assume that the numbers 1 and 2 are equally likely, as are the numbers 3 through 6, yet rolling a 1 is five times as likely as rolling a 3.

Question

If we have probability weights w1 through w6, one for each possible outcome, what is the the value of w3?

Answer 1

First off, all probabilities add to 1 so

w1+w2+w3+w4+w5+w6=1

 

Then we know w3=w4=w5=w6 so

w1+w2+w3+w3+w3+w3=1

 

Then we know w1 is five times w3 and w1=w2 so

5w3+5w3+w3+w3+w3+w3=1

Solve for w3!

 

Hope that helps!