Comments on: What are the odds that I'm such a loser?
http://ask.metafilter.com/38991/What-are-the-odds-that-Im-such-a-loser/
Comments on Ask MetaFilter post What are the odds that I'm such a loser?Fri, 26 May 2006 20:01:39 -0800Fri, 26 May 2006 20:01:39 -0800en-ushttp://blogs.law.harvard.edu/tech/rss60Question: What are the odds that I'm such a loser?
http://ask.metafilter.com/38991/What-are-the-odds-that-Im-such-a-loser
Probability filter: How do you determine the odds of being extremely unlucky? <br /><br /> I know I've seen this explained, but my google-fu fails me, as does my brute-force mathematical figuring. Here's my problem -- if there's a game in which you're supposed to win 12% of the time, what are the odds that after 150 plays you only win 3 times (i.e. 2% of the time)?post:ask.metafilter.com,2006:site.38991Fri, 26 May 2006 19:47:13 -0800TonyRobotsprobabilityoddsBy: neustile
http://ask.metafilter.com/38991/What-are-the-odds-that-Im-such-a-loser#602489
For the exact case (3 wins in 150 trials) you want the binomial distribution (or, M in N). <br>
<pre><br>
P(3 in 150) = ((150!)/((150-3)!*3!)) * 0.12^3 * (1 — 0.12)150 — 3 <br>
</pre><br>
<br>
I'll.. leave the math to you.comment:ask.metafilter.com,2006:site.38991-602489Fri, 26 May 2006 20:01:39 -0800neustileBy: MzB
http://ask.metafilter.com/38991/What-are-the-odds-that-Im-such-a-loser#602490
two words: <a href="http://en.wikipedia.org/wiki/Binomial_distribution">Binomial distribution</a><br>
p = 0.12<br>
n = 150<br>
x = 3comment:ask.metafilter.com,2006:site.38991-602490Fri, 26 May 2006 20:02:28 -0800MzBBy: neustile
http://ask.metafilter.com/38991/What-are-the-odds-that-Im-such-a-loser#602502
I screwed up the formatting. That last bit is (1-0.12)^(150-3). I tried computing it in bc and got 0.00000657, or 0.000657%comment:ask.metafilter.com,2006:site.38991-602502Fri, 26 May 2006 20:15:00 -0800neustileBy: iconjack
http://ask.metafilter.com/38991/What-are-the-odds-that-Im-such-a-loser#602616
The exact answer is<br>
<br>
129105822506510328235079035870803502997446159322335938209859903517778016592610656673611677829545693385926686940661856618755002678254299558255332801294973419393667922397147762715322224910896287172213604352<br>
<br>
over<br>
<br>
19636373861190906212383087819945102571900862049997798260446196468748421018886863425840391536149327809108734286260527514052671674441698875611230059296435853809012421857839447714155767243937589228153228759765625comment:ask.metafilter.com,2006:site.38991-602616Sat, 27 May 2006 04:29:38 -0800iconjackBy: klangklangston
http://ask.metafilter.com/38991/What-are-the-odds-that-Im-such-a-loser#602654
Ugh. Neither of those appears to be repeating...<br>
<br>
Google says the answer is 6.57483013 × 10-6.comment:ask.metafilter.com,2006:site.38991-602654Sat, 27 May 2006 06:47:52 -0800klangklangstonBy: TonyRobots
http://ask.metafilter.com/38991/What-are-the-odds-that-Im-such-a-loser#602978
Thanks guys. I've also found this binomial distribution calculator, which may be helpful to anyone else interested in this question:<br>
<br>
<a href="http://people.hofstra.edu/faculty/stefan_waner/RealWorld/stats/bernoulli.html">Binomial Disribution Utility (for Bernoulli Trials)</a>comment:ask.metafilter.com,2006:site.38991-602978Sat, 27 May 2006 18:02:01 -0800TonyRobotsBy: iconjack
http://ask.metafilter.com/38991/What-are-the-odds-that-Im-such-a-loser#602997
I just thought it would amusing to give the <b>exact</b> answer, which I did in my post above. The approximate answer is 6.57483013 × 10^-6.comment:ask.metafilter.com,2006:site.38991-602997Sat, 27 May 2006 18:27:56 -0800iconjack