Ask MetaFilter questions tagged with probability

Conditional Probability

chrisamiller — Wed, 16 Dec 2009 17:25:54 -0800

Stats-filter: Given a binary matrix, if I know the total number of ones in a given row and a given column, can I calculate the probability that a given position contains a one? I have a binary matrix, like so, where every value is either 1 or 0. So, if the first column contains 2 ones, and the first row contains 1 one, what's the probability that position A contains a one?

Example:



    2

    ________

 1 | A|  |  |

   |__|__|__|

   |  |  |  |

   |__|__|__|

   |  |  |  |

   |__|__|__|

   |  |  |  |

   |__|__|__|

(not homework-filter)

How many songs do I have to listen to before I start again?

jacquilynne — Sat, 12 Dec 2009 08:49:09 -0800

Can you help me figure out this math question related to probability? I feel like my college course in probability should have enabled me to figure this out for myself, but clearly, it did not.

I have an iTunes playlist with 500 songs on it. Once I've listened to a song, it disappears from the list, and is replaced by a new song from my library. Songs that have been on the list once can not reappear on the list, and for the purposes of this question, you can assume the library from which the replacement songs are drawn is infinite.

The list is selected randomly and then sorted in alphabetical order, and I start at the top and play my way through it alphabetically until I reach the 500th song. But because of the replacing of songs, which might appear in the list after the song I just listened to (and thus I will have to listen to them before I get to the end of list) or might appear in the list before the song I just listened to (and thus will not get listened to before I get to the end of the list) I actually listen to way more than 500 songs on the way through. What I'm trying to figure out is how many songs I'm likely to listen to in a single pass.

I know that when I'm listening to the first song, there's relatively high odds that it's replacement will belong later in the list. I assume this to be 499/500 -- if the replacement song gets slotted into the list pretty much at random, it could end up in any of 500 slots, and one of them is before the point I'm at in the playlist, while the other 499 after. And when I'm done listening to the 373rd song, then, the odds should be down to 227/500 that I'll have to listen to the replacement song. So, I can figure out the odds that I'll have to listen to any given replacement song.

But I can't get from there to an overall expected number, because I can't figure out how to account for new songs I've already had to listen to vs. the ones I haven't. Say I've listened to 100 songs, and 90 of their replacements ended up later in the list than where I was at the time. That means that while I've listened to 100 songs, I'm only at position 10 on my list -- a long way from the 100 I'd be at if I wasn't replacing the songs.

This is obviously not earth shattering, and I don't need to know this. But it's been nagging at me because I can't figure it out, and I feel like I should be able to.

How many grandmothers will die in two months?

Meatbomb — Tue, 08 Dec 2009 22:55:12 -0800

Actuarial / statistics geeks: a puzzle for you. You are on a two month training program with a group of Korean teachers of English.

There are 20 teachers, ranging in age between late 20s to early 50s.

How many of their grandmothers are likely to die during the course?

Consider the current months (October-November) as the timeframe. Assume the grandmothers are all residents of Daegu, Korea.

I probably cannot provide any more specifics than those, but if you want clarification I will try.

Teach me about this thing you call "math"...

incomple — Tue, 08 Dec 2009 05:03:16 -0800

Here's an obnoxious school question that's been nagging me for days: Suppose that the height (at the shoulder) of adult African bull bush elephants is normally distributed with µ = 3.3 meters and ∂ = .2 meter. The elephant on display at the Smithsonian Institute has a height 4 meters and is the largest elephant on record. What is the probability that an adult African bull bush elephant has height 4 meters or more? I know the answer is .0002, but how can I prove this? I'm not a math guy, and I can't find any "quick and easy"—or even vaguely comprehensible—formulas to help me pound this into my thick skull... Just having the answer doesn't matter to me—I wish to understand!

Other questions that have been sticking in my craw...

