Some more probability thoughts
Jul. 9th, 2010 10:30 am![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
peterbirksWeirdly, it isn't the counterintuitive 13/27 answer that caused the controversy (thus showing that you guys actually do understand numbers), but whether the starting point is a half or a third (thus showing that we don't really understand semantics, language and philosophy).
I've been thinking about this "two children" problem, and I think I've come up with an interesting, and probably solvable (given a few assumptions about certain likelihoods) poser.
Let's apply this to gambling!
A man walks in. Slaps his palm down on the table on top of two coins and looks at them. He says to you (Mr Aardvark) "There are two coins here. At least one of them is a head."
A second man (a Mr Birks) (who knows no more than you do about this state of affairs) now says: "I will offer you 3-to-2 on both coins being heads".
Do you take the bet?
Well, you, Mr Aardvark, are in the 50:50 camp. So obviously you do.
As it happens, Mr Birks wins the bet (the second coin is a tail), and you are ten quid down.
"Mr Aardvark", scoffs Mr Birks, "You are a fool! Everyone knows that the chance of the other coin being a head is one-third!"
But you, Mr Aardvark, are not non-plussed. "Double or nothing?" you say.
"Hah HAH!" says Mr Birks. "This is going to pay for my holiday!"
The first man then walks out of the door. Ten seconds later he comes back in. He repeats the exercise, and says:
"At least one of these coins is a tail."
At this point, Mr Birks (foolishly) says "ahh, it's exactly the same problem, so clearly the chance of the second coin being a tail is still 1/3." He offers you 3-to-2, and you absolutely lump on, because now you are certain (incorrectly, as it happens, but we can cover that later) that the chance of the other coin being a tail is 50%.
Let's take the sample of four.
(A) Coins are HH. Man says "At least one of these coins is a head"
(B) Coins are TT. Man says "At least one of these coins is a tail"
(C) Coins are TH. Man says "At least one of these coins is a tail"
(D) Coins are TH. Man says "At least one of these coins is a head"
The "law of restricted choice" in A and B causes the probability of the other coin to be the same to move up to 50% from 33%.
But, suppose the man comes in the second time, tosses the two coins and puts his palm down, and, once again, says... "At least one of the coins is a head". Once again, Mr Birks offers you 3-to-2 on the other coin being a head. Do you, Mr Aardvark, still lump on?
This is a tougher prospect.
Clearly, therefore, what matters here (in terms of making money) is the sample of "choices of what to say" held by the man who tosses the coin.
Suppose he comes in, tosses the two coins, looks at them. Says nothing. And then tosses the two coins again. At which point he says "At least one of the coins is a head".
Even you, Mr Aardvark, will deduce from the fact that the man said nothing after the first toss that there is a state of affairs that commits him to silence. Thus, when he makes the second toss and states "At least one of them is a head", you will induce (perhaps incorrectly, but it's a reasonable induction) that his previous silence was because he had tossed two tails.
At this point you, Mr Aardvark, will conclude that the chance of the other coin being a head is 1/3.
In real life, of course, Mr Birks would deduce as soon as the man said the second time "at least one of them is a tail" (having said the first time that "at least one of them is a head", that the chance of the other coin being a tail was, because of the law of restricted choice, 50%, not 33.3%.
The key, therefore, is what the man is instructed to say. Even the above is simplistic.
I have, in fact, assumed two possibilities (nos (1) & (4) below) when there are more than this. remember, we are toalking about real prop bets here.
1) The man is told (or has decided) to say either "at least one is a head" or "at least one is a tail". If it's a head and a tail, he can say either.
2) The man is instructed to say "at least one is a tail" ONLY if both are tails. Otherwise, he says "at least one is a head".
3) The man is instructed to say "at least one is a head" ONLY if both are heads. Otherwise, he says "at least one is a tail".
4) The man is instructed to say "at least one is a head" if at least one is a head. Otherwise, he stays silent and tosses again.
5) The man is instructed to say "at least one is a tail" if at least one is a tail. Otherwise, he tosses again.
There are many other combinational possibililties, and clearly it would be erroneous to apply an equal probability to all of them. So let's just stick to casees (1) and (4) for the moment.
Now, let's go back to the first coin toss again. He tosses it, and says "At least one is a head". What is the chance that the other is a head?
