coins

[Pjer d'Brt]

we throw 1000000 coins. what is the probability that one side has 0.1% advantage over another in that sample.

[pzhon]

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we throw 1000000 coins. what is the probability that one side has 0.1% advantage over another in that sample.

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The standard deviation of the count for one fair toss is 1/2. For n coin tosses it is squareroot(n)/2, so 1000/2=500 here. If the coin is fair, then 0.1%+ = 1000 tosses off is 2+ standard deviations away from the mean, which happens about 4.6% of the time (2.3% in each direction).

[BruceZ]

"One side has a 0.1% advantage over the other" means that one side comes up 1000 times more than the other, so one side comes up 500 more than the mean, while the other side comes up 500 less than the mean. This is +/-1 standard deviation, not +/-2, so the advantage will be that large or larger 32% of the time. In gambling, an advantage or "edge" of x% always means winning 50 + x/2% of the time while the opponent wins 50 - x/2%. Edge multiplied by dollars bet equals expected value EV.

This answer assumes that the coins are fair, but we learned awhile ago that coins in fact are not fair (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=562655&page=&view=&sb =5&o=), but instead they will come up the same way they started 51% of the time. We can get around this by starting exactly half the coins with heads up and half the coins with tails up. If we start them all with the same side up, then that side will essentially always have an advantage of more than 0.1% since that would be 20 standard deviations below the average advantage of 2%.

[pzhon]

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In gambling, an advantage or "edge" of x% always means winning 50 + x/2% of the time while the opponent wins 50 - x/2%. Edge multiplied by dollars bet equals expected value EV.

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Thanks for the correction.

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This answer assumes that the coins are fair, but we learned awhile ago that coins in fact are not fair (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=562655&page=&view=&sb =5&o=), but instead they will come up the same way they started 51% of the time.

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If the coin-tosses are in a sequence that is very weakly dependent like this, with each toss matching the previous 51% of the time, how much of an effect does this have on the probability that one side is ahead by a certain number of coin-tosses? I think this effect will be small.

Decompose the sequence of results into alternating streaks of heads and tails. Instead of stopping at 1,000,000 tosses, you could stop after the number of streaks that would have total length 1,000,000 on average. This is easier to analyze.

As a check, let's do this for independent tosses. The average length of a streak should be 2, with a variance of 2. The sum of the lengths of the first 250,000 streaks of heads has a variance of 500,000. The difference between the lengths of the first 250,000 streaks of heads and the first 250,000 streaks of tails has a variance of 1,000,000, so a standard deviation of 1000. This agrees with the earlier estimate that the standard deviation of the advantage after 1,000,000 tosses is 0.1%.

If each toss has a 51% chance of matching the previous, then the average length of a streak is 100/49, so we should be considering 245,000 streaks of each. The variance of the length of a streak is 5100/2401 ~ 2.124. The variance of the first 245,000 streaks of each is about 1,040,816, for a standard deviation of about 1,020. So, 1000 tosses is .98 standard deviations.

If each toss agrees with the next 51% of the time, it is slightly easier for one side to end up ahead than with ideal independent coin-tosses. Instead of a 31.73% chance to be 1000 tosses up, it looks like the probability should be roughly 32.70%.