The lifetime of a certain brand of tires is normally distributed with mean µ = 30,000 miles and standard deviation ∂ = 5000 miles. The company has decided to issue a warranty for the tires but does not want to replace more than 2% of the tires that it sells. At what mileage should the warranty expire?

Again, I know the answer is 19,750 miles, but I'm completely failing to grasp exactly how one comes to that answer.

In American roulette, the probability of winning when betting "red" is 9/19. What is the probability of being ahead after betting the same amount 90 times?

Answer is .2745. Grr!!

The probabilities of failure for each of three independent components in a device are .01, .02, and .01, respectively. The device fails only if all three components fail. Out of a lot of 1 million devices, how many would be expected to fail? Find the probability that more than three devices in the lot fail.

The first question is easy for me—0.01 x 0.02 x 0.01 = 0.000002, and 0.000002 x 1000000 = 2. But why is there a .1469 probability of more than three devices in the lot failing?

I flunked math in high school, and now I feel like everything in math is set up to make guys like me look dumb.

What is the chane of this event happening?

jackypaper — Fri, 06 Nov 2009 09:43:10 -0800

I have an event that has a 75% chance of happening. If I run the trial seven times, what is the probability of the event happening at least once? And what's the math behind it?

How to make this contest drawing fair?

smelvis — Mon, 26 Oct 2009 14:01:31 -0800

What's a fair way to chose winners for this ad-hoc giveaway? details inside... I have 100 items to give away to roughly 10,000 people. Folks can sign up for as many items as they want, as many times as they want, but can only win once (no money was involved).

Some items received hundreds of "votes", others received very few.
What order should I draw names to make it most fair for the most people?

Should I start with the high-number items first, since they represent the items with the most interest? Or should I start with the lowest interest items - or random?

Or does it even matter that Joe only signed up for item 1, while Bill signed up for every item, since the rules allowed it?

Not all random numbers are created equal

monkeystronghold — Sun, 25 Oct 2009 23:30:14 -0800

How do I get a controlled distribution of random numbers to fairly determine a start position. In a sporting event, start position is decided based on the last digit of your registration number. Each week, random numbers are drawn to decide the start order. For example, the random draw order for a single week is 4, 0, 3, 5, 1, 8, 7, 9, 6, 2. So everyone with a number ending in 4 starts first. Everyone with a number ending in 0 starts second. And so on. The next week the draw order is again random.

While this works, the distribution of of numbers can end up being unfair (one particular number can be "lucky" or "unlucky" for many weeks). Statistically, how would one generate a set of "random" start orders so that the value of each registration number was roughly equal over the course of a season (for ease of calculation, let's say 10 weeks).

I don't know anything about math or statistics, so my description of this situation probably uses lots of words incorrectly. I'd google this, but I don't even know how to start.

Basically, is it possible for value of all of the numbers to even out. But in a random order. So, for example, that 0's aren't always going after 4's. And one group doesn't always start in the middle.

What's the probability of multiple holes-in-one?

jeather — Mon, 21 Sep 2009 11:44:24 -0800

Some guy recently got 3 holes in one on the same course (different holes) within 5 days. My cousin contends that the odds against this are 5 billion to one, based on something he read. He's willing to have me prove him wrong, but this is well beyond my ability to calculate. I'm good with any time period -- the likelihood of someone doing this in a year, or 10 years, or whatever. Part of the problem is that he is convinced that the likelihood of a golfer getting even a single hole-in-one is minute. I found a statistic saying that it's around 12,000 to 1 for a round of golf, or possibly for each hole. But even given that, I have no idea how to estimate how many people play at least 3 games on the same course in any given 5 day period, especially since these periods can overlap.

I'd prefer to use stats for amateur only golfers, but really anything vaguely reasonable is fine -- if I understand the caculations I can sub in any other numbers that my cousin prefers.

What is the probability of one or both events occurring.

charlesminus — Fri, 18 Sep 2009 12:55:28 -0800

What is the probability of one or both events occurring. Seems like this should be a simple problem. but I have been unable to figure it out. Any help in the form of an answer or a pointer to somewhere that would help me solve it would be very nice.

