March 11th, 2021 at 7:18:46 AM
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I can't locate it just now but I remember blackjack hero Arnold Snyder giving a very simple method of calculating Standard Deviation

from googling I found out that trying to calculate it can be quite complicated

I would like to know if this is correct - what I remember

if you flip a coin 100 times and it ends up 60 heads and 40 tails then:

it varied from 50/ 50 by 20

to get the standard deviation of this series of 100 flips you get the square root of 20 - which is 4.47

you then divide that by 2 to get 2.235.............................and that is the standard deviation of the results of your 100 coin flip series if the results are 60/40 - if not exactly - a very good approximation

is this correct?

*

I can't locate it just now but I remember blackjack hero Arnold Snyder giving a very simple method of calculating Standard Deviation

from googling I found out that trying to calculate it can be quite complicated

I would like to know if this is correct - what I remember

if you flip a coin 100 times and it ends up 60 heads and 40 tails then:

it varied from 50/ 50 by 20

to get the standard deviation of this series of 100 flips you get the square root of 20 - which is 4.47

you then divide that by 2 to get 2.235.............................and that is the standard deviation of the results of your 100 coin flip series if the results are 60/40 - if not exactly - a very good approximation

is this correct?

*

Please don't feed the trolls

March 11th, 2021 at 9:21:04 AM
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Quote:lilredrooster................................

I can't locate it just now but I remember blackjack hero Arnold Snyder giving a very simple method of calculating Standard Deviation

from googling I found out that trying to calculate it can be quite complicated

I would like to know if this is correct - what I remember

if you flip a coin 100 times and it ends up 60 heads and 40 tails then:

it varied from 50/ 50 by 20

to get the standard deviation of this series of 100 flips you get the square root of 20 - which is 4.47

you then divide that by 2 to get 2.235.............................and that is the standard deviation of the results of your 100 coin flip series if the results are 60/40 - if not exactly - a very good approximation

is this correct?

*

The standard deviation for the proportion of heads in one hundred fair-coin spins is: 0.5 / sqrt(100), which is 0.05.

The mean, or average, number of heads for one spin is 0.5. And 0.5 is the standard deviation for one spin because 0 heads is 0.5 head below the mean, and 1 head is 0.5 head above the mean.

In your example of 60 heads and 40 tails, 60/100 = 0.6, which is 0.1 above the mean. This corresponds to 0.1 / 0.05 = 2 standard deviations above the mean.

If we approximate 100 spins with the “normal distribution,” we would figure the probability of having at least 60% heads, or 2 SDs above the mean, is about 2.5%.

So, be sure to use 0.5 (not 2) for the standard deviation for one spin, and to use the square root of the number of observations (not the excess heads.)

March 11th, 2021 at 10:19:40 AM
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.............................

thanks ChesterDog

thanks ChesterDog

Please don't feed the trolls

March 11th, 2021 at 1:13:34 PM
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The average is np and the standard deviation = SQRT(npq). For a coin flip p=q=1/2.

So 60 heads from 100 tosses is 2 SDs from the norm. Av = 100*(1/2) = 50. SD = SQRT ( 100 * 1/2 * 1/2 ) = 5.

https://ned.ipac.caltech.edu/level5/Berg/Berg3.html

So 60 heads from 100 tosses is 2 SDs from the norm. Av = 100*(1/2) = 50. SD = SQRT ( 100 * 1/2 * 1/2 ) = 5.

https://ned.ipac.caltech.edu/level5/Berg/Berg3.html