# Simple Probability? **URL:** \ **Category:** Factual Questions\ **Created:** [October 27, 2000, 2:05am UTC](https://boards.straightdope.com/t/simple-probability/38530 "2000-10-27T02:05:24Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![DeutschFox](https://avatars.discourse-cdn.com/v4/letter/d/49beb7/32.png) [@DeutschFox](https://boards.straightdope.com/u/DeutschFox)\ **Post date:** [October 27, 2000, 2:05am UTC](https://boards.straightdope.com/t/simple-probability/38530/1 "2000-10-27T02:05:24Z")
This is a probability problem which I am having difficulty with … maybe someone could answer it for me and there are supposely no tricks in it ! A cab driver is at an airport where there is a 50% probability that any given passenger will want to go to the East side and 50% to the West side. Two independent passengers get in the back of his cab simultaneously. The cabbie turns around and looks at one of the two passengers in the back seat and asks where does he want to go? The passenger he is looking at says the East side. What is the probability that he will be able to take both to the East side? Easy, 50% right? 50%/50% for the other passenger to the East side. Cabbie gets two tips, saves on tolls, gas, etc. … real affects … Then one night two independent passengers get in the back of his cab simultaneously. The cabbie does NOT turn around to look because he is working on his log book, but ask where do they want to go, East or West side? He hears from one of them, the East side. Now what is the probability that he will be able to take both to the East side? ---
**Author:** ![Greg\_Charles](https://avatars.discourse-cdn.com/v4/letter/g/839c29/32.png) [@Greg\_Charles](https://boards.straightdope.com/u/Greg_Charles)\ **Post date:** [October 27, 2000, 2:23am UTC](https://boards.straightdope.com/t/simple-probability/38530/2 "2000-10-27T02:23:16Z")
That would still be 50% I think. I’ve heard this problem in various forms, but never in this one. If the cabby were to ask, “Does either of you want to go to the east side?” and one said yes, then the answer would definitely be one in three. Why? Because there are four equally likely probabilities. They could both want to go east, both west, A west and B east, or B east and A west. The cabby has eliminated the “both west” possibility with his question, so there are three possibilities left. Only one of those has both passengers going east. The way that you worded it though is a little more confusing. He just asks the question and the first person who answers says east. If both people were going east, the first person to answer would of course say east, but if one is going west, there would be a 50% chance that the first answer would be west. That subtle difference is enough to bring the overall odds back to 50/50. ---
**Author:** ![Smeghead](https://avatars.discourse-cdn.com/v4/letter/s/f1d935/32.png) [@Smeghead](https://boards.straightdope.com/u/Smeghead)\ **Post date:** [October 27, 2000, 2:48am UTC](https://boards.straightdope.com/t/simple-probability/38530/3 "2000-10-27T02:48:07Z")
I think the way it was worded was meant to make the situation equivalent to your 1 in 3 situation, **Greg**. That is, we now know that one passenger wants to go east, but not which one. Therefore, the both east, A east and B west, and B east and A west possibilities are all still equally likely. You make a distinction between the first and second passenger to bring it back to 50/50. I think not looking at the passengers was meant to imply that the cabbie can’t distinguish between the first and second. If it had stated that they had identical voices, it would be more obvious. ---
**Author:** ![Race\_Bannon](https://avatars.discourse-cdn.com/v4/letter/r/5f8ce5/32.png) [@Race\_Bannon](https://boards.straightdope.com/u/Race_Bannon)\ **Post date:** [October 27, 2000, 2:59am UTC](https://boards.straightdope.com/t/simple-probability/38530/4 "2000-10-27T02:59:29Z")
I’d agree with Greg’s first paragraph, except for the first sentence, but I’d state it this way: the probibility of both passengers going East, given that one is going east is 1/3. This seems to cover DeutchFox’s first case. Let me say it a little more technically, the probibility of both passengers going East, given that _the cabbie selects an east-bound passenger to ask first_ (which I’ll assume is a 50/50 shot) is 1/3. But I don’t agree with Greg’s second paragraph. I don’t see why the second case is any different. You’re just asking for the probibility of both passengers going east given that _the first one to speak_ is going east. I’d assume (since you said “no tricks”) that the first one to speak is also a 50/50 shot. So the probibility should still be 1/3. ---