A procedure is about to occur. There is a 1/100 chance that outcome X will occur, and a 1/200 chance that Y will occur. The outcomes are independent of each other. What is the probability that one or both of the outcomes will occur?

Calculating Probability Over Several Attempts?

WCityMike — Wed, 16 Sep 2009 09:39:58 -0800

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%.)

A Strange Bucket

Blazecock Pileon — Thu, 06 Nov 2008 01:39:58 -0800

I have a bucket containing N marbles: M white marbles and N-M black marbles. I need to grab a handful of marbles (n) and figure out the probability of having picked up m white marbles. At first, I thought I could use the hypergeometric distribution. But there's a complication, namely that the white marbles are not equally distributed in my bucket.

In other words, if my handful of marbles contains one white marble, I'm more likely to have picked up one or more additional white marbles in my hand, and this probability is different depending on how many white marbles I may have picked up.

Is there a good approach to modeling or simulating this kind of situation?

Math and Odds and Blackjack

gummo — Tue, 28 Oct 2008 14:50:03 -0800

What are the chances, in blackjack, that I will start with 15, and the dealer will be showing a 10 (or face card worth 10)? Despite the fact that my buddy thinks the odds of this are near 100%, I've tried to calculate the actual odds (assuming a single deck).

I've come up with 2.62% -ish. Not really sure if I'm right.

I started with all the possible hands I could have.
2652 (52 * 51)

Then I figured out how many hands would give me 15.

64 hands with a 10 (or equivalent) and a 5.
64 hands with a 5 and a 10.
16 hands with a 9 and a 6.
16 hands with a 6 and a 9.
16 hands with an 8 and a 7.
16 hands with a 7 and an 8.
16 hands with an ace and a 4.
16 hands with a 4 and an ace.

These fall into two groups, those that have a 10, and therefore deprive the dealer of one, and those that don't.

128 hands have a 10.
96 hands are 15 some other way.

If I have a 10, she has 15 cards left, out of 50, to get a 10.
If I don't she has 16 out of 50.

So my odds of making a 15 using a 10 (or face card) are 4.8%.
Her odds of then having a 10 showing are 30%.

So we have a 1.44% chance of that happening.

-Plus-

Me making 15 some other way: 3.7%.
And her having a 10 showing: 32%.

Gives us 1.18% for that second scenario.

So 2.62% of my having 15 and the dealer showing a 10 or face card.

Two questions.
Assuming a single deck, is this correct?
Assuming multiple decks, what changes (if anything)?

If I'm wrong, where did I go wrong? (It would help my brain to have a combination of english and math to explain where I went wrong, rather than just something like "you should have used a factorial for possible hands 52!-4!", etc...)

Thanks.

Probability question (in need of code)

chrisamiller — Wed, 13 Aug 2008 22:04:42 -0800

It's late, I'm tired, and I have a probability and coding question that's fairly simple. Say I have 5 buckets (A,B,C,D,E), with different colored balls in them. The probability of removing a red ball is different for each bucket:

A: 1/3
B: 1/4
C: 1/16
D: 1/6
E: 1/9

If I draw 1 ball from each bucket, what is the probability (P) of drawing at least N red balls?

Doing it for small numbers of buckets is easy enough, but I have thousands of buckets here. Given that I know the probabilities for each bucket, what's the easiest way to calculate P? I suspect that there's an R function that makes this a breeze, but I'm having trouble tracking it down.

I can do basic operations in R, and I'm also open to functions or pseudo code from your favorite language. Ruby and Perl preferred, but I can use others if they get the job done.

For good measure, here's a preemptive "This isn't homework-filter"

How to calculate probability of being dealt certain cards?

b_thinky — Wed, 09 Jul 2008 17:44:11 -0800

You're dealt 8 cards face down from a 52 card deck. All values are determined by the number on the card, not the suit. What is the probability of each card being dealt? The first card you're dealt has a 4/52 chance of being any of the 13 values (2 thru Ace). The 2nd card you're dealt has a 4/51 chance of being any of the remaining 12 values, plus a 3/51 chance of being the same value as the first card.

