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The Time Everyone “Corrected” the World’s Smartest Woman (2015)

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Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#301

Post by amscanne »

A_No_Name_Mouse wrote: post #292
amscanne wrote: post #285

Earlier quoted context omitted.

I’m aware that it’s complicated. I intentionally phrased it in a way that makes it clear that the family moving in is not somehow selected from the set of families with at least one boy, but rather the observation is unrelated. This is the distinction. I believe that the 1/3 analysis may also be incorrect for the way you phrased your question. If you had said: “we select a second ball, and only observe the first ball…

I don't think this is correct. It matters whether the person removing the second ball can see the colors and choose accordingly. If you can, and you willingly remove a single blue ball, you have the Monty hall problem where you stick to your choice. It does not influence the chances, it is still 1/2. But the way I read the problem, you choose a ball at random, look at its color, it happens to be blue. Now this gives…

The sneaky part of this question is that it gives information about the outcome and then asks you again about the probability. Forget the extra balls, and just drive the point home directly: suppose I pick a ball at random, and show it to you. It’s blue. I ask: “what are the odds that the ball is blue?” (exact phrasing as original question).

There are obviously correct interpretations for 50% and 100%. It depends whether I’m asking: - What’s the probability of this outcome? - What’s the probability that the ball I’m holding in my hand in blue?

The second is effectively a “resampling” with a population of one. You are simply assuming the second interpretation and arguing for it, but I don’t dispute the logic. The original question is unclear whether it’s asking for the probability that you picked a blue ball initially (50%) or the likelihood that the ball is blue, given some information of the outcome. But we don’t normally speak of probabilities this way. The odds that you picked a blue ball initially were 50%, even if you picked a red one.

By giving only partial information, the question creates more ambiguity since the answer isn’t definite. (When there is ambiguity in a question, I believe most people will discard trivial interpretations over substantive ones, which is what pushes toward the “resample” here.)

Anyway, I’ll leave it there since I think it’s clear there are correct interpretations for both, depending on what the question is actually asking.

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#302

Post by jVinc »

Cpoll wrote: post #44

Another statistically unintuitive problem (which I've witnessed a lecture hall enter a state of uproar over): There are 2 red and 2 blue balls in a box. One ball is removed at random, what are the odds that the ball is blue? Now we repeat the problem, but before examining the ball, we remove a second ball. We observe that the second ball is blue. In this case, what are the odds that the first ball is blue?

I might be wrong here, but this is my understanding.

You'll always be able to show a blue ball as the second ball after the first is drawn. So arguing that it tells you something about the statistics of the state of the system after the first has been drawn is wrong. The chance that the first draw is blue is 50/50. The arguments that lead to 1/3 are trying to use the second event as a statistic for the state, however for that to be appropriate the problem would have to be phrased as:

You draw one ball and set it aside, then draw another, if the second ball is red, you start the entire thing over, if it's blue, then you continue the experiment.

Because otherwise the assumption that the second drawing can tell you the statistics of the underlying state is wrong.

The question of the odds really boils down to, did we randomly pick a ball and observe it? And if so, what would have happened if we didn't observe what we specifically stated for this instance.

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#303

Post by haberman »

someguydave wrote: post #162
haberman wrote: post #76

Earlier quoted context omitted.

When you choose the first door, you are partitioning the "board" into two parts: the door you chose, and "everything else". By opening a door, Monty lets you cover 100% of the "everything else" partition using only one guess. So now you get to choose between partition 1, which covers 1/3 of the board, and partition 2, which covers 2/3 of the board.

I think the confusing part for many people is the unstated fact that the host takes all prizes into account when revealing a door, and that the host is forbidden from revealing the car. If the host chooses randomly then his pick adds no information (other than the extra revealed prize)

I don't think this actually matters. Suppose Monty doesn't actually know which door has the prize, and just picks randomly between the two. Now the only difference is that Monty might accidentally pick a prize. Suppose he does. What are the options now?

1. The game goes on, and you can just switch your guess to the door he picked. (WIN)

2. The game resets due to Monty's error and you play again. (REDO)

3. Monty just decides that you lose since he picked the prize. (LOSE, NO CHOICE)

Options 1 and 2 don't deprive you of the car. But option 3 doesn't give you a choice, so there is no dilemma.

