Hi Panamajack … what is the outcome if one of them says “West”? Given the combinations of EW and WE … the lucky East guy would live … and maybe go flag down another cab? Thanks … Andi
The odds are 50/50. All you have done is reworded the OP. Rather then the first guy exiting the car, he gets killed by a death ray.
It’s still 50/50.
panamajack calls it correctly; there is a very subtle difference between “the cabbie … only knows that at least one of them said ‘East’” (which is what I said) and “both passengers speak up, but the cabbie only hears one of them (and he says east)” (which is what you said).
In the first case, the odds are 33/66. This is because there is no distinction between the two passengers, so the cabbie is not sure if he “didn’t hear quite hear the 2nd guy” (note that you said that, not me). Maybe he heard one, maybe he heard the other, or maybe he heard both, all he knows is that at least one said “East”.
In the second case, you are correct: the odds are 50/50. This is because there is a distinction between the two passengers, namely, that the cabbie hears one and doesn’t quite hear the other.
This difference between the two problem phrasings is the crux of the “A family has two children…” problem that both Arnold and I refer to on the first page (and Cecil refers to in the column I give a link to). This cabbie problem is exactly the same problem, it’s just (evidently) more difficult to phrase unambiguously.
madhatter: If you look closer, panamajack is agreeing with you that the odds are 50/50 in the case he gives (I agree too). Now you need to convince DeutschFox.
DeutschFox: I’ve posted a number of times in this thread, attempting to explain why you are incorrect. I’ve even given examples and asked if you agree or disagree. You, on the other hand, seem to be repeating the same (incorrect) conclusion over and over. If you are interested in understanding this problem, please reread my earlier posts and tell me what you disagree (or agree) with.
This is an interesting and tricky problem and, on the whole, an interesting discussion. However, it gets a little tiring explaining the same thing multiple times without any feedback.
Ahh, got it. I see where the misunderstanding is. Look at what you just said “The EW, WE, EE outcomes…” One was E so you can’t have both EW and WE as remaining options, now can you?
It could only possibly be one of those. So it can only be EW or EE, or if you want it the other way around, WE or EE. Either way you only get two options. Does this make sense?
The coin game where you were taking advantage of your “friend” 
would be similar to the taxi driver asking “Is anyone going East” and one of the passangers answering “Yes”.
Ok, then let’s say the question, from a wording viewpoint, were asked and answered this way? What would the answer be … 1/2 or 1/3 ?
But is this really different than a cab driver asking “Where do you want to go?” … and one answering “East” ?
Wow, I can’t believe this is still going on! If you phrased it that way, DrDoom, the answer goes to 1 in 3. That is if you assume that only one of the passengers anwers, and that if there was a westbound passenger, he would ignore the question entirely. I’d explain the difference, but it’s already been covered in gruesome detail by previous posts.
Is it really possible to get a 1 in 3 answer, even by rephrasing the question? Wouldn’t that still affect the cab driver’s passenger fares, tips, etc., as previously discussed?
Just like the death ray question, all you have done is re-worded the OP.
Once the first guy responds with “Yes” (East), then the second guy is going to respond with either “Yes” (East) or “No” (West).
The odds still remain exactly the same. 50/50
Yep. Think of there being three variables: where the first passenger wants to go, where the second passenger wants to go, and which passenger has the chance to speak before the other one. Call the possibilities WW1, WW2, WE1, WE2, EW1, EW2, EE1, & EE2.
So, when the cabbie turns, and looks at 1 passenger (call him passenger 1) then all the “2” possibilities are 0, so if passenger 1 says “East”, then there are 2 possibilities: EW1, and EE1. 50/50.
When the cabbie doesn’t turn and asks where the passengers are going, and one says “East”, then the possibilities are WE2, EW1, EE1, and EE2. Again, 50/50.
