Variance question
 

Hey,

I'm studying about variance and understand it mathematically, but don't quite understand it practically.

Take the simple example of flipping a coin for $1. EV=0. Variance=1. Standard deviation=.33. Now, what exactly do the variance and standard deviation numbers refer to? In this case do they refer to money, percentage, something else?

Thanks.

Re: Variance question
 

If the variance/flip is 1 , then the standard deviation/flip is the square root of 1 . (you have 0.1)

So for instance , if we flip 1 coin , 100 times , then the standard deviation is sqrt(n*p*(1-p) where n=100 ,p=1/2 .

s.d = 5

There is a ~68% chance that we will flip 45 heads to 55 heads . (or tails) . This is about 1 standard deviation/100 flips from the mean (50) .

There is a ~ 95% chance that we will flip 40 heads to 60 heads which is ~ 2 standard deviations/100 flips from the mean .

There is a ~ 99% chance that we will flip 35 heads to 65 heads which is ~ 3 standard deviations/100 flips from the mean .

This tells us that if we repeatedly flip this coin , then the number of heads should be equal to the number of tails .For a fixed number of trials , this is not always true , so we have some deviation from the mean . The mean tells us that when n is sufficiently large , then the number of heads approaches n/2 for p=0.5 . For a finite number of trials , we use the formula :

s.d = sqrt(n*p*(1-p)) for a binomial distribution , or when we have only two possible outcomes .

Re: Variance question
 

[ QUOTE ]
what exactly do the variance and standard deviation numbers refer to? In this case do they refer to money, percentage, something else?


[/ QUOTE ]
Mathematical variance is not an easy quantity to understand directly, since the units are the square of the units of the random variable. When you flip a fair coin for $1, the variance is 1 square dollar. When you flip a fair coin for $1,000, the variance is 1,000,000 square dollars.

The standard deviation is the square root of the variance, and it has the same units as the random variable. The standard deviation is a measure of how spread out a distribution is about its mean. A standard deviation of $x means that in some sense, the distribution is about as spread out as a fair coinflip for $x. This doesn't help you to understand the coinflip itself, but it lets you understand other distributions in terms of coinflips.

Re: Variance question
 

[ QUOTE ]
If the variance/flip is 1 , then the standard deviation/flip is the square root of 1 . (you have 0.1)

So for instance , if we flip 1 coin , 100 times , then the standard deviation is sqrt(n*p*(1-p) where n=100 ,p=1/2 .

s.d = 5

There is a ~68% chance that we will flip 45 heads to 55 heads . (or tails) . This is about 1 standard deviation/100 flips from the mean (50) .

There is a ~ 95% chance that we will flip 40 heads to 60 heads which is ~ 2 standard deviations/100 flips from the mean .

There is a ~ 99% chance that we will flip 35 heads to 65 heads which is ~ 3 standard deviations/100 flips from the mean .

This tells us that if we repeatedly flip this coin , then the number of heads should be equal to the number of tails .For a fixed number of trials , this is not always true , so we have some deviation from the mean . The mean tells us that when n is sufficiently large , then the number of heads approaches n/2 for p=0.5 . For a finite number of trials , we use the formula :

s.d = sqrt(n*p*(1-p)) for a binomial distribution , or when we have only two possible outcomes .

[/ QUOTE ]

Thanks for the reply. Yeah, I fixed the .1 to .33. What you said makes sense, but how did you determine the percentages? (68, 95, 99.)

Re: Variance question
 

[ QUOTE ]
[ QUOTE ]
what exactly do the variance and standard deviation numbers refer to? In this case do they refer to money, percentage, something else?


[/ QUOTE ]
Mathematical variance is not an easy quantity to understand directly, since the units are the square of the units of the random variable. When you flip a fair coin for $1, the variance is 1 square dollar. When you flip a fair coin for $1,000, the variance is 1,000,000 square dollars.

The standard deviation is the square root of the variance, and it has the same units as the random variable. The standard deviation is a measure of how spread out a distribution is about its mean. A standard deviation of $x means that in some sense, the distribution is about as spread out as a fair coinflip for $x. This doesn't help you to understand the coinflip itself, but it lets you understand other distributions in terms of coinflips.

[/ QUOTE ]

Thanks, that helps. Can you elaborate on this, though:

[ QUOTE ]
A standard deviation of $x means that in some sense, the distribution is about as spread out as a fair coinflip for $x.

[/ QUOTE ]

Re: Variance question
 

Sean , if the variance per flip is 1 , then the standard deviation per flip is 1 (now you have 0.33)

Remember that s.d(x) = sqrt(var(x)) ; and sqrt(1) = 1 .

Re: Variance question
 

[ QUOTE ]
Sean , if the variance per flip is 1 , then the standard deviation per flip is 1 (now you have 0.33)

Remember that s.d(x) = sqrt(var(x)) ; and sqrt(1) = 1 .

[/ QUOTE ]

LOL, duh, yeah. Sorry, long day. Thanks a lot for your help, this finally makes sense.

Re: Variance question
 

[ QUOTE ]
how did you determine the percentages? (68, 95, 99.)

[/ QUOTE ]
You can't do this by hand. You have to use a z-score table or a program (like Excel). It's common to memorize that for a normal distibution 68% of falls within 1 sd, 95% within 2 sd, and 99.7% within 3 sd.

Also notice that your distribution is not normal, it's binomial. To come up with these results a normal approximation was used. A binomial distibution can sometimes by approximated as a normal distibution. You can approximate a Binomial(100,0.5) as a Normal(50,5) (a Binomial(100,0.5) has a mean of 50 and a SD of 5).

You cannot approximate a Binomial(1,0.5) as a Normal(0.5,0.5) even though a binomial(1,0.5) has a mean of 0.5 and a SD of 0.5.

Re: Variance question
 

Normal distibutions are quite common because of the Central Limit Theorem. Even if the results of your individual hands do not follow a normal distibution, the results of your 100 hand sessions will.

Re: Variance question
 

[ QUOTE ]
[ QUOTE ]
how did you determine the percentages? (68, 95, 99.)

[/ QUOTE ]
You can't do this by hand. You have to use a z-score table or a program (like Excel). It's common to memorize that for a normal distibution 68% of falls within 1 sd, 95% within 2 sd, and 99.7% within 3 sd.

Also notice that your distribution is not normal, it's binomial. To come up with these results a normal approximation was used. A binomial distibution can sometimes by approximated as a normal distibution. You can approximate a Binomial(100,0.5) as a Normal(50,5) (a Binomial(100,0.5) has a mean of 50 and a SD of 5).

You cannot approximate a Binomial(1,0.5) as a Normal(0.5,0.5) even though a binomial(1,0.5) has a mean of 0.5 and a SD of 0.5.

[/ QUOTE ]

I got this all sorted out. Thanks for your help!