Probability is tricky.
If you are one of those people who likes to say “I was never any good at maths at school”, your response to this would be “of course!”. But probability seems to be tricky for mathematicians too, even for mathematicians who teach probability.
I have always loved probability paradoxes and clearly remember spending most of a train-trip to Edinburgh almost 20 years ago debating the famous Monty Hall problem (see below).
Now I’ve been thinking about paradoxes again. It all started with a passing reference to the two boys paradox in my recent post about the passing of Martin Gardner. Commenting on that post, Bob Walters (who was in fact my honours supervisor a long time ago) drew my attention to his own reflections on that paradox, which opened a can of worms for me. I’ve been obsessing on the topic ever since.
Rather than jumping straight to my conclusions, I thought it would only be fair to give readers a chance to think things through themselves first. So in this post, I will simply state a few well-known probability paradoxes and discuss my own thoughts on each of them in future posts. Feel free to share your thoughts in the comments.
Bob’s post alerted me to a variant of the two boys problem that has been generating a lot of discussion (here, here, here and here among other places). Appropriately enough, all of this discussion emerged from the most recent Gathering for Gardner, a conference on mathematical puzzles held in honor of Martin Gardner.
So here is the puzzle.
A man says to you “I have two children, one is a boy born on a Tuesday”. What is the probability the man has two boys?
This is a more complicated version of the original puzzle where the man simply says “I have two children, one is a boy”. Classically, the answer to this one is that the probability of two boys is 1/3. The argument is that there are four equally likely probabilities to start with: Boy-Boy, Boy-Girl, Girl-Boy and Boy-Boy*. The statement rules out Girl-Girl, and Boy-Boy is one of the three equally likely remaining possibilities.
I say “classically” because I no longer agree with this reasoning. I’ll explain why in a later post, and for now I’ll just pose another question. Is there any difference between the scenario in which you ask the man “do you have at least one boy?” and the scenario in which the man simply volunteers the information? I think there is a difference and understanding this difference is the key to the puzzle. As for Tuesday’s child, does the day of the week have any bearing on the probability?
This is perhaps the most famous of all probability paradoxes. It derives its name from the host of an old American TV game show called “Let’s Make a Deal”. Here’s how the problem was posed in a letter to the “Ask Marilyn” column in Parade magazine.
Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?
Should you switch? Does it matter? When the problem appeared in Parade, it generated an enormous volume of mail, arguing both ways. So, if you think the answer is obvious, it may be worth thinking again!
This is perhaps the trickiest problem of the three.
You are given the choice of two sealed envelopes and told that they both contain money, and that one envelope contains exactly twice as much as the other. You pick an envelope and open it, discovering that it contains, say, $20. You are then given the option of keeping the $20 or switching envelopes.
It is suggested that there is a 50% chance that the other envelope contains $10 and a 50% chance that it contains $40 and that your expected pay-off is therefore 0.5 x $10 + 0.5 x $40 = $25 if you switch. Is this reasoning correct and should you therefore change envelopes?
If that reasoning is correct, doesn’t it also mean that you should switch if the first envelope contains $30 or $500? In which case, why even bother opening the envelope if you know you are always going to switch?
* Here I’m assuming that human babies are equally likely to be boys or girls, which is probably not quite true.
Possibly Related Posts (automatically generated):
- The Monty Hall Problem (21 June 2010)
- Tuesday’s Child (8 June 2010)
- Eliminating the irrelevant (10 June 2010)
- Vale Martin Gardner (28 May 2010)
With the children question, the answer depends entirely on psychology and social context, putting this well outside the reach of probability. What are the circumstances in which someone (a stranger? a friend?) is likely to tell you that they have two children of various sexes (or birthdays)?
If you’re in a culture where boys are valued much more than girls, then a stranger telling you he – and that it’s not “she” may be significant – has one boy probably implies they don’t have two (and depending on the wording that may be the “ordinary English” interpretation anyway).
And a lot of cultures have superstitions about days of the week (though at least in Java and I think Japan these attach to older calendrical systems, not the European week).
Danny: While I agree that context is important, I don’t think that means it’s outside the reach of probability. I like to take a Bayesian approach and come up with the most reasonable prior you can. Obviously, if you knew something specific about the context, your prior would be different. As a simple example, if you thought a dice could be loaded but had nothing to suggest on which sides, your prior would still be to assign 1/6 to each face value.
Another great post, Mule.
