# probability question **URL:** \ **Category:** Factual Questions\ **Created:** [December 29, 2007, 7:54am UTC](https://boards.straightdope.com/t/probability-question/431579 "2007-12-29T07:54:25Z")\ **Posts on this page:** 19\ **Page:** 1
**Author:** ![D\_White](https://avatars.discourse-cdn.com/v4/letter/d/839c29/32.png) [@D\_White](https://boards.straightdope.com/u/D_White)\ **Post date:** [December 29, 2007, 7:54am UTC](https://boards.straightdope.com/t/probability-question/431579/1 "2007-12-29T07:54:25Z")
I made up the following probability question in hopes of explaining another probability question to someone, and now it’s got me confused: > [@](#): > > One million people, including yourself, purchase lottery tickets of which there will be three winners. If you could get someone with inside info to rule out 999,994 losers, other than yourself if you should be one, is your probability of being a winner 1/2? I explained how the answer is no and that you would still have a 1/1,000,000 chance of being a winner. Of course the other 5 people have an awesome chance of winning. My first instinct was that each one has a 1/2 chance of winning. Then it occurred to me that I must be wrong. Since one person has an only 1/1,000,000 chance of winning, the remaining five must each have a better than 1/2 chance of winning. How do I determine exactly what that probability is? ---
**Author:** ![Randy\_Seltzer](https://avatars.discourse-cdn.com/v4/letter/r/ee7513/32.png) [@Randy\_Seltzer](https://boards.straightdope.com/u/Randy_Seltzer)\ **Post date:** [December 29, 2007, 8:07am UTC](https://boards.straightdope.com/t/probability-question/431579/2 "2007-12-29T08:07:23Z")
What you’ve got is a variation on the Monty Hall problem. This is a favorite on this board. Check out these threads: [http://boards.straightdope.com/sdmb/showthread.php?t=426065&highlight=monty+hall](http://boards.straightdope.com/sdmb/showthread.php?t=426065&highlight=monty+hall) [http://boards.straightdope.com/sdmb/showthread.php?t=426065&highlight=monty+hall](http://boards.straightdope.com/sdmb/showthread.php?t=426065&highlight=monty+hall) [http://boards.straightdope.com/sdmb/showthread.php?t=411943&highlight=monty+hall](http://boards.straightdope.com/sdmb/showthread.php?t=411943&highlight=monty+hall) Short answer: Yes, your chances are still one in a million. But if you randomly trade with another of the remaining six ticket holders, your chance improves to 1/2. So counter-intuitively, everyone’s chances improve by trading tickets with someone else. (It doesn’t make sense at first glance, but trust me: submit your will to it before it drives you mad!) ---
**Author:** ![D\_White](https://avatars.discourse-cdn.com/v4/letter/d/839c29/32.png) [@D\_White](https://boards.straightdope.com/u/D_White)\ **Post date:** [December 29, 2007, 8:13am UTC](https://boards.straightdope.com/t/probability-question/431579/3 "2007-12-29T08:13:10Z")
[QUOTE=Randy Seltzer] Short answer: Yes, your chances are still one in a million. But if you randomly trade with another of the remaining six ticket holders, your chance improves to 1/2. [/QUOTE] That’s what I originally thought. But other five (there are five remaining ticket holders by the way) each have a slightly less than 3/5 chance of winning, not 1/2. I’m trying to figure out _exactly_ how much of a chance the other five have. ---
**Author:** ![D\_White](https://avatars.discourse-cdn.com/v4/letter/d/839c29/32.png) [@D\_White](https://boards.straightdope.com/u/D_White)\ **Post date:** [December 29, 2007, 8:18am UTC](https://boards.straightdope.com/t/probability-question/431579/4 "2007-12-29T08:18:09Z")
[QUOTE=Randy Seltzer] So counter-intuitively, everyone’s chances improve by trading tickets with someone else. (It doesn’t make sense at first glance, but trust me: submit your will to it before it drives you mad!) [/QUOTE] No, the only one who’s chances will improve after trading improve is the one who is currently at 1 in a million. ---
**Author:** ![Randy\_Seltzer](https://avatars.discourse-cdn.com/v4/letter/r/ee7513/32.png) [@Randy\_Seltzer](https://boards.straightdope.com/u/Randy_Seltzer)\ **Post date:** [December 29, 2007, 8:29am UTC](https://boards.straightdope.com/t/probability-question/431579/5 "2007-12-29T08:29:37Z")
I don’t usually participate in the Monty Hall threads, and upon reflection, I realized why not. I’m screwing it up. Anyway. > [@](#): > > That’s what I originally thought. But other five (there are five remaining ticket holders by the way) each have a slightly less than 3/5 chance of winning, not 1/2. I’m trying to figure out exactly how much of a chance the other five have. Either your conceptual problem is as follows, or I’m misunderstanding the question: Your odds do not change simply because you are acquainted with the cheater. Being buddies with the inside