Probability question that is driving me nuts!

[Mucman]

Here's the scenario...

There is a bus that is loaded with 20 people. At each stop, 1 person leaves the bus (so there are 20 stops). There is a camera on the bus and 2 men (lets call them A and B)are having a bet. A and B are betting on what side of the bus the last passenger is going to be on. If 19 men are sitting on the left side and 1 on the right side, how does that change the odds of their bet (does the word odds = probability?). Is it safer to bet that the last man on the bus will be on the left?

Another pair of main (C and D) are also having a bet, but they are waiting at the last stop, where the last man is to get off. They have the exact same bet, but have no prior knowledge of the seating arrangement when the bus left. Does that mean it reall is a 50:50 chance? Is the fact that 19 men are on the left in the beginning totally irrelevant?

Don't ask me why but I have been pondering this for the last two nights and can't seem to think clearly! Instinct tells me to say that the last man on the bus will be on the left, but it really is a 50:50 chance no matter what, correct?

 

[bizmark]

That's pretty hard, and I've always had problems with probability, but here goes.

Maybe it helps to think of this in a different way. Let's think of this as drawing balls from a pot. There are 19 red balls and 1 blue ball. You draw them one at a time without replacement. So what's the chance that the last ball you draw is the blue one?

It's 1 - (the chance that it's drawn on the first draw + the chance that it's drawn on the second draw + .... + the chance that it's drawn on the 19th draw).

Things are complicated by the fact that all of these chances are conditional probabilities. There's NO chance that it'll be drawn on the 2nd draw if it was already drawn on the 1st draw. So the stuff within the parentheses above becomes a huge headache, like

(1/20)+(19/20)(1/19)+(19/20)(18/19)(1/18)+.......

But notice that this collapses. 19/20*18/19*1/18=1/20. So it becomes 1/20+1/20+1/20.... 19 times, so the final probability that you'll draw the blue one last is then 1/20.

This also works via a symmetry argument. It is guaranteed that one of the balls will be drawn last. None of the balls are special in this regard or more likely to be drawn at any time than any one of the others. Then since they're all the same, the chance of any one of them being drawn last is 1/20.

Odds are not probability. The fair odds here would be 19:1.

I hate to post this because, as I've said, I've always had problems with thinking of probability problems in the right way. But this one seems to make sense to me for some reason.

[edit]:So what really is the question here? You didn't say for sure that the seats were taken as you said, 19 on one side and 1 on the other. Is this given, or is this something we're supposed to not know? Because if we don't know the initial distribution of the men sitting on the bus, then it's a 50/50 bet.

 

[Mucman]

Thanks for the reply... I did take a stats course and came to a similar conclusion as you (albeit I only got a C+ so what do I know .

My problem is this...

Men A and B know that 19 are on the right and 1 on the left.
Men C and D know nothing about the initial conditions but are betting
on the same bus. If man C bets that that last man will be on the right he has chosen poorly even though he does not know it?



<< Because if we don't know the initial distribution of the men sitting on the bus >>



I might have forgotten an important lecture I guess... Could the examples of calculating the probability of loaded dice work for thise question?

 

[Turkey]

The probability for C & D is the same as it is for A & B. Another way of describing the bet C & D are making is "scam" .

 

[bizmark]

<< Men A and B know that 19 are on the right and 1 on the left.
Men C and D know nothing about the initial conditions but are betting
on the same bus. If man C bets that that last man will be on the right he has chosen poorly even though he does not know it? >>



Well, since neither C nor D knows anything about the conditions on the bus, a 50:50 bet is fair. Either of them is equally likely to be wrong. (In expectation, the number of men on either side of the bus is the same. Because neither knows, they could make a bet with C saying Right and D saying Left, or they could swap and neither of them should care.) They'll never know what happened on the bus. They'll only see the last man get off, and they'll never know the chances and they'll think it was a fair bet. It WAS a fair bet since the information was symmetrical between the bettors. If D knew but C didn't, it wouldn't be a fair bet.

I know that I'm being kind of unclear, but I can't think of any better way to explain it.

 

[sygyzy]

Interesting.

