A question about Standard deviation
 

I looked at my 6-max SD, it is about 17BB/100 hands. Other posters report similar numbers (I saw 15-20BB/100)

Now what if I calculate SD/2 hands. The maximum I can win in a hand is 60BB, the maximum I can lose in a hand is 12BB.
So, a maximum SD in 2 hands would be 72BB/2 hands, right?
Now, to calculate maximum SD/100 hands, I would take max SD/2 hands and divide it by square root of 50, right?
If this is right, I come up with a maximum SD of less than 11BB/100. But it is less than my actual SD of 17BB/100. How can this be? Where did I make an error?

Re: A question about Standard deviation
 

(SD/2 hands)*sqrt(50)=SD/100 hands

so your SD/2 hands would be about 2.4, I believe. SD/hand would be 1.7BB.

BTW, I don't think my units are actually correct, but I don't remember how you correctly express the units of SD. Besides, you all know what I mean.