# Probability Basics **URL:** \ **Category:** Factual Questions\ **Created:** [November 26, 2004, 4:05pm UTC](https://boards.straightdope.com/t/probability-basics/276755 "2004-11-26T16:05:19Z")\ **Posts on this page:** 18\ **Page:** 1
**Author:** ![SiXSwordS](https://avatars.discourse-cdn.com/v4/letter/s/49beb7/32.png) [@SiXSwordS](https://boards.straightdope.com/u/SiXSwordS)\ **Post date:** [November 26, 2004, 4:05pm UTC](https://boards.straightdope.com/t/probability-basics/276755/1 "2004-11-26T16:05:19Z")
From this thread,[http://boards.straightdope.com/sdmb/showthread.php?t=288644](http://boards.straightdope.com/sdmb/showthread.php?t=288644) > [@doug-jensen](#): > > the classic probability theory question to the effect of "If there is a 60% chance of rain today, and it is twice as likely to rain tomorrow, what is the probability of rain tomorrow? Obviously not 120%.). You could explain the simple answer (80%) and you could mention how questions like that relate to various theories of probability Also, > [@C K Dexter Haven](#): > > [W]e’d not only have to explain how to deal with multiple occurrences in statistics, but in how to deal with assigning numbers to phenomena. … While Cecil might not want to deal with the basics in his column, I thought someone might care to take on ignorance single handedly and explain the basics of these concepts to me. I thought that the answer to the ‘classic thoery’ question would have been 90%. My understandiing of probability comes almost entirely from determining the odds of getting certain hands in poker. I am very bad at poker. ---
**Author:** ![spingears](https://avatars.discourse-cdn.com/v4/letter/s/ebca7d/32.png) [@spingears](https://boards.straightdope.com/u/spingears)\ **Post date:** [November 26, 2004, 5:46pm UTC](https://boards.straightdope.com/t/probability-basics/276755/2 "2004-11-26T17:46:02Z")
> [@SiXSwordS](#): > > From this thread,[http://boards.straightdope.com/sdmb/showthread.php?t=288644](http://boards.straightdope.com/sdmb/showthread.php?t=288644) > Also, > While Cecil might not want to deal with the basics in his column, I thought someone might care to take on ignorance single handedly and explain the basics of these concepts to me. > I thought that the answer to the ‘classic thoery’ question would have been 90%. My understandiing of probability comes almost entirely from determining the odds of getting certain hands in poker. I am very bad at poker. Try the local library first. Buy or order a good text book on “Probability” I have neither the time nor inclination to write a lengthy post of such an extensive topic! :rolleyes: ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [November 26, 2004, 6:14pm UTC](https://boards.straightdope.com/t/probability-basics/276755/3 "2004-11-26T18:14:14Z")
To me, doug’s question does not make any sense unless you define the meaning of ‘twice as likely to happen’ in some way other than the usual sense in which I think we all use it – that the probability of Y is double the probability of X. Probabilities greater that 1 are meaningless, and so doubling the probability of anything over 0.5 (50 percent) are also meaningless From the answer, I can guess that he is using it, at least some of the time, to mean “the probability of not-Y is half the probability of not-X”. But I don’t think that’s standard in lay probability at least… I’ve never delved really deeply into it. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [November 26, 2004, 6:23pm UTC](https://boards.straightdope.com/t/probability-basics/276755/4 "2004-11-26T18:23:28Z")
> [@chrisk](#): > > From the answer, I can guess that he is using it, at least some of the time, to mean “the probability of not-Y is half the probability of not-X”. But I don’t think that’s standard in lay probability at least… I’ve never delved really deeply into it. That was my first guess as well, but I wouldn’t want to try to make that into a general principle. ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [November 26, 2004, 6:31pm UTC](https://boards.straightdope.com/t/probability-basics/276755/5 "2004-11-26T18:31:42Z")
