$3$ identical Dice are tossed simultaneously, The probability that all dice shows same number.

If $3$ identical Dice are tossed simultaneously, The find probability that all dice shows

same number.

$\bf{My\; Try::}$ Let $A$ be the event in which upper face of all dice shows same number

and $S$ be the sample space

Now Here we have $3$ identical dice.

So $x_{1}$ be the number of times in which dice shows number $1$ on upper face.

Similarly $x_{2}$ be the number of times in which dice shows number $2$ on upper face.

...........

...........

...........

$x_{6}$ be the number of times in which dice shows number $6$ on upper face.

So here $x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6} = 3,$ Where $x_{1},x_{2},x_{3},x_{4},x_{5},x_{6}\geq 0$

So we get $\displaystyle n(S) = \binom{8}{2}=56$ and $n(A) = 6$

So required probability $\displaystyle P(A) = \frac{6}{56}$

although i have solve that question using that post

How many are the possible outcomes from throwing $n$ (identical) dice

but i did not understand what is the difference between identical dice and simple dice.

and why answer can not be equal to $\displaystyle \frac{6}{256}$

bcz whether dice are identical or not total number of possible outcome will remain same.)(Its my assumption.)

plz explain me, Thanks

• I think you made some typos - I think you meant $\binom{8}{3}=56$ and a final probability of $6/56$, not $6/256$. – Peter Woolfitt Mar 25 '16 at 13:38
• If I understand the problem correctly, the issue arises because not all of the $56$ combinations are equally likely (for example it is easier to roll two 1s and one 2 than it is to roll three 1s) – Peter Woolfitt Mar 25 '16 at 13:43
• Did you mean to ask why the answer cannot be equal to $6/216$? – N. F. Taussig Mar 25 '16 at 13:51
• If you meant to ask why the answer can't be $\frac6{216} (= \frac1{36})$, it is, in fact the correct answer. – true blue anil Mar 25 '16 at 15:06
• The crux is that outcomes can be counted in various ways, but for computing probability, if you are computing according to frequency of occurrence, the outcomes have to be equi-probable. E.g. if a family has two children, outcome can be all boys, all girls, or one of each, but P($2$ boys) is not $1/3.$ Similarly, here one way of counting outcomes is $56$, one way is $216$, but the first way doesn't give equi-probable outcomes. – true blue anil Mar 25 '16 at 20:16

Note that, as the outcomes of the three dice are independent, throwing them simultaneously is equivalent to throwing them after each other. Also note that, for all numbers to be equal, there is no restriction on the outcome of the first die. Assume the outcome of the first die is $n$, with $1\leq{n}\leq6$. Then the probability that all dice show the same number is the probability you throuw $n$ with both the second and the third die, i.e. $$\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{36}.$$

• Does someone care to explain the downvote? – Eric S. Mar 25 '16 at 13:33
• Not mine, but the question seems referred to not distinguishable dice (identical). If so, your answer is wrong. – Emilio Novati Mar 25 '16 at 13:38
• How so? Even if they're identical, their outcomes are still indepedent, right? As a thought experiment, just pick one of the non-distinguishable dice and throw it, after that, you want the outcome of the last two dice to match the outcome of the first die. Or am I missing something? Because I am not that familiar with identical dice. – Eric S. Mar 25 '16 at 13:45
• my downvote was not because I found the answer incorrect, but rather that I wanted an answer to address the issue with the OP's attempt (which to me is a more interesting question) – Peter Woolfitt Mar 25 '16 at 13:50

If the dice are indistinguishable (I suppose that ''identical'' means this) than all the outcomes that have the same tree numbers are identified, in the sense that we cannot distinguish, as example: $1,4,3$ from $1,3,4$ or $4,1,3$ and so on.

This means that the space of all possible outcomes is formed from the ''Combination with repetition'' of $6$ elements of order $3$. As you can see here, such $3-$multicombinations of $6$ elements are: $$\binom{6+3-1}{3}=\binom{8}{3}=\binom{8}{5}=56$$

since the outcomes with three same numbers are $6$ it seams that the answer is $6/56$, but this is wrong (as noted by @trueblueanil), because the $56$ outcomes are not equiprobable, so the correct answer is $1/36$.

• The dice are indistinguishable if we toss them together and we see the result without the possibility of distinguishing them. If we toss the dice separately, the dice become distinguishable. – Emilio Novati Mar 25 '16 at 14:52
• A little thought will show that even if we toss them simultaneously, $1,3,4$ will occur $6$ times more frequently than, say, 1,1,1. – true blue anil Mar 25 '16 at 14:57
• No. The six combinations $(1,3,4)$, $(1,4,3)$, $(3,1,4)$, $(3,4,1)$, $(4,1,3)$, $(4,3,1)$ are the same event. Or, in another way, we have six possible position of the same number $1$ in three places $(1_A,1_B,1_C)$ if we can distinguish the places. – Emilio Novati Mar 25 '16 at 15:09
• There are $56$ outcomes, but they aren't equi-probable. – true blue anil Mar 25 '16 at 15:30
• If the dices are identical (different colors are not admitted) and tossed all together, the space of all possible outcomes has $56$ distinguishable elements (it has no matter if the exit $1$ is on the first, or second or third die, because we can not identify the single die). Since the favourable case are $6$, the probability is, by definition, $6/56$. – Emilio Novati Mar 25 '16 at 15:54

For this problem, the dice being distinguishable or not is immaterial,
since they are all to show the same number.

The first die can show anything. The second and third must show the same number,

thus $Pr = \dfrac16\cdot\dfrac16 = \dfrac1{36}$

Even if they were distinguishable (say red, blue, yellow) the red die could show anything,
and the blue and yellow ones would have to show the same number with $Pr = \dfrac16$ each

When $3$ dice are rolled, or one die is rolled $3$ times, there will be $216$ outcomes, as under:
$3$ of a kind: $\binom61 = 6$, only $1$ permutation $\to 6$ ways
$2-1$ of a kind: $\binom61\binom51 = 30, \frac{3!}{2!} = 3$ permutations $\to 90$ ways
$1-1-1$ of a kind: $\binom63 = 20, 3! = 6$ permutations $\to 120$ ways
Combinations with repetitions counts $6+30+20 = 56$, ignoring frequency of occurrence,