However, when the cabbie asks if either of them are going east, and one says yes, there are six possibilities. WE1, WE2, EW1, EW2, EE1, and EE2. The reason that we can include WE1 and EW2 in this case, is that, while the west-bound passenger gets the opportunity to speak first, by the constraints of the question, he will remain silent, allowing the east-bound passenger to say “Yes.”
Now, you could make the argument that the west-bound passenger, given the opportunity, might say “No.” So, if a west-bound passenger that has the chance to speak first will do so, then you have to exclude WE1 and EW2, and you’re back to 50/50. But I don’t buy this argument. I feel the cabbie’s question was specific: only answer if you’re going east.
If the cabbie asks the question “Are either of you going east?” and only one passenger answers, psychological considerations make it almost silly to look at it from a purely mathematical perspective. Still, if all other probabilities remain the same, you’d have to admit that the likelihood goes to 1/3.
(By the way, MadHatter, I posted the death ray as a completely farcical rewording of the OP. I think you or someone else already set up the similar situation with the person leaving the cab, but it seemed to be getting nowhere).
That wasn’t why I posted, I’m actually here since I finally found the article that explained it for me. So if anyone still isn’t convinced, try reading :
“Some Teasers Concerning Conditional Probabilities” by Maya Bar-Hillel & Ruma Falk in Cognition, no. 11(1982) pp.109-122. You’ll probably need to check a university library to find the journal, but if all this discussion hasn’t convinced you and you still think you’re a reasonable person, try the article. It’s about 10 pages and covers a number of teasers, all along the same lines as this one.
And hopefully, nobody will ever the need to ask this question again.
I assume, with this example, you’re implying that the answer will be 33/66, rather than 50/50. Unfortunately, the way you word it, there are only four possibilities: WE2, EW1, EE1, and EE2, and thus a 50/50 chance that the other passenger is going east. By the constraints of the problem, WE1 and EW2 are eliminated; i.e., they didn’t happen. You can’t claim that they still did, but not quite in the same way.
The problem with this taxi-driver scenario is that it’s devilishly hard to word the scenario in such a way that the driver can’t make any distinction between the passengers. The “children” problem (A family has two children. One is a girl. What is the chance that the other is a boy? Answer: 66%) is easier to phrase and visualize, because you’re given minimal information, none of it differentiating.
In the taxi case, if the problem were phrased: A taxi has two passengers. One goes east. What is the chance that the other goes west?, then the solution (66%, natch) is more obvious. Once you start adding additional information, like “the one the driver asks goes east” or “the first one to speak goes east” or “the only one to speak goes east” then that additional information changes the problem by differentiating the passengers.
Given that, as per the OP, the cabbie has direct contact with the passengers it’s difficult to come up with a scenario where the cabbie doesn’t have any differentiating information. The best I can come up with (as above: The passengers both speak at once, but the cabbie can’t quite hear, and only knows that at least one of them said ‘East’) is a little tortuous in it’s wording.
If the wording were something like, the taxi driver asks “Is anyone going East” and one of the passengers answers “Yes” (as Dylan suggests) then the answer is a little tricky. I would argue that this is a case where the first passenger to answer says “east”, and thus the passengers are differentiated, and this the answer is 50/50. However, you could argue that the wording implies that the passenger speaking is answering for both of them (i.e., “at least one of us is going east”), and thus there is no differentiation, and thus the answer is 33/66. The problem here is not the probability theory, which, I think, everyone posting seems to grasp, but the ambiguity and interpretation of the wording.
Zut, I think Punoqllads’ does give a 1 in 3 answer. You can’t say WE1 and EW2 are eliminated; in these two cases, West stays silent, and East speaks up. The cabbie doesn’t know West would have spoken first given a different question. So in the six cases WE2, EW1, EE1, EE2, WE1, and EW2, the cabbie hears one of the passengers say “East”, and there is a 1 in three chance the other passenger is going East also.
I disagree. The original statement by Punoqllads was: *Think of there being three variables: where the first passenger wants to go, where the second passenger wants to go, and which passenger has the chance to speak before the other one. * So in the two cases above (WE1 and EW2) the West passenger has the chance to speak first.
