Hi All,
'A gambler rolls three fair six-sided dice. What is the probability that two of the dice show the same number, but the third shows a different number'?
The author goes on to explain the strategy to answer....
1. Set the result of the first die to be any number. * This will allow us to concentrate on matching the second and third die to the first, is that correct?
2. Prob of Second matching first, Third not matching = [tex]\frac{1}{6}.\frac{5}{6}[\MATH]
* There is a one in six chance of the second matching as there are 6 digits, is that correct? There is a five in six change of the Third not matching as [tex] 1 - \frac{1}{6} = \frac{5}{6}. [\MATH]Is that correct?
This is where it gets really confusing.
3. Prob of Third matching first, second not matching = [tex]\frac{5}{6}.\frac{1}{6} [\MATH]
* I can't figure out where this fractions came from. As above, matching = [tex]\frac{1}{6}[\MATH]
4. Prob of second and third matching each other = [tex]\frac{5}{6}.\frac{1}{6}[\MATH]
* as with 3. I can't figure out where these probabilities came from.
I understand the logic to this approach it's just the specifics I'm struggling with.
N.B. Latex does not seem to be compiling for me so forgive the [\MATH]. I hope it makes sense.
Thanks in Advance,
D
When rolling three dice, each outcome is a triple. Roll dice the outcome is a pair. Roll n dice the outcome is an n-tuple.
The outcome of rolling three dice and getting two 1’s and a different face can happen in three ways: (1,1,X), (1,X,1) or (X,1,1).
But X can have five different values: 2,3,4,5,6.
That makes fives 15 possible triples.
Any one of six values can be the double in such a triple.
6 times 15 is 90.