@AThousandYoung
Correct.
The way I did it was:
@athousandyoung said
What are the odds that each number will be equal to or less than the previous number?
@bigdogg saidYeah I wouldn't know how to approach such a question.
I never did find a formula for this. There probably is one, if you're 'mathy' enough.
I asked a 'mathy' person about this. Turns out there is a formula.
The sequence can be thought of as a combination of six digits and 9 markers. The markers reduce the digit value by 1.
For example, the number 998877 is shown as:
99-88-77-------
And the number 997777 is shown as:
99--7777-------
As Joe Shmo tried to teach us, before he got banned, the way of calculating the number of 6-digit possibilities spread out over 15 total positions is:
n! / [ k! * (n-k)! ]
15! / [6! * (15-6)! ]
= 15! / [6! * 9! ]
= 15*14*13*12*11*10 / 6*5*4*3*2*1
= 3603600 / 720
= 5005
5005 out of 1 million numbers is a probability of 0.5005%, which matches what I got from my program.
@bigdogg saidThanks for that.Just the sort of thing I like
I asked a 'mathy' person about this. Turns out there is a formula.
The sequence can be thought of as a combination of six digits and 9 markers. The markers reduce the digit value by 1.
For example, the number 998877 is shown as:
99-88-77-------
And the number 997777 is shown as:
99--7777-------
As Joe Shmo tried to teach us, before he got banned, the way of calc ...[text shortened]... 5005 out of 1 million numbers is a probability of 0.5005%, which matches what I got from my program.
I'll keep the formula for future reference.
I didn't know Joe was banned.
What was that all about?
@venda saidHe kept posting alt-right links in Debates after he was told to desist.
Thanks for that.Just the sort of thing I like
I'll keep the formula for future reference.
I didn't know Joe was banned.
What was that all about?
Hence the smattering of posts by "Removed" in this forum's history.