C and D don't know prior arrangements so you can place any sort of odds you want on them. It is not 50/50 like a coin flip because in a coin you know there really is a 50/50 chance. Who is to say all 20 people cannot be on the right? Common sense and experience would tell me that doors open only on the right side (right side of driver facing forward) and that most people in an empty bus start moving fwd and right anticipating their upcomming stop.

A and B definitely have weighted odds. It is not 50/50. Each time a guy gets off from the left, it decreases the odds that the last person will be on the left and the same goes for the right.

 

[bizmark]

<< C and D don't know prior arrangements so you can place any sort of odds you want on them. It is not 50/50 like a coin flip because in a coin you know there really is a 50/50 chance. >>



Well, if the only thing that they know is that there's 20 people on there, and that these 20 people could be arranged any way (20/0, 10/10, 13/7, whatever), then [I}in expectation[/i], there's 10 on the left and 10 on the right. Of course 10 on the left and 10 on the right is only one of many possibilities, but if you sum up all these possibilities and then divide by the # of possibilities, you come up w/ 10/10. That is, if all of these possibilities have symmetrical probabilties, as far as C and D know (ie. 0/20 is as likely as 20/0, 1/19 is as likely as 19/1, etc.)

The expected value of a (fair) coin toss is 1/2 (taking 0 to be tails and 1 to be heads). Of course, it's IMPOSSIBLE for a coin to land like that. But still, statistically the expected value is 1/2. Just as the expected value of the distribution of the people on the bus is 10 and 10.
Again, assuming symmetrical probabilities.



<< Common sense and experience would tell me that doors open only on the right side (right side of driver facing forward) and that most people in an empty bus start moving fwd and right anticipating their upcomming stop. >>



Heheh, common sense is thrown to the winds with these problems

 

[sygyzy]

Statistically, as you reach a large number n of tries, then it approaches the expected value very closely. Toss a coin 10,000,000 times and it will be very close to 50/50. Impossible, but close enough.

 

[gustavo]

Sorry: you all gave wrong answers !!

The real answer is that statistics is based on random events, so 2 conditions have to be met to answer the question:

- the events described are randomly distributed
- knowing the distribution of the events described

In this case none of the conditions is met !!!!

Regards Gustavo.-

 

[Agent004]

since you stated that 19 men are on left and 1 on the right, here is my go

For A and B, the probability of the last man is on the left hand side is 19/20 at the start. However, as time goes on (there is a higher probability that the man leaving the bus), the chance of the last man is on the left hand side is decreasing ( ie, from 19/20 ->18/19->17/18 and so on).

However this not change the fact that A and B, the probability of the last man on the bus (left or right) is the same through out (19/20 for left or 1/20 for right), since that's the probability when they started the bet (or subject to) if they can't change their decision during the process.

At the 19th stop (assming all the 18 men left are on the left hand side) then you have a 50/50 chance of the person leaving the bus is either right ot left. On the 20th stop, the last remaining person is also either on left or right, depending on the 19th. Since C and D knows nothing about it, the chance of getting the correct outcome is 50/50

So A & B is different from C & D

A similar and simpler example is the famou monty python game show. You are on the game show and the hosts offers you a choice of 3 doors, 2 of which has a goat behind, 1 of them is a car. First you picked a door, then the host ramdomly (not really, since he knowns what's behind each door) a door and reveals a goat. The host then offers you a single chance to change the door you previsouly picked. Questions

1) What's the probability of winning the car if you stick?
2) What's the probability of winning the car if you change?
3) So what's the preferred action if you wished to win the car?

If you managed all 3 questions with the correct answers, then you will have no problem with the bus scenario above

 

[CTho9305]

<< A similar and simpler example is the famou monty python game show. You are on the game show and the hosts offers you a choice of 3 doors, 2 of which has a goat behind, 1 of them is a car. First you picked a door, then the host ramdomly (not really, since he knowns what's behind each door) a door and reveals a goat. The host then offers you a single chance to change the door you previsouly picked. Questions

1) What's the probability of winning the car if you stick?
2) What's the probability of winning the car if you change?
3) So what's the preferred action if you wished to win the car?

