## A riddle

slipstick
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### Re: A riddle

Penn wrote:
Schultz wrote:I don't know how you can look at the purple boxes in the bottom row of Fred Barclay's diagram and still come to that conclusion.

[ I initially agreed with you guys, until I looked at that diagram. ]
Fine, you want pictures.

It starts with one alteration, I labeled the doors 1, 2 and 3.

So if you choose door 1, you will be shown door 2 and that option is removed. You can't change your mind AWAY from door 3 since you didn't choose that. So your CURRENT possibilities are

Not a thing erased affects your CURRENT choice and the resulting odds. That stuff is GONE.

This same logic is applicable if your first choice is 2 or 3.
But you are ignoring the fact that changing is a winner every time you initially choose a goat and your chances of initially choosing the goat is 2 out of 3. It's true that in any particular instance you don't know which door has the car, but the odds are only 50/50 if you ignore the knowledge that there are 2 goats and only 1 car when you initially choose.

Here's an analogy: say you roll two dice and the goal is to predict the total count on the two. Any number between 2 and 12 is possible, so there are 11 possibilities. At any particular roll of the dice, you have no way to know what the count will be, but to say that all possibilities are equal is to ignore the fact that there are more ways to get a 7 than to get a 2 or a 12. Similarly, there are more ways to win by changing than by sticking, because there are more goats than cars.
In theory, theory and practice are the same. In practice, they ain't.

samriggs
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### Re: A riddle

I would just choose door number 4 (the backdoor), put the goat in the car and drive home while you folks try to figure this out, I now have a new shiny car and a lawn mower called goat.

Winner winner chicken dinner.
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Schultz
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### Re: A riddle

slipstick wrote:
But you are ignoring the fact that changing is a winner every time you initially choose a goat and your chances of initially choosing the goat is 2 out of 3.
I tried to explain this in 2 out of 3 posts.

Penn
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### Re: A riddle

mintage wrote:Good Luck Penn, you're a real soldier.
Not so fast, I'm just trying to get people to explain in a simple and clear way the correct answer and I don't care if I appear to not know.
Schultz wrote:
The simple solutions above show that a player with a strategy of switching wins the car with overall probability 2/3, i.e., without taking account of which door was opened by the host.
https://en.wikipedia.org/wiki/Monty_Hall_problem
"Marilyn vos Savant is an American who is known for having the highest recorded IQ according to the Guinness Book of Records, a competitive category the publication has since retired." https://en.wikipedia.org/wiki/Marilyn_vos_Savant

She says by switching your odds are now 2 out 3 of winning the car. Is your IQ higher than hers?
Sure, try to take the fun out of it by trying to leave an explanation to others and deferring to a higher intellect (which isn't always logical since I have outsmarted both people I personally know and I can confirm have a higher IQ than me on more than one occurrence). What von Savant couldn't do is explain it in a simple and clear way. Both people more and less intelligent than her didn't agree with how she explained it (if she is the same person I think she is). Too many of the ways people explain can be argued. Since you went that route I guess i'll give my attempt in a moment even though others have already said similar things to what I would say.
Schultz wrote:
slipstick wrote:
But you are ignoring the fact that changing is a winner every time you initially choose a goat and your chances of initially choosing the goat is 2 out of 3.
I tried to explain this in 2 out of 3 posts.
You are right that the point by slipstick is probably the most relevant point anyone has made but he still didn't fully explain it.

My attempt to explain in a simple way - which will probably fail for those convinced.

You play the game a number of times divisible by 3 and the odds hit exactly. So 3 times you play and 2 times your first choice is wrong and only once correct. You stay all 3 times and the final result is you are right once.

And the debate is switching or staying every time so you can't combine those numbers switching.

Play again the same number of times and again you are wrong twice and right once but this time you switch all three times and you are now the final result is you are right 2 times and wrong once.

As I said, others have said it in a similar way. I still want a strong but simple explanation.

(this was my fifth attempt to post but three times changed my mind how to state that and just somehow lost the other)

rene
Level 14
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### Re: A riddle

Schultz wrote:I tried to explain this in 2 out of 3 posts.
And Wikipedia and an approximately six-figure Google hit count equally does. Guys, please note that this is probably the best known probability-calculus snafu around and gets basically everyone on the first encounter. But you should switch, and if you do not agree you do not have an alternate opinion but are wrong. Just simulate the game on your computer if you must: you will very quickly observe the win rate to converge to one-third if you stick, two-thirds if you switch. Mathematics is not an opinion.

https://en.wikipedia.org/wiki/Monty_Hall_problem

TooMuchTime
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### Re: A riddle

The initial odds are 1 in 3, or 33%. However, you know, that going into the game the host will ALWAYS show you one door that has a goat behind it. That is, you know beforehand that no matter what door you choose you will always get a second choice between the remaining two doors. So, the secondary choice is 1 in 2 or 50%. Because you can't choose the third door!!!! And remember, YOU DIDN'T choose it to begin with!! And now, it has been removed from the game and no longer exists as a choice.

