Should have deleted the phrase “to enter the cab” in that last post. Try again:
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).
Maybe my wording of the problem wasn’t clear enough. Not unusual so I’ve been told. But, then, could the answer be 1/3 if the cabbie couldn’t distinguish who said East and 1/2 if he could? This still seems to have an affect on the cab driver in terms of tips, tolls, gas, etc., which is still perplexing to me! Can the information available prior to an event have an affect on its outcome? … Andi
The information available prior to an event can’t affect the outcome, but it can affect the probability of an outcome. That’s the whole point of conditional probability.
As Greg Charles, ZenBeam, panamajack, and other have already said: the cabbie can distinguish between the two passengers. One of them answered and the other didn’t, and that’s all the distinction that needs to be made.
DeutscheFox, it depends (once again) on what you mean by ‘the cabbie can’t distinguish’ . The cabbie still knows that somebody spoke, unless the voices are in fact exactly identical (and this information is known to the cabbie). I can’t think of a way to put it so this situation works out to 1/3, since the passengers always distinguish themselves when they speak. Perhaps only if the cabbie is told by a 100% reliable psychic, “At least one of the two people who enter your car will want to go east” would it work out to 1/3.
The information gathered before the event affects the probability, not the event itself. If you have incomplete information, the probability simply does not map to the event. In this case, how you get your information is considered important as well.
Okay, okay, I’m changing sides on this. The probability for both cases in the OP is 1/2. I had prepared an example much like Greg’s 12:20 post, but he says it well enough.
I’ll try to contribute this - it doesn’t matter whether the cabbie can distinguish the two passengers or not. It goes like this: the whole population breaks down into quarters: pairs that look like WW, WE, EW, EE. Now let’s say the cabbie always looks at the guy on the left to ask. This means that none of the WW or WE pairs are going east - and all of the EW and EE guys are. Of the sub-population that goes east, half of that are EE. So the prob. is 1/2.
Now lets say he doesn’t distinguish the guys, and the E and W passengers have an equal chance to speak first (this isn’t given in the OP, but this has to be established). That means that none of the WW pairs are going east, and all of the EE pairs will (no change here). But among the EW pairs, half will say east first. Among the WE pairs, half will also go east. So in this case, we still have 1/2 * 1/2 = 1/4 of the original population going east in mixed pairs, and still have 1/4 going east together. Probability is 1/2.
Notice that in the two cases, different pairs of passengers go east first, but their number is the same.
You’d get (well, okay, I got ) probability of 1/3 by thinking that all of the WE and EW pairs are going east. That could happen if, say, a third party quizzed both guys, and then told the cabbie “I know at least one of these wants to go east, so let’s head that way”. That doesn’t fit the OP scenario. I’d also say that I don’t think there’s any trouble with the semantics of the OP either.
Now, sometime you folks should drop into rec.gambling.craps when there’s a discussion over whether laying odds on the don’t changes the house advantage. Hoo-boy!
Hey, if this cabbie is working St. Louis, I know some real neat places to visit on the East Side… but that’s another story.
Ok … here is the solution which all of you can actually try for yourself, you don’t have to trust me! By the way, do this with a “friend” for $5 a play … clearly I said one of the passengers said “East”, and I did not distinguish which one. The problem wording considerations, while enlightening, are only just that. The concepts of both having to have the same voice is not applicable, one could have been speechless. It would not matter so long as who said East was indistinguishable to the cabbie …
Take two coins … consider Heads as East and Tails as West.
Flip both coins … so they are hidden from the friend you are going to bet with.
Bet your friend even up … that if you tell him there one is a Head (East), he will win if both are Heads (East) … if they are both Tails (West) the coins get retossed …
Flip both coins … so they are hidden from the friend you are going to bet with.
If they are both Tails (West) display them to your friend … no play … since “a” passenger didn’t said West …
Still having the coins hidden … if there is a Head on either coin or both, tell him there is a Head (East) …
He wins if both are Heads … you win if both are not Heads (East)…
If you do this 1000 times at $5 a play with two Tails getting reflipped … you will milk your friend of $1,250 …
This is a classic three sided die problem … since WW or Tails/Tails is a no play …
Calculation of winnings on 1,000 flips … 250 tosses get reflipped (Tails/Tails) … 1/3 (250) will be Heads/Heads … your friend wins … 2/3 (500) will be either Heads/Tails or Tails/Heads … you win $5 x 250 … oh well … anyone want to play ?
I’m still sticking with my 1/3 answer, with the assumption that saying he didn’t look at the passengers means he can’t distinguish between them - again, semantics.
The OP is exactly equivalent to a problem that’s discussed in every college genetics class, kind of like the one Arnold mentioned: you have an organism that has a dominant phenotype, and you want to know the probability that it’s homozygous. I’ll skip the Punnet square, but the phenotype rules out the double recessive case, so the answer is 1/3. I’ve had this on enough tests to remember that answer for sure. That’s why I’m so adamant about this.