If we take the simple sample here (choices of (1) and (4) being the "actual state of affairs") , we could say: "He might have been told to stay silent if neither was a head, or he might have been told to say either "at least one is a head" or "at least one is a tail" (his choice) if it was a head and a tail. These are equally likely states of affairs (under the principle of 'in the beginning, everything was even money')".
Now, this is where it gets interesting. In effect, if we are talking about having a bet on this, we know that there's an evens chance that it's 33.3% (state of affairs 4 is true), and an evens chance that it's 50:50 (state of affairs 1 is true).
This gives us the intermediate real, I'm prepared to put my money on it, chance of 41.67% that the other coin is a head!
Once you add in the other "posible states of affairs", things become more complicated. I can return to that later.
Now, if the man goes out, comes back in, and says "At least one is a tail", I think that the odds shift directly to 50%.
However, if he once again says "at least one is a head", we have an interesting calculation.
It's my guess (and it is a guess -- I haven't seriously worked it out) that this would shift the best "prop bet" likelihood to about 37.75% that the other coin is a head.
For every time the man comes in and says "at least one is a head", rather than saying "at lest one is a tail" or saying nothing and tossing again, you halve the distance between your previous percentage and 33.33%.
If anyone wants to test this financially, I am happy to walk into a room and be the coin tosser and statement maker while two people bet against each other!.
PJ
I've been thinking about this "two children" problem, and I think I've come up with an interesting, and probably solvable (given a few assumptions about certain likelihoods) poser.
Let's apply this to gambling!
A man walks in. Slaps his palm down on the table on top of two coins and looks at them. He says to you (Mr Aardvark) "There are two coins here. At least one of them is a head."
A second man (a Mr Birks) (who knows no more than you do about this state of affairs) now says: "I will offer you 3-to-2 on both coins being heads".
Do you take the bet?
Well, you, Mr Aardvark, are in the 50:50 camp. So obviously you do.
As it happens, Mr Birks wins the bet (the second coin is a tail), and you are ten quid down.
"Mr Aardvark", scoffs Mr Birks, "You are a fool! Everyone knows that the chance of the other coin being a head is one-third!"
But you, Mr Aardvark, are not non-plussed. "Double or nothing?" you say.
"Hah HAH!" says Mr Birks. "This is going to pay for my holiday!"
The first man then walks out of the door. Ten seconds later he comes back in. He repeats the exercise, and says:
"At least one of these coins is a tail."
At this point, Mr Birks (foolishly) says "ahh, it's exactly the same problem, so clearly the chance of the second coin being a tail is still 1/3." He offers you 3-to-2, and you absolutely lump on, because now you are certain (incorrectly, as it happens, but we can cover that later) that the chance of the other coin being a tail is 50%.
Let's take the sample of four.
(A) Coins are HH. Man says "At least one of these coins is a head"
(B) Coins are TT. Man says "At least one of these coins is a tail"
(C) Coins are TH. Man says "At least one of these coins is a tail"
(D) Coins are TH. Man says "At least one of these coins is a head"
The "law of restricted choice" in A and B causes the probability of the other coin to be the same to move up to 50% from 33%.
But, suppose the man comes in the second time, tosses the two coins and puts his palm down, and, once again, says... "At least one of the coins is a head". Once again, Mr Birks offers you 3-to-2 on the other coin being a head. Do you, Mr Aardvark, still lump on?
This is a tougher prospect.
Clearly, therefore, what matters here (in terms of making money) is the sample of "choices of what to say" held by the man who tosses the coin.
Suppose he comes in, tosses the two coins, looks at them. Says nothing. And then tosses the two coins again. At which point he says "At least one of the coins is a head".
Even you, Mr Aardvark, will deduce from the fact that the man said nothing after the first toss that there is a state of affairs that commits him to silence. Thus, when he makes the second toss and states "At least one of them is a head", you will induce (perhaps incorrectly, but it's a reasonable induction) that his previous silence was because he had tossed two tails.
At this point you, Mr Aardvark, will conclude that the chance of the other coin being a head is 1/3.
In real life, of course, Mr Birks would deduce as soon as the man said the second time "at least one of them is a tail" (having said the first time that "at least one of them is a head", that the chance of the other coin being a tail was, because of the law of restricted choice, 50%, not 33.3%.
