I was wondering what 1 SD would be if you tossed a coin 100 times. Thanks to anybody who responds in advance.
Standard deviation = SQR{(N*p*(1-p)}
where p is the probability of 'success' and N represents the number of trials.
Suppose we toss a coin 100 times (N=100). The probability of heads (or tails) = 0.5. The standard deviation is SQR{100 * 0.5 * 0.5} = SQR(100 * .25) =SQR(25) = 5.
So, if we toss an unbiased coin 100 times then we would expect the number of heads to fall within 1 standard deviation approx' 2/3 of the time (68.2%).
In other words, 68.2% of the time we would expect to toss between 45 and 55 heads.
Hope this helps
'Switch'
If the coin cannot land on its side (i.e. no pushes), then the Variance for one toss is 1 � EV^2. For most coins, this will be close to 1. The SD is the square root of that.
For a unbiased coin, the SD is 1.
For a quick and dirty approximant, the SD is almost 1- EV^2/2. For a coin with a 10% edge, the SD is about 1-(10%)^2/2 = 0.995. It will be unusual for you to have a coin more biased than that.
If you have N coin tosses, then you multiply your SD by sqrt(N)
I remember a formula from a bj book ( CHambliss and Rojinsky, maybe) that said SD = sqrt ( # hands played) * bet size * 1.15. The 1.15 accounts for the extra cash due to splits, double downs and blackjacks.
This SD is quite a bit larger than the one you quote. Any comments?
