SD = % Randomness?

Isn't SD just a measurement of how accurate the closeness of true randomness was achieved?

Would or could any player win against true randomness?

jomoats

If cards where perfectly distributed through the deck every time, then there would never be any SD. The expected result would be the same each time. Because they aren't, this must be the reason for the SD. Even polls recognize their results aren't perfectly accurate etc.

jomoats

The casino would run out of chips very quickly.

SD is not % randomness. It is a number we look at that has useful properties.

Pay special attention to the verbiage leading up to the "Probability of Loss Table" at the end of the article. If this makes sense to me, it should make sense to anyone.

Jay Cobbson

If cards were perfectly distributed they wouldn't be random at all. And if they are not random they are predictable.And as Math Boy says the Casinos would bust very quickly.

Francis Salmon

The standard deviation is a measure of dispersion, more precisely a measure of how much the data differ from the mean. Suppose that the mean height and weight of adult males in the US are 5'10" (70 inches) and 170 lbs., respectively. No man is 50% taller than the average, i.e. no man is taller than 105 inches, which is 3 inches "shorter" than 9 feet. Many men are more than 50% heaver than the average, i.e. heavier than 255 lbs. Quite a few are heavier than 300 lbs. Thus, weights vary from the average more than heights at the right tail. Therefore, the standard deviation of weights is higher, relative to the mean, than the standard deviation of heights. Heights are more symmetrically distributed about the mean.

Suppose that the most common suit size of adult males in the US is 42. Suit size corresponds to chest girth. It is likely then that the mean weight of men who wear a size 42 suit is nearly 170 lbs. It is most likely less than the mean weight of all adult men because there are many more 240+ pounders, who will not fit into a size 42, than 100- pounders, who will also not fit into a size 42.

Suppose that the the standard deviation of weights of adult males in the US is 15 lbs. It is likely that the standard deviation of weights of men who wear a size 42 suit is less than 15 lbs, say 10 lbs or as little as 5 lbs. This is because very light men and very heavy men will not fit into a size 42 suit and the weights of men who wear a size 42 suit are less dispersed about 170 lbs. than the weights of all American adult males, which includes jockeys and feather weight boxers as well as linebackers and basketball players.

Sounds like I am being told to get into the clothing business.

It might make more sense if you say that ONE standard deviation of weights of men is 15 lbs. This would mean that roughly 68% of men weigh within 15 lbs of 170. Thirty-four percent of men would weigh between 155 and 170 (ONE standard deviation below) and another 34% between 170 and 185 (ONE standard deviation above). Assuming a normal curve, 95% would weigh within 2 standard deviations and 99% within 3 standard deviations.

Say your expected WinRate is $50/hr and the hourly standard deviation is $700. After an hour of play you will win between -$650 and +$750 68% of the time. In one hour you will lose more than $1350 (2 std dev) 2.5% of the time.