The key, therefore, is what the man is instructed to say. Even the above is simplistic.
I have, in fact, assumed two possibilities (nos (1) & (4) below) when there are more than this. remember, we are toalking about real prop bets here.
1) The man is told (or has decided) to say either "at least one is a head" or "at least one is a tail". If it's a head and a tail, he can say either.
2) The man is instructed to say "at least one is a tail" ONLY if both are tails. Otherwise, he says "at least one is a head".
3) The man is instructed to say "at least one is a head" ONLY if both are heads. Otherwise, he says "at least one is a tail".
4) The man is instructed to say "at least one is a head" if at least one is a head. Otherwise, he stays silent and tosses again.
5) The man is instructed to say "at least one is a tail" if at least one is a tail. Otherwise, he tosses again.
There are many other combinational possibililties, and clearly it would be erroneous to apply an equal probability to all of them. So let's just stick to casees (1) and (4) for the moment.
Now, let's go back to the first coin toss again. He tosses it, and says "At least one is a head". What is the chance that the other is a head?
If we take the simple sample here (choices of (1) and (4) being the "actual state of affairs") , we could say: "He might have been told to stay silent if neither was a head, or he might have been told to say either "at least one is a head" or "at least one is a tail" (his choice) if it was a head and a tail. These are equally likely states of affairs (under the principle of 'in the beginning, everything was even money')".
Now, this is where it gets interesting. In effect, if we are talking about having a bet on this, we know that there's an evens chance that it's 33.3% (state of affairs 4 is true), and an evens chance that it's 50:50 (state of affairs 1 is true).
This gives us the intermediate real, I'm prepared to put my money on it, chance of 41.67% that the other coin is a head!
Once you add in the other "posible states of affairs", things become more complicated. I can return to that later.
Now, if the man goes out, comes back in, and says "At least one is a tail", I think that the odds shift directly to 50%.
However, if he once again says "at least one is a head", we have an interesting calculation.
It's my guess (and it is a guess -- I haven't seriously worked it out) that this would shift the best "prop bet" likelihood to about 37.75% that the other coin is a head.
For every time the man comes in and says "at least one is a head", rather than saying "at lest one is a tail" or saying nothing and tossing again, you halve the distance between your previous percentage and 33.33%.
If anyone wants to test this financially, I am happy to walk into a room and be the coin tosser and statement maker while two people bet against each other!.
PJ
![[identity profile]](https://www.dreamwidth.org/img/silk/identity/openid.png){kind=small}
1 In 4, Baby, 4 In 1
Date: 2010-07-09 02:55 pm (UTC)For added stringency, case (1) needs a verbal toughening up -- something along the lines of the man tossing a coin whilst outside to predetermine his choice of words if the result inside is HT or TH. But it's fair to assume that some such aleatory procedure takes place.
I actually agree with everything you say after "Now, let's go back to the first coin toss again." I had to think quite hard about it, but I'm fairly sure I'd have reached each separate conclusion myself, given the wording. Of course that's easy for me to say, because you've already written it down ... Not entirely sure about all the preceding stuff, but then I haven't looked carefully enough at it.
The interesting thing, of course, is that the world changes the minute the man walks in, tosses the coins, and says nothing. At that point you've eliminated (1) and the odds permanently shift to 33%. If you expect Aardvark not to spot this and consequently not adjust his bet, there's a certain amount of arbitrage available on the previous bets to make sure he thinks he's winning...
To return briefly to the BB problem, I found this (http://mathforum.org/library/drmath/view/52186.html) interesting on the topic of "choosing" the candidate family. To simplify, if you pick a boy out of the pool to determine the family, the chances are 50:50. If you pick the family out of the pool and find a boy, the chances are 33:66.
Except ... if the set of all families with two children has a cardinality of one -- there's only one such family in the whole world -- then the chances are, apparently, simultaneously 50:50 and 33:66. Which just goes to show how difficult it is to formulate these problems.
Re: 1 In 4, Baby, 4 In 1
Date: 2010-07-09 03:19 pm (UTC)(If you're going to insist on specifying the gender, no matter what, then I believe that the Bayesian formulation in the wikipedia article is the way to go.)