Indeed, probability is tricky. No matter what definition you choose for conditionalisation, you can generate counter-intuitive situations — the ones offered in the Tuesday’s Child and Monty Hall problems show the problems with classical conditionalisation in particular. Of course, it is only convention that we choose classical conditionalisation (i.e. probability mass is redistributed uniformly over remaining possible states), we can choose a different definition if we like. So one could say that the troubles come from the (classical) definition of (conditional) probability.
The Envelope problem is a different matter. This is connected to the choice of measure (the space in which probabilities are calculated), which, perhaps counter-intuitively, impacts expectation calculations. In this case, the measure has not been fixed ex-ante, which leads to the paradox.
Anyway, you may be interested in the following seminal paper (doesn’t seem to be available online) if you haven’t seen it already:
David Lewis (1976) “Probabilities of Conditionals and Conditional Probabilities” The Philosophical Review, Vol. 85, No. 3 (Jul., 1976), pp. 297-315
Oh yes, under classical conditionalisation the answer to Tuesday’s child is 13/27, if my mental calculations are right.
Oh, I forgot to mention that in the Monty Hall problem, different conditionalisations correspond to different behaviours of the host. Furthermore, under all possible conditionalisations (resp. behaviours) you are no worse off, and may be better off, if you switch. Therefore a meta-argument says you should always switch.
Does the non-zero probability of twins – and identical twins – change the Tuesday’s Child probabilities?
Danny: it shouldn’t, even in the classical interpretation, which gives you 13/27, and certainly doesn’t in my interpretation.
But what if you want the goat?
Stilgherrian: I should not have picked such a fine looking goat: it really undermines the premise that you wouldn’t want the goat!
Doesn’t the possibility of identical twins make two boys slightly more likely given one boy? It should raise the probability of both BB and GG, but we can rule out GG…
I can get 13/27 easily enough, but in this case at least I think the classical approach makes no sense, as any kind of context at all will dominate. There may or may not be a difference between having “born on a Thursday” volunteered and having that information provided in response to a question — it depends on psychological assumptions.
I’m sorry Mule, but a non-zero identical-twinning rate does change the answer under classical conditionalisation. For example, if the identical-twinning rate is (the extremely high) 50% per pregnancy, then the probability of the other child being a boy is 5/7 (assuming identical twins always have the same sex, and as before sons and daughters are equally probable).
Mark: you are right, I missed the identical bit (which is silly…that was the point, after all!). Still, I’ll argue that you can make a case for the probability of Boy-Boy being 1/2 not 13/27 (or something a little higher because of the identical twin rate) and twins will not affect that argument.
Mark and Danny: of course the impact of a non-zero rate of identical twins is not just on the conditional probability, but the unconditional probability of having two boys, which risies above 1/4. As I understand it, the frequency of identical twins is under 0.5%, so for the purposes of the problem, I would be inclined to sweep it under the rug along with any differences in the boy/girl rate away from 50-50.
I liked that envelope puzzle even using the probability maths as you have to present the paradox.
The paradox I believe is caused by an error in logic with the problem.
The probability factors of 0.5 x$10 +(and) 0.5x$40 are mutually exclusive so the calculation should be 0.5x$10 or 0.5x$40 which is useless and when we try to treat it as an (and) the probability paradox occurs due to the logical non sequitur, hence also useless.
Further – All we can logically say is that there is a 50% probability that the amount in the envelope will be greater or less than the $20 in hand. The dollar amounts are irrelevant. It can be expressed as 0.5x-1(less) + 0.5×1(more)=0(no reason to pick the other)
The problem is interesting! It will be interesting to see what others think. Thanks for the teaser!
Mark: actually, identical twins would mess up my alternative version too. After all, if the twin rate was 100%, then as soon as you know there is at least one boy, you know there are two. So I’ll stick to ignoring twins (rude, I know!).
Firefly: glad you like the teaser…I’ll post a comment here when my discussion of the envelopes is posted.
On the two boys problem – leaving all the Tuesday preamble aside because it is about the boy child with sex already defined, one question remains – What is the sex of the child as yet un- revealed?
So at this point is the probability of this child being a male equals the probability of the man having two boys ? My answer is 0.5 or 50% assuming that we are going to use male to female ratio of 1 to 1.
You mean rises above 1/4, of course.
PS: Strangely enough, I have thought a fair bit about twinning :-).
Of course! I have now edited it.
I am really enjoying these puzzles – Thanks again:)
Further to my last “Tuesdays child” comment – If we ask the man if he has at least one boy? The problem may change completely depending on his answer!
If he answers “no” then the probability he has any boys becomes “zero” and the probability he has two boys equals “zero”.
