Probability, Paradox, and the Reasonable Person Principle
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When the contestant picked a box he was making a random choice between three boxes and had a probability of 1/3 of picking the car. The probability that one of the other two boxes was the car is thus 2/3. When Monty Hall opens a box it's still the case that the probability of the car being in the other two is 2/3 but the contestant can now eliminate the open box and knows that the probability that the car is in the other box is 2/3. Thus it's worth switching.
> Basically, you probably got a goat the first time, so switch.
It's so concise that I use more words to re-post it every time Monty Hall comes up than are in the explanation itself.
That helps people understand that their original probability of picking the car the first time was very small, and didn't change because of what Monty did.
You pick a door. Monty opens another door, revealing a goat. You now get the option to switch from your door to the other two doors, keeping the car if it's behind either one.
Whether Monty opens the door or not doesn't mean much.
Even in the case where your restatement is mathematically equivalent, it's obviously not at all equivalent as a puzzle because what is being requested by the puzzle is seeing that the original situation is mathematically equivalent to that one.
I thought he selected it at random.
That was always the confusing unstated assumption for me, that makes me think the problem is only confusing because of how the wording usually de-emphasizes that distinction.
If he chooses a door, you can benefit from the secret information he reveals sometimes by doing so.
If the door was chosen at random, he is not adding any information, so you can't act on it?
Edit: Since I have been downvoted, here is a citation: http://probability.ca/jeff/writing/montyfall.pdf (see Monty Fall problem)
Regardless, the problem as stated has Monty Hall open a losing door. Thus it's clear that you should switch after Monty Hall does so, and a very simple computer program can show that you tend to win by switching.
Another way to phrase the problem is that you choose one door, then Monty Hall (without revealing anything) gives you the option of taking the door you chose, or taking both the other doors. That's an equivalent problem (assuming the goat has zero utility), and it makes it very obvious that you should switch.
The problem is not always presented in a way that makes it clear that the game always progresses this way, and it's not just a description of the particular circumstance you find yourself in one particular play.
Consider:
"You have $X, your opponents have $Y and $Z. You select Potent Potables for $400 and it's a daily double. How much should you wager?"
I think you'll agree that is going to be read near-universally as a statement of a particular situation that could arise in Jeopardy, not a statement of how Jeopardy games always go.
I'm not sure how that matters. The question is whether, in this scenario, you choose to switch after Monty Hall reveals the door. The terms of the thought experiment dictate that Monty Hall will reveal a losing door. Regardless of what force is actually causing the revealed door to be a losing door, you should switch, because you're essentially being given the option to take your original one door, or to take both of the other doors (at least one of which is a losing door).
If you pick randomly between a fair coin and a double-headed coin, there's a 50/50 chance of picking either. If Monty then flips the coin and it comes up heads, that suggests you're more likely to be in a world where you've picked the biased coin. If, on the other hand, Monty deliberately takes your coin and places it so that it's heads (and he was gonna do that whatever coin you picked) then you have no new information.
But yeah, that the problem is moot if it were random is exactly my point.
I was deceived by the wording and it's frustrating to hear "answers" to this problem that ignore that deception because I got the wrong answer and so I feel "dumb" for being decieved so I am trying to defend how i'm not "dumb" for not seeing the probability, just dumb for not thinking through the fact that it would be unlikely for Monty to reveal a car at that point.
I guess a lot of people go "oh, he won't reveal a car".
I'm missing that intuition.
The problem isn't probabilistically difficult and most people's intuitions would be on point if it weren't set up to deceive.
That's all I want anyone to admit when they explain the answer.
Thank you.
Monty himself mentions that he would occasionally try hard to argue contestants into swapping. They would always refuse. He would even explain that 'his door' had the better chance - nobody would listen. But I suppose that's confounded by the suspicion that he was trying to trick them.
Again, the part I missed was "it would be silly for him to reveal a car".
I get that now, but that was not something that I intuited, and so the omission of that information I will continue to argue is at least a little bit deceptive.
Even if only to those like myself who have not seen the program.
I'm confident that there are lots of people who still fail to make the connection (though we're getting to the point where I think people are failing less due to a lack of statistical intuition and more due to a symbolic/physical model mismatch issue), but I think this problem wouldn't be as renowned as it is if it weren't for all of the even expert statisticians who are getting fooled.