If you managed all 3 questions with the correct answers, then you will have no problem with the bus scenario above >>



1. 1/3
2. 2/3
3. switch

you have a 2/3 chance of guessing wrong on the first guess, no matter what. when the host shows you the non-car, you STILL have a 2/3 chance of being wrong - the additional information does not change that fact. so switch. you will win 2/3rds of the time like that. not that I feel like thinking about the bus problem

 

[Shalmanese]

Methinks that A, B, C and D (If those are there real names!!!) need a serious visit to Gamblers Anonymous.

 

[Noriaki]

<< Sorry: you all gave wrong answers !!

The real answer is that statistics is based on random events, so 2 conditions have to be met to answer the question:

- the events described are randomly distributed
- knowing the distribution of the events described

In this case none of the conditions is met !!!!

Regards Gustavo.- >>

Be careful. While statistics and probability are often taught together at university they are distinct subjects.

Probability, we'll use some simple counting:
There are 20! (20 factorial) total ways for 20 people to leave a bus.

Single out the guy on the right. Call him x. If we want him to leave last we put him at the end, there is 1 way to do this. The other 19 can leave in any order and x will still be at the end. There are 19! ways for the other folk to leave the bus.
Thus there are 1 * 19! ways for x to be the last person on the bus.

Now that we have counted thse up, we simply divide them, number of ways for our event to occur divided by the total ways things could happen.
19! / 20! = 0.05 probability that x will be the last to leave (this is equal to 1/20 which is the inutitive answer, but the above is how you arrive at it).

Since all probabilities add up to 1 the chance that x isn't the last person to leave (ie someone on the left is the last) is .95 (19/20).

You can also express this as [20! - 19!] / 20! which also equals 0.95 (19/20) probability someone on the left is the last to leave.

Actually counting the ways in which x is not the last person to leave is much more complicated, it's much easier to just subtract the positive from the total to get the negative in this case.

Note: This assumes any given person is equally likely to leave the bus at any given stop.
Note #2: I use the mathematically notation that 1 is the absolute probabilty, but you can multiply by 100% and read as percentages if you are more comfortable with that. Thus the numbers above would be 5% and 95%.


As to the odds that is statistics so now gustavo's statement is exactly correct.

You can't work out odds on this, because it's not a random event. This is a specific bus ride.

If I told you that I have gathered information over the last year and determined that 95% of all people prefer the left side of the bus (that is the distribution of people on the bus).
Then we pick a random bus with 20 people on it, we can say there are 95% odds that the last person will be on the left side of the bus.
This is not probabiltiy. It's statistics. They are easy to get confused but they are very different.
Becuase in this situation you could still possibly have a situation where all 20 people sit on the right side. Very unlikely since 95% perfer the left, but it could happen. In this case there is a 0 probabiltiy that the last person will be on the left and there is a 1 probability that the last person will be on the right.

Statistics is sort of a weighted average of all possible probabilties so they are related. But when you know the specifics of a particular situation like this, you don't use stats.

If A and C both bet that the last person to get off will be on the left they both have a .95 probabilty of being correct.
If B and D both bet that the last person to get off will be on the right they both have a .05 probabilty of being correct.

The fact that A and B know where everyone is sitting before hand and that C and D don't is totally irrelevant to the probabilities.

As to the odds of the bet for C and D, you can't determine it without knowing the distribution as I mentioned. However it is likely that the distirbution is a Normal one (not the demented 95% one I described above), thus fair odds on a bet if you don't know the contents of the bus are 50/50.
If you do know the contents of the bus, odds are irrelevant, you'd use probability.

(My assumption that it's a normal distribution is just that an assumption, very likely a valid one, but still an assumption)

Hope that cleared things up a bit for you....

 

[Agent004]

<<

<< A similar and simpler example is the famou monty python game show. You are on the game show and the hosts offers you a choice of 3 doors, 2 of which has a goat behind, 1 of them is a car. First you picked a door, then the host ramdomly (not really, since he knowns what's behind each door) a door and reveals a goat. The host then offers you a single chance to change the door you previsouly picked. Questions

1) What's the probability of winning the car if you stick?
2) What's the probability of winning the car if you change?
3) So what's the preferred action if you wished to win the car?