I guess in some bizarro world where you might want a goat, you could always choose the door the host opens to reveal the goat. Not sure why you would since the object of the game is to win the car. So for sake of argument let's say that you are actually trying to win the car. Therefore, you cannot choose the goat behind the revealed door. Which means, there are only two doors remaining FOR YOUR SECOND CHOICE and you MUST choose either one of them. Either one -- as in a choice between ONLY TWO objects. No matter how hard you try to BS everyone, YOU ARE NOT CHOOSING BETWEEN 3 DOORS ON YOUR SECOND ATTEMPT. Removing one of the doors has changed the game. You know this because it was explained to you going into the game.

1st - 33.3%
2nd - 50%

Schultz
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### Re: A riddle

If a person who once held the distinction of having the highest IQ in the US can't convince you, then neither can I. I will say this though, everything you wrote proves the opposite of what you believe.
slipstick wrote:
But you are ignoring the fact that changing is a winner every time you initially choose a goat and your chances of initially choosing the goat is 2 out of 3.

TooMuchTime
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### Re: A riddle

The thing to remember is that YOU choose a door. THEN THE HOST CHOOSES A DOOR, NOT YOU! Then you choose again BETWEEN THE TWO DOORS REMAINING. There is NO WAY that choosing between two doors is a 66% chance. If THE HOST chooses door C, I now have the options between doors A and B. That is not 66%. There is no way in hell it is 66%. It is 50%. The exact same odds as if I flipped a coin. As a matter of fact, say door A is heads and door B is tails. Flip the coin. What were the odds of either one coming up? Yeah. 50%.

all41
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### Re: A riddle

TooMuchTime wrote:
1st - 33.3%
2nd - 50%
+1
That was my initial thoughts which are yet unchanged.
In real life:
After Monty reveals the goat door often a monkey wrench was thrown into the mix.
Monty would often offer up yet another choice--something to the effect of "or you may give up your door completely and choose the envelope being held by the lovely Jennifer.

Schultz
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### Re: A riddle

Numbers doen't lie.
In her (i.e., Marilyn vos Savant) final column on the problem, she gave the results of more than 1,000 school experiments. Nearly 100% found it pays to switch. Of the readers who wrote computer simulations of the problem, 97% reached the same conclusion. Most respondents now agree with her original solution, with half of the published letters declaring their authors had changed their minds.

https://en.wikipedia.org/wiki/Marilyn_v ... ll_problem
I'm out of here.

all41
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### Re: A riddle

If I entered the studio after the goat was revealed, with no knowledge of what transpired before that time,
and the contestant let me choose between the remaining doors what would my odds be of choosing the car?

rene
Level 14
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Joined: Sun Mar 27, 2016 6:58 pm

### Re: A riddle

What most explicitly TooMuchTime is failing to note is that two possibilities only make for 50 percent chance each if either is equally probable. In this case the initial door still has its original 33 percent whereas the single remaining alternative door has obtained the summed probability of the original alternative two. Monty after all opened a wrong door from among those original alternative two.

Find below a bash simulation of the game. For those that can read it: note the if-then-else structure to nicely lay bare WINS_WHEN_FAITHFUL and WINS_WHEN_SWITCHED to necessarily be complementary. With the former moreover obviously 33% thereby the answer to be evident. So evident in fact that I wondered for a bit if I had not basically coded the answer rather than the question but if you feel so: do be explicit about what you find to be incorrect.