In their case it's definitely a matter of them misinterpreting the situation and the way that it's worded, for someone unfamiliar with the show, is at least a little "tricky".
'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?'
Given "the host, who knows what's behind the doors", I think the original phrasing was less ambiguous than many rephrasings I encounter. It certainly could have been still clearer.
Also, her answer - which was presumably read by those writing in to asset her answer was wrong - plainly states that the host always picks a losing door.
At the point when I read it, I didn't realize Monty was always choosing the door with pre-knowledge of which one the car wasn't behind.
It wasn't until somebody wrote in with a computer program example that showed the benefits of switching, in which I inspected the source code, that I understand what Monty was doing.
personally, I think this has to do with ambiguous way word problems are stated, and how people model the word problem mentally.
This is so detached from normal events in life that the paradoxical nature is much less impactful, so it's nowhere near as enticing of a thought experiment. It is, however, similar (equivalent, methinks) to the "God's Coin Toss" problem, which is also popular and which also has gotten some attention on Hacker News: http://www.scottaaronson.com/democritus/lec17.html
1) contestant wins (notion being the contestant retains the option to switch to any door, and now knows where the car is)
2) contestant loses (notion being Monty picked right and "won" in place of the contestant)
3) Round is aborted, things are repositioned, and the round is replayed (doesn't require any drugging, aborted rounds may or may not be aired but player learned nothing relevant to future rounds).
"Have you ever noticed when you're driving that anybody driving slower than you is an idiot, and anyone going faster than you is a maniac?"
I'll include the canonical quip by Laplace:
"The theory of probabilities is at bottom nothing but common sense reduced to calculus; it enables us to appreciate with exactness that which accurate minds feel with a sort of instinct for which ofttimes they are unable to account." -Introduction to Théorie Analytique des Probabilitiés
Its this warping of the probability curve that is responsible for major facets of human society: war, witchcraft, gambling, marriage, on and on.
But if you think in an evolutionary sense, if each individual in a tribe acted rationally, we'd get a sort of Tragedy of the Commons: the enemy tribe wins the war, and slaughters your men. So it may happen that if each individual acts rationally the group, each individual included, suffers; so it makes sense individual sense to develop mechanisms to go to war, which by symmetry must include yourself.
Even less intuitively even if the probability of death is lower by not going to war (no tragedy of the commons in traditional sense), which means no individual would rationally choose war or develop mechanisms to force going to war, it would be rational for the tribe to choose war, in an evolutionary sense, provided many more of your men are killed than the enemy tribe (so they do better in evolutionary terms).
In particular, when selecting for war, there's a clear individual level selection pressure: if you don't cooperate with your group, you likely die. This is a just so story, I have no idea how irrationality actually developed, but say you have small groups of hunter gatherers and one individual develops a preference for cooperating in raids against their best interest. If this means that on average you now have two versus one whenever this individual participates, there's a clear competitive advantage which could allow that gene to propagate. The key here in this hypothetical is that the mutation occurs at the individual level in a lone individual, who then cooperates with someone else who wouldn't necessarily follow them, but is happy to have the help. This results in a disproportionate gain (100%) in effectiveness vs lone opponents. As time evolves, the gene becomes more widespread by its early disproportionate effectiveness and groups that fail to cooperate are killed on average, thus eliminating individual competitor genes.
Again, I have no idea how war actually evolved, but it seems easy to believe that when a trait influences whether one group kills another, that it would cause the killers to have a reproductive advantage. I'd love someone more educated on this topic to send me up though.
"That guy" was Pierre-Simon Laplace, and the book he was introducing was his own seminal Théorie analytique des probabilités!
