Comments on: Probability calculations
http://ask.metafilter.com/10079/Probability-calculations/
Comments on Ask MetaFilter post Probability calculationsFri, 10 Sep 2004 17:53:06 -0800Fri, 10 Sep 2004 17:53:06 -0800en-ushttp://blogs.law.harvard.edu/tech/rss60Question: Probability calculations
http://ask.metafilter.com/10079/Probability-calculations
If the likelihood of event A occurring is 15%, and the likelihood of independent event B occuring is 5%, what is the likelihood of EITHER event occuring? <br /><br /> [StatisticsFilter...]So, I know the answer isn't 20%, and I'm pretty sure it's not 15%. I'm sure there's a formula that could be applied, but I was an english major.<br>
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I reasoned as far as thinking of two coin tosses - the likelihood of EITHER coin toss yeilding HEADS is greater than 50% but less than 100%. But that's where I stopped.<br>
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Incidentally, this came up from reading the normally stats-savvy <a href=http://p086.ezboard.com/fsonsofsamhornbostonredsox.showMessageRange?topicID=14026.topic&start=1&stop=20>Sons of Sam Horn</a> discussion board. In it someone opines that there's a 5% chance of Ichiro hitting .400 this year and a 15% chance of Bonds hitting .400 this year. The author then makes an aside of there being a "1 in 5" chance of there being a .400 hitter this year, which I think is incorrect.<br>
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Unfortunatly, much like Metafilter, as far as I can tell SOSH is a closed forum at this point, so I can't ask directly. Anyone wanna get me in?post:ask.metafilter.com,2004:site.10079Fri, 10 Sep 2004 17:41:53 -0800sohcahtoamathprobabilityBy: shepd
http://ask.metafilter.com/10079/Probability-calculations#184396
How do you know it's not 20%? Seems that mathematically, all possibilities should add up to 100%, so it makes sense that you would just add the possibilities. It works the same in lotteries. Having a 1 in a million chance and getting 2 tickets doesn't mean you get twice the chance of winning (1/2), it means you have 2 / 1,000,000 chances of winning.<br>
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In the case you have, though, your problem is compounded. Your looking at chances OF a chance, not chances WITHIN a chance.<br>
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x = 15 / 100 for possibility z<br>
y = 5 / 100 for possibility z<br>
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z = 1 / 5 for possibility w<br>
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w = 100 / 100 for possibility w<br>
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v = x + y<br>
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soo, I get:<br>
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z = (1 / 5) * (100 / 100) = 1 / 5<br>
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then:<br>
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x = (15 / 100) * (1 / 5) = 3 / 100<br>
y = (5 / 100) * (1 / 5) = 1 / 100<br>
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and:<br>
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u = 3/100 + 1/100 = 4/100<br>
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Or, I get 4% that of the 0.400 batting averages either event will occurr.<br>
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Anyone want to tell me where I went wrong? I barely passed stats.comment:ask.metafilter.com,2004:site.10079-184396Fri, 10 Sep 2004 17:53:06 -0800shepdBy: ROU_Xenophobe
http://ask.metafilter.com/10079/Probability-calculations#184397
For two independent events A and B, the probablity of (A or B) is pr(A) + pr(B) - pr(A and B). It's the last part that keeps you from nonsense like there being a 150% chance of getting at least one heads in three coin-tosses -- you get that by counting the same probability mass two (or more) times.<br>
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So it's 0.15 + 0.05 - (0.15*0.05). Which seems to be 0.1925 if one can trust google calculator.comment:ask.metafilter.com,2004:site.10079-184397Fri, 10 Sep 2004 17:54:18 -0800ROU_XenophobeBy: LimePi
http://ask.metafilter.com/10079/Probability-calculations#184398
