Thread: Multiple Probabilities

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

Originally Posted by dumluck

'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'?

That is making entirely too much of this problem.
Consider, (1,1,X),(1,X,1), (X,1,1) where X is not 1.
That means we can have 15 triples with two ones and another number.
So there are 90 ways to have two of one kind and one different.
There are a total triples.

Originally Posted by Plato

That is making entirely too much of this problem.
Consider, (1,1,X),(1,X,1), (X,1,1) where X is not 1.
That means we can have 15 triples with two ones and another number.
So there are 90 ways to have two of one kind and one different.
There are a total triples.

Hi Plato,
I would agree. However, the answer is . Is it possible to derive this from your approach?

Originally Posted by dumluck

Hi Plato,
I would agree. However, the answer is . Is it possible to derive this from your approach?

(90)/(216)=(15)/(36)=(5)/(12)

Originally Posted by Plato

That is making entirely too much of this problem.
Consider, (1,1,X),(1,X,1), (X,1,1) where X is not 1.
That means we can have 15 triples with two ones and another number.
.

Plato,
Thanks again for the response. What is the meaning of triplet here? I've only heard it in reference to geometry. I understand '(1,1,X),(1,X,1), (X,1,1) where X is not 1' but am failing to get where the 15 came out of this?

Sorry.
D

Originally Posted by dumluck

Plato,
Thanks again for the response. What is the meaning of triplet here? I've only heard it in reference to geometry. I understand '(1,1,X),(1,X,1), (X,1,1) where X is not 1' but am failing to get where the 15 came out of this?

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.

Originally Posted by Plato

There are a total triples.

Ok I've got it up to this point now. Where did the come from? We have 6 'matching pairs' (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). By why to the power for 3?

Thanks,

Originally Posted by dumluck

Ok I've got it up to this point now. Where did the come from? We have 6 'matching pairs' (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). By why to the power for 3?

There are triples made from .
Three places to put six different numbers: 6x 6x 6=6^3.