Maggie’s Probability- question

[hermosabeach]

So I’m listening to a book on Math

The book covers the game shows on TV

The contestant is shown 3 doors and told that behind 2 doors is a crap prize and the One door conceals a new car

Contestant picks door #2

Host then builds excitement by opening a different door- say they open door #3 revealing a crap prize.

There are now two doors left- one had crap one has a new car

Door #1 and
The one the contestant picked- door #2.

Host asks-do you want to stay with your original pick of Door #2?
Or change you mind and select door #1?


mathematically / probability say you should change your mind and select door #1

You should now select door #1 to increase your odds of wining.


without googling the scenario -

agree / disagree?

why?

 

[earthtrekker1775]

Your odds in scenario 1 are 33% of choosing the prize. Remove a door and your odds are now 50/50. Strange math to look at the original odds and say if you now change your choice you have a higher odds of winning. I would just ignore the door picked by the host as between my original choice and the one left my odds remain 50%.

However, the host picked a door which he knew was not a winner suggesting the last door may be the winner. On the flipside you may have easily picked the winner to begin with. Statistically I don't think your odds change mathematically, but psychologically you may have an edge by changing your choice at the last stage. I also likely have no idea what I am talking about.

 

[Jaron3]

“I would obviously want the iokane powder as far from me as possible so I can clearly not choose the glass in front of you!”
Or something like that.

 

[ah1356]

Your odds double if you switch to the other door. It is counter-intuitive until you think backwards.

The odds are always 1/3 behind each door. The only way you can switch and lose is by having guessed right on the first try. Your odds of guessing right on the first try are 1/3. Therefore if you switch, your odds double to 2/3.

Your odds are never 50/50, but I have had long screaming matches with people who cannot be convinced otherwise.

 

[Waorani]

This ^^^^^

 

[Renegade]

So I’m listening to a book on Math

The book covers the game shows on TV

The contestant is shown 3 doors and told that behind 2 doors is a crap prize and the One door conceals a new car

Contestant picks door #2

Host then builds excitement by opening a different door- say they open door #3 revealing a crap prize.

There are now two doors left- one had crap one has a new car

Door #1 and
The one the contestant picked- door #2.

Host asks-do you want to stay with your original pick of Door #2?
Or change you mind and select door #1?


mathematically / probability say you should change your mind and select door #1

You should now select door #1 to increase your odds of wining.


without googling the scenario -

agree / disagree?

why?

Let's Make A Deal - ALWAYS switch.

 

[theLBC]

there is no advantage to changing your pick, unless there is a pattern of behavior to follow for how the game plays out.

 

[Jscb1b]

I read an article about this. The article said you have a better chance to win if you switch.

 

[theLBC]

Your odds double if you switch to the other door. It is counter-intuitive until you think backwards.

The odds are always 1/3 behind each door. The only way you can switch and lose is by having guessed right on the first try. Your odds of guessing right on the first try are 1/3. Therefore if you switch, your odds double to 2/3.

Your odds are never 50/50, but I have had long screaming matches with people who cannot be convinced otherwise.

sorry, but statistically the prior flip of the coin is not a predictor for the next flip. it is always 50/50.
if you flip a coin and it lands on heads 50 times in a row, the chance of it landing on tails on the next flip is 50/50 (unless the coin or flip is rigged).

 

[Jscb1b]

sorry, but statistically the prior flip of the coin is not a predictor for the next flip. it is always 50/50.
if you flip a coin and it lands on heads 50 times in a row, the chance of it landing on tails on the next flip is 50/50 (unless the coin or flip is rigged).

Google Monty Hall problem.

 

[ah1356]

sorry, but statistically the prior flip of the coin is not a predictor for the next flip. it is always 50/50.
if you flip a coin and it lands on heads 50 times in a row, the chance of it landing on tails on the next flip is 50/50 (unless the coin or flip is rigged).

You are solving the wrong problem (correctly, but still the wrong problem). Coins have no memory, but this game does. The car goes behind a door once. It is wherever it is no matter how many doors are opened. It is not a sequence of independent coin flips.