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3845016/
> Studies have shown that reward uncertainty rather than reward per se, will magnify mesolimbic DA, both in monkeys (Fiorillo et al., 2003; de Lafuente and Romo, 2011) and healthy human participants (Preuschoff et al., 2006). In PG, accumbens DA is maximal during a gambling task when the probability of winning and losing money is identical—a 50% chance for a two-outcome event representing maximal uncertainty (Linnet et al., 2012). Although non-dopaminergic neurons might also be involved in the coding of reward uncertainty (Monosov and Hikosaka, 2013), these results based on electrophysiological and neuroimaging techniques indicate that DA is crucial for the coding of reward uncertainty. This suggestion is corroborated by a large number of behavioral studies, showing that mammals and birds respond more vigorously to conditioned cues predicting uncertain rewards (Collins et al., 1983; Anselme et al., 2013; Robinson et al., under review) and tend to prefer an uncertain food option over a certain food option in dual-choice tasks (Kacelnik and Bateson, 1996; Adriani and Laviola, 2006), sometimes despite a lower reward rate (Forkman, 1991; Gipson et al., 2009). According to Greg Costikyan, an award-winning game designer, games cannot hold our interest in the absence of uncertainty—which can take many forms, occurring in the outcome, the game's path, analytical complexity, perception, and so on (Costikyan, 2013).
(And yes, I'm sure your interpretation of the joke was always consistent with the above, but some people consider it an actual insight in addition to being funny.)
To be sure, the general point is sound, but that is a really bad, false example. A better version might be "Ever notice how when you're angry, it's because of a legitimate injustice, but when someone else is angry, it's because they're just in a bad mood?"
Also, the interview with Peter Norvig in the book "Coders at Work" [3] is great - one of my favorites in the book (actually, the whole book is great).
[1] http://norvig.com/SET.html
[2] http://henrikwarne.com/2011/09/30/set-probabilities-revisite...
[1] https://en.wikipedia.org/wiki/Bertrand_paradox_%28probabilit...
Normally, if you assume triangle does not move, and then draw a random chord, then depending on relative position of the chord and triangle you use one of three methods described by Bertrand to judge whether it is longer or shorter than triangle side.
Then you need a way to count probability of each of those cases to happen and then count the weighted average of all three "paradoxical" cases.
EDIT: in fact I would just bet on Method 3 as a correct solution, because it is the most general: it does not require rotating the triange.
If you are rotating the triangle then you are in fact changing problem definition during solving it, which leads to some hidden assumptions (i.e. Method 2 is only good for chords which are parallel to one side of triangle, and Method 1 is only good for chords that do not cross with triangle sides)
The final step of the St. Petersburg paradox, which he took far further than I had ever seen before, might be implementing the Kelly Criterion, where the amount you bet is related to the size of your own personal bankroll, but I don't know offhand how to relate that to a probability distribution since the Kelly Criterion is normally calculated with one odds value and one payoff value. At the least, I'd like to see a calculation of how many times you'd have to place the bet before you could be reasonably sure of a positive payout - you could do that easily enough with a monte carlo simulation I guess.
If you feel comfortable that the answer to the Monty Hall problem is "switch", you might want to test your understanding to be sure you reach the correct answer here as well. As Wikipedia states, Three Prisoners "is mathematically equivalent to the Monty Hall problem with car and goat replaced with freedom and execution respectively". But mathematically equivalent may not make it intuitively equivalent, and the answer may feel like it contradicts the correct Monty Hall answer.
Wikipedia points out that it's originally from a 1950's Martin Gardner column, but I came across it in the textbook "Stastistical Inference" by Casella and Berger. Here's C&B's phrasing:
Three prisoners, A, B, and C, are on death row. The governor
decides to pardon one of the three and chooses at random the
prisoner to pardon. He informs the warden of his choice but
requests that the name be kept secret for a few days. The
next day, A tries to get the warden to tell him who had been
pardoned. The warden refuses. A then asks which of B or C
will be executed. The warden thinks for a while, then tells A
that B is to be executed.
Warden’s reasoning: Each prisoner has a 1 in 3
chance of being pardoned. Clearly, either B or C must be
executed, so I have given A no information about whether A
will be pardoned.
A’s reasoning: Given that B will be executed, then either A
or C will be pardoned. My chance of being pardoned has
risen to 1 in 2.
Who is right?
While it's true that the warden's reasoning is correct, leading off with "it should be clear" seems cruel. Here's a slight variation of the Monty Hall problem, one of the most famous "paradoxes" of popular statistics, and you are going to start with a paraphrase of "it should be obvious to the reader" without the slightest sense of irony, even though this variation produces an answer superficially incompatible with the better known problem? Ah, the strange humor of textbook authors! The remainder of their answer (which is solid) can be found here on Section 1.3 page 22: http://people.unica.it/musio/files/2008/10/Casella-Berger.pd...
