## A Conversation for What do Probabilities Mean?

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### Degree of information

Vandervecken Started conversation May 31, 2001

I think I'm kind of a subjectivist on this one, but I'd go further, and say that the probability of an event's occurrence, as seen from a particular person's point of view, must depend on how much information is available to that person about circumstances determining the event's occurrence, or lack of it.

I think one interesting side-effect of this is that probability could almost be said to depend on your intelligence, in that the more capable you are at maths, the more use you can make of various clues available to you about the abovementioned circumstances. Of course, this isn't really the case, in my view. Instead, I would say that someone with perfect probability-calculating abilities must by definition come up with the correct answer, and anyone else is making an estimate that is partly calculated and partly guessed - and which must be considered an approximation.

This is well-illustrated by a number of popular probability puzzles with highly-unintuitive answers. Here's an example:

You are a contestant on a game show and the host shows you three closed doors, telling you that behind one of them is a prize, and behind the others is nothing. You are asked to choose a door, which you do. The host then opens one of the other two doors, showing that there is nothing behind it. At this point he now gives you the option to change your choice to the third door. If you change your choice, will you be:

a) better off,

b) worse off, or

c) No different

from a probability point of view?

### Degree of information

Martin Harper Posted May 31, 2001

It's an interesting question, actually - consider the question "What is the probability that Fermat's Last Theorem is valid?". Someone from the future, when it has been proved one way or the other, would get the answer one or zero; what would someone from the past get it as? Is there any way to put a meaningful figure on it?

The answer to your puzzle is "it depends" - on things like whether the host *always* opens one of the other two doors - and whether the host *never* opens a door with a prize behind it - and whether the host is malicious or benevolent. I should really go and write an entry on that puzzle...

### Degree of information

Vandervecken Posted May 31, 2001

Interesting. I had always assumed that the host always opens a door, and it never has a prize behind it - under which constraints it becomes irrlevant whether he is malicious. In such a situation, I think the answer normally given is a reasonable one, i.e. you are better off switching, as your probability of winning then increases from one-third to two-thirds. However, I see what you mean about how hard it is to pin probability down. There are so many factors in reality that might affect the outcome, and it's just so *hard* to determine which ones are to be counted from a given perspective...

By the way, I believe Fermat's Last Theorem actually *has* been proved now

### Degree of information

Martin Harper Posted May 31, 2001

That is somewhat open to debate - the 'proof' is long and tortuous, and not everyone is convinced it doesn't have any holes in it...

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Zathras (Unofficial Custodian of H2G2 Room 101. ACE and holder of the BBC Pens) Posted Jun 1, 2001

You should indeed change your opinion. I could show the probability workings to demonstrate it but nobody beleives them.

This analogy helped me get to grips with why you should change.

Suppose there are a million boxes and you choose one.

The host then opens 999,998 of the other boxes showing that they are empty leaving just one.

Most people would then accept that they should switch to the one he has not opened.

This is exactly the same idea as the 3 box version (although the game show would probably be really tedious to watch).

Z

### Fermat's Last Theorem

Vandervecken Posted Jun 4, 2001

OK, as I do not have the knowledge to get anywhere near understanding the proof, I'll have to concede that one...

I was at least aware that it was a very unorthodox proof, as it relied on a problem from a different branch of mathematics being proved first. I think this is quite quaint, myself - a bit like taking a great work of poetry and translating it into another language. The greatness is still somehow "present" in the new version, but is unlikely to make fully as satisfying reading as the original.

### Degree of information

Vandervecken Posted Jun 4, 2001

I'm sorry to have to say this, but unfortunately the 999,998 boxes don't do anything for me I'm not sure why people would be swayed by that argument at all. After all, no matter whether you have chosen correctly or not, the host will still be able to open 999,998 boxes, provided he knows which one contains the prize. So what does it prove if he does it?

P.S. What's so unusual about a game show that's really tedious to watch?

### Fermat's Last Theorem

Martin Harper Posted Jun 4, 2001

Using results from multiple different branches of mathematics isn't unorthodox - maths is a richly interconnected subject, and that kinda thing happens a lot...

### Degree of information

Zathras (Unofficial Custodian of H2G2 Room 101. ACE and holder of the BBC Pens) Posted Jun 5, 2001

The point of the example with lots of boxes is that in the three box example people tend to say that they wouldn't change their mind since they'd have a fifty fifty chance of winning either way. In the million box example most people are more willing to accept that given the hosts choice of a box not to open the chance is more than 50% that that box contains the prize.

The alternative demonstration (going back to the three boxes) is to say that when you make you first choice you have a 1 in 3 chance of getting the box with the prize. Since the host will definitly open a box that is empty you are given no information by his doing so. Since the probability of an event is unchanged when no new information is provided the probability that you have the correct box is still 1 in 3. Therefore the probability that the other (unopened) box contains the prize must be 2 in 3 (since probabilities sum to unity).

Someone can write out the Bayesian analysis of this if they want.

Z

### Degree of information

Vandervecken Posted Jun 5, 2001

I understand the argument about 1/3 becoming 2/3, and I agree with your analysis.

