Can anyone help me on this probability

Reply Mon 13 Jul, 2015 11:42 am
The "happy" outcome means I have to do a leg day (workout).
Which isn't really happy... :/
Thanks to anyone who can help.
Say I have a single fair 100-sided die.
I am going to roll it 5 times each morning, recording the result of each test.
The die is numbered 1 through 100.

If among the five discrete rolls, it lands on any number between 1 and 20 [inclusive] and it does this TWICE OR MORE (among those 5 rolls) then I have to do a leg day.

Here is an example of a leg day: 11 / 47 / 99 / 20 / 66
Here is another example of a leg day: 12 / 33 / 13 / 97 / 20
Here is an example of a day in which I am released from legs (!) :
21 / 22 / 23 / 24 / 25

Many thanks!
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Reply Mon 13 Jul, 2015 12:45 pm
Simplify, Effectively you have a five sided die.

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

Whatta the odds.

Reply Mon 13 Jul, 2015 01:59 pm
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.

Reply Tue 14 Jul, 2015 09:51 am
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.

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