I threw three identical dice on the table and looked at the outcome. If I throw them again, what is the probability I get the same outcome? Post solutions in the comments section, below.

March 26, 2007 am31 9:47 am 9:47 am

No matter how you define “outcome,” if there are $N$ different outcomes, and the probability of the $i^{\rm th}$ outcome on the first roll of the dice is $P_i$, the probability of getting the same outcome twice in a row is: $\Sigma_{i=1}^N P_i^2$

But actually figuring out all the $P_i$‘s the hard part…

March 26, 2007 pm31 8:18 pm 8:18 pm

For the second throw there are $6^3$ possible outcomes,
111, 112, … 666. Now if your first throw was of the form

(i) a,b,c with a,b,c different from each other, then the probability
for throwing the same dice “state” is $p=3! \frac{1}{6^3}$

(ii) a,a,b with a,b different from each other, then the probability
for throwing the same dice “state” is $p= {3 \choose 2} \frac{1}{6^3}$

(iii) a,a,a, then the probability for throwing the same dice “state” is $p=\frac{1}{6^3}$

March 26, 2007 pm31 9:40 pm 9:40 pm

The total number of possible outcomes in the first throw is 6x6x6=216. The number of different combinations in the first throw is 6!/3!x3! = 20. So the probability of getting one combination in the first throw is 20/216. The probability of getting this exact same combination in the second throw is the number of permutations of the compniation of the first throw divided by the total possible outcomes which is 3!/216=6/216. So the probability of the two steps together is 20/216×6/216=5/1944.

March 26, 2007 pm31 10:05 pm 10:05 pm

The probability of getting the same outcome with (i) 3 different numbers is 6/216; expressed as 3/6 for the first die, 2/6 for the second die and 1/6 for the third die.
The probability of getting the same outcome with (ii) 2 similar numbers is 3/216 and the probability of getting the same outcome with (iii) 3 similar number is 1/216.

Combining the above 3 options, the probability of getting the same outcome in the second throw is 6/216 + 3/216 + 1/216 = 10/216 = 5/108.

March 27, 2007 am31 2:04 am 2:04 am

Nonna, your answer is less than 1/216, which can’t possibly be right: after all, the probability of each die landing exactly as it did before is 1/216, so the probability that we get the same outcome must be at least this large.

Samy Sadat, your answer is larger than 6/216 = 1/36, and this can’t possibly be right, either: each arrangement can appear in at most 6 different ways, so the probability of duplicating a given arrangement is surely bounded above by 6/216.

Edgardo has provided a key piece of the answer to this version of the problem. I’ll answer a different version, where the “outcome” is the sum of the three numbers shown. In this case, the number of ways of getting a 3, 4, 5, 6, 7, 8, 9, 10 is 1, 3, 6, 10, 15, 21, 25, 27, respectively, and then they go back down in the same way (so there are 27 ways to get 11, 25 ways to get 12, etc.). So, for example, the probability that I rolled a total of 8 and that I now roll a total of 8 again is $\left(\frac{21}{216}\right)^2$, and so on for the other terms. The probabilities that I arrive at are the probabilities of the disjoint, independent events, “I rolled an $n$ the first time, and then I roll an $n$ again.” Thus, to get the total probability of “I rolled something, then I rolled the same thing again,” I do exactly as rdt suggested and add up these probabilities. The final answer I get is 361/3888, or just a shade less than 1 in 10.

Incidentally, one good way to get the sequence I wrote above is through the use of generating functions, which have a lot of nice combinatorial uses. The definitive text on the subject is generatingfunctionology, by Herbert Wilf. The second edition is available for free as a .pdf file on his website. (Just search for generatingfunctionology.)

March 27, 2007 am31 3:55 am 3:55 am

As for the other interpretation with the sum:
If the dice yielded a sum of s, then the probability of
throwing three dice with the same sum s is: $\frac{(s-2) \cdot (s-1)}{2} \frac{1}{6^3}$

In your example you gave that would be:
s=6 $\frac{(6-2) \cdot (6-1)}{2} \frac{1}{6^3} = \frac{10}{6^3}$

7. March 27, 2007 am31 8:47 am 8:47 am

I notice lots of formulas. But rdt makes it easy: find the P for each sum, square it, and add them up. There are only 18 – 3 ( + 1) = 16 sums, and by symmetry we can go from 3 to 10, and double the sum of the squares. This is what JBL does (and we get the same answer).

BTW,
P(3) = 1/216
P(4) = 3/216
P(5) = 6/216
P(6) = 10/216
P(7) = 15/216
P(8) = 21/216
P(9) = 25/216
P(10) = 27/216

I’ll take explanations of the apparent break in pattern, if anyone wants to try. Why aren’t the last two numerators 28 and 36 respectively (the next two triangular numbers)?

8. March 28, 2007 pm31 9:04 pm 9:04 pm

JD – because with numbers 9 and higher, 11x is no longer a valid dice roll, elimnimating 3 of the possibilities. 10 and higher changes further because 12x won’t work either (eliminating 6 more).

March 28, 2007 pm31 9:13 pm 9:13 pm

Excellent! Now, we know why limiting our dice to 6 sides makes it deviate from the triangular numbers, but that doesn’t explain why it was the triangular numbers to begin with. So, suppose we had “sufficiently-many-sided dice.” Why is the resulting sequence the triangular numbers?

10. March 29, 2007 pm31 2:38 pm 2:38 pm

Consider 8. How many ways can we add 3 counting numbers to make 8?

Let’s substitute that question with this one: How many ways can we divide ******** (8 stars) into 3 groups?
*|*|******
*|**|*****
*|***|****
etc. These placements correspond to 1+1+6 (look at the groups the stars are divided into), 1+2+5, 1+3+4…. Will every sum be used up by this count? Sure. Pick one, and we can draw it. 4+1+3 ? ****|*|***

Now the math: we have 8 stars, but 7 gaps to put dividers into. Of those 7, we will fill 2 of them (not 3. Notice that we need one less divider than we need group). so we have $\binom{7}{2}$.

In general, if we have d dice we have d-1 dividers. If we have a sum of s we have s – 1 spaces. So for a sum of s with d dividers we get $\binom{s-1}{d-1}$

(If you haven’t seen it before, that “fraction” without the fraction bar represents the number of combinations of s-1 things taken d-1 at a time, or $_{s-1}C_{d-1}$

Back to above, when we have three dice, we have numbers of the form $\binom{s-1}{2}$, or $\frac{(s-1)(s-2)}{2}$, which are triangular numbers. (Alternate explanation refers to the 3rd diagonal of Pascal’s Triangle.)

the probability of the first landing on a “different” number is $\frac{6}{6}$. The probability of the second landing different than the first is $\frac{5}{6}$. And the third? $\frac{4}{6}$. So we multiply $\frac{6}{6} \times \frac{5}{6} \times \frac{4}{6} = \frac{120}{216}$