# Lottery probability question **URL:** \ **Category:** Factual Questions\ **Created:** [April 28, 2010, 2:45am UTC](https://boards.straightdope.com/t/lottery-probability-question/537594 "2010-04-28T02:45:09Z")\ **Posts on this page:** 20\ **Page:** 2
**Author:** ![Chessic\_Sense](https://avatars.discourse-cdn.com/v4/letter/c/7c8e57/32.png) [@Chessic\_Sense](https://boards.straightdope.com/u/Chessic_Sense)\ **Post date:** [April 28, 2010, 3:30pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/21 "2010-04-28T15:30:10Z")
I’m going to weigh in on the side of “depends on the combinations”. Once you know that _someone_ had the winning combination, your chance becomes 1/X where X=combinations covered, NOT tickets sold. After all, if you and your brother buy the same number ticket, and no one else plays at all, then you know you both won. Your probability is 1, not 1/2. ---
**Author:** ![Maserschmidt](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/maserschmidt/32/18829_2.png) [@Maserschmidt](https://boards.straightdope.com/u/Maserschmidt)\ **Post date:** [April 28, 2010, 3:37pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/22 "2010-04-28T15:37:50Z")
Did the last three of us say the same thing? I think we did. At SDBM that counts as consensus! ---
**Author:** ![Munch](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/munch/32/5281_2.png) [@Munch](https://boards.straightdope.com/u/Munch)\ **Post date:** [April 28, 2010, 3:46pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/23 "2010-04-28T15:46:31Z")
Without knowing who won, I think the question that should be asked is “has my chance for winning been eliminated?”. The answer of “no” doesn’t reduce them to 0, but it also doesn’t raise them because no one else’s chances have been eliminated. ---
**Author:** ![Darth\_Panda](https://avatars.discourse-cdn.com/v4/letter/d/ee7513/32.png) [@Darth\_Panda](https://boards.straightdope.com/u/Darth_Panda)\ **Post date:** [April 28, 2010, 4:25pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/24 "2010-04-28T16:25:05Z")
Just because you cannot calculate the new odds, doesn’t mean that you can’t know something about them. When I see someone pour a bucket of water into a pool, I may not know the volume of the water in pool before they poured it in or after they pured it in, but I know that it is greater now than it was before. Similarly, you may know that your probability of winning is greater or lesser after a given event occurs than prior, without knowing what the probability either before or after was or is. Prior to discovering that there was a winner, a reality existed in which many realities could unfold, with many of those possible futures having events play out such that no one won the lottery. In none of those realities could you have won the lottery. After finding out that there was a winner, all of the possible realities where no one would have won the lottery have been eliminated - they may no longer come to pass. However, all of the possible realities where you would have one before are still possbile - the new piece of information has not made them impossible. So if your odds of winning are equal to the number of possible realites in which you win divided by the number of possible realities overall, then you can tell that your odds have increased. Before it was x/y. Now it is x/(y-n). And given that x,n, and y are all positive numbers, x/(y-n) is greater than x/n. ---
**Author:** ![CurtC](https://avatars.discourse-cdn.com/v4/letter/c/ce73a5/32.png) [@CurtC](https://boards.straightdope.com/u/CurtC)\ **Post date:** [April 28, 2010, 4:54pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/25 "2010-04-28T16:54:33Z")
Darth Panda nailed it - if most of the time no one wins, then by finding out that someone _has_ won, your odds have gone up. You can break down that 1 in 200,000,000 number this way: 1/200,000,000 = (prob that someone will win) \* (prob that if someone wins, it will be you) For the example in the OP, you’re just trying to find that last term. If the prob that someone will win is less than one, then the prob that it will be you has to be _more_ than 1/200,000,000. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [April 28, 2010, 4:58pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/26 "2010-04-28T16:58:02Z")
> [@Bryan\_Ekers](#): > > I don’t understand why later posters don’t understand the situation as described in the OP. The only part of the OP I made use of was that, if you were to learn that there wasn’t a winner, the probability that you are a winner will drop to zero, upon conditioning upon this information. From this, I concluded that if you were to learn that there was a winner, the probability that you are a winner will rise, upon conditioning upon this information. I gave the mathematics to support this, as well, though it’s all rather simple. If I have misappreciated what was meant when the OP said the probability will drop to zero upon learning that there isn’t a winner, please clarify the situation for me and the others in what would be the same confusion. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [April 28, 2010, 5:34pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/27 "2010-04-28T17:34:44Z")
