Can anyone help me on this probability

Simplify, Effectively you have a five sided die.

You throw two ones (1) in five rolls, you have a leg day.

Whatta the odds.

RAP

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.

p(e)=1/5*1/5*4/5*4/5*4/5=4^3/5^5=64/3125

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)

P(E0)=1*1024/3125

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)

P(E1)=5*256/3125=1280/3125

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

(3125-2304)/3125=821/3125=0.2627

So a leg day in this case will occur about one day in four.

Rap

raprap, Thanks to you so much for the explanation!
I had deduced by experiment that it had to be a little bit more than 20%*, but I just wasn't sure how to take the counting of multiple ways (to end up with at least two 1's).
You're so nice to help me. Thank you.
*I have actually been doing this for a while and had noticed it was pretty much between [once every 4th or 5th day]. I was really curious how to figure out a more exacting odds! I use an Excel spreadsheet RAND() function (for the value between 0.01 and 0.20) over 5 rows of the spreadsheet.