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...
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...
