Yes you are right, We need the #times the event occurs over the sample space.

No of possible outcomes for 4 throws is 6*6*6*6=1296

No of these times which your number appears is ( 1 for when your number appears 5 when it doesnt appear)

1 *5*5*5 = 125

1*1*5*5 = 25

5*1*1*5 = 25

1*1*1*5 = 5

1*1*1*1 = 1

1*5*1*1 = 5

1*1*5*1 = 5

5*1*1*1 = 5

5*1*1*5 = 25

5*5*1*5 = 125

5*5*1*1= 25

5*5*5*1 = 125

5*1*5*5 = 125

Adding these we get 621

So 621/1296 = 0.4791

Maybe my arithmetic is off somewhere but I think this method is right??

Maybe this is better

Cheers Paul