This scenario has inspired arguments for decades. It will inspire arguments for more decades. It is unintuitive. But I assure you, my original explanation is the correct one.

 

[Jscb1b]

Is this still available? Asking for a friend.

 

[Renegade]

sorry, but statistically the prior flip of the coin is not a predictor for the next flip. it is always 50/50.
if you flip a coin and it lands on heads 50 times in a row, the chance of it landing on tails on the next flip is 50/50 (unless the coin or flip is rigged).

It is probability problem that was solved 50 years ago.

Basically your goal is NOT to pick the winning door and keep it, but to pick a losing door and let Monty offer you to switch to the winning door. Pick the right door and keep, 1/3, pick the wrong door and switch, 2/3.

 

[Renegade]

Google Monty Hall problem.

It literally has a web site of its own.

The Monty Hall Problem

 

[Barneybdb]

Depends on the guy behind the doors switching the prizes around......

 

[hermosabeach]

So much for not googling the problem

if a prize was behind one of 1,000 doors

would you pick 1 door or 999?

we would all pick to open 999 doors


—
in this scenario, we picked one door. Door # 2

the host opened 998 doors

the odds are that we did not pick the right door

the 1:1000 makes it more obvious than the one in 3


but yes- you should charge your answer

your success rate goes from 33.33%
to
66.66%

 

[SanPatHogger]

“I would obviously want the iokane powder as far from me as possible so I can clearly not choose the glass in front of you!”
Or something like that.

I was thinking the same part of the same movie. Princess bride

 

[eklarsen]

Yes, but you have a 50/50 shot of picking the wrong door on the second pick nerds

 

[theLBC]

You are solving the wrong problem (correctly, but still the wrong problem). Coins have no memory, but this game does. The car goes behind a door once. It is wherever it is no matter how many doors are opened. It is not a sequence of independent coin flips.

This scenario has inspired arguments for decades. It will inspire arguments for more decades. It is unintuitive. But I assure you, my original explanation is the correct one.

hmm
when there were 3 doors, you had a 1 in 3 chance of getting the right one.
but eliminating the 3rd door does not carry over to the next question.

there are now 2 doors, the car can still be behind either door.
eliminating the 3rd door earlier is not part of the problem anymore.
it is a new "bet" for door 1 or door 2 because he is allowed to change his guess.

now it comes down to how the question is worded (imo).
if you ask "would his odds of having chose the right door improved if he chooses to switch to door 1?"
the answer to that is YES, because he chose BOTH door 1 and door 2 - but he didn't win yet.

but the OP says:
You should now select door #1 to increase your odds of winning.

BUT with the NEW bet, he only has ONE choice between TWO doors, so his chance of WINNING is 50/50 no matter which he chooses.

he won the first bet at 1/3 odds.
but he would always win that bet because no matter what is behind the door he chose, there is always another crap prize to be revealed.

 

[buffalowinter]

This was brought up and explained on the tv series "Numbers"

 

[JbBooks]

So it’s kind of like farting after 70.

 

[Waorani]

It wouldn't be a brain teaser that's been around for years if it was obvious.

 

[theLBC]

So it’s kind of like farting after 70.

dust comes out?

 

[JbBooks]

dust comes out?

No, a 50/50 chance whether you’re gonna shit or not

 

[hermosabeach]

Yes, but you have a 50/50 shot of picking the wrong door on the second pick nerds

Wrong. Problem still has 331/3 vs 66 2/3 probability

 

[theLBC]

Wrong. Problem still has 331/3 vs 66 2/3 probability

wrong. it was 2 games or bets.
he wins the first game at 1/3 odds, because there is always another crap prize behind another door to reveal first, regardless of what is behind the door he chooses. it is rigged.
that bet is over.

now he has a choice between 2 doors that is 1/2 no matter what door he goes with.

 

[Jscb1b]

Is the goat still available? Asking for a muslim.

 

[theLBC]

Is the goat still available? Asking for a muslim.

