Comments on: Calculating Probability Over Several Attempts?
http://ask.metafilter.com/133006/Calculating-Probability-Over-Several-Attempts/
Comments on Ask MetaFilter post Calculating Probability Over Several Attempts?Wed, 16 Sep 2009 09:46:35 -0800Wed, 16 Sep 2009 09:46:35 -0800en-ushttp://blogs.law.harvard.edu/tech/rss60Question: Calculating Probability Over Several Attempts?
http://ask.metafilter.com/133006/Calculating-Probability-Over-Several-Attempts
How do you multiply probability across multiple chances? Let's say that every time you roll a die, you have a one-in-six chance of having five come up. What math would you perform to come up with the probability of five coming up at some point with the dice being rolled two times? Three? Five? Ten? (I'm using dice as a shorthand here: the actual probability figure I'm working with is 24%.)post:ask.metafilter.com,2009:site.133006Wed, 16 Sep 2009 09:39:58 -0800WCityMikeprobabilitymathematicsmultipleresolvedBy: unixrat
http://ask.metafilter.com/133006/Calculating-Probability-Over-Several-Attempts#1900116
It's probability of it happening times the rolling percentage of times it hasn't happened.<br>
<br>
That's a terrible description, sorry. Let me clarify.<br>
<br>
If there's a 50% chance of rain for today and tomorrow, the overall chance that it'll rain once is 75%.<br>
<br>
100 outcomes (original) * 50% (today) + (50 dry outcomes * 50% (tomorrow)) = 75%<br>
<br>
For three days, it's 87.5%:<br>
<br>
100 * 50% + 50 * 50% + 25 * 50% = 87.5comment:ask.metafilter.com,2009:site.133006-1900116Wed, 16 Sep 2009 09:46:35 -0800unixratBy: Obscure Reference
http://ask.metafilter.com/133006/Calculating-Probability-Over-Several-Attempts#1900118
This sounds like a <a href="http://en.wikipedia.org/wiki/Binomial_distribution">binomial distribution</a>, Something either "comes up" or does not, with a fixed probability in repeated trials.comment:ask.metafilter.com,2009:site.133006-1900118Wed, 16 Sep 2009 09:47:15 -0800Obscure ReferenceBy: kmz
http://ask.metafilter.com/133006/Calculating-Probability-Over-Several-Attempts#1900121
If you want the change that at least one five will be rolled at <i>some</i> point, what you want is the inverse of the change that no five ever is ever rolled. For n rolls, that probability is (5/6) to the nth power. Subtract that from 1 and you have the original probability you want.comment:ask.metafilter.com,2009:site.133006-1900121Wed, 16 Sep 2009 09:47:39 -0800kmzBy: unixrat
http://ask.metafilter.com/133006/Calculating-Probability-Over-Several-Attempts#1900123
Argh, and of course that's "it'll rain *at least* once". Things get complicated quickly if we're not specific enough. The probability that it'll rain is still 50%.comment:ask.metafilter.com,2009:site.133006-1900123Wed, 16 Sep 2009 09:48:44 -0800unixratBy: kmz
http://ask.metafilter.com/133006/Calculating-Probability-Over-Several-Attempts#1900124
Gah, don't know why I kept typing "change" instead of "chance".comment:ask.metafilter.com,2009:site.133006-1900124Wed, 16 Sep 2009 09:48:46 -0800kmzBy: FishBike
http://ask.metafilter.com/133006/Calculating-Probability-Over-Several-Attempts#1900126
In the example you gave, I find it much easier to start by calculating the probability of NOT rolling a 5 across multilple throws, because these probabilities can be just multiplied together. For example:<br>
<br>
1 roll: 5/6 (83.333%) probability of NOT rolling a 5<br>
2 rolls: (5/6) x (5/6) (69.444%) probability of NOT rolling a 5<br>
3 rolls: (5/6) x (5/6) x (5/7) (57.870%) probability of NOT rolling a 5<br>
<br>
To find the probability of rolling a 5, just subtract the percentage of not rolling it from 100%, e.g. for 3 rolls, 100% - 57.870% = 42.13% probability you'll roll a 5 in at least 1 of those 3 throws.comment:ask.metafilter.com,2009:site.133006-1900126Wed, 16 Sep 2009 09:49:50 -0800FishBikeBy: Orange Pamplemousse
http://ask.metafilter.com/133006/Calculating-Probability-Over-Several-Attempts#1900130
You could work from the reverse. Ie. what is the chance that there are NO fives in 2 dice rolls? 5/6 * 5/6 = 25/36. The chance of a 5 is therefore 1 - 25/36 = 1/36.<br>
<br>
You could also just sum up each individual option. 1/6 * 5/6 (1st roll is 5, second roll isn't) + 5/6 *1/6 (the reverse) + 1/6*1/6 (2 fives).<br>
<br>
