what is the variance of playing 3 box/hands simultanoeously, as well the SD ?
if var =1.26 / 1 hand
If n is the number of simultaneous spots, then:
V(n) = 1.26n + .50n(n-1).
So, for n = 3, V = 1.26(3) + .50(3)(2) = 6.78, and since s.d. is the square root of variance, s.d. = 2.60.
Don
the Var. of 3 hands seems too huge.
i don't have the exact figure but it suppose to be less than 3.78
if u bet 1 unit x 3 spot.
most likely between 2.5 - 3.0 by est.
don't u thk so?
I think you are falling for the common fallacy that variance per hand goes down as hands per round increases. Actually, variance per hand goes up.
The constants in the formula, 1.26 and 0.50, can vary a little depending on the rules, but it would be unusual for the total variance, to be less than 6.5, for three hands of one unit apiece.
ETF
i confused the concept of var. & SD.
sorry to Don,
pls ignore my previous post.
Maybe you are just asking about the math of playing 3 hands however it is not recommended to do so. Very seldom do you ever see a player taking 3 hands. Even high rollers will take a max of 2 hands. Taking 3 hands will make you stand out like a sore thumb. Too much attention. Don't do it.
1) Don is right, thks. look like simple but not so.
2) thks to Thet. as well, i don't want to raise a red flag on my head.
many mentors r here, fortunately.
thx in advance
if anyone can answer.
