 - .9025 = .0975\$

These kinds of statistical shortcuts are pretty cool, and can save a lot of time!

On Physical Dice

As you note in your question, your dice don't seem to match with reality here (you fumbled twice as often as we would expect!) there are a few explanation for this.

  1. Not all dice are fair. Many dice are produced on the cheap in factories with minimal quality control. Most of the time these dice aren't too far off of fair, but sometimes they can be heavily weighted, affecting outcomes.
  2. 100 die rolls simply isn't a very large sample size. It's a good start, but if you look in the Math section of my answer, you'll notice that the odds of rolling a natural 20 with disadvantage is only .25%, that's a 1 in 400 chance that means that it would take 400 rolls to expect to see a single natural 20 with disadvantage! Even at 400 rolls you're not guaranteed to see one! Most statistical simulations are done with thousands or millions of samples for this reason. The more samples you have, the better you can trust your measurements!
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  • \$\begingroup\$ Thanks! As for why I fumbled twice as often, that's me. I was trying to prove to my DM by rolling in front of him how disadvatage was huge in terms of probabilities so trying to roll a 1. I did roll a 20 on one of the dice several times. One of my dice was one I call the dice of chaos which seems weighted to come up with either 1 or 20 about twice as often as normal. Which is fine by me, either way something interesting happens. \$\endgroup\$ – Billy the baby goat. Mar 31 '17 at 6:39
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Neither, it's 9.75%. The odds of rolling at least one 1 are the odds of not rolling a 1 twice, or 1 - (.95^2).

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  • \$\begingroup\$ For completeness sake, the chance of rolling a nat 20 is 1 in 400. \$\endgroup\$ – BlueMoon93 Mar 30 '17 at 13:38
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    \$\begingroup\$ Your answer is correct but would be greatly improved if you explained how you came to your conclusion. "The odds of rolling at least one 1 are the odds of not rolling a 1 twice" is not intuitive to someone who isn't familiar with how probabilities work. \$\endgroup\$ – LegendaryDude Mar 30 '17 at 13:41
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It’s \$9.75\%\$. To calculate this, you need to do some double-negation. First, you calculate the odds of not rolling a 1. The probability of rolling a 1 is \$5\%\$, so the probability of not rolling a 1 is \$\left( 1 - 5\% \right) = 95\%\$. With disadvantage, though, you roll twice. The odds of doing something twice in a row is the individual probability squared, so in this case \$\left(95\%\right)^2 = 90.25\%\$ is the probability of not rolling a 1 on both dice. Then to get the odds of rolling a 1 on at least one of them, we negate this again: \$1 - 90.25\% = 9.75\%\$.

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Neither, it's 9.75%

The odds of rolling a 1 are, as you correctly identified, 5% or 0.05.

The odds of an event (rolled "1") occuring at least once are complementary (100% - ... or 1-...) to the odds of the event not occuring at all (which means that for a given number of independent attempts X, each attempt will result in "not 1").

P("1" >= 1) = 1 - P("1" = 0) = 1 - P("not 1" = X)

The probability of an event occuring X times in a row is the individual probability to the power of the number of consecutive occurences:

P("not 1" = X) = p("not 1")*p("not 1")...p("not 1") = p("not 1")^X

The odds for rolling anything but a 1 are 95% or 0.95. Which brings us to the final result for X = 2 attempts:

P("1" >= 1) = 1 - P("1" = 0) = 1 - P("not 1" = 2) =
            = 1 - p("not 1")^2 = 1 - 0.95^2 = 0.0975 = 9.75%

See also this question in math.SE.

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