man does not change the fact that you are in the same position as the other five “surviving” ticket holders. Each of the six survivors have an equal chance. That is, 3/1,000,000. You would insist that this does not add up to a probability of 1. You would be correct. This shocks the intuition, I know, because to create your whole probability of one, you need to re-introduce the “eliminated” tickets. If you want the _individual_ probabilities to add up to 1 post-elimination, everyone needs to trade tickets. Then, everyone will have a 3/6 chance, which does, in fact, add up to one. For an interesting perspective on counterintuitive probability puzzles, read _The Curious Incident of the Dog in the Nighttime_, a novel which discusses the Monty Hall problem briefly. ---
**Author:** ![D\_White](https://avatars.discourse-cdn.com/v4/letter/d/839c29/32.png) [@D\_White](https://boards.straightdope.com/u/D_White)\ **Post date:** [December 29, 2007, 8:40am UTC](https://boards.straightdope.com/t/probability-question/431579/6 "2007-12-29T08:40:53Z")
[QUOTE=D\_White] No, the only one who’s chances will improve after trading improve is the one who is currently at 1 in a million. [/QUOTE] I mean 3/1,000,000 ---
**Author:** ![D\_White](https://avatars.discourse-cdn.com/v4/letter/d/839c29/32.png) [@D\_White](https://boards.straightdope.com/u/D_White)\ **Post date:** [December 29, 2007, 8:43am UTC](https://boards.straightdope.com/t/probability-question/431579/7 "2007-12-29T08:43:00Z")
[QUOTE=Randy Seltzer] Each of the six survivors have an equal chance. That is, 3/1,000,000. You would insist that this does not add up to a probability of 1. You would be correct. [/QUOTE] Each person DOES NOT have a 3/1,000,000 chance of winning. Five of the men have an almost 3/5 chance of winning. The only one who has a 3/1,000,000 chance of winning is the one you’re calling the cheater. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [December 29, 2007, 9:29am UTC](https://boards.straightdope.com/t/probability-question/431579/8 "2007-12-29T09:29:19Z")
[QUOTE=Randy Seltzer] Anyway.Either your conceptual problem is as follows, or I’m misunderstanding the question: Your odds do not change simply because you are acquainted with the cheater. Being buddies with the inside man does not change the fact that you are in the same position as the other five “surviving” ticket holders. [/QUOTE] I think you’re misunderstanding the question. **D\_White** has set up the problem so that “Monty” _always_ picks “you” to be in the final six. The other two (or occasionally three) losers in the final six are randomly picked, and the three winners are _guaranteed_ to be present in the final six. That means “you” _really are_ different from the other people. Probability is based on what you _do_ and _don’t_ have knowledge of. In this case, you know that “you” are guaranteed a spot in the final six, so your odds of winning are still 3/1,000,000, because appearing in the final six tells you nothing about your chance of winning. However, you also know that all three winners are guaranteed to be in the final six, so appearing in the final six _does_ affect the chance that any of the other five people are winners. Their chance of winning is now about 3/5. The exact odds for the other five people are 3/5 times the probablity that “you” _aren’t_ a winner plus 2/5 times the probablity that “you” _are_ a winner. That’s (999,997/1,000,000)_(3/5) + (3/1,000,000)_(2/5) = 2,999,997/5,000,000. Also, remember that probability is based on what you _do_ and _don’t_ have knowledge of. That means this calculation is from _your_ perspective, taking into account what _you_ know. The odds would be different from the perspective of “Monty” (since he already knows who the winners are) or from the perspective of one of the other five finalists (since, presumably, they don’t know Monty has rigged the final). ---
**Author:** ![D\_White](https://avatars.discourse-cdn.com/v4/letter/d/839c29/32.png) [@D\_White](https://boards.straightdope.com/u/D_White)\ **Post date:** [December 29, 2007, 9:41am UTC](https://boards.straightdope.com/t/probability-question/431579/9 "2007-12-29T09:41:30Z")
Thanks, zut. I just got done calculating and I came up with a slightly different answer than yours. You came up with 2,999,997/5,000,000 which equals .5999994 I came up with 5,999,998/10,000,000 which equals .5999998 .5999998 x 5 = 2.999999 The guy who has a 1/1,000,000 chance = .000001 ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [December 29, 2007, 5:33pm UTC](https://boards.straightdope.com/t/probability-question/431579/10 "2007-12-29T17:33:24Z")
You’re mixing up your odds. The guy with the 1/1,000,000 chance doesn’t really have a 1/1,000,000 chance; he has a **3** /1,000,000 chance, since there are three prizes. If you factor that into your calculations, yours should match mine. ---