This is where my retard brain ceases to work. How do I know the odds of the 3rd card, given that the 2nd card may or may not be the same as the 1st card? If the 2nd card has a new value, the third card is 4/50 of the 11 remaining cards, plus 3/50 it's the same value as the 1st card and 3/50 it's the same as the 2nd card. But if the 2nd card is a repeat of the 1st value, there is a 4/50 chance of it being one of the 12 remaining values and a 2/50 chance of it being the same as the 1st and 2nd card.

So you see the dilemma I have calculating the odds for the 3rd card to be dealt until the 8th. I know I'm probably looking at this entirely the wrong way but I've always been a math retard beyond basic arithmetic. Anyone care to explain how this works?

21 judges give 4 contestants a 4, 3, 2, or 1. How often is there a first place tie?

23skidoo — Tue, 24 Jun 2008 11:22:02 -0800

Imagine a contest, with 21 judges and 4 contestants. Judges must give one contestant 4 points, one contestant 3 points, one contestant 2 points, and one contestant 1 point. The winner is decided by summing the scores of all 21 judges. Mathematically speaking, what percentage of the time would you expect there to be a tie for first place after all 21 scores have been added?

Oh, boy, parallel universe #57339! That's where I'm a Viking.

Rhaomi — Sat, 24 May 2008 17:39:43 -0800

How would the many-worlds interpretation work on the human level, if at all? (I'm no quantum physicist, so please forgive me if the following is woefully simplistic, ridiculously naive, and/or hopelessly wrong)

The way I've heard it explained, the many-worlds interpretation of quantum mechanics holds that for every situation in which multiple outcomes are possible, each one of those outcomes does happen -- albeit in its own universe. That the universe we perceive is just part of an inconceivably large multiverse of infinitely branching possibilities, and that every interaction between every atom everywhere in the universe creates another one, or multiple ones, all the time.

I've also heard that because every possible outcome occurs, even the most bizarrely improbable event has happened in at least one universe. This makes sense if the ideas in the above paragraph are true.

For instance, it is incredibly unlikely that a fair coin could come up heads twenty times in a row -- the odds are about 1,048,576 to 1. But if each coin toss branches into a universe where it lands heads and a universe where it lands tails, then at the end of the line one of the million+ universes would see it land heads all twenty times. Of course, most of the rest of the million branches would see a mixed outcome, so from the point of view of a single universe the odds are still very unlikely. But the many-worlds theory says it does happen somewhere.

But when you think about it, it wouldn't be that simple. For starters, each coin toss would make more than two universes. A lot more. For example, a single coin toss could have two universes where the coin lands tails, but one universe sees it land one centimeter further to the right than the other. And there would be a small set of worlds where the coin landed perfectly on its side. And an even smaller minority where all the molecules of the coin spontaneously evaporated at the same time. A colossally improbable event, but possible.

And hey, if this holds true for molecules and small objects like coins, could you not also extend it to the rest of the world (which is just a collection of 10^huge molecules)? Must there be a universe out there where, say, every person who bought a Florida State lottery ticket happened to pick the same number, which was the winning one? Or where every building on Earth suffered simultaneous structural failure? Or where everyone spontaneously decided to break into Broadway-style song and dance? And a trillion variations on these and other scenarios, each slightly different from the other? And that the only reason we (most of our selves?) don't experience these things is because the infinity of "normal" universes where probable things happen outnumbers the infinity of universes where "impossible" things happen?

It feels absurd, like I'm talking about the Infinite Improbability Drive instead of a theory of physics, but I'm not seeing why it shouldn't be true. I've tried finding answers, but most of the literature out there (with the occasional exception) deals with quantum physics in a dry, academic context that limits the discussion to the atomic level. And of the material that imagines crazy outlier universes like the ones I described, I don't have a good way of telling if the physics involved is real or just taking artistic liberties for the sake of interesting fiction.

