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I am stuck at the following question.

What is the probability that two throws with three dice each will show the same configuration if (a) the dice are distinguishable (b) they are not.

Solution.

(a) $P(\text{Two throws of three dice s.t. each show the same configuration})=6^{3}/6^{6}=1/216$.

(b)Firstly from the second set of throws, the $3$ dies with the same face value as the first ones can be chosen in $3!=6$ ways.

Place $r=3$ dies in $n=6$ cells, such that order does not matter with repetition, can be done in ${{3+6-1}\choose{3}}={8\choose 3}=56$

${\displaystyle P(\text{Two throws of three indistinguishable dice show the same configuration})=\frac{56\times6}{6^{6}}}$

However, my answer to the second part of the problem is incorrect. Could someone help me think correctly about the problem.

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    2 throw of 3 dice is equals to one throw of 6 dice. You want same number on all six dice? – 2017-01-16
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    My understanding is, the problem asks for the probability of getting the same configuration - for e.g. if the first throw results in (5,6,6), what is the likelihood that the second throw also results in, perhaps (6,5,6)? – 2017-01-16
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    You mean if in first throw we have (1,2,3) then in next throw we have exact (1,2,3) same pattern. – 2017-01-16
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    after the first throw, the probability of then getting the same configuration is effected by the first configuration - if you first get 6-6-6 then there is only a 1/216 of getting that again. Therefore you can calculate the prob of getting 3 of a kind, 2 of a kind , or 3 unique - then multiply by the prob of repeating that - the chaces of repeating is different for each initial roll type – 2017-01-16
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    1-2-3 is more likely to be repeated than 6-6-6 - can you see why? Then can you use this to calculate the probability? (this is in the indistinguishable case) – 2017-01-16
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    what you've got wrong in b) is that if you roll 6-6-6 then you have only 1 way of doing this again, but if you roll 1-2-3 then you have [how many?] ways of doing this again. – 2017-01-16
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    @Cato, when all three faces are distinct, the probability of rolling an ace, deuce and trey again are $6/216$. – 2017-01-16
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    @KanwaljitSingh, in the next throw I must have (1,2,3) in any order, is what I understand. – 2017-01-16
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    @Cato, thanks for leading me in the right direction! Cheers. :) – 2017-01-16
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    @quasar - cool! – 2017-01-16

2 Answers 2

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(b)

Three distinct faces :

$$P(\text{getting the same configuration}) = \frac{(6\cdot5\cdot4)\times(3\cdot2\cdot1)}{6^6}$$

One repetition:

$$P(\text{getting the same configuration}) = \frac{{3\choose2}(6\cdot5)\times{3\choose2}}{6^6}$$

Two repetitions:

$$P(\text{getting the same configuration}) = \frac{6}{6^6}$$

Thus, the required probability is,

$$P(\text{getting the same configuration}) = \frac{996}{6^6}$$

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Simulation: Two successive throws of three fair dice with faces 1 through 6. Probability of same three faces showing: (a) if dice are indistinguishable, (b) distinguishable.

In the distinguishable case, it may help to imagine the dice are colored red, green, and blue, respectively. The probability is pretty clearly $1/6^3 = 0.00462963.$ In simulating for the indistinguishable dice, we sort the outcomes before comparison because color does not matter.

m = 10^6;  same.i = same.d = numeric(m)
for (i in 1:m) {
  roll.1 = sample(1:6, 3, repl=T)
  roll.2 = sample(1:6, 3, repl=T)
  same.d[i] = sum(roll.1==roll.2)
  same.i[i] = sum(sort(roll.1)==sort(roll.2)) }
mean(same.d==3);  mean(same.i==3)
## 0.004632
## 0.021579

With a million iterations, probabilities should be accurate to three places (a little more for very small probabilities), so the first simulated answer is consistent with $1/63.$ And, the second answer is consistent with with @Quasar's $996/6^6 = 0.02134774.$ (+1)

For a little more detail, my same.d and same.i are numbers of faces that agree for distinguishable and indistinguishable dice, respectively. here are the histograms of their respective distributions. (The probabilities mainly at issue here are the ones for 3 matches.)

enter image description here

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    I'm getting my hands wet at R, it's a lot of fun to simulate these games and see astonishing results. I liked your sort() trick in the indistinguishable case. – 2017-01-17
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    I see you are fairly new to this site, and welcome. If you will do a search on my user number, you will see that I have posted a _lot_ of R simulations (too many for the tastes of some users, I suppose). These tend to use 'for loops' which is not the greatest style in R (and they don't run as fast as they might), but some of the more sophisticated object oriented R structures would not be as easy for non-R users to understand. (So 'steal' whatever of my tricks you like, but try to develop a more sophisticated style than I feel comfortable using here.) – 2017-01-17
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    @BruceET, Your R programming is good . But where did you get $\frac{1}{63}$.Answer to the first question where distinguishable dice are used is $\frac{1}{216}$. – 2018-10-02
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    It's $1/6^3 = 1/216.$ In R, `1/6^3` returns 0.00462963, `6^3` returns 216 and `1/216` returns 0.00462963. – 2018-10-02