But: I still think that if most people believe that the host's opening 999,998 empty boxes makes it more likely to be the one he didn't open, as opposed to the one they have chosen, then I am definitely NOT most people. It just doesn't make sense to me at all. It seems obvious that if the host knows where the prize is, he can always open 999,998 boxes and show them to be empty, so it proves absolutely nothing. Maybe this is because of my mathematical training, but I simply think it's common sense!!

### Degree of information

Martin Harper Posted Jun 5, 2001

> " if the host knows where the prize is, he can always open 999,998 boxes and show them to be empty, so it proves absolutely nothing"

Precisely so - so it makes absolutely no difference to the probability that your first choice is right: it's still one in a million, because the opening of the boxes can always, will always, has always been done, and is always possible.

But that means that the unopened box is now 999,999 in a million, so a prety safe bet

### Degree of information

Vandervecken Posted Jun 6, 2001

Aaaaah! Light bulb just appeared above my head. Thanks, I get it now. I think what I was forgetting was that the chance of my first choice being correct with 1,000,000 boxes was miniscule..

I'd better not go on any game shows

### Degree of information

Zathras (Unofficial Custodian of H2G2 Room 101. ACE and holder of the BBC Pens) Posted Jun 6, 2001

So do you see now that when there are only three boxes you should also change your mind?

Z

### Degree of information

Vandervecken Posted Jun 6, 2001

I believe I've already said this, but I never actually had a problem with the answer to the problem.

The thing I couldn't work out at first was why the 1,000,000 boxes example made the problem any easier for a "lay person" to understand. In fact, although the combination of your example and Lucinda's further commentary on it has helped me a little, I still feel uncomfortable with the whole thing. *I* can see that you should change your opinion, but I still don't understand what is the best way of explaining it to a non-mathematician without using what they would consider to be obscure theorems. In other words, how do you make the answer to this problem easy to understand? My personal view is that the 1,000,000 boxes don't really achieve this. When I referred to the light bulb above my head, I was saying that I understood the comparison, but that doesn't mean that I am convinced that *everyone* will understand it (hopefully without sounding arrogant).

I'm not sure how clear I'm making myself here, but hopefully I've explained things better than I did before.

Anyway, going back to the 3-boxes setup, how about this for an alternative explanation, which I prefer for some reason:

Imagine that, instead of opening one of the boxes you didn't choose, the host says that you may either stick with your current choice, or switch to BOTH the other boxes. If you switch, they will both be opened, and provided that one of them contains the prize, it's yours. In this situation, it seems obvious that one should switch, as two chances are better than one. But now for the clever part: If the host had opened an empty box and given you the opportunity to switch to the other unopened one, the situation would have been exactly the same as the 'Stick, or switch to the remainder as a group' option mentioned above. Because the host knows where the prize is, and therefore could never accidentally open the prize box, the characteristics of the two unchosen boxes, as a group, have not really changed. If that group previously contained the prize, it will still contain it. If it did not, it will not. The probabilities are unchanged.

### Degree of information

Zathras (Unofficial Custodian of H2G2 Room 101. ACE and holder of the BBC Pens) Posted Jun 6, 2001

Yes thats another good way of explaining it. I might try that next time somebody doens't understand.

Z

### Degree of information

Bez (arguaby the finest figure of a man ever found wearing Bez's underwear) <underpants> Posted Sep 13, 2002

Maybe I'm being thinck here, but how does the other box become 2/3? You've chosen one out of the 3, and another has been opened and shown to be empty, no mateer which box you choose. This means that there are only two boxes that are important for whether you change or not. One has the prize, the other doesn't. Surely this means that you have a 1/2 chance of choosing the right one whther you change or not?

Now your chances of choosing the correct one initially are 1/3, and the chances of it being the other one that is not opened are 2/3 before you start, but each box has until more data is recieved an equal chance, therefore if one of the boxes is known to not contain it but no new information is provided about the other two they still have an equal probability, thus 1/2.

Bez

### Degree of information

manolan Posted Sep 20, 2002

BTW, that was a real game show in the States.

One thing about the 3 doors is there are so few combinations, you can always write them all out and it becomes totally clear that you're better off switching. Doesn't do anything to rationalise the result, however.

The trick is that the host has two constraints. He cannot open the door you have chosen and he cannot open the door with the prize. That means that the only time he has a real choice of which door to open is when you picked the right door first time. If you didn't, then there is only one door he can open and the other _must_ be the prize.

You can see it here. "x" shows the prize. "1" is your first choice. "2" is your second choice. "r" is revealed. You only lose if you picked right first time and there's only a 1 in 3 chance of that.

Location / Outcome

x1 -r -2 or x1 -2 -r / lose

x2 -1 -r / win

x2 -r -1 / win

-1 x2 -r / win

-2 x1 -r or -r x1 -2 / lose

-r x2 -1 / win

-1 -r x2 / win

-r -1 x2 / win

-r -2 x1 or -2 -r x1 / lose

### Degree of information

Iago Posted Sep 21, 2002

I think you are all to be congratulated. In typical mathematician's fashion, you have turned the simple solution into a problematic nightmare. I would not have you on the end of a phone to help me!

