If a fair die which six sides is tossed twice what is the probability
that the lagest face value obtained is twice the smallest face value
obtained?

(1/6)^6

Heres my work:

Let x = the value of the first roll
Let y = the value of the second roll

so if x = 1,2,3,2,4,6
then y = 2,4,6,1,2,3
then the pairs are (1,2), (2,1), (4,2), (2,4), (3,6), and (6,3), which is 6 pairs

the probability of obtaining each value is 1/6
So I get (1/6)^6

2. Hello,

Originally Posted by shogunhd
If a fair die which six sides is tossed twice what is the probability
that the lagest face value obtained is twice the smallest face value
obtained?
Hm, I don't think it's the right reasoning

Your pairs are OK. Each pair has an equiprobability to be gotten.

The probability that the first dice yields X (for example 1) is 1/6. The probability that the second dice yields Y (for example 2) is 1/6.
Hence, the probability of getting the given pair (X,Y) is (1/6)*(1/6)=1/36

Among the 36 possible pairs, only 6 are the ones you're interested in.

Therefore, the probability of getting one of the 6 given pairs is 6/36=1/6.

Do you understand ?

3. So I get (1/6)^6
What you did here corresponds to the probability of getting one of these pairs every time for the 6 tosses.
Multiplying probabilities this way corresponds to a repetition of events, independently. It's like tossing 6 times the dice (twice).

4. Yes, I understand it much better now. Thank you