The answer is 50/50. What are the odds? 1/2. The % chance? 50%. Expressed as a number? 0.5.
This isn’t brain surgery. The second guy in the cab either wants to go east, or he wants to go west. Therefore, 50/50. Problem solved.
But I have this to add:
The first guy who speaks up says “East”. So the cabbie drives east and this very same guy who responded first hops out of the car. The cabbie turns to the remaining guy and says “East or West?”.
50/50 chance.
REGARDLESS of who spoke up first, why they spoke first, how they spoke first, or the accent they said it in.
It’s 50/50.
Incorrect. The problem is stated as «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.»
(This applies also to what panamajack has been saying.) The way the problem is phrased, we (and the driver) have no idea which person has spoken. So this situation is exactly similar to the case I cited above, where a couple has two children and you know one of them is a girl. The odds in that situation: 2/3 vs. 1/3 that the second child is a boy.
MadHatter, all I can say is that your statement : «The second guy in the cab either wants to go east, or he wants to go west. Therefore, 50/50. Problem solved.» is definitely incorrect. Take the example of the two children. If you know one is a girl, then that means that there is an equal likelihood of the three possibilities:
- oldest child girl, youngest child boy
- oldest child boy, youngest child girl
- oldest child girl, youngest child girl
You will notice that 2 out of 3 equally likely possibilities have a boy as the second child.
ZenBeam said:
«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 wants to go west, assuming the odds of east vs. west travel are equal? (Answer: 1/2).»
The problem does not say “the first person to answer the cabbie”, and does not say that the cabbie knows which person spoke. The way you are phrasing the problem, the answer would always be 1/2!
Let’s take the example of the children again. «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.)»
ZenBeam, following your lead above, I could rephrase it to say «The child for which you have been given certain information is a girl.» Therefore I have a way to distinguish the children, therefore the odds are 1/2. That is incorrect, the point being that I don’t know to which child the information applies.
Still 50%. Turning around has no effect.
Why should you think it would?
Wow! … Arnold where have you been? … Kudos … Andi
It’s true that the second passenger either wants to go East or West, but that does not imply that the odds are 50-50. Consider another problem: What are the odds that a meteor is going to land on my head and kill me? Well, either it’ll hit, or it won’t, so it’s 50-50, right?
The point is that although the second passenger can go either direction, the probability is not equal for either direction.
Wow, just when we seemed to turn the corner more people have weighed in on the incorrect 1 in 3 case!
Andi, I’m glad to hear you’re a published author in probability. That means I can do a formal proof.
Given: 50% of passengers go east and 50% go west.
Given: two people have entered the cab simultaneously.
Given: cabbie has asked the destination and heard from only one passenger, who has said east.
Assumed: There is nothing about going east that would make a passenger more likely to answer first.
Bayes rule: The probability of event A occurring given that event B has occurred equals the probability of both A and B occurring divided by the probability of event B occurring.
Let event A be both passengers going east.
Let event B be the first passenger to answer says east.
The probability of A is 25%. (So far there’s been no argument about that.)
The probability of B is 50%. (This follows from the first given and the assumption.)
Since when A occurs, B will always occur, the probability of A and B both occuring is equal to the probability of A occuring. That makes the probability of A and B occurring also equal to 50%.
Therefore the probability of A given B equals 50%. QED
So no more comparing this problem to other problems, saying that you had it on a test, or published an article about it. (Impressive though that all is.) I challenge anyone who still insists on clinging to the 1 in 3 answer to show the flaw in my proof, or in my assumption.
To be fair, let me point out the flaws in the other arguments for 1 in 3. Chronos is correct that just because two events are possible doesn’t mean they have an equal chance of occuring, but then of course that also means they might. I’ve proved above that in this case they do.
Arnold is right that this problem is exactly the same as the two daughters problem, and it has the same pitfalls. How you find out that the woman has one daughter makes all the difference. Did you ask her if she had at least one daughter? Then it’s 1 in 3 that she has two daughters. Did you seen a picture of a daughter? Then it’s 50/50 that the other one is a daughter. Did you see a Britney Spears poster in her house? Then it’s 1 in 3 that the second is also a daughter, or her husband is a pervert.
