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re: Let's discuss the Monty Hall problem (probabilities, odds)
Posted on 8/12/17 at 4:04 pm to Rebel
Posted on 8/12/17 at 4:04 pm to Rebel
quote:I got it by thinking about a thousand doors. If he shows you 998 goats, you probably should switch to the door you didn't pick because your chances of picking the car were 0.1% on your first door
The odds are 50/50.
Since it is known 1 goat will be revealed (not necessarily door 3) all the "theater" before choosing between the final 2 doors is nothing more than a distraction.
1
Posted on 8/12/17 at 4:09 pm to TigerCoon
quote:
if you are asking about Monte's chance of selecting the car with a random pick, it is 1 in 3. The fact that he picked second, and your pick eliminated one door makes no difference, Correct?
Posted on 8/12/17 at 4:10 pm to Rebel
I see someone picked up where I left it this morning.
Lol
Lol
Posted on 8/12/17 at 4:13 pm to SlapahoeTribe
Is the one picking a Muslim, and how pretty are the goats?
Posted on 8/12/17 at 4:16 pm to chalmetteowl
One more post and I am out.
Let's play the game 3 times, and assume the outcomes play out exactly per the probabilities.
2 of 3 doors haves goats . You pick a goat door with your first pick 2 of 3 games.
Monte picks a goat all 3 games (game show drama rules, not random), leaving the car as the unpicked door 2 times.
If you switch all 3 games, you win twice. If you don't switch, you win once.
Not 50-50 on the switch. Always switch.
Let's play the game 3 times, and assume the outcomes play out exactly per the probabilities.
2 of 3 doors haves goats . You pick a goat door with your first pick 2 of 3 games.
Monte picks a goat all 3 games (game show drama rules, not random), leaving the car as the unpicked door 2 times.
If you switch all 3 games, you win twice. If you don't switch, you win once.
Not 50-50 on the switch. Always switch.
Posted on 8/12/17 at 4:22 pm to Korkstand
quote:
If Monty knows where the car is, and he reveals a goat intentionally, then your odds of winning are 1/3 if you pick and stay, and 2/3 if you pick and switch.
If Monty does NOT know where the car is, then your odds of winning are 1/3 if you pick and stay, 1/3 if you pick and switch, and 1/3 that Monty reveals the car.
Yes, I understand that Monty creates a new game by opening a door when he picks at random. And yes, this NEW GAME has odds of 50/50. I'm just saying that when considering the game as a whole, there is no scenario where you have a 50% chance of winning from the outset. If there are 3 doors and you switch, you're trading 1/3 for 1/3. With 100 doors, you're trading 1% for 1%.
I understand that odds change throughout a random game. I'm just saying that nothing the player does can change the odds in a random game.
I just wanted to illustrate your point.
If Monty does not know where the car is and I pick first.
I will pick correctly ~333 out of 1000 times.
In those games, Money will pick wrong every time.
In the 666 I pick wrong, Monty will pick correctly half the time.........IE.....333 times.
So, this means if I approach from the outset with a "pick and switch strategy,
Switch out of victory 333 times.
Switch IN to victory 333 times(the times Monty picks a goat
and just lose 333 times(the times Monty gets the car.
This is why him knowing makes the "pick switch" strategy happen.
In that case. I'll STILL pick right 333 times and wrong 666 times.
BUT.
So. I'll switch OUT of victory 333 times
BUT
I'll switch IN to victory 666 time because the knowing Monty will NEVER show me the car.
Posted on 8/12/17 at 4:27 pm to ShortyRob
Posted on 8/12/17 at 4:30 pm to ShortyRob
What if the contestant has built up an immunity to iocane powder?
Posted on 8/12/17 at 4:31 pm to SlapahoeTribe
What if my heart desires a goat? Win-win for me.
Posted on 8/12/17 at 4:32 pm to Rebel
depends on if the host is Sicillian. What are those odds?
Posted on 8/12/17 at 4:33 pm to SlapahoeTribe
How I understand the problem:
Let's take it to the extreme. You have a jar of 1 million pennies. One of the pennies has a golden star on it. You're allowed to take one penny out, and if the penny you select has the golden star, you win a new sports car.
You take out your penny, and then the host removes 999,998 pennies that DON'T have the golden star.
He then asks you: do you want to trade with the one penny that's left?
You obviously say yes. Why? Because your odds are now 999,999/1,000,000 instead of one in a million.
In the Monty Hall problem, the host asks you to choose your original odds of guessing right (1/3) with your odds of guessing wrong (2/3). So you switch, and say thank you.
Let's take it to the extreme. You have a jar of 1 million pennies. One of the pennies has a golden star on it. You're allowed to take one penny out, and if the penny you select has the golden star, you win a new sports car.