 

[P-Squared]

How about playing the ANTIFA move and burn the bit%^ down.

 

[Snuby642]

The host stopped the 1/3 game after a partial elimination.

He now offers a new game of 1/2 because it's more dramatic or knows you have won, knows you have lost or doesn't really gaf it's not his car and just wants ratings.

At this point there is no damn reason to change at all.

It's like a fucking troll thread on moa versus millard.

It has no winners just loosers with the exception of the op that is laughing at everyone.

 

[Bolo]

Wrong. Problem still has 331/3 vs 66 2/3 probability

Ah. But... having studied the three prisoners problem a long tome ago, there is a key fact that's left out of your description.
Does the game show host already know which door the car is behind?

That changes the rules. If the host's (warden's) selection is random with respect to the prize behind the door, it destroys the interdependence of the second and first round probabilities. The second round becomes a 50/50 shot.
This is also the source of the logical fallacy.. failure to recognize that the procedure unveiling a zonk in Let's Make a Deal is planned, will always eliminate a zonk, and makes round 2 dependent on round 1.

Otherwise, Monty would have accidentally revealed the car in the first round 1/3 of the time and NBC/ABC would have to figure out four minutes of filler for the timeslot.

 

[theLBC]

Ah. But... having studied the three prisoners problem a long tome ago, there is a key fact that's left out of your description.
Does the game show host already know which door the car is behind?

That changes the rules. If the host's (warden's) selection is random with respect to the prize behind the door, it destroys the interdependence of the second and first round probabilities. The second round becomes a 50/50 shot.
This is also the source of the logical fallacy.. failure to recognize that the procedure unveiling a zonk in Let's Make a Deal is planned, will always eliminate a zonk, and makes round 2 dependent on round 1.

Otherwise, Monty would have accidentally revealed the car in the first round 1/3 of the time and NBC/ABC would have to figure out four minutes of filler for the timeslot.

yup, exactly what i said.
actually, the odds are never 1 out of 3 because they always reveal one of the booby prizes first and let him pick again from the remaining 2 doors.

 

[hermosabeach]

But your pick is 1/3

so you are only 33.33% likely to win

If the host said you can pick one door or 2 doors- we would want to pick 2. 66% chance

So picking the other door increases the probability of winning

Picture 10,000 doors
You pick one
Host opens 9,998 doors


would you take your first pick or switch to the door that was part of the 9,999 out of 10,000?

 

[Barneybdb]

You changed your mind and I didn't, which one of us now has a new car????

 

[G-Dog]

Beautiful counter intuitive stats problem. And yes, switching doors increases the probability of winning from 1/3 to 2/3.

There are a few classic stats problems like this, one of my favorites was in regard to WWII airplanes and Abraham Wald. He studied planes that made it back to base, and looked at where the most damage was. He then told engineers to add more armor to the planes where there was the *least* amount of damage. Brilliant man knew about selection bias.

 

[Steel+Killer]

You guys are all wrong he never had a 1/3 chance even in the first scenario he had 0/0 chance because have you ever seen a damn game show where they actually let someone win on the first door opening? They don't put the car behind any damn door until it comes down to the last two doors so he always had a 50/50 chance........can't believe you guys don't know this........LOL

 

[pmclaine]

What if the Gimp is behind one of the two remaining doors?

Winning means you get a car.

Losing means you get Gimped.

Are you than playing to win or in a serious contest to avoid losing because you really don't care about getting a new car but the thought of having the Gimp run you down is really disturbing?

 

[Barneybdb]

You guys are all wrong he never had a 1/3 chance even in the first scenario he had 0/0 chance because have you ever seen a damn game show where they actually let someone win on the first door opening? They don't put the car behind any damn door until it comes down to the last two doors so he always had a 50/50 chance........can't believe you guys don't know this........LOL

See post #15

 

[Jaron3]

And this is why statistics can be manipulated to say anything you want them to. Both cups were poisoned with iokane. You lost.