> [@doubled](#): > > Here’s the thing I don’t get – most of the time, there is not a winner. […] most drawings go without a winner. […] After all, most of the time no won wins the lottery This is important information that wasn’t present in the OP. To put some numbers on it, suppose on average there’s a winner only every twenty drawings. Prior to learning whether someone won, your odds were 1 in 200 million. 19 times out of twenty, after learning whether someone won, your odds drop to 0. That one time out of twenty you learn that there’s a winner, your odds have increased to 1 in 10 million. > [@](#): > > Is this a rational feeling that you think you’re better of now? Is this because we’re implictly making the assumption (without acknowleding it) that fewer than 200 million tickets were sold? Or is there some other reason why people assume the knowledge that somebody won increases their personal odds? It’s not strictly the number of tickets, since there can be duplicates. It’s how often there’s at least one winner. Depending on how likely duplicates are, having at least one winner every twenty drawings may mean there’s about ten million tickets sold per drawing (in the case of no duplicates), or twenty million, or whatever. ---
**Author:** ![panamajack](https://avatars.discourse-cdn.com/v4/letter/p/47e85d/32.png) [@panamajack](https://boards.straightdope.com/u/panamajack)\ **Post date:** [April 28, 2010, 6:06pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/28 "2010-04-28T18:06:44Z")
> [@doubled](#): > > Yeah, that happened the last time I asked this question, too. 🙂 > Once you find out _someone_ won, you feel happier than you did before the drawing. You think, “Well, now at least I have a chance!” After all, most of the time no won wins the lottery, so you just survived a big hurdle. Is this a rational feeling that you think you’re better of now? Is this because we’re implictly making the assumption (without acknowleding it) that fewer than 200 million tickets were sold? Or is there some other reason why people assume the knowledge that somebody won increases their personal odds? In the first place, the psychological feelings around probability (even when they are correct, as in this case) are often not rationally proportionate to the numbers involved. In other words, you feel better than you “ought” to. Despite my use of “ought”, however, I want to stress that there is nothing wrong with feeling that way. Indeed, about the only good reason for playing the lottery is that the hope or dream of winning - the psychological boost - is worth the $1 to you - it’s almost never worth it “rationally” to buy a ticket. What if there was a lottery where you never even knew the odds involved? Suppose that you simply buy a ticket, and sometimes there’s a winner, and sometimes there isn’t. On any week where there is a winner, you are justified in feeling that it could be you, even though you don’t know the mechanism. But that, too, could be changed by additional knowledge. Suppose this unknown lottery is presided over by an old woman in a room, who picks a winner (or not) based on a name she likes, or a photograph. But she also really dislikes redheads, and will rarely pick them. If you have red hair, then you now know you’re less likely to win at any time. But you’re still more likely to win on a week that she picks a winner. ---
**Author:** ![rbroome](https://avatars.discourse-cdn.com/v4/letter/r/838e76/32.png) [@rbroome](https://boards.straightdope.com/u/rbroome)\ **Post date:** [April 28, 2010, 6:58pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/29 "2010-04-28T18:58:15Z")
> [@doubled](#): > > Yeah, that happened the last time I asked this question, too. 🙂 > > Here’s the thing I don’t get – most of the time, there is not a winner. So right before I check to see whether there was a winner, my odds are 1 in 200million and I also know that empircally, most drawings go without a winner. I don’t know for certain that the number of tickets sold was less than 200 million; but it’s certainly possible (and indeed likely based on past results). It seems like simply learning that SOMEONE won, must give me more hope now than before I knew that. > Yes, it’s possible that every possible ticket was bought, in which case my odds haven’t changed at all – there HAD to be a winner and my odds are still 1 in 200 million. But it’s also possible less than 200 million tickets were sold, in which case my odds have increased to 1/X. > Once you find out _someone_ won, you feel happier than you did before the drawing. You think, “Well, now at least I have a chance!” After all, most of the time no won wins the lottery, so you just survived a big hurdle. Is this a rational feeling that you think you’re better of now? Is this because we’re implictly making the assumption (without acknowleding it) that fewer than 200 million tickets were sold? Or is there some other reason why people assume the knowledge that somebody won increases their personal odds? I don’t understand lotteries, but why the focus on “sold”? At the beginning of the experiment, there are 200 million possible outcomes. At the end of the experiment there is one outcome. Whether you or anyone picked one of the outcomes doesn’t change those two facts. Prior to the winning number being selected each possible outcome has a 1 in 200 million chance of occurring. After the winning number is selected, the chances of any other number being selected drops to zero while the chance of the winning number being selected (it just was) jumps to 1. Seems straigthforward to me. Whether you also picked one of those numbers, or whether anyone did, doesn’t change the above argument. Your selection of a number doesn’t effect the odds at all. So the odds that you picked a winning number is one in 200 million, before and after you selected the number. At least that is how I understand this game. ---