**Author:** ![scr4](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@scr4](https://boards.straightdope.com/u/scr4)\ **Post date:** [October 27, 2000, 3:22am UTC](https://boards.straightdope.com/t/simple-probability/38530/5 "2000-10-27T03:22:42Z")
I think Greg is correct in verything he said. As described by the OP, the first person who happens to answer says “east.” So the probability that person A answers first is 50%, same as turning randomly to one passenger and asking where he wants to go. Here’s another way to think about it. There are four possible configurations of passengers with equal probability: Person A goes to east, person B to west (call that EW), or EE, or WE, or WW. If you ask “do either of you want to go east?” and one answers “yes”, you have eliminated WW. So you are left with EE, WE and EW. Keep in mind that these three are still equally likely, because they were equally likely to begin with. So you have 33% chance of having the EE combination. However, if you ask where they want to go and one answers, there are 8 possibilities. If the combination is WE, person A may answer first (call that We) or person B may answer first (call that wE). So the eight possibilities are Ee, Ew, We, Ww, eE, eW, wE, wW, and these are equally likely. Now, if the first person to answer says “east,” you have eliminated all the combinations with an upper case W. So you are left with Ee, Ew, wE, and eE. So the chance of having eE or Ee is 50%. ---
**Author:** ![Smeghead](https://avatars.discourse-cdn.com/v4/letter/s/f1d935/32.png) [@Smeghead](https://boards.straightdope.com/u/Smeghead)\ **Post date:** [October 27, 2000, 3:29am UTC](https://boards.straightdope.com/t/simple-probability/38530/6 "2000-10-27T03:29:35Z")
But my point is that the cabbie doesn’t know which passenger answered first - that’s why the problem bothered to specify that he wasn’t looking at the passengers. Therefore, the cabbie’s only knows that at least one of the passengers wants to go east. Any explanation that leads to a 50/50 answer requires a distinction between the person that answered and the other. I am interpreting the OP to mean that that distinction is not possible. ---
**Author:** ![scr4](https://avatars.discourse-cdn.com/v4/letter/s/59ef9b/32.png) [@scr4](https://boards.straightdope.com/u/scr4)\ **Post date:** [October 27, 2000, 3:35am UTC](https://boards.straightdope.com/t/simple-probability/38530/7 "2000-10-27T03:35:00Z")
> [@](#): > > Any explanation that leads to a 50/50 answer requires a distinction between the person that answered and the other. I don’t think that’s the case. All that is required is that there is a 50% chance that person A answers first, and 50% chance that person B answers first. This is in contrast to the “do either of you want to go east” problem, where person A answers first only if he is going east, etc. I know my explanation was hard to follow but that’s the best I could do without tables. Read it again - when I eliminated 4 of the 8 possible configurations, it didn’t require any knowledge of which person answered the question. If the first answer (from whichever person) is “east”, you can immediately eliminate the 4 choices, and the remaining 4 choices are all still equally possible. ---
**Author:** ![Greg\_Charles](https://avatars.discourse-cdn.com/v4/letter/g/839c29/32.png) [@Greg\_Charles](https://boards.straightdope.com/u/Greg_Charles)\ **Post date:** [October 27, 2000, 3:54am UTC](https://boards.straightdope.com/t/simple-probability/38530/8 "2000-10-27T03:54:16Z")
You did a good job of explaining that scr4. The form in which I’ve usually heard this problem concerns a woman who has two children and somehow you learn one is a daughter. (I think Cecil himself covered this at some point.) Anyway, the probability that the other one is a daughter is either 50/50 or 1 in 3 depending on some very subtle differences in how you learn that one kid was a daughter. In the cabby problem, there’s no fundamental difference between the cabby choosing a passenger to ask the question and the passenger choosing himself by simply answering first. ---
