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The Goober
12-10-2004, 05:00 AM
Suppose you play a game where you bet for even money on a fair coin toss. Say you play at a rate of 100 bets an hour. Obviously your EV is exactly 0 no matter how long you play, but what's your variance?

cepstrum
12-10-2004, 09:59 AM
each coin flip is a bernoulli trial with p=0.5, assuming a fair coin. repeated bernoulli trials are modeled with the binomial distribution. the variance of the binomial distribution over n trials is np(1-p). in this case, your variance is 100 * 0.5 * 0.5 = 25 bets per hour.

The Goober
12-10-2004, 03:47 PM
[ QUOTE ]
each coin flip is a bernoulli trial with p=0.5, assuming a fair coin. repeated bernoulli trials are modeled with the binomial distribution. the variance of the binomial distribution over n trials is np(1-p). in this case, your variance is 100 * 0.5 * 0.5 = 25 bets per hour.

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I'm a little confused now. Using your formula, it looks like if I was to play twice as many flips per hour, then my variance would be 200 * 0.5 * 0.5 = 50 bets per hour. Shouldn't increasing the number of flips actually lower my variance? If I flipped the coin a billion times an hour, shouldn't my actual result be very, very close to the expected result and my variance be really low?

topspin
12-10-2004, 04:35 PM
[ QUOTE ]
Shouldn't increasing the number of flips actually lower my variance? If I flipped the coin a billion times an hour, shouldn't my actual result be very, very close to the expected result and my variance be really low?

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Nope. Total variance increases linearly since your trials are independent. Variance for two coin flips is twice the variance for a single coin flip.

You're probably thinking of standard deviation per coin flip, which does decrease with the number of coin flips: sqrt(p*(1-p))/sqrt(n). One way to view this is that the percentage deviation from your total expected earn decreases as the number of flips goes up.

jason1990
12-11-2004, 12:58 PM
The net change in bankroll after each flip is not a Bernoulli random variable. Bernoulli's take the values 0 and 1. This variable takes the values -1 and 1. So the variance of the outcome of each flip is 1, and the variance after n flips is n.

Paul2432
12-12-2004, 01:36 PM
[ QUOTE ]
[ QUOTE ]
each coin flip is a bernoulli trial with p=0.5, assuming a fair coin. repeated bernoulli trials are modeled with the binomial distribution. the variance of the binomial distribution over n trials is np(1-p). in this case, your variance is 100 * 0.5 * 0.5 = 25 bets per hour.

[/ QUOTE ]

I'm a little confused now. Using your formula, it looks like if I was to play twice as many flips per hour, then my variance would be 200 * 0.5 * 0.5 = 50 bets per hour. Shouldn't increasing the number of flips actually lower my variance? If I flipped the coin a billion times an hour, shouldn't my actual result be very, very close to the expected result and my variance be really low?

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I think you are confused about the definition of variance. Suppose you expressed your results as a win-rate. For a fair coin your expected result is zero. The more coins you flip the closer this win-rate is likely to be to zero. I think this is what you are thinking of when you say variance gets smaller with large trials.

However, variance refers to the absolute value of the difference. This amount gets increasingly larger with the number of trials. For example, after ten flips if you have 6 wins and 4 losses your win rate will be 0.2/flip but you are only ahead by two units. After 1000 flops you might be ahead by 30 units, but your win rate has dropped to 0.03.

Paul

blank frank
12-12-2004, 07:30 PM
[ QUOTE ]

However, variance refers to the absolute value of the difference. This amount gets increasingly larger with the number of trials. For example, after ten flips if you have 6 wins and 4 losses your win rate will be 0.2/flip but you are only ahead by two units. After 1000 flops you might be ahead by 30 units, but your win rate has dropped to 0.03.

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No, variance has to do with the square of the difference, not the absolute value. If you want something in the same units as the thing you're observing, you want to take the square root of the variance and get the standard deviation. Even then it's not really related to the absolute difference.

So say you flip a coin 100 times, you'll have an expected value of 50 with a variance of 25 (100*.5*.5). But your standard deviation is 5. About 70% of the time the actual result will be within one standard deviation of the mean, and about 95% of the time it will be within two.

Now at 10,000 trials you'll have a variance of 2500, for a standard deviation of 50. While this is bigger, the coefficient of variance (measuring a percentage with standard deviation divided by the mean) is 0.5%, compared to 5% for 100 trials.

cepstrum
12-13-2004, 12:43 PM
Correct; I didn't follow through by noting that a 25-flip variance in trial successes implies a \$100 variance in bankroll.

You can still use the bernoulli/binomial principle, however. A bernoulli variable is just an indicator variable; it can mean anything you want. So the bankroll change for a single toss is not itself a bernoulli random variable, but it is a function of a bernoulli random variable; likewise the bankroll chance for a sequence of tosses is a function of a binomial random variable.

cepstrum