Probability Issue

Philosophy Forums Science, Technology, and Math
1

I don’t know much about probabilities, but as far as I know, two random events don’t affect each other.

Take two identical sets of numbers, {1, 2, 3, 4, 5} and {1, 2, 3, 4, 5}. You choose one random number from each set, a and b.

The probability of a+b to be 2 is 1 to 25: (1;1)
The probability of a+b to be 3 is 2 to 25: (1;2) (2;1)
The probability of a+b to be 4 is 3 to 25: (1;3) (2;2) (3;1)
The probability of a+b to be 5 is 4 to 25: (1;4) (2;3) (3;2) (4;1)
The probability of a+b to be 6 is 5 to 25: (1;5) (2;4) (3;3) (4;2) (5;1)
The probability of a+b to be 7 is 4 to 25: (2;5) (3;4) (4;3) (5;2)
The probability of a+b to be 8 is 3 to 25: (3;5) (4;4) (5;3)
The probability of a+b to be 9 is 2 to 25: (4;5) (5;4)
The probability of a+b to be 10 is 1 to 25: (5;5)

So a+b is most likely to be 6.
So if a is n, b is most likely to be 6-n.

I need to know if I made myself understood, and, if so, where I did wrong.

2

What you’re doing is confusing the probability of the sum of a+b with the value of a and b. Neither rely on each other. Ach, I’m not very good at explaining it, but I understand where you’ve gone wrong. Maybe this example will help.

Say a is 1. Then b is just as likely to 1, 2, 3, 4 or 5. It is not most likely to be 5. (from your own diagram, the probs of (1;1), (1,2) (1,3) (1,4) (1,5) only appear once, there is no greater prob of (1,5)).

You could express b as 6-n, where n could be (1,2,3,4,5) as all you are saying is b is (6-1, 6-2, 6-3, 6-4, 6-5) which is the same as (5,4,3,2,1), which was what you defined b as in the first place.

When you change n to be defined as a it doesn’t work any longer though.

3

Thanks, Matt. Now I think I understand…