Must I mourn for the Earth somewhere out there that suffered Total Existence Failure? (And the one that suffered it five minutes later, and the other one that suffered it sixteen years later, and the one that suffered it partially, losing the western hemisphere, and the one that lost the eastern hemisphere, and the one that suddenly split into two planets, and the one where Australia turned into gelatin, and etc.)

Oh, and I know that we still aren't sure about which interpretation of quantum mechanics is correct, and that even if we knew the many-worlds theory were right, we'd have no way to observe other universes. I just want to know if the things I described are allowable in the context of the theory as it's understood today.

What is a good sample size for determining whether dice are truly random?

Bugbread — Wed, 30 Apr 2008 11:41:41 -0800

How many times would I have to roll a standard 6-sided die to get a statistically representative view of whether it was truly random or not? I have a bunch of dice that I haven't used in years. The other day, I was playing with one, and I noticed that the 5 came up fairly often. I started rolling the die and writing down the results, to see if it was just a short term statistical fluke, or observer bias, or if the die really favored the 5.

I think I ended up rolling it around 200 times, and 5 definitely had a significant edge.

Now, I'd like to check my other dice. However, I'm not a statistician. I understand that, obviously, the larger your data set (the more times you roll each die and write down the results), the better your analysis of non-randomness will be. However, I know that it isn't necessary to roll the die 1 billion times to check for randomness, that there is some generally accepted statistical minimum, below which the margin of error is too large, and above which the margin of error is generally considered acceptable.

How many die rolls is that point?

Probability and Truth

binturong — Sat, 12 Apr 2008 12:59:35 -0800

Suppose you take a test for a rare type of cancer that affects 0.01 percent of the population. The test is 98 percent reliable. You get a positive reading. What are the chances you have the cancer? I read this probability puzzle today and the writer said the statistical chances of you having the cancer in this scenario are less than half a percent. I don't get it. Isn't the rarity factor irrelevant compared with the test reliability? Please explain.

Do you know my friend?

Gungho — Tue, 01 Apr 2008 13:37:58 -0800

Mathematics filter: What are the odds of a pair of my friends meeting another couple who are also my friends, but they have never met before? Bonus. This happened on vacation in another city where neither of them, nor I, have ever lived... Is this an example of six degrees of separation?

what's the probability distribution, kenneth?

russm — Fri, 18 Jan 2008 19:20:25 -0800

for the statisticians: what probability distribution do I want to fit to my data, and how? I have a collection of widgets. they age, and as they age some of them get tweaked. it's been suggested to me that given enough time every widget will eventually be tweaked, but I'm not convinced. the data I have covers ~64000 widgets, and for each widget is either the age at which it was tweaked, or alternatively the current age and the fact that it has never (yet) been tweaked.

as I understand it (I'm a database programmer with an interest in maths but very little working stats) I want to fit some probability distribution to my data and see whether the area under the probability mass function is < 1.

the 2 things that make this different from everything I've been able to find on the 'net are

1) I specifically don't expect the cumulative distribution to asymptote to 1, which appears to count out obvious candidates like a Poisson distribution

2) my data contains negative examples - those widgets which have not yet been tweaked, and may never be. I'm assuming these are relevant to the shape of the final distribution, but I can't work out how to get tools like R to take them into account.

the best advice I've had so far is from an engineer friend who said to chuck the data on only the tweaked widgets into R, fit a Poisson distribution and be done with it. unfortunately this fails on both of the above points. (then again, all he normally cares about is failure rates in aircraft parts, and you know that given enough time every lug will eventually break).

so, if there are any stats geeks who understand what I'm trying to do and can point me at tools/docs/info on how to do it you'd rock my world.

complex dice probability make head esplode

DetonatedManiac — Fri, 21 Dec 2007 13:52:56 -0800

I need a tool to help me analyze the probability of certain events occurring regarding a pool of X dice for a game I am creating. See my examples below. I am looking for a dice statistics program or, baring that, a set of formulas I could input into excel, that can answer the following questions. Graphing capabilities are a plus.