**Author:** ![milquetoast](https://avatars.discourse-cdn.com/v4/letter/m/258eb7/32.png) [@milquetoast](https://boards.straightdope.com/u/milquetoast)\ **Post date:** [December 29, 2007, 7:00pm UTC](https://boards.straightdope.com/t/probability-question/431579/11 "2007-12-29T19:00:10Z")
Why are you guys overthinking this? > [@](#): > > One million people, including yourself, purchase lottery tickets of which there will be three winners. If you could get someone with inside info to rule out 999,994 losers, other than yourself if you should be one, is your probability of being a winner 1/2? If the guy with inside inside info rules out 999,994 losing tickets, there’s only 6 possible tickets left, of which I have one, that can be a winner. If three of the six remaining tickets are winners, I have a 1 in 2 chance of being a winner. Though the pool started with 1,000,000 tickets. It’s not down to 6. The original number is irrelevant. Half of the remaining “unknown” tickets are winners. That’s all that matters. The guy eliminating the losers is not Monty Hall. He’s just making sure that you are in the final 6. Period! Question over. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [December 29, 2007, 7:21pm UTC](https://boards.straightdope.com/t/probability-question/431579/12 "2007-12-29T19:21:20Z")
[QUOTE=milquetoast] Why are you guys overthinking this? If the guy with inside inside info rules out 999,994 losing tickets, there’s only 6 possible tickets left, of which I have one, that can be a winner. If three of the six remaining tickets are winners, I have a 1 in 2 chance of being a winner. [/QUOTE] The reason this is being “overthought” is because your answer is completely incorrect. As you say, the guy eliminating the losers _makes sure that you are in the final 6._ That means that “being in the final six” tells you _nothing at all_ about your chances of winning, and thus your chances of winning stay the same as they were at the beginning: 3/1,000,000. ---
**Author:** ![milquetoast](https://avatars.discourse-cdn.com/v4/letter/m/258eb7/32.png) [@milquetoast](https://boards.straightdope.com/u/milquetoast)\ **Post date:** [December 29, 2007, 7:49pm UTC](https://boards.straightdope.com/t/probability-question/431579/13 "2007-12-29T19:49:49Z")
**zut** , You’re right. I’m wrong. The ace of spades analogy in [Cecil’s original column](http://www.straightdope.com/classics/a3_189.html) helped me see the error of my ways. My apologies, milquetoast ---
**Author:** ![D\_White](https://avatars.discourse-cdn.com/v4/letter/d/839c29/32.png) [@D\_White](https://boards.straightdope.com/u/D_White)\ **Post date:** [December 30, 2007, 5:19am UTC](https://boards.straightdope.com/t/probability-question/431579/14 "2007-12-30T05:19:03Z")
[QUOTE=zut] You’re mixing up your odds. The guy with the 1/1,000,000 chance doesn’t really have a 1/1,000,000 chance; he has a **3** /1,000,000 chance, since there are three prizes. If you factor that into your calculations, yours should match mine. [/QUOTE] Doh! I don’t know why I keep doing that. Thanks again ---
**Author:** ![chorpler](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chorpler/32/2888_2.png) [@chorpler](https://boards.straightdope.com/u/chorpler)\ **Post date:** [December 30, 2007, 12:23pm UTC](https://boards.straightdope.com/t/probability-question/431579/15 "2007-12-30T12:23:16Z")
I don’t know about anybody else, but it would probably be simpler for me to think about if the original question was rephrased so that there was only one prize, and you eliminated all but two of the lottery ticket holders, yourself and some random guy. Then the question can still be asked “Is your probability of winning 1/2?” Cecil tackled a problem very similar to this one during the whole [Monty Hall debacle](http://www.straightdope.com/classics/a3_189.html), where he said: > [@](#): > > Finally, this one from a friend. Suppose we have a lottery with 10,000 “scratch-off-the-dot” tickets. The prize: a car. Ten thousand people buy the tickets, including you. 9,998 scratch off the dots on their tickets and find the message YOU LOSE. Should you offer big money to the remaining ticketholder to exchange tickets with you? (Answer: hey, after all this drill, you figure it out.) To which reader Jim Balter of Los Angeles replied: > [@](#): > > … If you think the answer is “yes,” you are wrong. If you think the answer is “no,” then you are intentionally misleading your readers … And Cecil replied: > [@](#): > > Do you think I could possibly screw this up twice in a row? Of course I could. But not this time. Cecil is well aware the answer to the lottery question is “no”–if there are only two tickets left, they have equal odds of being the winner. The difference between this and the Monty Hall question is that we’re assuming Monty knows where the prize is, and uses that information to select a non-prize door to open; whereas in the lottery example the fact that the first 9,998 tickets are losers is a matter of chance. I put the question at the end of a line of dissimilar questions as a goof–not very sporting, but old habits die hard. I think the only difference is that in **D\_White** ’s question, you cheat and find out that you’re one of the possible winners ahead of time. Of course there’s probably some other subtlety I’m missing. ---