So conditional on having a game where no prize has been revealed yet and you get a choice to switch, I think it still pays to always switch.

Another way of putting it: you get to choose between partition 1 covering 1/3 of the board where you get 1 guess, or partition 2 covering 2/3 of the board where you get two guesses. It works even if all of the guesses are random.

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#304

Post by [deleted] »

leephillips wrote: post #158
failwhaleshark wrote: post #99

This reminds me of the time my entire family screamed and shouted that I was wrong that buying two different lottery tickets slightly more than doubles the total (infinitesimal) odds of winning over just one. 1 in a zillion vs slightly greater than 2 in a zillion because eliminating one choice reduces the pool by one for the next choice.

If you are talking about a simple lottery with N tickets and a fixed probability of any one ticket being the winner = 1/N, then the screamers were correct. To see this, imagine that there are a total of two tickets (N = 2). You buy one ticket. The probability that it is the winner = 1/2. After buying the second ticket, is your probability of winning somehow > 1? Or has it exactly doubled?

[deleted]

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#305

Post by [deleted] »

jonasenordin wrote: post #291
failwhaleshark wrote: post #99

This reminds me of the time my entire family screamed and shouted that I was wrong that buying two different lottery tickets slightly more than doubles the total (infinitesimal) odds of winning over just one. 1 in a zillion vs slightly greater than 2 in a zillion because eliminating one choice reduces the pool by one for the next choice.

I think if you buy a first ticket, and somehow found out (but lotteries usually don't allow this) that it wasn't a winner, and only _then_ buying a second one, your reasoning would be correct.

[deleted]

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#306

Post by failwhaleshark »

qot wrote: post #159
failwhaleshark wrote: post #99

This reminds me of the time my entire family screamed and shouted that I was wrong that buying two different lottery tickets slightly more than doubles the total (infinitesimal) odds of winning over just one. 1 in a zillion vs slightly greater than 2 in a zillion because eliminating one choice reduces the pool by one for the next choice.

But you are wrong: it's exactly double the odds. If you bought all the tickets, do you think you'd have over 100% chance of winning? Edit: looks like leephillips beat me to it

Mea culpa. Instead of arguing with the shifting sands of ambiguous and imprecise English, it's better to use equations and evidence.

P(A) = first ticket wins

P(B) = second ticket wins

P(A|B) + P(A|^B) + P(^A|B) = 1 - P(^A|^B) = 1 - (1 - 1/N) * (1 - 1/(N-1)) = 2 / N

It is slightly > 2 / N if the ticket is revealed before the next one is chosen because then it does throw away that possibility.

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#307

Post by failwhaleshark »

feoren wrote: post #253
failwhaleshark wrote: post #99

This reminds me of the time my entire family screamed and shouted that I was wrong that buying two different lottery tickets slightly more than doubles the total (infinitesimal) odds of winning over just one. 1 in a zillion vs slightly greater than 2 in a zillion because eliminating one choice reduces the pool by one for the next choice.

There are still N possible outcomes, and you have 2 of them instead of 1. Your chance of winning is 2/N instead of 1/N and has exactly doubled. And most lottery games allow numbers to be re-used, so you haven't "eliminated" anything. If you exhaustively bought all N number combinations, you have a 100% chance of winning, but you also have a decent chance to split the pot with someone else who also bought the winning…

Piling-on and screaming isn't a rational argument for anything. No listening can occur. None of them had above a high-school education.

Your piling on isn't helpful either, so can it.

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#308

Post by failwhaleshark »

aaplok wrote: post #160
failwhaleshark wrote: post #99

This reminds me of the time my entire family screamed and shouted that I was wrong that buying two different lottery tickets slightly more than doubles the total (infinitesimal) odds of winning over just one. 1 in a zillion vs slightly greater than 2 in a zillion because eliminating one choice reduces the pool by one for the next choice.

I may misunderstand you, but replace the lottery with a coin toss. Buying a ticket gets you a 50% chance of winning. This reduces the pool of possible outcomes by 1 for the next choice, so there is precisely one choice left. However buying a second (different) ticket does not give you a 150% chance of winning.

