Chance of a double in three dice
What is the chance of rolling doubles in three six sided dice?
The answer I have is: $$ \frac{1}{6}•\frac{1}{6}•\frac{3}{2} = \frac{1}{24} $$
$\endgroup$7 Answers
$\begingroup$The number of elements in the sample space = 216 or (6*6*6)
Of the three numbers two of them to be same is (1,1) (2,2) (3,3) (4,4) (5,5) (6,6) = 6 doublets. Every one of this can happen in 3 different ways, that is (1,1,x) (1,x,1) (x,1,1) = 3 ways. Now x can be any of the remaining 5 numbers. So total number of ways two of them can be same = 6 x 3 x 5 = 90
So probability of the event = 90/216 = 5/12
$\endgroup$ 1 $\begingroup$I will assume that "doubles" does not include triples. Imagine the dice are coloured blue, white, and red. Record the outcome as a triple $(b,w,r)$, where $b$ is the number on the blue, and so on. There are $6^3$ possible outcomes, all equally likely.
How many doubles are there? The number we have two of can be chosen in $6$ ways. For each choice, the number we have one of can be chosen in $5$ ways, and its location (colour) can be chosen in $3$ ways, for a total of $(6)(5)(3)$. the required probability is therefore $\frac{(6)(5)(3)}{6^3}$.
Another way: The probability all the tosses are the same is $\frac{1}{36}$, since whatever we get on the blue, we must get on the white and on the red.
The probability the tosses are all different is $\frac{5}{6}\cdot\frac{4}{6}$. So the probability of not getting a double is $\frac{1}{36}+\frac{20}{36}$. It follows that the probability of a double is $\frac{15}{36}$.
$\endgroup$ $\begingroup$$\displaystyle{% \mbox{No triplets} \quad\Longrightarrow\quad \left[\vphantom{\Huge A^A}% 3\times\left(\vphantom{\Huge A}{1 \over 6}\times{1 \over 6}\times{5 \over 6}\right) \right] \times 6 = \color{#ff0000}{\large{5 \over 12}} }$
$\endgroup$ $\begingroup$Between the first and the second dice there is a 1/6 chance of getting doubles, which is 36/216 possible outcomes. Between the first and third dice there is also a 1/6 chance of getting doubles but 6 of the 36 doubles in this set of outcomes have already been accounted for in the combination of the first and second die so that's another 30/216 possible outcomes. Between the second and third dice there is also a 1/6 chance of getting doubles but 6 of the 36 doubles in this set of outcomes have already been accounted for in the combination of the first and second die so that's another 30/216 possible outcomes. This leaves the total probability that you get doubles (including triples) to be 36/216+30/216+30/216 = 96/216 which simplifies to 4/9. Excluding triples makes the probability 96/216-6/216 = 90/216 which simplifies to 15/36.
*note the 6 duplicate doubles from the second and third dice outcomes and the first and third dice outcomes are all from the triples of each respective number so they could be counted in any of the three combinations of dice when looking at the question this way
$\endgroup$ $\begingroup$Probability of all three dice the same = $1 \times 1/6 \times 1/6 = 1/36$. Probability of no dice the same = $1 \times 5/6 \times 4/6 = 20/36$. Probability of a pair = $1-1/36-20/36 = 15/36$.
$\endgroup$ $\begingroup$- Another way to approach this problem to calculate the compliment i.e. no of outcomes in which there are no doubles.
Step 1:Calculate the cardinality of sample space #S = 6 ^ 3 = 216
Step 2: Calculate #outcomes of not getting doubles
#of outcomes of not getting doubles = (P(6,3) + 6) = 126
Note: 6 is added to count the triples i.e (1,1,1).....(6,6,6)
Step 3 : Calculate probability of getting doubles:
If #outcomes of not doubles is 126, then # outcomes with doubles = 216-126 = 90
Therefore,P(getting doubles) = 90/216
Note : # represents ( number of)
$\endgroup$ $\begingroup$There are 36 possible permutations when rolling the first and second die. 6 of these produce doubles. Therefore the probability of rolling doubles with the first and second die is 6/36, or 1/6. There is also a 1/6 chance of rolling doubles with the first and third die. And a 1/6 chance of rolling doubles with the second and third die. Therefore the probability of rolling doubles with all three dice is 1/6 + 1/6 + 1/6 = 3/6 or 1/2 (including triples). The probability of rolling triples is 1/36 so the probability of rolling doubles with no triples is 18/36 - 1/36 = 17/36.
$\endgroup$ 1