Code: Select all

``````#!/bin/bash

GAMES=1000

WINS_WHEN_FAITHFUL=0
WINS_WHEN_SWITCHED=0

RANDOM=\$(date +%-N)
for ((I=0; I < GAMES; I++)) do
DOOR_WITH_AUTOMOBILE=\$((RANDOM % 3))
DOOR_SELECTED=\$((RANDOM % 3))
if ((DOOR_SELECTED == DOOR_WITH_AUTOMOBILE)); then
((WINS_WHEN_FAITHFUL++))
else
((WINS_WHEN_SWITCHED++))
fi
done

PRCT_WHEN_FAITHFUL=\$(((100 * WINS_WHEN_FAITHFUL) / GAMES))
PRCT_WHEN_SWITCHED=\$(((100 * WINS_WHEN_SWITCHED) / GAMES))

echo \
In \$GAMES iterations, we won \$WINS_WHEN_FAITHFUL times, \$PRCT_WHEN_FAITHFUL%, \
when staying faithful to our originally selected door and \$WINS_WHEN_SWITCHED \
times, \$PRCT_WHEN_SWITCHED%, when we switched.
``````
A specific comment: note that \$RANDOM is a magic bash variable; the assignment seeds it from the current time, reads from it provide a pseudo-random integer. Shall leave it at that unless there's specific comment and end on a quote from the above linked Wikipedia article,
Pigeons repeatedly exposed to the problem show that they rapidly learn always to switch, unlike humans (Herbranson and Schroeder, 2010).
I rather enjoy that quote.

sphyrth
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### Re: A riddle

The problem of intuition usually arises when one sees "switching vs staying" as a separate game from "choosing one of the doors".

Here's a video trying to explain it.
I have a favorite game. It's on my Youtube Channel.

Fred Barclay
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### Re: A riddle

rene wrote:Find below a bash simulation of the game.
Nice work! You beat me to it - I was working on an interactive Python version.
Still may finish it...

"Once you can accept the universe as matter expanding into nothing that is something, wearing stripes with plaid comes easy."
- Albert Einstein

all41
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### Re: A riddle

You could simulate it with three cards--one ace, two others
Let someone be your Monty (monty knows where the ace is but will never reveal)

greerd
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### Re: A riddle

Looks to me that the script only shows the original odds, before a choice is removed.

What I see argued above is that since your original choice has a 1/3 chance of the car, when Monty opens an unpicked goat door your odds stay frozen at 1/3, yet the other un-picked door's odds jump to 2/3, doesn't sound right.

Hypothetically: I pick the car door, Monty opens both goat doors, are my odds still frozen at 1/3?

sphyrth
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### Re: A riddle

greerd wrote:Hypothetically: I pick the car door, Monty opens both goat doors, are my odds still frozen at 1/3?
Your odds were still 1/3 if you stayed with the door. The fact that you happened to pick the car door didn't increase the probabilities.

When you roll 2 dice and betted for 12, and the dice happened to fall on 12, it didn't increase the probability of getting the same result.

You just got lucky aqainst the odds.
I have a favorite game. It's on my Youtube Channel.

TooMuchTime
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### Re: A riddle

Monty after all opened a wrong door from among those original alternative two.
MONTY opened the door. NOT THE CONTESTANT. The contestant KNEW, going into the game, that one of the doors would be opened to reveal the wrong choice and that he would then get A SECOND CHOICE TO KEEP THE ORIGINAL DOOR OR CHOOSE THE ONLY REMAINING DOOR OF TWO. What part of that tells you it's anything other than 50%? As I said earlier, you choose door A, Monty reveals door C, you are then given the option of keeping door A or choosing door B. Take a coin and write an A on heads and a B on tails and flip it; what are the odds of hitting either A or B? If you say anything other than 50% then you must live in a world were the laws of probability don't exist.

In other words, THE CHOICE OF THE CONTESTANT WAS ALWAYS GOING TO BE BETWEEN ONLY TWO DOORS!!! It doesn't matter that a third door ever existed. The third door is just part of the showmanship of the game.
If I entered the studio after the goat was revealed, with no knowledge of what transpired before that time,
and the contestant let me choose between the remaining doors what would my odds be of choosing the car?
BINGO!!! We have a winner! The player ALWAYS knows that one door will be taken from the game and that only two will remain. This is no different than if there were only two doors and you had to choose. The mere fact that you got a first choice and then a second choice does not change the fact that you are only really picking between two doors. You may start with three doors but YOU ARE NOT CHOOSING BETWEEN THREE DOORS ALL OF THE TIME. In fact, the instant that Monty chooses a door, means that you didn't. Could the contestant have chosen door C? Of course. Then MONTY CHOOSES door B. The effect is the same. The contestant picks a door, the host picks a door, the contestant then chooses between two doors.

50%.

TooMuchTime
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### Re: A riddle

Question: What would be the odds for the contestant if Monty opened a door BEFORE the contestant made a choice?

Ummm...uhhhh...wait a minute...I know this...it's uhhhh...

Yeah. 50%.

Again. The third door is just something to give the game some extra spice. The choice was always only going to be between two doors.

rene
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### Re: A riddle

the-long-fight-goes-on-300x185.jpg