<a href="http://www.ruf.rice.edu/~lane/hyperstat/A136602.html">p(A or B)</a> = p(A) + p(B) - <a href="http://www.ruf.rice.edu/~lane/hyperstat/A127969.html">p(A and B).</a><br>
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soo... p(a or b) = <a href="http://www.google.com/search?q=.15+%2B+.05+-+(.05+*+.15)&ie=UTF-8&oe=UTF-8">.15 + .05 - (.05 * .15)</a><br>
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on preview: ROU beat me to it.comment:ask.metafilter.com,2004:site.10079-184398Fri, 10 Sep 2004 17:56:42 -0800LimePiBy: sohcahtoa
http://ask.metafilter.com/10079/Probability-calculations#184402
ROU and LimePi - Thanks!comment:ask.metafilter.com,2004:site.10079-184402Fri, 10 Sep 2004 18:01:52 -0800sohcahtoaBy: mosch
http://ask.metafilter.com/10079/Probability-calculations#184404
0.15 = chance of event A happening<br>
0.05 = chance of event B happening<br>
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0.15 * ( 1 - 0.05 ) = chance of only event A happening<br>
0.05 * (1 - 0.15) = chance of only event B happening<br>
0.15 * 0.05 = chance of both event A and event B happening<br>
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(A && !B) + (B && !A) + (A && B) =<br>
(0.15 * (1 - 0.05)) + (0.05 * (1 - 0.15)) + (0.15 * 0.05) = <br>
.1425 + .0425 + .0075 = 0.1925<br>
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19.25% chance of one or more events happening.<br>
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And on preview, everybody beat me to it.comment:ask.metafilter.com,2004:site.10079-184404Fri, 10 Sep 2004 18:04:28 -0800moschBy: lbergstr
http://ask.metafilter.com/10079/Probability-calculations#184405
The lottery ticket analogy only works if they're two tickets to different lotteries (otherwise they're not independent events.)comment:ask.metafilter.com,2004:site.10079-184405Fri, 10 Sep 2004 18:09:15 -0800lbergstrBy: shepd
http://ask.metafilter.com/10079/Probability-calculations#184407
YAY. I lose.comment:ask.metafilter.com,2004:site.10079-184407Fri, 10 Sep 2004 18:10:32 -0800shepdBy: LairBob
http://ask.metafilter.com/10079/Probability-calculations#184411
I just want to point out that "sohcahtoa" is a pretty math-oriented nickname to pick, for an english major.comment:ask.metafilter.com,2004:site.10079-184411Fri, 10 Sep 2004 18:17:49 -0800LairBobBy: sohcahtoa
http://ask.metafilter.com/10079/Probability-calculations#184412
Yes. I didn't miss that irony (or that of the misspellings in my above, for that matter).<br>
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It was a remnant from my rock-and-roll days.comment:ask.metafilter.com,2004:site.10079-184412Fri, 10 Sep 2004 18:24:03 -0800sohcahtoaBy: sfenders
http://ask.metafilter.com/10079/Probability-calculations#184428
huh. My instinct is to calculate this by figuring "p(A or B) = 1 - p(not A) * p(not B)". <br>
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So, 1-(.85 * .95) = 0.1925. <br>
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Yet another way of getting the same answer. I figure it's worth mentioning since it might sometimes be easier to estimate. Also note that the smaller the probabilities of each event, the closer p(A)+p(B) is to the right answer. So 1 in 5 isn't far off in this case.comment:ask.metafilter.com,2004:site.10079-184428Fri, 10 Sep 2004 19:36:48 -0800sfendersBy: Kwantsar
http://ask.metafilter.com/10079/Probability-calculations#184434
FWIW, you ought to check out <a href="http://www.baseballthinkfactory.org/files/primer/therapy/">Sox Therapy</a>. SoSH suxks.comment:ask.metafilter.com,2004:site.10079-184434Fri, 10 Sep 2004 20:09:15 -0800KwantsarBy: ikkyu2
http://ask.metafilter.com/10079/Probability-calculations#184560
To be extremely pedantic: the original questioner used the word 'likelihood' where 'probability' would be the correct term. 'Likelihood' is a term used to refer to conditional probabilities (non-independent events).<br>
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<a href="http://plato.stanford.edu/entries/bayes-theorem/">A discussion of conditional probabilities and Bayes' Theorem</a> (extra credit).comment:ask.metafilter.com,2004:site.10079-184560Sat, 11 Sep 2004 11:09:26 -0800ikkyu2