If you managed all 3 questions with the correct answers, then you will have no problem with the bus scenario above >>



1. 1/3
2. 2/3
3. switch

you have a 2/3 chance of guessing wrong on the first guess, no matter what. when the host shows you the non-car, you STILL have a 2/3 chance of being wrong - the additional information does not change that fact. so switch. you will win 2/3rds of the time like that. not that I feel like thinking about the bus problem >>



Spot on

The analogy applies to the bus scenario

 

[Mucman]

Wow! I am glad I posted this here and not Off-Topic... I probably would have been told to get a life and stop thinking up stupid things . Thanks for the responses. I am familiar with the 3 door problem since my prof explained that one really well in my discrete math class.

I think Noriaki made a good point of the differences of probability and statistics... Heck my text book was called Probability and Statistics! I still get confused
about the differences but Noriaki's explain really helped me out.

 

[Noriaki]

Oh, one minor other addition.

I don't know if the term "odds" is the proper one. I think statistical probability is probably the more "correct" term. But I wanted to differentiate cleanly between probability probability and statistical probability.

I picked on the word odds becuase that was mentioned already, and when gambling you base it off statisitcal probability and call it "odds". Because of course gambling is a random event and thus fits the criteria

But anyway, despite that I can't remember exactly what it's called, there is a clear distinction between mathametical calculated probability and statisical probability. Glad I could help you understand it

 

[Evadman]

Ow! I have a headache on one side...

 

[gustavo]

Noriaki posts deserve some annotations:

1- Statistics has branches like:
Descriptive statistics: Not to apply in this situation, not intended to infer anything
Experimental design: not to apply in this situation, researchers use this branch.
Sampling Theory: not to apply in this situation, lets assume you allready got the sample.
Probability: the one intended to apply here, assign probability of occurrence to certain events, inference.

Other terms like the one you used ?odds? are not technical and being I Spanish speaking cannot tell what you mean by using them.


2- Probability is based on studying random distribution models, like the one you assumed (the normal distribution). Then you can assign those distribution models to variables and you could finally predict likelihoods of occurrence for certain events.


3- Your assumption of the Normal distribution model is a tremendous mistake, that model is for continuous variables and the variable we are talking about is a discrete one, models for discrete variables could be for example The binomial model or the hypergeometrical model


To all the others: Probability is not affected in any way by previous occurrences of events, the doors example does fall into the probability scene, the events are random ones, the distribution is binomial with 0.5 probability of success and 0.5 probability of failure (strictly speaking this is false because of psychological influence on taking decisions, but lets leave this outside of the argument) so the decision of what door to choose is irrelevant statisticaly speaking keeping or changing door has a half probability of success.


Regards Gustavo.-

 

[woolmilk]

i would walk through the bus and ask all the people when they will leave. Then I would make my bet

 

[Mucman]

Gustavo - I rememember learning Normals approximations with discrete values. It is a very stron approximation, a good example of this is the IQ levels. I would have to hit the text books again to explain it any better though.

I am going to re-iterate my conclusion of the problem :

Men A and B are betting on a 1-case probability problem which can be computed. Since they know the initial conditions. It's like two men playing with
loaded dice, but they both know it.

Men C and D are assuming the bus has a Normal distribution with the mean value being 10/10. Since value of 19/1 is just an anomily and if C and D
were to keep on betting on infinite amount of buses the probability will be around 50:50. Since they are only betting on the one bus, one of them
is getting screwed in the bet because of this statistical anomoly.

If people tended to favour the door side of the bus (right side) would that mean that you could still have a Normal distribution but the mean will perhaps
be centered around a 8/12 mean?

Sorring for the persons heads that are getting hurt by this .

 

[Noriaki]

<< Other terms like the one you used ?odds? are not technical and being I Spanish speaking cannot tell what you mean by using them. >>

Yeah I know. I said that in my second post. I wasn't trying to write a formal specification. I was trying to explain to him the difference between a particular bus, and a random bus. I used some colloquial language to help get my point across, so he would understand. To be honest I don't know what the proper terminology is, I don't have much of a stats background. In fact just barely enough to know the difference between stats and combinatorical probabilty.



