When rolling 3 dice, is the probability of rolling the same number three times the same as rolling three different numbers?

Question

Is the outcome 111 as probable as e.g. 123?

Answer 1

No. To see this, consider that for the first, second, and third die to be the same, you can only roll a $1$ on all three dice. On the other hand, consider that the first die you roll can be $1,2,3$, that the second die can be either of the two remaining numbers, and that the third die must be the last number. The actual probabilities are $\frac1{6^3}=\frac1{216}$ for the same and $\frac1{6^3}*3!=\frac1{36}$ for the three different dice.

Answer 2

No, if you are interested in 111 vs. any outcome with three different numbers.

The probability of getting same outcome thrice is $(\frac{1}{6})^3 $ but that of getting 3 different numbers is $(\frac{1}{6}).(1-\frac{1}{6}).(1-\frac{2}{6})$ i.e. (Probability of getting say 1) * (Probability of not getting 1) * (Probability of not getting 1 AND the number you got on second throw) .

Yes, if you are interested in 111 vs the specific outcome 123. If you want the probability of outcomes specifically 1,2,3 then it is same as 1,1,1 because probability of getting 1 = probability of getting 2 = probability of getting 3 = $(\frac{1}{6})$. Therefore the result is same as that of three simultaneous 1s = $(\frac{1}{6})^3$.

Answer 3

What do you mean by the words "rolling the same number three times"?

What do you mean by the words "rolling three different numbers"?

If "rolling the same number three times" means "choose a number, then roll it three times in a row" then the probability of success is $\left(\frac16\right)^3.$ If it means merely that the three rolls should all be the same as each other, then the probability is much higher: $\left(\frac16\right)^2.$

If "rolling three different numbers" means "choose three numbers $x, y, z,$ all different, and roll $x$, then roll $y$, then roll $z$" the probability again is $\left(\frac16\right)^3.$ If it means "choose three numbers $x, y, z,$ all different, and roll all three numbers, but it does not matter which of them you roll first or which you roll last" the the probability rises to $3! \times\left(\frac16\right)^3 = \left(\frac16\right)^2.$ If it means, "roll three times, and and the end of the three rolls none of the numbers you rolled should be the same as either of the others," then the probability is even higher: $\frac56\times\frac23.$

In summary, the probability of rolling $1,1,1$ is the same as the probability of rolling $1,$ then $2,$ then $3.$ But that is not what most people would think you mean when you ask about "rolling the same number three times" or "rolling three different numbers."