Problem Solving

A gambler rolls three fair 6 sided dice. What is the probability that 2 of the dice show the same number but the third shows a different number?

I don't see the flaw in my reasoning:

Lets say the gambler played the dice the first time: 1/6 is the probability for any number to come up, then for the second to match there is 1/6 again and the third not to match, its 5/6.

So the probability for the first chain of events becomes 1/6 * 1/6 * 5/6.

Now to figure out how many such chains exist:
Either the 1st and 3rd are same or 1st and the 2nd are the same or 2nd and the 3rd are the same, which gives us 3 possibilities.
This is also confirmed by the distinguishability logic: So if we are looking for all possibilities of arranging the alphabets: S, S and D => its 3! / 2! = 3.

So the total probability becomes: 1/6 * 1/6 * 5/6 * 3. But the answer is 15/36.

I don't understand the flaw in this reasoning. Please help! Thanks...

your logic is almost correct,just one point.two of the dices has same number ,so probability for the second and third dice depends on the first dice face

first dice can have any value,second one has to have same value as the first so 1/6 and third has to have different value than the first so 5/6

total 1/6*5/6=5/6

now as you have said,there will be 3C2=3 ways of piling the same number dice pair.So total probability is 3*5/36=15/36

btw,if it has been mentioned that what is the probability that two dice have value '6' and the other something else ,then your ans would have been correct.Hope this will help.

liferocks wrote:your logic is almost correct,just one point.two of the dices has same number ,so probability for the second and third dice depends on the first dice face

first dice can have any value,second one has to have same value as the first so 1/6 and third has to have different value than the first so 5/6

total 1/6*5/6=5/6

now as you have said,there will be 3C2=3 ways of piling the same number dice pair.So total probability is 3*5/36=15/36

btw,if it has been mentioned that what is the probability that two dice have value '6' and the other something else ,then your ans would have been correct.Hope this will help.

Got it! Your last statement helped. Thats when I realized the difference between the 2 scenarios...Thanks!

prob of rolling any number*prob of rolling same number as first number*prob of rolling different number*number of arrangements=

6/6*1/6*5/6*3=
15/36