**Author:** ![XT](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/xt/32/456_2.png) [@XT](https://boards.straightdope.com/u/XT)\ **Post date:** [April 28, 2010, 7:22pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/30 "2010-04-28T19:22:55Z")
I guess I’m totally spacing on this one. I would have thought that if you got your number before the drawing then your probability of having the winning number would be exactly the same whether you knew if there was a winner or not, until you actually find out what the numbers for the lottery were (obviously if you won, then your probability is 1, while if you lost it’s 0). Granted, I haven’t even considered probability statistics in \, but the formulas I remember for calculating the probability didn’t have variables for if someone won or not, or the number of winners. I’ll take it as a given, though, that I’m simply looking at the problem incorrectly, so I’ll apologize to the OP for my totally incorrect answer up thread and slink off… -XT ---
**Author:** ![Voyager](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/voyager/32/133_2.png) [@Voyager](https://boards.straightdope.com/u/Voyager)\ **Post date:** [April 28, 2010, 7:33pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/31 "2010-04-28T19:33:46Z")
> [@Indistinguishable](#): > > The only part of the OP I made use of was that, if you were to learn that there wasn’t a winner, the probability that you are a winner will drop to zero, upon conditioning upon this information. From this, I concluded that if you were to learn that there was a winner, the probability that you are a winner will rise, upon conditioning upon this information. I gave the mathematics to support this, as well, though it’s all rather simple. > > If I have misappreciated what was meant when the OP said the probability will drop to zero upon learning that there isn’t a winner, please clarify the situation for me and the others in what would be the same confusion. As others have mentioned, it depends on the combinations covered. If you knew that all combinations had been covered, learning that there is a winner does not increase your probability of winning at all. If however p/n combinations had been covered then knowing there is a winner makes your probability of winning 1/p \> 1/n of p \< n. If p = n, then the probability doesn’t change. If however there is no winner (assuming a number was picked) then it must be in the n-p unpicked combinations, and the probability of anyone with a ticket winning is now 0. ---
**Author:** ![Maserschmidt](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/maserschmidt/32/18829_2.png) [@Maserschmidt](https://boards.straightdope.com/u/Maserschmidt)\ **Post date:** [April 28, 2010, 7:48pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/32 "2010-04-28T19:48:31Z")
> [@Voyager](#): > > As others have mentioned, it depends on the combinations covered. If you knew that all combinations had been covered, learning that there is a winner does not increase your probability of winning at all. If however p/n combinations had been covered then knowing there is a winner makes your probability of winning 1/p \> 1/n of p \< n. If p = n, then the probability doesn’t change. If however there is no winner (assuming a number was picked) then it must be in the n-p unpicked combinations, and the probability of anyone with a ticket winning is now 0. Yes, this is what many of us are saying. And it’s the right answer, I swear. ---
**Author:** ![Munch](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/munch/32/5281_2.png) [@Munch](https://boards.straightdope.com/u/Munch)\ **Post date:** [April 28, 2010, 8:04pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/33 "2010-04-28T20:04:48Z")
> [@CurtC](#): > > You can break down that 1 in 200,000,000 number this way: > > 1/200,000,000 = (prob that someone will win) \* (prob that if someone wins, it will be you) No - this is false logic. (prob that if someone wins, it will be you) is entirely independent from (prob that someone will win). Powerball odds are, for one ticket, 1 in 195,249,054 ([cite](http://www.powerball.com/powerball/pb_prizes.asp), but the OP changed it to 200,000,000 for simplicity). Those odds are the same for each ticket whether you bought 1 ticket or 1,000,000 tickets. Those odds are also the same if every single lottery number was taken, or if your ticket was the only one sold. The only information we have is that there is a winning ticket - and the only difference that means to your odds is that they aren’t reduced to zero. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [April 28, 2010, 9:38pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/34 "2010-04-28T21:38:38Z")
> [@Munch](#): > > No - this is false logic. (prob that if someone wins, it will be you) is entirely independent from (prob that someone will win). “(prob that if someone wins, it will be you)” was poor wording. Rather, what was meant was “given that someone wins, prob that it will be you”; i.e., the conditional probability p(you win | someone wins). It is by definition true that p(you win) = p(you win | someone wins) \* p(someone wins). On this point, **CurtC** is correct. ---