Andi, you’re right. There are four equally likely possibilities WW, WE, EW, and EE. We know we can toss out the WW case, leaving three possibilities. Are all three equally likely? No! There was a 50% chance of throwing out the WE and EW cases as well. Given the knowledge we have, the EE case is twice as likely as either the WE or EW case. That’s what changes the nature of this problem. I’d be interested in reading your articles sometime though!
Hi Greg, geez, I don’t really have the basic education to understand something as complicated as you have just outlined … sorry … and
gosh, did say I was a published author in probability? … sorry, I thought I said that if I were, maybe I could win you over to my side? … But, anyway, who knows, maybe I might get lucky flipping coins and someday, like Bernoulli, get a chance to write something …
All I know, is that I keep flipping the two coins as previously outlined, eliminating the Tails/Tails outcome as a no play, and I am winning a lot of $ betting against the Heads/Heads outcome, 1/3 on a play. Want to play against me over the internet? Maybe one of the moderators would be nice enough to flip the two coins and tells us if there are two Tails, or a Heads (in play)? And if you notice, even an independent third party or group of people could look at the flipped coins to see if there is a heads or two tails. Furthetmore, we don’t even have to be in the same room with them, they could even send us a smoke signal, one puff if there is a Head, two puffs if there are two Tails. Andi
“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.”
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Wrong. But if you expressed it as a hypothetical question like DeutschFox did and specifically stated that the odds were 50/50, then the answer would be yes. Therefore, both of your arguements are flawed.
The problem is, this is not the same situation as in the OP.
Here’s another experiment. Flip two coins, and look at one of them. Do not look at the other, yet. If the coin that you’ve looked at came up tails then it’s a “no play” in your words. But if the coin that you’ve looked at came up heads, then go ahead and bet on whether or not both coins are heads. In this case the bet is even money; both coins will be heads with probability 50%.
That is what’s happening in the original problem. The cabbie finds out where one of the passengers wants to go, but not the other; that’s exactly the same as looking at one of the coins but not the other. It doesn’t matter if the cabbie asks the first passenger to get into the cab or the first passenger to speak up, the effect is the same: he learns where one of the passengers wants to go.
In your coin-flipping experiment, there’s no way to eliminate just the tails-tails outcomes unless somebody looks at both coins. If you only look at one coin, sometimes you will eliminate a heads-tails or tails-heads outcome, which means that the heads-heads outcome becomes more likely, just as Greg has said repeatedly. And in the case of the cabbie and his passengers, no one person in the problem knows the destinations of both passengers. (That is, if you assume the passengers are acting independently and without consulting each other, but that’s a psychological point and not a mathematical one.) We and the cabbie know that one passenger is going east, and we know nothing about the second passenger, so the probability that both passengers are going east is 50%.
(I’m not really saying anything that hasn’t been said before–twice–by Greg, scr4, panamajack, and others, but I can support Greg’s position at the very least. While I haven’t published anything about probability either, I am a graduate student in mathematics and have taught conditional probability to undergrads.)
Sorry, Andi. I guess I read too much into that. Too bad though. It would have been nice to have a published author here, for the prestige of the board and all. Still, welcome to the SDMB. Your first post has certainly been memorable, at least for me. Keep them coming!
As for the proof, it’s really pretty simple as proofs go. Give it a read-through. The only hard part is Bayes Rule, which I’ve defined. Even with your lack of education, you should be able to follow it. 
[QUOTE]
*Originally posted by Math Geek *
**
Math Geek … I wish I was smart enough to develop and instruct classes at a university … but all I have to rely on is my simplistic brain … but thanks …
in your proposed experiment, e.g. you are essentially flipping just one coin, since you want to look at only one of them first, you might as well flip a $1 worth, what’s the difference if you are only going to look at one of them first? This helps clearly show the distinction between the way you, and others, want to construct the problem and the real problem … 1) both passengers get in at the same time … 2) the cabbie asks both at the same time, since he is the only one speaking … 3) it is a random event as to which passenger answers first … and in fact one of the passengers might have been unable to talk but did answer in sign language, or maybe he was deaf … all that matters is that the cabbie heard “East” and was unable to distinguish who said East between the two …
Also, for you trying the 2 coins flip experiment … the problem can be simply stated by “Is there a Heads” (East)? If not … no play … flip the coins again … this simulates the three sided die problem … maybe the real problem is who can visualize a three sided die?
Alright, are you saying that in this part of the problem, the cabbie actually hears answers from both of the passengers? But that, for some arcane reason, we (the probability debaters) are not given this information?