Years ago, Peter Norvig was a tech editor for an AI book I wrote. I realized two things: he has a clearer way of analyzing things than I do, and also that I was very fortunate that he ended up spending several hours helping me with his review material.
Most paradoxes disappear when the problem is strictly described.
Say you had a bunch of families with two children each. The children are evenly distributed in terms of gender and the the days of the week on which they were born. If you pick one parent from the crowd, the chance that they have at least one boy is 3 / 4, the chance that they have two is 1 / 4, and the chance that they have none is also 1 / 4:
| B| G|
--|--|--|
B |BB|BG|
--|--|--|
G |GB|GG|
Now, for the day-of-the-week problem. If you ask a parent if they have a male child born on Tuesday, it is not equiprobable that they're in any of those three squares. In the BG group, they all have male children, so the chance that any parent chosen has a male child born on Tuesday is 1 / 7. Similarly in the GB group. However, in the BB group, either one of their children may be born on a Tuesday to satisfy the condition. The chance that either child is born on a Tuesday is the same as the inverse of neither child being born on a Tuesday, or
1 - (6/7 * 6/7) = 13/49
(13/49) / (1/7 + 1/7 + 13/49) = 13/27Also, for interpretation of problem 2a there is one fundamental flaw that I can see: The sample space listed as:'BB', 'BG', 'GB' is wrong, because 'BG' and 'GB' are in fact the same sample. Or, to put it in a different way: if you decide that ordering does not matter then you should list either 'BG' or 'GB', not both. And if order does matter then you should list 'BB' twice, for all the possible orderings. Both of which will make probability equal to 1/2.
EDIT: Moral of the story is: "If it looks like paradox you are doing it wrong". Which I think was the Norvig's point from the start.
Flip two coins. Exclude the case where both are tails. What is the probability the two coins are different? If 'BG' and 'GB' are "the same sample", is it also the case that 'HT' and 'TH' are the same sample?
The variances are due to how people interpret the outcome space .
It is right that probability = favaroble out comes / total possible outcomes.
In problem 2, in my opinion the possible outcomes are not how it was suggested in the post but as below. When a family has 2 kids , the below are the only possible outcomes
1) Potential Outcome 1: Both are boys 2) Potential Outcome 2: One boy and one girl 3) Potential outcome 3: Both are girls
The 3rd one is not a legal potential outcome in our particular constraint of problem 2, since problem #2 statement already states 'at least one is a boy'
So Total possible legal outcomes = 2 Favorable out comes for our event (both boys) = 1
So probability for Problem 2 = 1/2
Similarly for problem #3, I think the post unnecessarily complicates the calculation of problem space. The fact that 'Tuesday' is mentioned is irrelevant in my opinon, if you state the problem #3 in a different way that is more clearly understood.
There are 14 baskets labelled as follows "Sunday Boy", "Sunday Girl", "Monday Boy", "Monday Girl",....."Saturday Boy", "Saturday Girl". A stork came and dropped 2 babies. One baby was dropped in "Tuesday Boy" basket. What is the probability that both are boys?
Now the total outcomes and favarable outcomes are :
Total possible outcomes = Number of ways second baby could have been dropped = 14 possible baskets = 14
Favorable outcomes = second baby dropped in 'boy' basket = 7 possible baskets = 7
Probability that both are boys = 7 / 14 = 1/2
> It is right that probability = favaroble out comes / total possible outcomes.
No, not actually: It is only right if all outcomes are equally likely! (There's an old joke about the guy who has a 50% chance of winning the lottery, since either he will win it or he won't.)
In particular, you make that mistake here:
> 1) Potential Outcome 1: Both are boys 2) Potential Outcome 2: One boy and one girl 3) Potential outcome 3: Both are girls
These three outcomes are not all equally likely. Outcome 1 has probability 1/4, outcome 2 has probability 1/2, and outcome 3 has probability 1/4. (This is if you assume that each child has a half chance each of being a boy/girl.)
Norvig gets rid of this problem by listing out all four possible outcomes, which are all equally likely.
1. First child boy, Second child boy
2. First child boy, Second child girl
3. First child girl, Second child boy
4. First child girl, Second child girl