Silicon Investor (SI) -- The First Internet Community

STOCKTALK

We've detected that you're using an ad content blocking browser plug-in or feature. Ads provide a critical source of revenue to the continued operation of Silicon Investor. We ask that you disable ad blocking while on Silicon Investor in the best interests of our community. If you are not using an ad blocker but are still receiving this message, make sure your browser's tracking protection is set to the 'standard' level.

Politics : Formerly About Advanced Micro Devices -- Ignore unavailable to you. Want to Upgrade?

To: Ali Chen who wrote (128125)	11/10/2000 7:40:47 PM
From: andreas_wonisch	Read Replies (1) \| Respond to of 1583733

Ali, Re: With 20 attempts I expect to have 10 heads with probability of 1-0.5/sqrt(20), or 94%. Sorry, but that's just not right. Let's look at an easy example. We have four flips and want to get the probability of having at least ten heads. Let's say h is head and t is tail. Then we have the following possibilities for head: 1. hhhh 2. thhh 3. hthh 4. hhth 5. hhht 6. tthh 7. thth 8. thht 9. htth 10. htht 11. hhtt We have 2^4 = 16 possibilities overall. Since all outcomes are equally probable we have p = 11/16 = 0.6875. This is exactly in line with my solution: p(n=2) = (4 over 2) * 0.5^2 * 0.5^2 = 0.375 p(n=3) = (4 over 3) * 0.5^2 * 0.5^2 = 0.25 p(n=4) = (4 over 4) * 0.5^2 * 0.5^2 = 0.0625 => p(n>=2) = p(n=2) + p(n=3) + p(n=4) = 0.6875 Your formula gives: 1-0.5/sqrt(4) = 0.75 This can't be right. Let's look at an even simpler example: You flip only two times. With your formula the outcome for one or more heads would be 1-0.5/sqrt(2) = 0.646 when in fact it's 0.75 (three out of four possibilities). It's very evident that we have a binomial distribution here, just look in any book about stochastics. For great n (2000) we can approximate this distribution with the gaussian distribution. Here in Germany you can find a picture of it on every 10 DM note. ;) Andreas