**Author:** ![Race\_Bannon](https://avatars.discourse-cdn.com/v4/letter/r/5f8ce5/32.png) [@Race\_Bannon](https://boards.straightdope.com/u/Race_Bannon)\ **Post date:** [October 27, 2000, 3:58am UTC](https://boards.straightdope.com/t/simple-probability/38530/9 "2000-10-27T03:58:00Z")
I think I follow you, scr4, but I’m still sticking with 1/3, like smeghead ( you guys are posting faster than I can think this thru!) I’m not really confident here, but I’d say that the issue is that your cases eE and Ee are not distinct. That is, they are the same case as far as the driver can tell, so you can’t count them twice, they are only one case, so your list still comes down to 1/3. Look at it this way, with two craps dice (a nasty hobby of mine) there are two ways to roll 11, I could roll (5,6) or (6,5). But there’s only one way to roll 12: (6,6). The “other” way to roll 12 is still (6,6) and I can’t tell 'em apart (just like the cabbie can’t tell the passengers voices apart). Even if you have one blue die and one red die, and count(B6,R6) as different from (R6,B6) then there are still twice as many ways to get 11: (B6,R5),(R5,B6),(R6,B5),(B5,R6). ---
**Author:** ![Smeghead](https://avatars.discourse-cdn.com/v4/letter/s/f1d935/32.png) [@Smeghead](https://boards.straightdope.com/u/Smeghead)\ **Post date:** [October 27, 2000, 4:10am UTC](https://boards.straightdope.com/t/simple-probability/38530/10 "2000-10-27T04:10:58Z")
> [@](#): > > \*Originally posted by scr4 \* > \*\*All that is required is that there is a 50% chance that person A answers first, and 50% chance that person B answers first. \*\* That’s what I mean by there’s a distinction being made. You’re narrowing it down to where one of the two passengers answers first, and while that is a true statement, we can never know which of those two cases is true, so that consideration drops out. This is a really subtle difference, but I still think I’m right. All the cabbie knows, or can ever know, is that at least one of them wants to go east. He can never know _which_ one answered. Man, what a crappy cabbie. The OP better cough up an answer to this sooner or later! ---
**Author:** ![Smeghead](https://avatars.discourse-cdn.com/v4/letter/s/f1d935/32.png) [@Smeghead](https://boards.straightdope.com/u/Smeghead)\ **Post date:** [October 27, 2000, 4:15am UTC](https://boards.straightdope.com/t/simple-probability/38530/11 "2000-10-27T04:15:26Z")
I just looked at the forum page - you’ve been a busy little thread-starter, **Fox**. Just thought I’d mention I’m done for the night. I’ll check this thread again tomorrow evening. I’ll probably be up all night thinking about it now. Grumble. ---
**Author:** ![Greg\_Charles](https://avatars.discourse-cdn.com/v4/letter/g/839c29/32.png) [@Greg\_Charles](https://boards.straightdope.com/u/Greg_Charles)\ **Post date:** [October 27, 2000, 5:29am UTC](https://boards.straightdope.com/t/simple-probability/38530/12 "2000-10-27T05:29:19Z")
kelly, kelly, kelly … Smeg, Smeg, Smeg … Now _I’m_ going to go crazy until I can convince you of this. kelly is right that the driver cannot distinguish between scr4’s cases eE and Ee, but this does _not_ mean that they aren’t distinct cases. There are two people and one person answers first. In fact the cabby can’t distinguish between any of the four remaining possibilities, eE, Ee, wE, Ew. (In scr4’s notation, the letter is the destination of each person, and the capital indicates who answered the question.) That’s the whole point of the problem. You’re also right about the dice roll, but you’re not applying it correctly to this problem. I’ll tell you what, if you still have lingering doubts by tomorrow, I’ll write a computer program to prove it to you empirically. ---
**Author:** ![DeutschFox](https://avatars.discourse-cdn.com/v4/letter/d/49beb7/32.png) [@DeutschFox](https://boards.straightdope.com/u/DeutschFox)\ **Post date:** [October 27, 2000, 7:56am UTC](https://boards.straightdope.com/t/simple-probability/38530/13 "2000-10-27T07:56:58Z")
Jeez … I couldn’t remember, but knew it was contrary to logic and seemed to actually affect the cabbie’s earnings. It seems that most of you are saying there are 4 outcomes … WW … EW … WE … EE … and once one of the passengers says East, there would only be three possibilities remaining since WW is not possible … EW … WE … EE … a three sided die! Only a 1 out of 3 probability versus 1 out of 2 (WE … EE) if he looked at the passenger who said East … Thanks … ---
**Author:** ![dylan\_73](https://avatars.discourse-cdn.com/v4/letter/d/ecccb3/32.png) [@dylan\_73](https://boards.straightdope.com/u/dylan_73)\ **Post date:** [October 27, 2000, 2:32pm UTC](https://boards.straightdope.com/t/simple-probability/38530/14 "2000-10-27T14:32:28Z")
I don’t see this 1/3 thing. Think of it as coins, instead of people. The odds of throwing 2 heads (East!) are 1/4, yes? But if you throw one coin, and it lands heads, then ask yourself what is the chance that the second coin will land heads when you throw it? ---