The first method is probably easiest. 1 - (5/6)^n, where n is the number of rolls. On preview, what fishbike said.comment:ask.metafilter.com,2009:site.133006-1900130Wed, 16 Sep 2009 09:51:28 -0800Orange PamplemousseBy: FishBike
http://ask.metafilter.com/133006/Calculating-Probability-Over-Several-Attempts#1900135
<i>On preview, what fishbike said.</i><br>
<br>
... although kmz said it even earlier. And that last (5/7) of mine is a typo, that should of course be (5/6) just like all the others.comment:ask.metafilter.com,2009:site.133006-1900135Wed, 16 Sep 2009 09:53:27 -0800FishBikeBy: WCityMike
http://ask.metafilter.com/133006/Calculating-Probability-Over-Several-Attempts#1900141
So, with my actual probability figure, where the figure is 24% -- I'd essentially do this?<br>
<br>
1 - (76/100)^(er, let's say 5)<br>
1 - 0.253553<br>
<br>
Chance of the event occurring once in five runs: 74.6%<br>
<br>
Yes?comment:ask.metafilter.com,2009:site.133006-1900141Wed, 16 Sep 2009 09:55:34 -0800WCityMikeBy: kmz
http://ask.metafilter.com/133006/Calculating-Probability-Over-Several-Attempts#1900150
At least once, yes.comment:ask.metafilter.com,2009:site.133006-1900150Wed, 16 Sep 2009 10:00:28 -0800kmzBy: justkevin
http://ask.metafilter.com/133006/Calculating-Probability-Over-Several-Attempts#1900156
More correctly, "at least once in five runs", but yes.comment:ask.metafilter.com,2009:site.133006-1900156Wed, 16 Sep 2009 10:02:57 -0800justkevinBy: jeather
http://ask.metafilter.com/133006/Calculating-Probability-Over-Several-Attempts#1900165
No, that's actually the chance of it occurring at least once in five runs, not exactly once. To find exactly one (or exactly two, or at least 3, or whatever) in five (thirty, etc) trials, you need to use the binomial distribution.comment:ask.metafilter.com,2009:site.133006-1900165Wed, 16 Sep 2009 10:08:10 -0800jeatherBy: ROU_Xenophobe
http://ask.metafilter.com/133006/Calculating-Probability-Over-Several-Attempts#1900173
So you know, you can also do this for "What's the probability of five coming up exactly three times in five rolls?" or "What's the probability of five coming up at least twice?" and so on.<br>
<br>
The reference distribution is indeed the binomial, which applies when you have:<br>
<br>
(1) A series of <i>k</i> trials, each of which is totally independent of the others. Repeated rolls of fair dice or repeated flips of a fair coin are good examples. Not to nitpick, but unixrat's rain example is less good -- right now the probability of rain Saturday and Sunday might be 0.5 each, but whether or not it rains Sunday will be influenced by whether or not it actually rains Saturday.<br>
<br>
(2) Some number of interesting things you want to count.<br>
<br>
(3) That interesting thing happens with probability π in each trial, and that probability never changes.<br>
<br>
There are lots of combinations of rolls such that five comes up three times in five rolls. What the binomial distribution's ugly formula does is multiply the probability of any one of those different combinations by the number of different combinations that provide that number of "hits."<br>
<br>
The easy method mentioned here for calculating the probability of at least one "hit" works because there is one and only one way to not get at least one five, which is that five never comes up. So you figure out the probability of that branch and multiply it by one.<br>
<br>
Using your second example, the probability of getting one "hit" -- no more no less -- in five trials is 0.4003.comment:ask.metafilter.com,2009:site.133006-1900173Wed, 16 Sep 2009 10:12:23 -0800ROU_XenophobeBy: Justinian
http://ask.metafilter.com/133006/Calculating-Probability-Over-Several-Attempts#1900186
Yes, as others have said, if you want the probability of it happening <i>at least</i> once it is trivial and straightforward. If you want the probability of it happening <i>exactly</i> once, or twice, or three times, or whatever it is a little more complex.comment:ask.metafilter.com,2009:site.133006-1900186Wed, 16 Sep 2009 10:21:28 -0800JustinianBy: WCityMike
http://ask.metafilter.com/133006/Calculating-Probability-Over-Several-Attempts#1900200
Thanks, guys ... all I need is actually just the "at least once" calculation, so you guys gave me my answer. I am much obliged. :)comment:ask.metafilter.com,2009:site.133006-1900200Wed, 16 Sep 2009 10:34:20 -0800WCityMike