> [@ultrafilter](#): > > That was my first guess as well, but I wouldn’t want to try to make that into a general principle. Yeah, one thing that might make a good general principle, (though it would be a pain to work with,) might be: Y is twice as likely as X =\> Y / ~Y = 2 \* (X / ~X) That is, the ratio of Y happening to not happening is double the ratio of X happening to not happening. so, if X has a probability of 0.3 and Y is twice as likely to happen as X, then the Y/not Y ratio would have to be about 2 \* (0.3 / 0.7) or 6/7 , which makes it only about 46% . The good thing about this definition is that no matter how likely or unlikely some event is, (as long as it isn’t 0 or 1 - never or certain,) you can calculate ‘X times more/less likely’ as a meaningful probability. Oh, let’s see how that works with X = 0.6 Y/~Y = 2 \* (0.6 / 0.4) Y / ~Y = 3 0.75 / 0.25 = 3 So I get 75 % with this system… not 80, darnit. ---
**Author:** ![nivlac](https://avatars.discourse-cdn.com/v4/letter/n/3bc359/32.png) [@nivlac](https://boards.straightdope.com/u/nivlac)\ **Post date:** [November 26, 2004, 6:35pm UTC](https://boards.straightdope.com/t/probability-basics/276755/6 "2004-11-26T18:35:13Z")
> [@SiXSwordS](#): > > I thought that the answer to the ‘classic thoery’ question would have been 90%. My understandiing of probability comes almost entirely from determining the odds of getting certain hands in poker. I am very bad at poker. How you came up with 90% is beyond me. The question that you’re answering is meaningless. The probability of an event occurring is always based on a relative frequency, i.e., from past situations with the same conditions, how many times have the event occurred? That’s basically the answer in Cecil’s original reply. It’s the same in poker. The probability of getting a specific hand is the number of times that the hand can be obtained divided by the number of possible poker hands that can be dealt to you (details at [http://mathworld.wolfram.com/Poker.html](http://mathworld.wolfram.com/Poker.html)). ---
**Author:** ![C\_K\_Dexter\_Haven](https://avatars.discourse-cdn.com/v4/letter/c/b2d939/32.png) [@C\_K\_Dexter\_Haven](https://boards.straightdope.com/u/C_K_Dexter_Haven)\ **Post date:** [November 26, 2004, 6:36pm UTC](https://boards.straightdope.com/t/probability-basics/276755/7 "2004-11-26T18:36:23Z")
One of the probability problems is the question of terminology: what does it mean, really, that “X is twice as likely as Y”? **SiXSwords** , I’m not exactly sure what you’re asking (i.e., how basic is basic?) but… - If you’re looking at multiple events occuring, and you want both, you multiple the probabilities - If you’re looking at multiple events occurring, and you want either, you add the probabilities. However, sometimes the “real” probability inovlves the negative, not the positive. Example: What’s the chance of getting heads on the first or second toss of a coin? WRONG: Chance of heads on first toss is 1/2, chance on second toss is 1/2, chance of either one is 1/2 + 1/2 = 1, so certainty. RIGHT: Chance of heads on first toss is 1/2; chance of not-heads on first toss but yes-heads on second toss is 1/2 \* 1/2 = 1/4, so chance of heads on first toss or second toss is 1/2 + 1/4 = 3/4. This is all pretty basic. ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [November 26, 2004, 6:40pm UTC](https://boards.straightdope.com/t/probability-basics/276755/8 "2004-11-26T18:40:31Z")
> [@C K Dexter Haven](#): > > - If you’re looking at multiple events occurring, and you want either, you add the probabilities. That’s a little misleading, though you do explain a little more in detail. I would say: - If you want either of two outcomes to the same event, you can add the probabilities. - If you want either of two independent events… (ie, two events that could both happen,) then calculate the odds of each event NOT happening, multiply those to get the odds of both events not happening, and subtract that from 1 How’s that for a rule of thumb? 🙂 ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [November 26, 2004, 7:46pm UTC](https://boards.straightdope.com/t/probability-basics/276755/9 "2004-11-26T19:46:45Z")