If he answers yes then the probability he has two boys is once again equal to the probability that the 2nd child is a boy or 0.5 because one child is already defined as a boy and only one is unknown.
Another very nice post Mule. Unfortunately however I had just managed to get that train trip out of my mind! Edinburgh or Glasgow?!
Evo: you’re right, it was Glasgow!
Haven’t read the comments here.
The envelope problem is not defined. Either the person who is asking the question is making an assumption or the person trying to answer. There cannot be a uniform probability distribution amongst all envelope values. Its not normalizable. Here’s a link discussing that http://blogs.discovermagazine.com/cosmicvariance/2010/05/17/non-normalizable-probability-measures-for-fun-and-profit/
Ramanan: you are correct that at the heart of the problem is the impossibility of having a uniform probability distribution on all dollar values. Two further questions then arise: (1) why does this create a problem for the attempt to calculate an expected value of the other envelope and (2) is there anything sensible that can be said about the problem?
Hi again Mule:)
I am back here again to play on your blog.
In the Monty Hall Puzzle the two player choice episodes are separate and should be treated as such, rather like two separate tosses of a coin are separate events each having a 50% probability of either heads or tails. Each one is irrelevant to the others unless you wish to treat them as a series which this puzzle avoids.
So unless we are going to attempt to bring into the calculation conditions such as the game show hosts propensity to assist or mislead the players based on some as yet undefined data, and other factors like mood effect on the host and player of the weather and time of day I believe the players should regard the choice they are offered as simple mathematical exercise as the other conditions are not defined. If any of these conditions could be defined there could as in any human condition be a variance to the probabilities. So what do we know from the puzzle as presented?
There is a car and a goat behind two doors. We don’ t know which door hides the car or the goat. So the probability of the car being behind either door is 0.5 the same is true for the goat. Therefore there is no advantage to the player either way! Pick the currently chosen door or pick the other as it makes no difference to the probability of a car win. After that the player can do as they please based on any reasoning they like to choose either door as it won’t matter.
firefly: OK, now let’s imagine we can assume the following:
1. The host has no particularly propensity to help or hinder you.
2. The host knows which door the car is behind.
3. The host will always open a door with a goat (to string out the tension).
Would those assumptions change your reasoning?
Interesting story – Pigeons beat humans in Monte Hall :)
Hello Mule :)
No my reasoning for a resolution of the puzzle would like most humans incorrectly remain the same.
Well done mule you beat me with that one. I suspect it might take me a while to correctly understand this puzzle as the correct result seems counter intuitive for us mere humans.
Oh well reality is often weird and it seems I need to do a bit more work on correct reasoning :) Still I enjoyed the puzzle so thanks for the fun!
Hello Mule :)
That will teach me NOT to go playing games with Mathematicians specialising in probabilities – I just wandered off and checked all the puzzle resolutions and I have not reasoned a correct solution for any of your wonderful puzzles he he :) These puzzles are great and indeed probability is tricky ! Thanks again:)
At that I better leave this fun to those better equipt to deal with it he he :)
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Here’s the first of two posts about Tuesday’s child.
Ramanan: that pigeon post is brilliant!
Hello Stubborn Mule:)
Regarding the puzzle “Tuesdays Child”
“A man says to you “I have two children, one is a boy born on a Tuesday”. What is the probability the man has two boys?”
Please allow me to ask a question or two….
So with classical reasoning the probability we are after is 13/27. or not quite 0.5. When we change the puzzle to – “”A man says to you “I have two children, one is a boy born on a Tuesday”. What is the probability the man has a boy and a girl?
Is as I believe the classical probability solution the same(13/27) as I now believe?
If the answer is yes – how well does this probability solution describe our reality? Why? And what is the sex of the 1 child in 54 that does not fit either probability? Hermaphrodite? Just kidding :) Seriously though why does the information of the known child’s birthday effect the result?
Is this a really profound and mysterious property of this universe we live in or merely an artefact of the mathematics we have used to calculate the probability? As you can see I am still here enjoying your blog and your probability puzzles but if I make a pest of myself please feel free to ask me to move on.
firefly: you are onto something…it is reasoning along these lines that led me to rethink the classical approach. I really like this line of thinking. If you are told by Mr Smith he has at least one boy you will (classically) conclude there is a 1/3 chance that he has two boys. If he says the boy is born on Tuesday, you will revise that to 13/27. But, you would do the same if he said the boy was born on Wednesday, or indeed any day of the week. So if he said he had a boy born on *$$*@day (his voice was muffled and you couldn’t quite catch which day he said), wouldn’t you then revise your probability to 13/27 too. After all, you’d make that revision regardless of what day he actually said!