**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:** [October 27, 2000, 2:42pm UTC](https://boards.straightdope.com/t/simple-probability/38530/15 "2000-10-27T14:42:49Z")
> [@](#): > > \*Originally posted by DeutschFox \* > It seems that most of you are saying [that there is] Only a 1 out of 3 probability versus 1 out of 2 Well, no. It looks to me like **kelly** and **smeghead** are saying there’s a 1/3 probability, while **Greg** and **scr4** are saying the probability is 1/2. Therefore, 50% say that the answer is 50%. However, I’d like to weigh in by agreeing with **Greg** and **scr4** that the answer is still 50% (so now most of us are saying that there is a 1 out of 2 probability versus 1 out of 3). Admittedly, the key to this problem is semantics. It looks similar to [the infamous Monty Hall problem](http://www.straightdope.com/classics/a3_189.html) that tripped up even Cecil (this is where **Greg** remembers the son/daughter problem from). However, I think in this case the intuitive answer (50%) is correct, because, as **scr4** points out, \*a distinction can be made between the passengers \* (i.e., first to answer/second to answer). If this distinction could not be made, the answer 1/3 would be correct. I’ll admit this is a fine line. In fact, I think that to be truly convincing, the OP ought to be reworded one way or the other to clearly state that the cabby can or cannot distinguish between the passengers. It appears that the OP was _intended_ to be phrased in such a way as to carry the meaning that the cabby _couldn’t_ distinguish between the two, but that it wasn’t worded correctly. If you search for some of the many Monty Hall threads on this board, you’ll see that most of them wind up boiling down to semantical arguments. ---
**Author:** ![Greg\_Charles](https://avatars.discourse-cdn.com/v4/letter/g/839c29/32.png) [@Greg\_Charles](https://boards.straightdope.com/u/Greg_Charles)\ **Post date:** [October 27, 2000, 5:20pm UTC](https://boards.straightdope.com/t/simple-probability/38530/16 "2000-10-27T17:20:07Z")
OK, I don’t have time to write the program right now, but here’s how it would do its caculations: We agree that there is 25% chance that both passengers are going east, right? So in 1000 trials, 250 should show both passengers going east. Now, the hard part. What are the odds that the first person who answers the destination question says east? We know 50% of the passengers are going east, so it follows that 50% of the time the first person to answer will say east. In 1000 trials, 500 will result in the answer east. Now to determine the probability of the passengers both going east given that the first person to answer said east, we devide the cases where both are going east (250) by the cases where the first person answered east (500) and the answer 50% pops out. It’s not 1 in 3 because there are trials where the passengers are going in opposite directions, but the one going west answers first. In 1000 trials, that should happen 250 times. Those trials are thrown out, just as are the trials where both passengers are going west, because we know the first answer was east. If you can phrase the question so that the only trials we can throw out are the ones where both passengers are going west, then we’ll get 250/750 for 1 in 3. That’s about as well as I can explain it without falling back on formulas. I’ve got them written out here, but posting them would get ugly. I’m not sure they would really convince anyone anyway. ---
**Author:** ![ElvisL1ves](https://avatars.discourse-cdn.com/v4/letter/e/5daacb/32.png) [@ElvisL1ves](https://boards.straightdope.com/u/ElvisL1ves)\ **Post date:** [October 27, 2000, 5:30pm UTC](https://boards.straightdope.com/t/simple-probability/38530/17 "2000-10-27T17:30:21Z")
This thread has gotten buried in math. Try this: If one passenger answers “East” and the second says nothing, then the second one either also wants to go East or he’s deaf. Human nature would be that he’d say “West” if he wanted to go west, but might not bother to say anything if the first passenger had already given the answer he also wanted. So the probability that both want to go east is pretty close to 100%. ---