Re: 1 In 4, Baby, 4 In 1
Date: 2010-07-10 01:51 pm (UTC)PJ
Re: 1 In 4, Baby, 4 In 1
Date: 2010-07-10 05:12 pm (UTC)"Thus, when he makes the second toss and states "At least one of them is a head", you will induce (perhaps incorrectly, but it's a reasonable induction) that his previous silence was because he had tossed two tails."
Induction in the mathematical sense does pretty much what it says on the tin. You start with the assumption of no historical knowledge, and you build up from one or more minimalist preconditions. (Usually it's one, but as long as it covers the entire problem space for preconditions, that's fine.) The reason you use induction is because (a) it eliminates all things that you do not know -- or else it fails, by definition -- and (b) you can use etablished rules to build from a lemma to a proof.
In this case, you are clearly (and correctly) using deduction, which is mathematically a process that involves historical knowledge. I'm not sure I've got that definition completely right, so let me assert that proof by induction is a recursive proof, whereas proof by deduction is linear.
I'm prepared to consider that it is possible to present a "proof by deduction" as a "proof by induction." Within the set of proofs (given axioms blah blah don't forget Goedel blah blah) there may be proofs that are, idempotently, both inductive and deductive.
Let's say you found one of those.
In that case, using a common verb that reminds people of using mechanical and/or pharmaceutical means for causing a birth to occur is almost certainly an unfortunate alternative to using another common verb that reminds people of Sherlock Holmes.
Re: 1 In 4, Baby, 4 In 1
Date: 2010-07-10 05:54 pm (UTC)The latter is right (and induce is wrong) when you say "It's a number from 1 to 8, but it isn't 8". I therefore deduce that it's "a number from 1 to 7".
I "induce" the general case that the silence was because he had two tails, because his statement "at least one of them is a head" after his previous silence acts in a different logical way on the previous silence after a coin toss from the way the statement "but it isn't 8" acts on the previous statement "it's a number from 1 to 8".
It was possible, for example, that the man was operating under a rule of which we were unaware (but which was not the rule that, if he threw two tails, he remained silent and tossed again) that required him to toss the coins again.
Therefore I felt that I was not using mathematical deduction in the Holmesian sense (eliminating all cases until whatever remained, must, however improbable, be the truth).
I was, instead, inducing a state of affairs in the world from a single instance.
The unfortunate fact that "induction" has other medical associations is an irritation, but I don't think I should allow that to discourage me from using it in the sense which does exist.
PJ
Re: 1 In 4, Baby, 4 In 1
Date: 2010-07-10 06:36 pm (UTC)Or, in other words, and God knows I've been reading you for 30 years because you obviously believe in this, language is about clarity.
(It's still a side-issue, though.)
I can't think of a single instance of a logical or mathematical framework that allows you to "induc[e] a state of affairs in the world from a single instance in the future."
Well, I can, actually. That single instance (and as I say, it doesn't have to be a single instance -- as long as it covers the problem domain) is by necessity the single instance from which you infer the subsequents. Given my assumption that there is a part of the proof space where deduction is exactly equivalent to induction (and don't forget, I'm not sure there is one; I'm just allowing you that possibility), then, I believe, you are taking a deductive proposition such as
D <- C <- B <- A
and, in your words, "inducing [actually inducting, I think] a state of affairs in the world from a single instance."
I wish to put forth the proposition (in a Jim Morrison sort of way) that you cannot induce D from A. You cannot. Dum de dum de soft machine dum de dum ...
Your single instance is in the historical present. Your conclusions are based upon analysis of the historical past.
That would be deduction -- not induction.
Induction is normally represented as a recursive process. There is no recursion present in your argument.
On the side: No, you weren't using Holmesian elimination. You weren't even suggesting that you were. I used that as a metaphor, and you are throwing it back as an accusation. What you were doing was to use a set of (reasonably assumed) preconditions and arguing based on those preconditions. That still makes it a deduction when you are talking about the set of results based upon those preconditions.
Basically, I think what I'm saying is that you can inductively generate a statistical spread on "first principles," without history. If you are going to use (perfectly validly) a set of historical results (which already represents a statistical spread), then it's deduction. If only because you know more. (Which is what I thought was the point at issue in the first place.)
Of course, this might only go to prove that I was right to do two A-levels in Maths & Mechanics rather than one in pure maths and one in stats.
Can't say I feel the loss after all these years.