Here’s where the 1/3 answer and the 1/2 answer differ: the 1/2 answer says that the two cases (i) person A answers and person B also wants to go east and (ii) person B answers and person A also wants to go east are different, leading to 4 possibilities, 2 of which result in both going east. The 1/3 answer says that the two cases are the same, resulting in three possibilities, 1 of which results in both going east.
Again, the reason I think the two cases are really the same from a probability standpoint is because you can’t tell the passengers apart. In both cases, both passengers want to go east and one of them says so. They’re the same case. Which one spoke is irrelevant, because you can’t pin down the speaker to a person.
And to answer a previous question, the amount of available information most certainly does affect probability. In the phenotype example above, if you knew that the dominant phenotype came from the mother (or the father), the answer changes to 1/2. Similarly, if you know the answer came from person A (or B), it also becomes 1/2.
That’s my answer and I’m sticking to it. I’ll also add that I spent all day thinking about this instead of studying for my tests. Thank you, SDMB!!
SDMB … you are absoulutely correct the probability is 1/3, period! semantics or not! The real question here was how information can change n outcome, as illogical as it may seem or not. This is a common problem faced regarding entropy, e.g., are particles distiguishable or not. If we have a tank of pure nitrogen vs a tank of nitrogen in which one separate half of the nitrogen molecules have a white dot on them, there a difference in their entropy. But geez, I wish I could help you with you studies . 
Andi
DutschFox, your coin flipping game is interesting, but It’s not equivalent to the cab scenario. (Do you think it is? that wasn’t clear). The difference is that in your step 6 you look at both coins to determine if there is one head. In the cab problem, the driver asks, or the rider says, only one of their destinations.
Smeghead, this problem is not equivalent to your genetics example. In that case you can’t tell the EE, WE, EW apart (to use the cabbie notation) so the answer is 1/3. But the when you determine the destination of one of them (no matter how) then you rule out half of the WE EW set, boosting the probability to 1/2.
Come over to the bright side…
In the coin toss experiment, Step 6, only states that one of the coins is a Heads (East), but both coins could be a head, the left one could be head, or the right coin could be a head. Just as the cabbie hears that one (“a”) of the passengers says East (Heads). Let’s pretend that I’m that psychic as previously mention on this issue, and then it was ok, what difference does it make? This is not a semantics problem or issue … it is a simple probability problem … the solution is well established and I have provided an experimental methodology to prove it … there are only 4 outcomes ( the keyword here is outcomes, there are not 8 or 16 or … ) WW, EW, WE, EE. As soon as one or “a” passenger says East … there are only 3 outcomes left, the WW out … EW, WE, EE. Thus, the probability that both will be going to East or even more generally, in the same direction is 1/3!..
Word, Fox.
Wanna arm-wrestle for it, kelly?
All right kelly! If I’ve helped fight the ignorance of just one person, then it’s all been worth it. As for Smeg and Fox, well, they may come around eventually. The 50% probability has already been proven. If you want to point out a flaw in the proofs, I’m ready listen. Meanwhile, I can show how your analogies differ.
In Smeg’s genetic example, the observer see a dominant characteristic. This only eliminates the possibility of both genes being recessive. That is, when you see a recessive trait, you toss out that sample and don’t use it to calculate the probability. That makes 1/3 the right probability for homozygotes, or whatever Smeg said.
In Fox’s toin coss example, the only possibility being eliminated is the two tails case. That makes the chance of two heads 1/3. You will win money from your friend. Right again.
Now read this part carefully. It’s the crushing point, but so far it’s failed to crush. In the cabbie problem, you eliminate the case where both passengers are going west AND the case where they are going in opposite directions, but the one going west is the one who has answered. It’s completely immaterial whether the cabbie can or can’t distinguish the voices. It’s the fact that you’ve eliminated this extra possibility that changes this to a 50/50 probability.
For the probability to be 1 in 3, you have to assume that a person going east would always answer first, or that the cabbie can’t hear the word west and only knows both passengers are going west when he hears nothing. Psychology could come into play if we started to speculate why the second passenger hasn’t responded. Does that in some way indicate that he is satified with what the first passenger has said? Psychology has no place is a probability problem though. I assumed one person had answered and in the interim before the other person says, “Hey, me too!” or “Too bad, I’m going west,” we were asked to compute the odds.
DeutschFox, I’d like to win you over. I can always say Smeghead is just a smeghead, but Germans are known for their precision and foxes for their cunning. We need you on our team!
I still don’t buy it, but I don’t have much new to add, except this minor nitpick:
What “always”? This is a one-time thing. It’s not like we’re repeating the same experiment over and over.
Man, I should have something better to do on a Friday night.
Let me try this. I need to be clear - in the OP we must agree that the east- and west- bound speakers have an equal chance of speaking up, right? You’re not assuming that the East-bound guy will always be the first to speak, right? Otherwise, that’s were the problem is.