**Author:** ![The\_Seventh\_Deadly\_Finn](https://avatars.discourse-cdn.com/v4/letter/t/6bbea6/32.png) [@The\_Seventh\_Deadly\_Finn](https://boards.straightdope.com/u/The_Seventh_Deadly_Finn)\ **Post date:** [December 30, 2007, 1:42pm UTC](https://boards.straightdope.com/t/probability-question/431579/16 "2007-12-30T13:42:44Z")
This seems to get a little fogged because of some ambiguity about what the “inside man” is up to. This could be one of two things: a) he’s only talking to you in the first place because he’s got a list of 999,994 losers and you’re not on it, in which case yours odds really are at 3 in 6, or b) he’s got a list of 5 potential winners and he adds you just for the hell of it to make you feel special till the drawing, in which case, yeah; it’s still 3/1,000,000. I feel like the question is phrased so you could take it either way. Whether the inside guy is a Monty Hall-type figure depends on why he made you one of his special 6. I’m sure I missed something obvious, though. Go ahead and rip me a new one, I deserve it. ---
**Author:** ![The\_Seventh\_Deadly\_Finn](https://avatars.discourse-cdn.com/v4/letter/t/6bbea6/32.png) [@The\_Seventh\_Deadly\_Finn](https://boards.straightdope.com/u/The_Seventh_Deadly_Finn)\ **Post date:** [December 30, 2007, 1:46pm UTC](https://boards.straightdope.com/t/probability-question/431579/17 "2007-12-30T13:46:36Z")
Yup. I did; I missed the phrase “other than yourself if you should be one.” Although why any sane human being would torture him or herself in this fashion is beyond me. You’d have to have some kind of disappointment fetish. ---
**Author:** ![zut](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/zut/32/2875_2.png) [@zut](https://boards.straightdope.com/u/zut)\ **Post date:** [December 30, 2007, 5:27pm UTC](https://boards.straightdope.com/t/probability-question/431579/18 "2007-12-30T17:27:49Z")
[QUOTE=chorpler] I think the only difference is that in **D\_White** ’s question, you cheat and find out that you’re one of the possible winners ahead of time. Of course there’s probably some other subtlety I’m missing. [/QUOTE] What **thetruewheel** said. I’m reasonably confident that the intention of **D\_White** ’s scenario is that the “inside man” is Monty Hall. In other words, “Monty” sets it up ahead of time that _you_ will be part of the final six, whether or not you’re a winner. Since that is set up from the start, your chances of winning never change. _If_ you were to alter the scenario, so that there is no precondition that _you_ will be part of the final six, then learning you are part of the final six _does_ alter your chances. This scenario is more akin to the one you quoted from **Cecil**. ---
**Author:** ![chorpler](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chorpler/32/2888_2.png) [@chorpler](https://boards.straightdope.com/u/chorpler)\ **Post date:** [December 30, 2007, 8:19pm UTC](https://boards.straightdope.com/t/probability-question/431579/19 "2007-12-30T20:19:00Z")
[QUOTE=zut] What **thetruewheel** said. I’m reasonably confident that the intention of **D\_White** ’s scenario is that the “inside man” is Monty Hall. In other words, “Monty” sets it up ahead of time that _you_ will be part of the final six, whether or not you’re a winner. Since that is set up from the start, your chances of winning never change. _If_ you were to alter the scenario, so that there is no precondition that _you_ will be part of the final six, then learning you are part of the final six _does_ alter your chances. This scenario is more akin to the one you quoted from **Cecil**. [/QUOTE] Okay, I see what you and **thetruewheel** were getting at – so in the first case, “Monty” essentially just happens to find almost all the losers while not checking on you at all, which tells you nothing about whether or not you are also a loser. Whereas if you were to go in to Monty and say “Look up all the losing tickets with your cheatin’ computer,” and you didn’t happen to be among them, _then_ you would have a much better chance of having a winning ticket. Because in the first scenario, you’ve basically been artificially excluded from the analysis, and the overwhelming likelihood is that _if Monty had included you in the analysis, you would have been among the losers_, so you’re still overwhelmingly likely to have a losing ticket.