Standard Deviation Question
 

Let's say you expect to win a certain bet 60% of the time. You would lose 40% of the time. To find out your SD for 10 bets you would take the square root of (10 x .6 x .4)= 1.549

You would then multiply this times 3 for for 3 SDs = 4.67

Now to find your range of winners you would subtract 4.67 form 6 and add 4.67 to 6 getting 1.33 and 10.67.

Within 3 SDs you could expect to win this bet anywhere from 1.33 to 10.67 times out of ten.

My question is how can you get a number over 10? I'm taking this straight from a book. It works out OK for larger sample sizes, but when I tried it for smaller ones this is what happens. I guess I don't understand what is happening here. Did I do the calculations correctly?

Thanks for any explanation.

Re: Standard Deviation Question
 

That's because the approach you use assumes the game (all 10 trials at once) has a normal distribution (with an ev of 6 and an sd of 4.67) of winners where in fact it's a binomial distribution. Now the binomial distribution converges to a normal distribution for a large number of trials, but 10 just isn't large enough.
If this were game normal distributed it would be possible (although unlikely) to get values bigger than 10 or smaller than 0.

Re: Standard Deviation Question
 

http://www.stat.wvu.edu/SRS/Modules/...malapprox.html

Don't think the binomial converges to the normal.
The normal is just a good approximation of the binomial.

Re: Standard Deviation Question
 

But the distribution of sums of enough bernoulli-distributed random variables converges to a normal distribution due to the central limit theroem. And the sum of bernoulli-distributed random variables has a binomial distribution.

Re: Standard Deviation Question
 

[ QUOTE ]
But the distribution of sums of enough bernoulli-distributed random variables converges to a normal distribution due to the central limit theroem. And the sum of bernoulli-distributed random variables has a binomial distribution.

[/ QUOTE ]

wow...is this English, ...i feel so dumb...but truly i am impressed


(although it does sound sarcastic, and i don't mean it to)

Re: Standard Deviation Question
 

[ QUOTE ]
http://www.stat.wvu.edu/SRS/Modules/...malapprox.html

Don't think the binomial converges to the normal.
The normal is just a good approximation of the binomial.

[/ QUOTE ]
eh, doesn't everything converge to the normal?

Re: Standard Deviation Question
 

[ QUOTE ]
wow...is this English, ...i feel so dumb...but truly i am impressed

[/ QUOTE ]

i has no clue, i is german [img]/images/graemlins/smile.gif[/img]

Re: Standard Deviation Question
 

[ QUOTE ]
[ QUOTE ]
http://www.stat.wvu.edu/SRS/Modules/...malapprox.html

Don't think the binomial converges to the normal.
The normal is just a good approximation of the binomial.

[/ QUOTE ]
eh, doesn't everything converge to the normal?

[/ QUOTE ]

The Poisson is used for expectation of wins in tourneys if your name isn't Stu Unger.
Maybe for very large n it converges. Only your skill differential from the field doesn't remain constant over time.

Re: Standard Deviation Question
 

[ QUOTE ]
[ QUOTE ]
http://www.stat.wvu.edu/SRS/Modules/...malapprox.html

Don't think the binomial converges to the normal.
The normal is just a good approximation of the binomial.

[/ QUOTE ]
eh, doesn't everything converge to the normal?

[/ QUOTE ]

No.

Re: Standard Deviation Question
 

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
http://www.stat.wvu.edu/SRS/Modules/...malapprox.html

Don't think the binomial converges to the normal.
The normal is just a good approximation of the binomial.

[/ QUOTE ]
eh, doesn't everything converge to the normal?

[/ QUOTE ]

No.

[/ QUOTE ]
okay, what doesn't?