<< 3- Your assumption of the Normal distribution model is a tremendous mistake, that model is for continuous variables and the variable we are talking about is a discrete one, models for discrete variables could be for example The binomial model or the hypergeometrical model >>

That's a good point, I forgot about that. I was never as good at the statisitcal parts as I am at calculated probability for a given situation. I've only got one course in stats whereas I have several in combinatorics.

i withdraw my specific statements about stats, I'm not an "expert" on that subject.

But none the less Mucman I hope my post helped you understand the difference here.

- Jon.

 

[gustavo]

Mucman: you did not understand, no statistical probability is possible to apply here, this is just betting on behavior of the bus passengers and statistics dont work with non-random samples.

Thats my point, to apply statistics the man to go down the bus should have to be selected at random, thing that is not stated by you, but for the sake of reaching a conclusion and come to an end now we will assume:

- the man to go down the bus is selected by some randomization process.
- the process has some probability of selecting a man from the left, call it "p" and some probability of
selecting a man from the rigth, call it "q", the sum of p + q equals 1 => (p + q = 1)

Now assuming this, please get a statistcs book, look for the chapters called Binomial and Hypergeometrical Distribution inside a section on Discrete variables, read it and chose between binomial & hypergeometrical the one described as the no-reposition model (to be honest, dont remember which of both models is that neither have a book at hand). Then make your calculations choosing values to assign to p & q.

Noriaki: Your attitude shows us what an Elite Member should be, capable of recognizing a mistake, you deserve your title. Surely you know many more things than me and many other members of this forums in other fields

Regards Gustavo.-

 

[Mucman]

gustavo - Ahh, now I see! In fact I do have a statistics book handy and and I am going to check it out... I sort of remember binomial, but I do not recall hypergeometrical.

 

[Mucman]

OK, I reviewed Hypergeometric Disctributions and it seems to fit the bill!



<< The assumption leading to the hypergeometric distribution are as follows:

1. The population or set to be sampled consists of N individuals, objects, or elements (a finite population).
2. Each individual can be characterized as a success (S) or a failure (F), and there are M successes in the population.
3. A sample of n individuals is selected without replacement in suck a way that each subset of size n is equally likely to be chosen.

The random variable of interest is X = the number of S's in the sample. The probability distribution of X depends on the parameters n, M, and N, so we
wish to obtain P(X = x) = h(x; n,M,N). >>

p.128 of Jay L. Devore; Probability and Statistics 5th Ed.

So the probability will be 1 (one possibility of 19 on the right leaving first) / 2^20 (the total number of ways they can leave the bus)

Hmm, that seems like an extremely low probability! Is that correct?

 

[bizmark]

<< So the probability will be 1 (one possibility of 19 on the right leaving first) / 2^20 (the total number of ways they can leave the bus)

Hmm, that seems like an extremely low probability! Is that correct? >>



No, that's not correct. No need to get fancy with the distributions. A simple symmetry argument works.

Let's number ALL of the men, 1, 2, 3, ... , 20. Number 20 is the one on the right side, and numbers 1-19 all sit on the left side.
Now looking at each man individually, is the chance that any one of them (say #13) gets off last any greater than the chance that any other (say #12) gets off last?

No. They all have the same probability of getting off last. The probablity that one of them gets off last is 1. (SOMEBODY's gotta get off last, right?) The probability of any single man gets off last is 1/20. This same probability applies the same to #1, #2, ..., and #20.

However, now looking back at your calculation it would appear that you calculated wrong. The answer would probably work if you used hypergeometric (unfortunately I don't have my prob. book with me right now so I can't look it up), but you must make sure to calculate the results correctly. In particular, the number of possible ways that 20 men can get off a bus is 20!, not 2^20. And there isn't just one possible way of 19 on the right getting off first; there's 19! ways that the 19 men on the right can get off first. And 19!/20!=1/20.