**Author:** ![CurtC](https://avatars.discourse-cdn.com/v4/letter/c/ce73a5/32.png) [@CurtC](https://boards.straightdope.com/u/CurtC)\ **Post date:** [April 28, 2010, 9:50pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/35 "2010-04-28T21:50:38Z")
> [@rbroome](#): > > Whether you also picked one of those numbers, or whether anyone did, doesn’t change the above argument. Your selection of a number doesn’t effect the odds at all. So the odds that you picked a winning number is one in 200 million, before and after you selected the number. I can point out why this must be wrong by taking it to an extreme. What if only ten tickets had been sold, you have one of them, and you read in the next morning’s paper that one of those ten had won? What are your odds then? I would definitely buy that ticket from you for way more than you paid for it. Because it has a one in ten chance of being the powerball winner. Therefore, finding out that there was a winning ticket _does_ change the probability of yours being a winner. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [April 28, 2010, 9:59pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/36 "2010-04-28T21:59:20Z")
> [@Indistinguishable](#): > > “(prob that if someone wins, it will be you)” was poor wording. Rather, what was meant was “given that someone wins, prob that it will be you”; i.e., the conditional probability p(you win | someone wins). It is by definition true that p(you win) = p(you win | someone wins) \* p(someone wins). On this point, **CurtC** is correct. Actually, I should word the tenses on this a little differently, to better reflect the situation and stave off misunderstandings: the quantity under discussion is “give that someone has won, the probability that you have won”, aka p(you’ve won | someone has won). p(you have won) = p(you have won | someone won) \* p(someone has won). ---
**Author:** ![Darth\_Panda](https://avatars.discourse-cdn.com/v4/letter/d/ee7513/32.png) [@Darth\_Panda](https://boards.straightdope.com/u/Darth_Panda)\ **Post date:** [April 28, 2010, 10:59pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/37 "2010-04-28T22:59:55Z")
The answer has been provided several ways alredy, but consider one more exmple. On any given day, your mom has a chance of dying. Someone calls you up and tells you that one of your parents died. Has the probability that your mom died increased? Ok, not actually the same thing - but seriously, information changes probabilities, it just does. It alters what is possible and what is not. Finding out that a car that you previously knew nothing about is not orange increases the odds that it is green. ---
**Author:** ![Bryan\_Ekers](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bryan_ekers/32/183_2.png) [@Bryan\_Ekers](https://boards.straightdope.com/u/Bryan_Ekers)\ **Post date:** [April 28, 2010, 11:25pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/38 "2010-04-28T23:25:10Z")
> [@Indistinguishable](#): > > From this, I concluded that if you were to learn that there was a winner, the probability that you are a winner will rise, upon conditioning upon this information. Well, more accurately, it won’t _drop_. The probably of you winning, once is it known that _someone_ won, may stay the same or it may rise. It would stay the same in a raffle-type lottery in which each participant has a unique number (and all tickets were sold), or (far less likely) in a powerball-type lottery if every possible combination was chosen by an equal number of people. > [@](#): > > If I have misappreciated what was meant when the OP said the probability will drop to zero upon learning that there isn’t a winner, please clarify the situation for me and the others in what would be the same confusion. I don’t understand what _could_ be confusing about this. If he knows nobody won, he knows the odds are zero that _he_ won. ---
**Author:** ![Chipacabra](https://avatars.discourse-cdn.com/v4/letter/c/e19b73/32.png) [@Chipacabra](https://boards.straightdope.com/u/Chipacabra)\ **Post date:** [April 28, 2010, 11:50pm UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/39 "2010-04-28T23:50:53Z")
I think the fundamental misunderstanding is that the OP is treating probability like it’s some objective thing that exists on its own merit. Probability (in this context, we’re not going to delve into quantum stuff) isn’t a thing, it’s a measurement of uncertainty. When you first buy a ticket, the only meaningful uncertainty to measure is whether or not the numbers you chose will be the numbers drawn. That’s your odds of winning. After the drawing, you can measure the uncertainty of whether the numbers you chose _were_ the numbers drawn. That’s the exact same odds of winning. Now, if you know that there was a winner, you can start talking about your odds of whether or not you are that winner. But this doesn’t mean the probability changed, it just means you’re measuring something else! If 100 people bought tickets, and there was 1 winner, then you have a 1 in 100 chance of being that winner. This is a new and _different_ probability measurement. If you don’t know how many people bought tickets, then you haven’t actually gained any information. The fact that you have purchased a lottery ticket, and that _someone_ chose the correct numbers are completely independent events. Without knowing how many other tickets were purchased, the fact that someone had a winning ticket isn’t enough information to make a new measurement. ---
**Author:** ![Darth\_Panda](https://avatars.discourse-cdn.com/v4/letter/d/ee7513/32.png) [@Darth\_Panda](https://boards.straightdope.com/u/Darth_Panda)\ **Post date:** [April 29, 2010, 12:58am UTC](https://boards.straightdope.com/t/lottery-probability-question/537594/40 "2010-04-29T00:58:32Z")
You don’t need to be able to measure x to know that it is greater than y - if n is positive then x plus n is greater than y, even though I have no idea what either x or y are. [Previous page](https://boards.straightdope.com/t/lottery-probability-question/537594.md?page=1) [Next page](https://boards.straightdope.com/t/lottery-probability-question/537594.md?page=3)