Each of five, standard, six-sided dice is rolled once. What is the probability that there is at least one pair but not a three-of-a-kind (that is, there are two dice showing the same value, but no three dice show the same value)?

RektTheNoob Feb 5, 2018

#1**+2 **

**I'll answer RektTheNoob but I want a response of you.** If you think my answer is wrong then I want an explanation of how you think it should be done, or the answer that you think is correct with the reason.

Each of five, standard, six-sided dice is rolled once. What is the probability that there is at least one pair but not a three-of-a-kind (that is, there are two dice showing the same value, but no three dice show the same value)?

Number of possible outcomes in total = 6^5 = 7776

Number of outcomes with 5 of a kind = 6

Number of outcomes with 4 of a kind and one different = (6*1*1*1*5)*5 = 150

Number of outcomes with 3 of one kind and one pair = (6*1*1*5*1)*5C2 = 30*10=300

Number of outcomes with 3 of a kind and 2 different ones = (6*1*1*5*4)*5C2=120*10=1200

(At first I wanted to multiply this last one by two but that results in double counting.)

Number of outcomes where they are all different = 6*5*4*3*2=6!= 720

I think this represents all the ones that do not have just have one or even 2 pairs.

Add them up

6+150+300+1200+720 = 2376

Total with at least one pair and no triples = 7776-2376 = 5400

So the prob of at least one pair and no tripples is \(\frac{5400}{7776} = \frac{25}{36}= 0.69\dot4\)

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Alan also ran a MonteCarlo simulation on this and consistenly got 0.695 so there may be some small error here but if there is I can't find it.

Thanks Alan

.Melody Feb 5, 2018