 

[Jaron3]

I was thinking the same part of the same movie. Princess bride

Correct sir

 

[Mike Islander]

The first explanation that overcame my intuition is this: on your first pick, you have a 1/3 chance of being right. Switching after Monty opens a zonk door is mathematically the same as Monty saying, "You can keep your 1st pick 1/3 chance of being right. Or, I will let you switch to both of these other doors, and I'll open one that is a zonk." Choosing 2 doors gives you a 2/3 chance of winning.

Imagine there were 100 doors. First pick gives you 1/100 chance of being right. Monty then says, I'll give you the remaining 99 doors in trade for your door, and I'll open 98 of them, all zonks. Whether you switch before or after Monty opens the other doors, it's 99/100 chance you choose the car.

 

[Mordamer]

The beauty of this problem is that it is very easy to test.

Get yourself 3 Dixie cups and an object underneath them and do the test with someone about 20 times. I guarantee you they'll win more than 10 times if they switch each time and they'll lose more than 10 if they don't switch.

It becomes very obvious that it is not 50/50 if you test it.

 

[theLBC]

The beauty of this problem is that it is very easy to test.

Get yourself 3 Dixie cups and an object underneath them and do the test with someone about 20 times. I guarantee you they'll win more than 10 times if they switch each time and they'll lose more than 10 if they don't switch.

It becomes very obvious that it is not 50/50 if you test it.

unfortunately, reality and probability rarely match with a low number of attempts.
flip a coin 50 times. no matter how many times you try, you will probably never get 25 head and 25 tails, even though that is the statistical probability.
the answer is that the 3rd door is never part of the game, because they always reveal one of the booby prizes.

 

[Mordamer]

You left all three dixie cups in play.

In the monty hall scenario, there is always a 100% chance that one of the wrong choices will be eliminated. Therefore it doesn't statistically figure into the problem.

The very framework of the problem is misleading . It isn't even really a statistics problem, but a critical thought problem.

I didn't leave all 3 in play. You assumed that.

I said do the exact scenario. Get a friend, ask them to pick a cup, then turn over one of the two cups that doesn't contain the object. Then have them switch their guess to the other cup. They will win more than half the time.

If they never switch it becomes obvious real quick that they can only win 33%.

 

[Mordamer]

unfortunately, reality and probability rarely match with a low number of attempts.
flip a coin 50 times. no matter how many times you try, you will probably never get 25 head and 25 tails, even though that is the statistical probability.
the answer is that the 3rd door is never part of the game, because they always reveal one of the booby prizes.

Agreed, but if you flipped the quarter 50 times you have an incredibly low chance of it being 30/20 or more lopsided.

 

[Waorani]

 

[theLBC]

Agreed, but if you flipped the quarter 50 times you have an incredibly low chance of it being 30/20 or more lopsided.

you'd have a hard time hitting on 17 at roulette 4 times in a row, but i have seen it.
the odds of hitting 17 x4. 2,085,136

 

[PDXGS]

Doubletap the host and open both doors......take what's rightfully yours by the rules of conquest

 

[HeavyAssault]

It's not math but a psych game. The concept IS to make you think your chances are now 50/50....It's always 33.3334%.

It's no different than multiple choice questions and answers on tests. One will always be 100% wrong, two others are close, only one right answer. What are your chances of getting the right answer on the first guess??? The two "close" answers are distractors.

In the end by being allowed to change your decision you show you are easily manipulated thru the suggestion you now have a
"better chance".

I'm sure that's my answer and I'm not changing it. If you want to try please do so.

 

[PDXGS]

Watch the last video posted...post #46
The decision for the unknowing player to switch doors is entirely dependent on whether Monty knows or doesn't know where the goats are. Also stated in post #31.
In the original scenario/post#1 we don't know if the host knows where the prize is.
In the video's description it is stated clearly that he knows.
Given that he knows, switch.
All that said....I'm still sticking with double tap... This and the fact that goats are awesome: eat poison oak, are social, their milk makes excellent cheeses and they taste great on tacos if cooked long enough with the right spices... Cars have brought me nothing but trouble and frustration and don't taste good on tacos.