If you buy heads and tails, then you have a 100% chance of winning. :) Unfortunately, each ticket cost 5 cents.

(See above.)

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#309

Post by dllthomas »

tpush wrote: post #270
dllthomas wrote: post #56

Earlier quoted context omitted.

Yes it does. Or more precisely, it matters whether the host could be counted on to do so reliably; the mechanism for that doesn't matter. There's a difference here, that our language obscures, between procedure and hypothetical.

The intention of the host only matters if the contestant would have to choose the subsequent action (switching or not) before the hosts opens a door. If the host has revealed a goat door, and the contestant then has to decide what to do, the intentions of the host for having chosen the door are irrelevant.

(Noting again that the mechanism doesn't matter, what matters is the odds of various behavior by the host, but - I think reasonably - using "intentions of the host" as a proxy for that.)

The intentions of the host do matter.

Imagine the host picks the correct door by the following procedure: 1) picks an available door at random; 2) if that door has a goat, opens it; 3) if that door has the car, opens the other door.

I hope you will agree that this is equivalent to the problem as originally intended - Monty can be relied on to reveal a goat, and exactly why doesn't matter.

Breaking it down into equally likely cases, assuming the contestant picks door 3:

    A) The car is behind door 1, Monty picks door 1, Monty corrects.
    B) The car is behind door 1, Monty picks door 2
    C) The car is behind door 2, Monty picks door 1
    D) The car is behind door 2, Monty picks door 2, Monty corrects
    E) The car is behind door 3, Monty picks door 1
    F) The car is behind door 3, Monty picks door 2
When Monty reveals the goat behind (say) door 2, we know we're in case A, B, or F. All remain equally likely, and switching wins in A and B.

If Monty would not have corrected, then revealing the goat behind door 2 eliminates (the new) A as well, leaving us with only B and F, again equally likely.

If all of this remains unconvincing, I encourage you to write a simple simulation of the problem.

Re: The Time Everyone “Corrected” the World’s Smartest Woman (2015)

#310

Post by nzealand »

listenallyall wrote: post #202
mkl wrote: post #185

Earlier quoted context omitted.

The 1990 Parade Magazine description is almost identical (and actually more explicit, since "say #3" is a removable parenthetical): "the host, who knows what’s behind the doors, opens another door, say #3, which has a goat" -- https://web.archive.org/web/20130121183432/http://marilynvos... The host's knowledge is explicitly mentioned, and the only purpose this could have is that so he can use it to avoid giving the g…

It's a tv show - a valid purpose is that he needs to stretch by 30 seconds before going to commercial. Another purpose is that the audience thinks seeing a goat is funny. Another purpose is to prove the show uses two different goats and doesn't do a switcheroo behind the scenes. What the Parade article does NOT say is that the first door opened is ALWAYS not the contestant's choice and ALWAYS reveals a goat.

The problem the reader originally submitted clearly implied both those things, at least clearly enough for Marylin

> and the host, who knows what’s behind the doors, opens another door ... which has a goat

The explanation Marylin originally wrote makes the second assumption explicitly clear to avoid any possibility of confusion

> Suppose there are a million doors, and you pick door #1. Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You’d switch to that door pretty fast, wouldn’t you?

People still wrote in claiming she was wrong.

> Since you seem to enjoy coming straight to the point, I’ll do the same. You blew it! Let me explain. If one door is shown to be a loser, that information changes the probability of either remaining choice, neither of which has any reason to be more likely, to 1/2. As a professional mathematician, I’m very concerned with the general public’s lack of mathematical skills. Please help by confessing your error and in the future being more careful.

She then explained the solution a second time with a table clearly illustrating the assumptions and the odds.

People continued to write in claiming she was wrong.

> I’m receiving thousands of letters, nearly all insisting that I’m wrong

> 92% are against my answer, and and of the letters from universities, 65% are against my answer.

Lack of clarity did not generate all the letters.

I run a small website with logic puzzles. I get thousands of emails. Some tell me the problem is misleading. Some tell me the answer is wrong.

The one thing I have learned is that once someone's mind is made up, it will never change.

The other thing I have learned, is that a good problem is succinct.

The smartest people in the room will ask clarifying questions to test out their assumptions before they even attempt an answer.

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