Wasn't it Bayes himself, by the way, who considered that the postulate on which he had argued might not perhaps be looked upon by all as reasonable? I confess to not understanding your own lack of understanding.

### Degree of information

Iago Posted Sep 23, 2002

Light Bulbs

Q: How many scientists does it take to change a light bulb?

A: None. They use them as controls in double blind trials.

Q: How many academics does it take to change a lightbulb ?

A: None. That is what their students are for.

A: Five: One to write the grant proposal, one to do the mathematical modelling, one to type the research paper, one to submit the paper for publishing, and one to hire a student to do the work.

Q: How many senior researchers does it take to change a lightbulb?

A: Five; one to change the lightbulb, the other four to stand around arguing whether he/she is taking the right approach.

Q: How many research technicians does it take to change a lightbulb?

A: One, but it'll probably take him/her three or four tries to get it right.

Q: How many post-doctoral fellows does it take to change a lightbulb?

A: One, but it'll probably take three or four tries to get it right because he/she will probably give it to the technician to do.

Q: How many students does it take to screw in a light bulb?

A: Only one, but it may take upwards of five years for him to get it done.

A: It all depends on the size of the grant.

A: Two and a academic to take credit.

A: 1/100. A student needs to change 100 lightbulbs a day.

Q: How many general relativists does it take to change a light bulb.

A: Two. One holds the bulb, while the other rotates the universe.

Q: How many quantum physicists does it take to change a lightbulb ?

A: One. Two to do it, and one to renormalise the wave function.

(Explanation - Renormalising the wave function is something that has to be done to a lot of quantum physics calculations to stop the answer being infinity and makes the answer always come out as one.)

Q: How many quantum mechanicians does it take to change a light bulb?

A: They can't. If they know where the socket is, they cannot locate the new bulb.

Q: How many Heisenbergs does it take to change a light bulb?

A: If you know the number, you don't know where the light bulb is.

Q: How many astronomers does it take to change a light bulb?

A: None, astronomers prefer the dark.

Q: How many radio astronomers does it take to change a light bulb.

A: None. They are not interested in that short wave stuff.

Q: How many particle physicists are necessary to change a light bulb?

A: Two hundred: 136 to smash it up + 64 to analyse the tiny pieces.

Q: How many statisticians does it take to change a lightbulb?

A: 1-3, alpha = .05

A: This should be determined using a nonparametric procedure, since statisticians are NOT NORMAL.

A: One -- plus or minus three (small sample size).

(Notes: Someone has been asking this as a bonus question on statistics exam papers for quite a while. Judging from some of his own students' exam answers, it depends on whether the lightbulb is negatively or positively screwed.)

Q: How many numerical analysts does it take to screw in a light bulb?

A: 0.9973 after the first three iterations.

Q: How many topologists does it take to change a light bulb?

A: It really doesn't matter, since they'd rather knot.

Q: How many mathematicians does it take to screw in a lightbulb?

A: None. It's left to the reader as an exercise.

A: Just one, once you've managed to present the problem in terms he/she is familiar with.

In earlier work, Wiener [1] has shown that one mathematician can change a light bulb.

If k mathematicians can change a light bulb, and if one more simply watches them do it, then k+1 mathematicians will have changed the light bulb.

Therefore, by induction, for all n in the positive integers, n mathematicians can change a light bulb.

Bibliography:

[1] Weiner, Matthew P., , "Re: YALBJ", 1986

Enjoy!

Iago

Key: Complain about this post

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### Degree of information

- 1: Vandervecken (May 31, 2001)
- 2: Martin Harper (May 31, 2001)
- 3: Vandervecken (May 31, 2001)
- 4: Martin Harper (May 31, 2001)
- 5: Zathras (Unofficial Custodian of H2G2 Room 101. ACE and holder of the BBC Pens) (Jun 1, 2001)
- 6: Vandervecken (Jun 4, 2001)
- 7: Vandervecken (Jun 4, 2001)
- 8: Martin Harper (Jun 4, 2001)
- 9: Zathras (Unofficial Custodian of H2G2 Room 101. ACE and holder of the BBC Pens) (Jun 5, 2001)
- 10: Vandervecken (Jun 5, 2001)
- 11: Martin Harper (Jun 5, 2001)
- 12: Vandervecken (Jun 6, 2001)
- 13: Zathras (Unofficial Custodian of H2G2 Room 101. ACE and holder of the BBC Pens) (Jun 6, 2001)
- 14: Vandervecken (Jun 6, 2001)
- 15: Zathras (Unofficial Custodian of H2G2 Room 101. ACE and holder of the BBC Pens) (Jun 6, 2001)
- 16: Bez (arguaby the finest figure of a man ever found wearing Bez's underwear) <underpants> (Sep 13, 2002)
- 17: Martin Harper (Sep 13, 2002)
- 18: manolan (Sep 20, 2002)
- 19: Iago (Sep 21, 2002)
- 20: Iago (Sep 23, 2002)

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