# Can I randomly determine the outcome of an event with 5 outcomes with a d6?

Could I use a d6 to find which event would happen given there were 5 events, lets say named A, B, C, D, and E? I am thinking that for 1–5, you could have the events assigned to their respective numbers, so:

1 → A
2 → B
3 → C
4 → D
5 → E
6 → Reroll the d6

Would this work? I don't have much experience in tabletop games, but it makes sense that each event A–E would have an equal chance assuming the d6 was fair.

You have it -- roll the D6, if you roll a 6, roll again. You'll get a flat probability curve. Alternatively, you could roll a D10 or D20, and integer divide the result by 2 or 4 respectively (that is, 1 or 2 on the D10 become 1, 3 or 4 become 2, etc.) -- whatever is comfortable for you.

Yes, this works: in statistics, it's known as rejection sampling.

The very slight problem is that you could roll a large number of consecutive sixes so you can't guarantee that the procedure will finish within any particular amount of time. However, the average number of dice rolls needed is just 1.2 and there's less than a 0.5% chance of needing more than three rolls.

• To be fair, you were already going to have to re-roll that die if it landed cocked or in someone's drink. Re-rolling sixes isn't that much different. Sep 24, 2015 at 0:11
• Ah, the craps theorem. Either that or the OP's 'by symmetry' will do it. (Both ways are predicated on not rolling sixes indefinitely, but that has probability 1.) Sep 24, 2015 at 3:13
• @GMJoe Exactly. And this procedure requires very few rerolls -- just 0.2 on average. Sep 24, 2015 at 8:34
• If you keep rolling 6's .. remember which die it is and use it next time you cast fireball ... don't roll 10 dice, just roll that die 10 times .. :) (you'll be the most popular person there ... rolling a single die ... 10 times .. O.o ) Sep 25, 2015 at 15:24

Yes

The 6th "reroll" option doesn't impact the weights of each of the other events, so by giving a specific face a reroll, it's like reducing the number of faces of the die.

In the scenario you set up, each event would have a 20% chance of occuring—or in mathiness: P(1/5).

• mathiness :o)
– Tas
Sep 23, 2015 at 21:40

This would work. Some alternatives would be to use a d10 or d20 as they are evenly divisible by 5. You could then either use the result divided by 2 for a d10 or 4 for a d20, or the result modulus 5.

• Thanks! I didn't have any situation in mind, just wondered in case something like this ever happened.
– OSG
Sep 24, 2015 at 4:44

If you're interested in the mathematical formula (who isn't??), you can see how each outcome has a 1/5 chance of happening using your method, which is what you want.

The chance of rolling a 1 under this circumstance is 1/6 (rolling a 1 first time) + 1/6 ✕ 1/6 (chance of rolling a 6 followed by a 1) + 1/6 ✕ 1/6 ✕ 1/6 (chance of rolling 2 sixes then a 1) + ...

This can be rephrased as (1/6 + 1/36 + 1/216 + ...)

Which, from the sum of a geometric sequence, is equal to

$\dfrac{1 \div (1- \frac{1}{6})}{6} = \dfrac{1 \div (5/6)}{6} = \dfrac{6/5}{6} = \dfrac{1}{5}.$

And the same for rolling a 2,3,4,5.

• Don't you mean (1/6 + 1/36 + 1/216+...)? I do like thinking about the math behind things, this is part of why I originally asked.
– OSG
Sep 24, 2015 at 4:47
• IMO it's simpler to get the probability it this way: $P(X=n) = 1/6 + P(X=n)*(1/6)$ (to get result n you can roll either get it on the first dice roll or roll a 6 and repeat the process). That equation is easy to solve and you also get $P(X=n) = 1/5$. Sep 24, 2015 at 9:20

You can, but you can't guarantee that it would happen after N throws of the die for any finite value of N. If you agree to re-roll (and discount the roll) when the die falls on 6, the probability of never getting anything but 6 is 0. But you may get a 6 an arbitrarily large number of consecutive times. However, if you ignore all the 6's, then each roll that remains has a 1 in 5 probability of occurring.