Some of the events I want to analyze:

1) The probability that, from a pool of X dice, Y dice can be chosen by a player such that the sum of the Y dice is equal or lesser than a number N.

2) Related to the previous question; of the pools that succeed what is the distribution of how close they are to hitting N, (for instance, if the sum of Y dice is 10 and the number the player was aiming for was 12 then they are only off by 2).

3) In a pool of X dice How likely is it that there will be Z sets of doubles (say two 6s or snake eyes) and, more specifically, what are the chances of getting a specific set of doubles (say just snake eyes).

I am pretty statistics theory inept (never took it in school) but I am average to above average with math in general... basically please keep any advice in laymen terms as much as possible. Thanks

How many rolls 'till n values in a row.

lucasks — Wed, 19 Dec 2007 15:01:26 -0800

Given a fair die with s sides, what is the expected number of rolls r before I get o outcomes with the same value v in a row? [Note this is emphatically NOT a homework problem; I just really want to know, feel like I should be able to figure it out, can't, and it is driving me nuts.] For clarity a sample case would be, how many times do I need to roll a 6 sided die before I get 5 1s in a row. Please help.

What are the odds that I can solve this problem?

reformedjerk — Tue, 19 Jun 2007 07:46:07 -0800

Probability/stat question: I'm looking for patterns in a protein sequence, and I've found a few that occur quite frequently. How do I know these are actual patterns and not just an artifact of random amino acid distribution? I have a roughly 1000 amino acid sequence, and I've used a sliding window to chop it up into overlapping 6-mers. Some of these 6-mers occur much more frequently than others and I suspect they have some sort of biological significance. Unfortunately, I don't know to test whether these are true pattern in the biological sense, or if they could just as easily have been the result of random distribution.

I've tried comparing the expected frequency of these 6-mers based on the amino acid distribution with the observed frequency; but the chance of getting any given 6-mer randomly is so low that almost anything I observe (even the ones that only show up once) seem really significant. I'll be happy to clarify things if this post is a bit messy.

Modal verbs can versus could

stavrosthewonderchicken — Tue, 12 Dec 2006 18:20:35 -0800

English grammar: 'could be Xing' versus 'can be Xing' -- how can we explain why one is correct and one isn't? (Somebody fire up the languagehat signal!)

I'm trying to figure out a reasonably succinct structural way explain why one of these is just fine, but one isn't.

We're using modals here to express probability or guessing. Scenario: you call your wife at home, but there's no answer. You guess that she might be in the bathroom, and express it by saying "She could be taking a shower." I say this is correct.

Saying "She can be taking a shower" in this situation, I call incorrect, but I can't seem to hit on any plausible explanation for why, other than that it's just common usage. That's a good enough answer, perhaps, but I'd like to actually understand what's happening here, structurally. It could be that my brain is just wobbly today, and I'll figure it out as soon as I post this, but it's driving me a bit nuts.

Infinite vs Infinite

777 — Tue, 21 Nov 2006 13:35:13 -0800

ProbabilityFilter: Calling all mathematicians... I am not a math person, but this is something I've always wondered/had trouble wrapping my head around:

Imagine you are standing on a football field, there is a trash can about 30 yards away, and you are holding an average sized stone.

I submit that there are _infinite_ ways (different trajectories) of tossing the stone such that it lands in the trash can.

I also submit that there are _infinite_ ways of of tossing the stone such that it doesn't land in the trash can.

Surely the probability of missing is higher than making the shot? Yet if both are infinite, are they not the same probability? Are there not, in fact, infinite ways to achieve the former? Or is there something else I'm not getting?

Extra Credit: Does the math change at all if I'm _trying_ to make the shot vs. randomly throwing the stone in any direction?