When 4 dice are thrown what is the probability that the same number appears on each of then?

It's worth noting another approach to this solution that imagines each die being rolled individually and looks at the odds of the complimentary event that there are no matching dice.

The first rolled die can be anything, doesn't matter. Imagine its a 1 if it helps.

The second rolled die has a $\frac{1}{6}$ chance of matching the first (which we're not interested in), and a $\frac{5}{6}$ chance of differing from the first.

Given that the first two dice were different, the third rolled die has a $\frac{1}{3}$ chance of matching one of the first two, and a $\frac{2}{3}$ chance of differing from the first two.

Finally, given that we have no matches by the fourth die, there's a $\frac{1}{2}$ chance that the fourth die matches, otherwise, there are no matches. Given that this is the only way to not get a match, we can compute the probability of not having a match as $\frac{5}{6} \cdot \frac{2}{3} \cdot \frac{1}{2}$. This simplifies to $\frac{5}{18}$, and the compliment $\frac{13}{18} \simeq 0.7222$ agrees with the other approaches.

Let #E# be the event 'All four dice show different numbers' and #S# be the sample space of the experiment. Then #absS=6^4# and #"P"(E)=absE"/"absS.#

In order to find #absE#, we can consider the four dice rolls independently. On the first roll, there is no risk of duplicating any previous roll (since there aren't any), so there are 6 acceptable outcomes. For the second roll, we want to avoid whatever the first roll was, so there are 5 acceptable outcomes.

If we keep going, assuming we keep rolling different numbers, then on roll three, there are 4 acceptable outcomes, and on roll four, there are 3 acceptable outcomes. (As the roll counter goes up, the number of possible unique rolls goes down.)

Assuming independence of rolls, we get #absE=6 xx 5 xx 4 xx 3,# which gives #absE=360.#

Since #"P"(E)=absE/absS#, we get

#"P"(E)=360/6^4=60/216=5/18,#

which is approximately 27.78%.

Answer:

The total number of elementary events associated to the random experiments of throwing four dice simultaneously is: = 6*6*6*6=64 n(S) = 64 Let X be the event that all dice show the same face. X = { (1,1,1,1,), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)} n(X) = 6 Hence required probability = n(X)/ n(S) = 6/ 64*4 = 1/ 216