Suppose I have 3 dice. Each has some mechanism that can prevent other dice being in 2 if itself is 2 when they are rolled together. Now I roll the 3 dice at the same time. Then what is the probability of having a 2 appearing?

Is the answer 1/6 + 1/6 + 1/6 = 1/2 since the event are mutually exclusive and each has a probability of 1/6?

But without the mechanism the probability of at least one 2 appearing is 1 - (5/6)^3 = 91/216 < 1/2?

Why should the probability decrease?

  • $\begingroup$ How does your "mechanism" not affect the probability of an individual die roll being $2$? $\endgroup$ Sep 1, 2014 at 23:36
  • $\begingroup$ To make an even more absurd example, suppose you rolled $12$ dice with the mechanism. Your computation would make the probability $2$. $\endgroup$ Sep 1, 2014 at 23:37
  • $\begingroup$ @RobertIsrael So you mean because of the mechanism, the event of 2 appearing for the single die is not mutually exclusive? Then how should I calculate the probability? $\endgroup$
    – Allitee
    Sep 1, 2014 at 23:45
  • $\begingroup$ It depends on what this "mechanism" does. If it does leave the individual probabilities at $1/6$ and makes them mutually exclusive, then your calculation was correct. $\endgroup$ Sep 2, 2014 at 0:17
  • 2
    $\begingroup$ Allitee, it seems that it's not possible to give a complete answer to this question unless you say more about the mechanism. For example, the results will be different if the mechanism deals with multiple twos by: (a) rolling all the dice again, until an acceptable combination is obtained; or (b) selects one of the two twos (or two of the three twos) at random, and rolls only those dice again, repeating the process as necessary if one or two twos are obtained again. There are other conceivable possibilities besides (a) and (b), and the results would be different again. $\endgroup$
    – Dave
    Sep 2, 2014 at 4:00

3 Answers 3


If the mechanism only applies to 2, then they are just like ordinary dice until a 2 appears. Hence the probability of no 2s in three rolls is still $\left(\frac56\right)^3$. So the probability of at least one 2 in three rolls is $\frac{91}{216}$ just like ordinary dice.

  • $\begingroup$ Then the probability of getting a 2 is 1/6? But then since the event is mutually exclusive, we can get 3 * 1/6 which is 1/2 for the probability? $\endgroup$
    – Allitee
    Sep 2, 2014 at 2:40
  • $\begingroup$ @Allitee: Actually the probability of a 2 on roll 2 is $\frac16\cdot 0+\frac56\cdot \frac 16 = \frac{5}{36}$ (because in the case that the first die was 2, the second die has $0$ probability of coming up $2$). Similar reasoning leads to the probability of the third die being 2 is $\frac{25}{216}$. The three events are mutually exclusive, and so the probability of a 2 is $\frac16+\frac{5}{36}+\frac{25}{216}=\frac{91}{216}$. $\endgroup$
    – paw88789
    Sep 2, 2014 at 5:22

It is reasonable to suppose by default that the mysterious mechanism does not apply when a number other than $2$ is tossed.

The probability of not getting a $2$ is then as usual $\left(\frac{5}{6}\right)^3$, so the probability of getting (at least) one $2$ is the same as usual.

Note that the probability of multiple $2$'s is now $0$, so for example the probability of at least one $6$ is greater than usual.

Remarks: $1.$ The problem now is to produce a physically reasonable model, to put flesh on the bones of the mysterious mechanism. I have not succeeded in doing so.

One possibility that is contrary to the wording is that the dice are rolled sequentially. If the first toss is a $2$, the machinery kicks in, and we end up with one $2$, probability $\frac{1}{6}$. If we don't get a $2$, the probability of a $2$ on the next toss is $\frac{1}{6}$, and then the machinery kicks in and we again get a success. And of course we can have success if we get a non-$2$ twice, then a $2$. That gives probability $\frac{1}{6}+\frac{5}{6}\cdot\frac{1}{6}+\frac{5^2}{6^2}\cdot\frac{1}{6}$. That gives probability $\frac{91}{216}$.

$2.$ Below, verbatim, is my earlier (wrongly analyzed) suggestion. The error was pointed out by Did.

We supply an explicit non-magical mechanism that satisfies the OP's condition. Toss three dice. If we get more than one $2$, toss again until we get one or fewer $2$'s. In any round of this game, the probability of getting no $2$'s is unchanged from the usual independent fair dice assumption.

  • $\begingroup$ The probability that the "non-magical mechanism" presented in the Remark produces $0$ twos is not $(5/6)^3$. To see why, let $p=(5/6)^3$ the probability to get $0$ twos in a toss of the three dice and $q=3(1/6)(5/6)^2$ the probability to get $1$ two in a toss of the three dice. Then, in any round of the game, the probability of getting $0$ twos is $p/(p+q)=5/8$, not $p$. $\endgroup$
    – Did
    Sep 2, 2014 at 0:43
  • $\begingroup$ @Did I don't understand why you say the probability is not (5/6)^3 and then let p = (5/6)^3 where p is the probability of get 0 twos. $\endgroup$
    – Allitee
    Sep 2, 2014 at 3:17
  • $\begingroup$ @Allitee This p is the probability to get 0 two when one throws three dice once. The procedure suggested in the answer was to throw the three dices anew when one got more than 1 two, until one got 0 or 1 two. The probability to get 0 two when one stops this procedure is indeed p/(p+q). $\endgroup$
    – Did
    Sep 2, 2014 at 3:32
  • $\begingroup$ @AndréNicolas You suppressed the paragraph explaining a so-called non-magical procedure. Let me suggest to restore this paragraph, first because it describes an interesting mistake, second because without it my comment debunking it becomes difficult to understand. $\endgroup$
    – Did
    Sep 2, 2014 at 3:34
  • $\begingroup$ I will restore it, it will indeed be useful. $\endgroup$ Sep 2, 2014 at 3:35

Due to this mechanism, there's only two possibilities for how many twos appear. Either no twos appear or one two appears. Let $A$ be the event that no twos appear. Let $B$ be the event that one two appears.

So $P(B) = 1-P(A)$.

Since $P(A) = {125 \over 216}$,

$P(B) = 1 - {125 \over 216} = {91 \over 216}$.

The trick is to disregard events where more than one two appears since that is impossible in this scenario.


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