> [@C K Dexter Haven](#): > > **SiXSwords** , I’m not exactly sure what you’re asking (i.e., how basic is basic?) but… > > - If you’re looking at multiple events occuring, and you want both, you multiple the probabilities > - If you’re looking at multiple events occurring, and you want either, you add the probabilities. I’m not **SiXSwords** , but I don’t see how p = .8 is the correct answer to “If there is a 60% chance of rain today, and it is twice as likely to rain tomorrow, what is the probability of rain tomorrow?” ---
**Author:** ![ftg](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/ftg/32/2801_2.png) [@ftg](https://boards.straightdope.com/u/ftg)\ **Post date:** [November 26, 2004, 7:58pm UTC](https://boards.straightdope.com/t/probability-basics/276755/10 "2004-11-26T19:58:38Z")
If the probability is 60%, the odds are 6 out of ten or 6:4 which is the same as 1.5:1. Doubling 1.5:1 gives 3:1. And 3:1 is 75%. ---
**Author:** ![chrisk](https://avatars.discourse-cdn.com/v4/letter/c/6de8d8/32.png) [@chrisk](https://boards.straightdope.com/u/chrisk)\ **Post date:** [November 26, 2004, 8:00pm UTC](https://boards.straightdope.com/t/probability-basics/276755/11 "2004-11-26T20:00:30Z")
> [@ftg](#): > > If the probability is 60%, the odds are 6 out of ten or 6:4 which is the same as 1.5:1. Doubling 1.5:1 gives 3:1. And 3:1 is 75%. Yeah, I went through similar math a little earlier on the thread, though I didn’t use the colon ratios notation. Still haven’t gotten any closer to 80% than 75%… guess maybe nobody but doug jensen knows how he arrived at the 80% figure. ---
**Author:** ![LSLGuy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lslguy/32/5813_2.png) [@LSLGuy](https://boards.straightdope.com/u/LSLGuy)\ **Post date:** [November 26, 2004, 11:57pm UTC](https://boards.straightdope.com/t/probability-basics/276755/12 "2004-11-26T23:57:15Z")
60% chance of rain is a 40% chance of non-rain. if rain is twice as likely tomorrow, it’s also half as unlikely. So the chance of non-rain drops from 40% to 20%. And if the chance of non-rain is 20%, then the chance of rain is 80%. As **chrisk** & **C K Dexter Haven** said, you’ve got to be careful about your terms. There are many, many cases where you’ve got to subtract your data point from 1 to get the probability of the opposite outcome, then perform some calcs on that, then subtract that result from 1 again to convert back to the probability of the outcome you’re interested in. There’s always exactly one right answer, but these aren’t problems yuo can do by rote; yuo need to understand not only what you’re doing but why at every step of the way. ---
**Author:** ![1010011010](https://avatars.discourse-cdn.com/v4/letter/1/ea666f/32.png) [@1010011010](https://boards.straightdope.com/u/1010011010)\ **Post date:** [November 27, 2004, 12:12am UTC](https://boards.straightdope.com/t/probability-basics/276755/13 "2004-11-27T00:12:57Z")
Look up [Bayesian](http://everything2.com/index.pl?node_id=1441299) Statistics. That’s usually what’s used for weather predictions rather than straight frequency based statistics. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [November 27, 2004, 12:24am UTC](https://boards.straightdope.com/t/probability-basics/276755/14 "2004-11-27T00:24:24Z")
> [@LSLGuy](#): > > if rain is twice as likely tomorrow, it’s also half as unlikely. That’s not obvious to me. What interpretation are we dealing with here? ---
**Author:** ![Sample\_the\_Dog](https://avatars.discourse-cdn.com/v4/letter/s/8797f3/32.png) [@Sample\_the\_Dog](https://boards.straightdope.com/u/Sample_the_Dog)\ **Post date:** [November 27, 2004, 12:57am UTC](https://boards.straightdope.com/t/probability-basics/276755/15 "2004-11-27T00:57:12Z")
> [@LSLGuy](#): > > 60% chance of rain is a 40% chance of non-rain. > > if rain is twice as likely tomorrow, it’s also half as unlikely. So the chance of non-rain drops from 40% to 20%. And if the chance of non-rain is 20%, then the chance of rain is 80%. > > As **chrisk** & **C K Dexter Haven** said, you’ve got to be careful about your terms. There are many, many cases where you’ve got to subtract your data point from 1 to get the probability of the opposite outcome, then