P.S. You should definitely not move on….stick around for more puzzling (and the rest)!
“Is there any difference between the scenario in which you ask the man “do you have at least one boy?” and the scenario in which the man simply volunteers the information? I think there is a difference….”
Yes, there’s a diff. Why? There’s no diff probability wise, between boy:girl and head:tail outcomes in coin tossing. So re-cast the paradox in the latter.
Thus, the man says, “I tossed two coins, one landed heads”.
It becomes clear then, the paradox is one of semantics. Why? Because you can interpret the statement as either:
1. Of the two coins that could have landed heads, ONLY one coin did.
2. One coin landed heads, and nothing is implied about the other.
Semantically, the first interpretation is the most appropriate.
Bruce: would you interpret the semantics any differently if the man said, “I tossed two coins, at least one landed heads.”?
“At least one landed heads” is unambiguous.
“One landed heads” is ambiguous.
Bruce: so, what would you conclude if the puzzle was phrased as:
“A man says to you “I have two children, at least one is a boy born on a Tuesday”. What is the probability the man has two boys?”
Seems to me the punters are easily confused by paradoxes!
1. The man with two boys paradox is about timing. Before he started procreating, the man had a 1/2 chance of a son first up, and a 1/2 chance of a son second time around – total chance of 2 boys is 1/2 x 1/2 = 1/4. But now we know he has produced 2 kids, one of whom was a boy, the only question left is what is the probability of the second being a boy. Since none of the other facts affects this outcome it’s 1/2 (and the Tuesday bit is irrelevant).
2. The quiz show host job is also only about timing. Before the contestant picks a door he/she has 1/3 chance of the car (note that some might prefer a goat).
After the host reveals the goat behind one door, the chances of picking a car from the other 2 doors is 1/2, but there is no information for the contestant which makes one door more likely to hold the car than the other, unless it be that the host is subtly trying to steer the contestant away from the door already chosen, which probably means that’s where the car is … Never trust a game show host!
3. Doesn’t matter what was in the first envelope – all it does is tell you how much you stand to gain or lose. The chances are still 1/2.
Now as your devoted followers seem to love a paradox, you might want to hit them with a conundrum:
Three cannibals and three missionaries are travelling together. They come to a river which they all must cross. They only have a 2-man canoe, which is their only means of crossing the river. Only one of the cannibals can row, but all of the missionaries can row. At no time or place can the missionaries be outnumbered by cannibals, or they will be eaten. How do they cross the river, and how many trips does it take?
Maybe a prize of a chaff bag of mule droppings to the solution with the least number of trips?? Second prize, 2 bags, etc…
“A man says to you “I have two children, at least one is a boy born on a Tuesday”. What is the probability the man has two boys?”
Why is the solution not thus?
1. P(1st is b) = .5 * P(2nd is b) = .5 -> .25
2. P(1st is b) = .5 * P(2nd is g) = .5 -> .25
3. P(1st is g) = .5 * P(2nd is b) = .5 -> .25
4. P(1st is g) = .5 * P(2nd is g) = .5 -> .25
Therefore, P(2 boys) = 0.25/0.75= 0.333?
Because the information presented changes all possible outcomes to:
1. P(1st is b) = 1 * P(2nd is b) = .5 -> .5
2. P(1st is b) = .5 * P(2nd is b) = 1 -> .5
3. P(1st is b) = 1 * P(2nd is g) = .5 -> .5
4. P(1st is g) = .5 * P(2nd is b) = 1 -> .5
And these reduce to
P(2 boys) = 0.5
Bruce: did you also see these posts: Tuesday’s Child and Eliminating the irrelevant?
Plunko: does having one cannibal and one missionary on a bank while the canoe is stopped at the same bank with a cannibal in it count as having more cannibals than missionaries in the same place?
Yes, of course…since when do self-respecting cannibals let a chance go by?
And while we’re at it, don’t forget that the GFC proved that most banks (especially the Yank ones) are heavily populated with cannibals! Err, apologies and greetings to the memsahib!!
In which case, the best solution I can come up with so far has 13 trips, which seems like a lot. Not sure whether I can do better.
SM: did you also see these posts: Tuesday’s Child and Eliminating the irrelevant?
The answer is obviously dependent on the probabilities associated with the interpretation of statements, and as you rightfully show using symmetry and exclusion (in eliminating the irrelevant), the probabilities vary.
I understand the derivation of your formula, but am not confident of a proof (35 years since I did senior math).
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