**Author:** ![Arnold\_Winkelried](https://avatars.discourse-cdn.com/v4/letter/a/3d9bf3/32.png) [@Arnold\_Winkelried](https://boards.straightdope.com/u/Arnold_Winkelried)\ **Post date:** [October 27, 2000, 5:40pm UTC](https://boards.straightdope.com/t/simple-probability/38530/18 "2000-10-27T17:40:45Z")
Come on people! Read the article at the Monty Hall link provided by **zut** , in which Cecil Adams has the following example: > [@](#): > > There is a family with two children. You have been told this family has a daughter. What are the odds they also have a son, assuming the biological odds of having a male or female child are equal? (Answer: 2/3.) Changing the conditions to fit the OP: There are two passengers in a taxi. You have been told that one wants to go east. What are the odds that the other wants to go west, assuming the odds of east vs. west travel are equal? (Answer: 2/3). Therefore the odds of the second wanting to go east are 1/3. Now, to slightly change the problem: There is a family with two children. The oldest child is a daughter. What are the odds the the youngest child is male, assuming the biological odds of having a male or female child are equal? (Answer: 1/2.) or, to change it to taxicabs: There are two passengers in a taxicab. The first passenger to enter the cab wants to go east. What are the odds that the second passenger to enter the cab wants to go west, assuming the odds of east vs. west travel are equal? (Answer: 1/2). The difference is that in the two last examples the passengers are differentiated in some way. ---
**Author:** ![panamajack](https://avatars.discourse-cdn.com/v4/letter/p/47e85d/32.png) [@panamajack](https://boards.straightdope.com/u/panamajack)\ **Post date:** [October 27, 2000, 6:59pm UTC](https://boards.straightdope.com/t/simple-probability/38530/19 "2000-10-27T18:59:25Z")
There’s a very good article that’s already explained this much better than I can, but I can’t find the citation at the moment (there is actually a page on a website out there devoted to Marilyn vos Savant’s handling of this problem). This may not go over well, but I’ll try to say what I can for now. Greg Charles was correct in the very first response, though one clarification might be needed about the second statement. (see below) It is indeed often a matter of semantics and I think zut explains that well. In _both_ cases as stated by the OP the answer is 50%. I’ll try to explain this by comparing the problems that Arnold presented to the OP. Arnold Winkelreid wrote : > [@](#): > > Changing the conditions to fit the OP: There are two passengers in a taxi. You have been told that one wants to go east. What are the odds that the other wants to go west, assuming the odds of east vs. west travel are equal? (Answer: 2/3) The key phrase is “You have been told that one …” This is usually understood to mean “given that” in a ‘Bayesian’[sup]\*[/sup] sense, meaning an unknown one of a group. If this is true, the answer of 2/3 is correct. But this interpretation of the phrase “you have been told that one” is not really the situation in the OP : DeutschFox wrote: > [@](#): > > The cabbie does NOT turn around to look because he is working on his log book, but ask where do they want to go, East or West side? He hears from one of them, the East side. Here’s the subtle part. It is immaterial whether the cabbie turns around to look or not. What we know is that he hears from one of them. A specific person, not an unknown one of a group. (This is the semantics : the passenger isn’t ‘known’ in the sense that the cabbie knows who spoke, but she is ‘known’ in a probabilistic sense because she spoke. ) We can in fact label her as the passenger who spoke, thus she is distinct. The answer still works out to 50%. This case is in fact covered in Cecil’s column, near the very end (Len Ragozin’s contribution.) The clarification I mentioned about Greg’s first post has to do with this. He says that if the cabbie asks “Do either of you want to go to the East side?” and one person responds yes, the answer is 1/3 only if the passenger who spoke knows the intention of the other passenger, and speaks for her to cover that case. Of course, this situation likely fails under a psychological explanation. Why would only one speak when a question like that is asked? [sup]\*[/sup] A little surprised this hasn’t been brought up yet. Bayes’ Rule is a formula for conditional probabilities (“A given B”) like these . By ‘Bayesian’, I mean that the analysis following this rule will work. ---
**Author:** ![ZenBeam](https://avatars.discourse-cdn.com/v4/letter/z/3ab097/32.png) [@ZenBeam](https://boards.straightdope.com/u/ZenBeam)\ **Post date:** [October 27, 2000, 7:05pm UTC](https://boards.straightdope.com/t/simple-probability/38530/20 "2000-10-27T19:05:34Z")
Taking Arnold’s example one tiny step further: There are two passengers in a taxicab. The first passenger to _answer the cabbie_ wants to go east. What are the odds that the second passenger to enter the cab wants to go west, assuming the odds of east vs. west travel are equal? (Answer: 1/2). [Next page](https://boards.straightdope.com/t/simple-probability/38530.md?page=2)