perform some calcs on that, then subtract that result from 1 again to convert back to the probability of the outcome you’re interested in. > > There’s always exactly one right answer, but these aren’t problems yuo can do by rote; yuo need to understand not only what you’re doing but why at every step of the way. But why is halving the remainder any better than doubling the percent? Do we do the latter for values under 1/2 and the former for values over 1/2? Or do we always halve the remainder because it’s always possible? I like **chrisk’s** solution of finding a value w/ a doubled X/~X ratio. That said, I don’t deal w/ probabilities like this, so I have no idea what the standards are. Anyone got any professional standards you use, or know of? ---
**Author:** ![Cabbage](https://avatars.discourse-cdn.com/v4/letter/c/f07891/32.png) [@Cabbage](https://boards.straightdope.com/u/Cabbage)\ **Post date:** [November 27, 2004, 4:10am UTC](https://boards.straightdope.com/t/probability-basics/276755/16 "2004-11-27T04:10:52Z")
My two cents: The only interpretation of “twice as likely” that makes sense to me is that it means the probability is simply doubled. In other words, if there’s a 30% chance of rain today, and for tomorrow rain will be twice as likely, then there’s a 60% chance of rain tomorrow. Of course, under that interpretation, it doesn’t make sense to speak of “twice as likely” when the probability is greater than 50%. That doesn’t look like a problem to me. If you’re taking a class and score 90% on the final exam, and a friend says, “Ha! I did twice as well as you!”, you’d say (rightly), “That’s impossible!” Similarly, if on a given day the chance of rain is 60%, it’s a logical impossibility to say that rain will be twice as likely on another day. I don’t see a problem here–it’s just like division by zero: It simply can’t happen. Re: Twice as likely to happen=half as likely to not happen. This makes no sense to me. Under this interpretation, if there’s a (virtually) 0% chance of rain Sunday, and rain is twice as likely on Monday, then we have a (virtually) 50% chance of rain on Monday. Probability 50% means twice as likely as probability 0%? I don’t think so. If a rare event (i.e., probability near zero) suddenly becomes “twice as likely”, it’s still a rare event (i.e., not a 50% chance of happening). Understand that, to my knowledge, there’s no formal definition of “twice as likely” in probability. This is simply how I would define it. ---
**Author:** ![OldGuy](https://avatars.discourse-cdn.com/v4/letter/o/3bc359/32.png) [@OldGuy](https://boards.straightdope.com/u/OldGuy)\ **Post date:** [November 27, 2004, 5:25am UTC](https://boards.straightdope.com/t/probability-basics/276755/17 "2004-11-27T05:25:42Z")
> [@LSLGuy](#): > > 60% chance of rain is a 40% chance of non-rain. > if rain is twice as likely tomorrow, it’s also half as unlikely. So the chance of non-rain drops from 40% to 20%. And if the chance of non-rain is 20%, then the chance of rain is 80%. Uh no. What if the change of rain tomorrow were 40%? Then twice as likely would mean 80%. Half as unlikely would mean 1- (1-.4)/2 = 70%. Clearly not teh same at all. ---
**Author:** ![nivlac](https://avatars.discourse-cdn.com/v4/letter/n/3bc359/32.png) [@nivlac](https://boards.straightdope.com/u/nivlac)\ **Post date:** [November 27, 2004, 5:40am UTC](https://boards.straightdope.com/t/probability-basics/276755/18 "2004-11-27T05:40:14Z")
> [@ftg](#): > > If the probability is 60%, the odds are 6 out of ten or 6:4 which is the same as 1.5:1. Doubling 1.5:1 gives 3:1. And 3:1 is 75%. Doubling the odds is not the same as doubling the probability. The original question had to do with “twice as likely”, with “likely” quantified as a probability. If you toss a fair coin, the probability of heads is .50 with the odds of getting heads 1:1. If you have a coin with double the chances of getting heads, you would have a coin that always comes up heads. Doubling the odds to 2:1 will not achieve that. > [@cabbage](#): > > … if on a given day the chance of rain is 60%, it’s a logical impossibility to say that rain will be twice as likely on another day. This is spot on.