The way I interpret the statment “He hears from one of them, the East side” is that he hears from one and only one of them. Otherwise, from the cabbie’s point of view, the problem becomes trivial.
Maybe (and I don’t think this is likely, but here goes) you 1/3’ers are thinking that this sentence means that he hears from both of them, one of whom says, “the East side”. This means that the cabbie has some information that he’s not sharing with us.
Which is it?
Thanks to all of you who have participated … could we ever have acheived a consensus on this real life problem?
For those who think 1/3 is the answer, including me, we have a real dilemma … this would infer that the cab driver’s tip income would be affected by if he knows who said “East” or not. I believe the saying is “what you don’t know, won’t hurt you”. Well for the 1/3’rs, it does hurt the cab driver. If he would get a $10 tip per passenger per trip and the situation in the problem happens 12 times a day, he would realize by “knowing” who said “East”, 6 single fare tips and 6 double fare tips, a total of $180 in tips. If he doesn’t pay attention, never “knowing” who said “East”, then he would only receive 4 double fare tips and 8 single fare tips … a total of $160 in tips. Would this really happen?
For the 1/2’rs, I understand your reasoning … but is there a situation, such as this, where “information”, knowing or not knowing, could actually affect the outcome of an event?
Onwards …
DeutchFox, I don’t have a clue what you’re talking about now. The fares are determined by how many people want to go to which side, and maybe how they pair up. Assuming the cabbie learns where everybody wants to go, and actually takes them there, his tips have nothing to do with whether he looks at them or not.
Now, if I wanted to go West, and he didn’t listen to me and kicked me out on the East side, what kind of tip do you think he’d get?
Probablity (and in our discussion, conditional probability) does not “change” an outcome - it’s only an attempt to quantify our expectations, based on the information we have at various stages of an experiment.
Kelly, I don’t know for sure, but I guess the cab driver could ask if one of the passengers (if it is EW or WE), wanted to go on a long ride in the opposite direction to his destination from the airport, instead of getting out and flagging another cab at the airport? Never considered that probability after the event was completely defined? Thanks
That’s it, that’s f^%$ing it.
I’m gonna write a software program simulating this scenario so I can stick it up all you little 1/3 PATHETIC LOSERS!!! (Sorry, got a bit carried away there.)
I’ll write it in Visual Basic, under these assumptions:
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The 2 passengers will have a 50/50 chance of wanting to go east or west.
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Each passenger will have an equal chance of speaking up first when asked where they want to go.
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If the guy who speaks up first answers “EAST”, the simulator will then check to see if both dudes want to go east. If one of them answers “WEST”, the scenario will ignore this trial and reset itself, until it gets the answer “EAST”.
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I will then upload the program to my site, either today or tomorrow.
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I will then come back on here and SHOVE IT UP YOUR ASSES!! (Again, carried away. My apologies.)
- MadHatter
OK, here’s my program, written in IDL but I think you can follow it. The square brackets indicate array indices. The randomu function returns a floating point between 0 and 1, so multiplying by 2 and taking the floor (integer) of the number gives either 0 or 1 with 50% probability. The $ sign somewhere indicates the line continues to the next line; otherwise each line contains a separate command.
PRO cab_driver, n
first_east = 0
both_east = 0
for i=0,n-1 do begin
dest = floor(randomu(seed,2) * 2) ; 2-dim. array containing destination for each passenger
ask = floor(randomu(seed) * 2) ; which passenger answers first
if (dest[ask] eq 0) then begin ; if first answer is 'east'
first_east = first_east + 1 ; increment counter
if (dest[0] eq 0) and (dest[1] eq 0) then $ ; do both of them go east?
both_east=both_east+1 ; if so, increment counter
endif
endfor
return
end
I ran this with n=10000 and came up with first_east=4980 and both_east=2507. I think that settles it, or at least help point out where the disagreement is.
I should add that the assumpsions used in my program is exactly the same ones MadHatter suggested. Also, 0 indicates east, 1 indicates west.
To present a similar question, intended to have a similar answer as the OP:
Say that the custom in that part of the world was to announce your sex upon entering a taxicab. Male & female passengers are equally likely. 2 passengers enter simulataneously; one announces that she’s female. What are the chances that the other passenger (who hasn’t yet spoken) is male?
Is there anything about this question that makes it essentially different from either the OP or the 2 children probability question?