Punoqllads complicates the problem by adding an additional variable: Does the passenger who had the chance to speak first take that chance. So in the four cases EW1, EW2, WE1, and WE2, sometimes a passenger will say “east”, and sometimes a passenger will say “west”, and you haven’t properly accounted for all the possibilities.
So, to rephrase the italicized quote above: *Think of there being four variables: where the first passenger wants to go, where the second passenger wants to go, which passenger has the chance to speak before the other one, and whether or not he takes the opportunity to speak. * Call the possibilities WW1Y, WW2Y, WE1Y, WE2Y, EW1Y, EW2Y, EE1Y, EE2Y, WW1N, WW2N, WE1N, WE2N, EW1N, EW2N, EE1N, & EE2N.
Since the cabbie hears “east”, the remaining possibilities are WE2Y, EW1Y, EE1Y, EE2Y, WE1N, EW2N, EE1N, & EE2N. By inspection, you can see that there is a 50/50 chance that the other passenger chooses east.
My take on Punoqllads case is that the passenger who has the chance to speak before the other one will do so if he knows the answer to the question, and will remain silent if he does not know the answer. The cabbie asks if either of them are going east. If the passenger who has the first chance to speak is going east, he will speak up and say “yes”. If the passenger who has the first chance to speak is going west, he doesn’t know the answer, and so remains silent, long enough for the other passenger to answer “yes” if the other passenger is going east. In this case, the fourth variable (whether or not he takes the opportunity to speak) is a dependent variable. Your cases EE1N, & EE2N will never occur. When the cabbie hears “yes”, the possibilities are WE2Y, EW1Y, EE1Y, EE2Y, WE1N, EW2N.
Oho, I see your point. If you assume (as you do) that the answer to “Is anyone going east?” is either “yes” or silence, then you are correct (33/66 are the probabilities). If you assume (as I did) that the answers are either “yes” or “no”, then the probabilities are 50/50. Dashed upon the rocks of semantics, yet again.
I dislike this problem because it’s too difficult to phrase in a way that is completely unambiguous. I guess that’s real life, though.
Thanks Punoqllads. Still hard to comprehend!
– grumble – appears to be a real paradox?
zut, ZenBeam:
In my mind the issue is which way you interpret the statement “He hears from one of them, the East side”.
I interpret that statement to mean “We are being told that one of the passengers is going east, with no way to differentiate which passenger said that.” With that interpretation we get the 1/3 vs. 2/3 probability.
You interpret it to mean “After hearing the answer, the cabbie has enough information to differentiate the passengers”. Then of course the probability is 1/2 vs. 1/2.
The way the problem is worded, we can have a never-ending argument about what the problem creator intended. Well of course not really never-ending since our time on this earth is finite, but long enough anyways to make it futile IMHO.
Let’s get down to brass tacks here…
First of all We all agree that there is four possible combinations, EE, EW, WE, WW. We all also agree that the WW is out after one of the people says he is going East. So that leaves three of which only one is EE, so the first blush may be 1/3, like some are saying.
but then the naysayers come in and are essentially saying EW and WE are the same and/or they are saying that once one says something one of those two must be elimnated, therefore it is 1/2.
But let’s step back a step, suppose the odds are 1/2 that each will speak up first. And since the cabbie can’t distinguish who is was, then you have new scearios.
(50%)A speaks…EE EW are equally likely
(50%)B speaks…EE WE are equally likely
So EE happens .5 x .5 + .5 x .5 = .5
the other two happen equally at .25.
See there are still three possible scenarios, but they are not all equally likely anymore.
So I vote for the 1/2 side of the room.
these are all our remaining scarios. Since
I tried the two coin trial as previously outlined. The outcome was about 1/3 for EE when WW’s were reflipped. While EW and WE are equal in outcome, they outnumber the EE 2 to 1 when a WW is reflipped. Wouldn’t this suggest that EW and WE are, in fact, really distinct and different outcomes?