Now, we don’t seem to have any problem with the WW or EE pairs, so let’s leave those out for now. That leaves the WE and EW group. You seem to think that all of these pairs are headed for the east side, because one of them wants to go there. I say, since they have an equal chance to be heard first, only half of the WE pairs are going to go east (i.e. only those in which the E speaks first), and only half of the EW pairs are (again, for the same reason). This means that only half of the mixed pairs are headed east. This is the same number of EE pairs that are headed east, hence, the probability is 1/2.
In the coin toss example, you’re willing to tell me (or let the psychic tell) that at least one of the coins is heads. Fine. But in the cab example we know more information. We know that one guy wants to go east, but we also know that the west-bound guy had an even chance to speak up and that did not happen. That’s the difference.
The equivalent situation would be if your coins had an even chance of speaking up for themselves. You could then be looking at a head and tail (in either order)and truthfully say “I have at least one head”. But then half the time I’d hear a coin chirp up and say “I’m a tail!” Then I’d know you’re not looking at two heads. This would eliminate half of the HT combos.
Now, let me step back and take a shot at something else DFox said. You said you offered an “experimental methodology to prove it”. Well, you really did no such thing. This is a mathematics problem, and and cold hard logic and axioms rule the proofs. Mathematicians do not perform expriments to form proofs. Descriptions of experiments may be illistrative and useful, but there’s no way to experimentally prove (for example) the irrationality of Sqrt(2) or the theorem of Pythagorus.
You also wondered about how information can change an outcome. Well, think about this, the real outcomes of how many cabs go in which directions are not changed at all by the information we’re talking about. What we are discussing is conditional probability, and, of course, the conditional probability of where a given cab will go is dependent on what we know about the guys in the cab. The outcome doesn’t change, but our expectation of the outcome in cases where we have incomplete information is what probability tries to quantify.
While I was typing a reply, Smeghead asked:
What we’re asking is whether the east and west bound guys have an even chance of speaking first (as I and Greg are asuming) or whether you think that the eastbound guy is certain to speak first. This could make all the difference in our views.
Friends, please … who speaks first is immaterial, so long as it is random … please don’t make this problem more complicated than it is … again there are only 4 outcomes … WW, EW, WE, EE. Even if the passenger said West, the probability of both going the same direction is 1/3 … think about it ! … once one says one of the two directions possible the probability is some y/3 … lets get this behind us so we can move on to the more important issue … can information change the outcome of an event ?
I think we’ve hit the nail on the head. I’m not worrying about who could have spoken first. I’m looking at after the speech has been spoken - that is, there is a probability of 1 that the person who spoke said East. It’s already happened, so no other option is possible.
[sub]And I still think I’m right.[/sub]
OK, here’s yet another way to think about it. Without any prior knowledge, the probability that both want to go east is 25%, and I think everyone agrees on that. The driver then asks “does either one want to go east?” - what’s the probability that the answer is “yes”? There’s a 25% chance that both want to go west, so a 75% chance that at least one wants to go east. So the probability is 75% (3/4). So given the “yes” answer, what’s the proability that the other also wants to go east? It’s 1/3. So the probability that both want to go east is (3/4)*(1/3)=1/4, and it works out.
Now, what if the driver asks the other question instead - i.e. “which way do you guys want to go” and listen to the first answer. What’s the probability the answer is “east”? It’s 1/2, right? Because it’s equally likely that the first answer is “west”. So, the probability that the other also wants to go east is 1/2, because (1/2)*(1/2)=1/4. If the probability was 1/3, the probability that both is going east would be 1/6, which clearly isn’t right.
For those more comfortable with equations: Let P(A1) be the probability of answer “yes” to first question, and P(A2) be the probability of “east” to second question. Let P(B) be the probability that both wants to go east. As explained above, P(B)=1/4, P(A1)=3/4, and P(A2)=1/2. What we want to know are P(B|A1), the probability of both going to east given answer A1, and P(B|A2). Now, P(A1)*P(B|A1)=P(B), and P(A2)*P(B|A2)=P(B). So you can immediately see that P(B|A1)=1/3 and P(B|A2)=1/2. I hope you don’t think I’m trying to blind you with math; for those of us used to the notation it’s a lot more clear than explaining in words.
Greg … you are certainly a gentleman. I wish you could win me over, however, I guess, that even if I told you that I was a published author in the concepts of probability on this very subject, I couldn’t win you over? Andi
DeutchFox:
If you don’t understand that the probibility of one passenger speaking first drives this answer, we’ll never reach agreement. If you truly think it’s immaterial, then we’re done here.
A probability may be assigned to a random event, and is usually given or evident in the problem statement. Here, some of us assumed that the E passenger speaking first with respect to a W passenger is 50%, which drives us to the 1/2 answer. Others, like you, think it doesn’t matter, and advocate the 1/3 answer. The 1/3 answer is only correct if you assume that the E passenger speaks first with 100% probability.
Now, if that prob. were, say 90%, the answer to this problem would be a number between 1/2 and 1/3. There’s an exercise for you.