OK a solution--if a leg day is 2 ones, then
a solution is to calculate the probability P(E) of throwing 2 ones p(e) times the possible ways n to throwing them.
now there are a lot of way to throw them (counting) there 1,1,x,x,x; 1,x,1,x,x ; x,x,x,1,1 so you have to count them. Fortunately 15th century gamesmen answere this question its n!/(c!*(n-c)!) where n is the number of throws, c is the number of targets (# of 1's).
so this is 5!/(2!*3!)=5*4/2=10
so the probability of a leg day (exactly 2, 1's) is 10*64/3125=640/3125=-.2048
or about one day out of five is a leg day.
But wait---if I throw more then two one's is that a leg day too?
A quicker way to do this one is to count the number of non leg days
so zero 1's is not a leg day
p(e0)=4^5/5^5=1024/3125 and there is only one combination (5!/5!/1!=1)
now 1 1's is also not a leg day
p(e1)=4^4/5^5=256/3125 but there is (5!/4!/1!=5 ways to roll 1 one)
So the probability of no leg days is 1024/3124+1280/3125=2304/3125
So if a leg day is 2 or more ones then a leg day has a probability of
So a leg day in this case will occur about one day in four.