You take out your penny, and then the host removes 999,998 pennies that DON'T have the golden star.
He then asks you: do you want to trade with the one penny that's left?
You obviously say yes. Why? Because your odds are now 999,999/1,000,000 instead of one in a million.
In the Monty Hall problem, the host asks you to choose your original odds of guessing right (1/3) with your odds of guessing wrong (2/3). So you switch, and say thank you.
Posted on 8/12/17 at 4:37 pm to LordoftheManor
quote:
In the Monty Hall problem, the host asks you to choose your original odds of guessing right (1/3) with your odds of guessing wrong (2/3). So you switch, and say thank you.
Yeah.
The 3 doors is what fricks people up.
That's why ALL variations of Monty hall type games use 3. Shell games, door game etc.
No one would EVER fail in the shell game if it had 10 shells and after they picked, the house revealed 8 shells with no pea under them and then asked if they'd like to switch
Posted on 8/12/17 at 4:41 pm to ShortyRob
I'm going to throw a little monkey wrench into this thing.
In Let's Make a Deal..........did Monty ALWAYS do a door reveal in the doors game?
Or, did he just do the door reveal sometimes?
Cause, ya know, that fricks up the problem.
Monty could frick things up if he ONLY did door reveals when he knew you picked the car.
In Let's Make a Deal..........did Monty ALWAYS do a door reveal in the doors game?
Or, did he just do the door reveal sometimes?
Cause, ya know, that fricks up the problem.
Monty could frick things up if he ONLY did door reveals when he knew you picked the car.
Posted on 8/12/17 at 4:43 pm to ShortyRob
That made me start thinking:
I always thought the people that did 50/50 on Who Wants to be a Millionaire were idiots. The obvious strategy is to say that you think one answer is correct that you KNOW is wrong, because that is the one that is GUARANTEED to appear, along with the correct answer. The game masters always included the contestant's thoughts into the "randomization" of the 50/50.
When people were torn between two answers, lo and behold, the two answers would always appear (assuming one of the two was correct)! And when they were leaning towards one answer, that answer would always appear even if it was wrong. It wasn't a random 50/50, the game was fixed.
I always thought the people that did 50/50 on Who Wants to be a Millionaire were idiots. The obvious strategy is to say that you think one answer is correct that you KNOW is wrong, because that is the one that is GUARANTEED to appear, along with the correct answer. The game masters always included the contestant's thoughts into the "randomization" of the 50/50.
When people were torn between two answers, lo and behold, the two answers would always appear (assuming one of the two was correct)! And when they were leaning towards one answer, that answer would always appear even if it was wrong. It wasn't a random 50/50, the game was fixed.
This post was edited on 8/12/17 at 5:00 pm
Posted on 8/12/17 at 4:46 pm to SlapahoeTribe
I knew a guy named Monty. Had a beagle. Borrowed a cup of Brown sugar once. Hell of a nice guy.
Posted on 8/12/17 at 4:51 pm to LordoftheManor
Yes it in the that game there is always a correct answer. You either know it or dont
Posted on 8/12/17 at 5:01 pm to ShortyRob
The 3 doors seems to imply the ""Big Deal" game at the climax of Let's Make A Deal.
For that Big Deal game, your pick was your pick. You couldn't swap doors. Monte just opened the doors to reveal the prizes in ascending order of value. On the plus side, there were no "zonks" (goats) behind doors in the Big Deal. But, you had to give up your original prize for your shot at the Big Deal, so if you picked the wrong door, you may have traded down to something less valuable. At least, that is how I remember it. I am not much of a statistician, but I am one hell of a couch potato.
The rules are changed to support this probability exercise.
For that Big Deal game, your pick was your pick. You couldn't swap doors. Monte just opened the doors to reveal the prizes in ascending order of value. On the plus side, there were no "zonks" (goats) behind doors in the Big Deal. But, you had to give up your original prize for your shot at the Big Deal, so if you picked the wrong door, you may have traded down to something less valuable. At least, that is how I remember it. I am not much of a statistician, but I am one hell of a couch potato.
The rules are changed to support this probability exercise.
Posted on 8/12/17 at 8:44 pm to LordoftheManor
quote:
How I understand the problem:
Let's take it to the extreme. You have a jar of 1 million pennies. One of the pennies has a golden star on it. You're allowed to take one penny out, and if the penny you select has the golden star, you win a new sports car.
You take out your penny, and then the host removes 999,998 pennies that DON'T have the golden star.
He then asks you: do you want to trade with the one penny that's left?
You obviously say yes. Why? Because your odds are now 999,999/1,000,000 instead of one in a million.
It all depends on whether or not the host purposely revealed the goats or revealed them by chance.
If the host deliberately revealed the goats, you should switch. That's the Monty Hall problem. At the end of the game, with 2 doors left, your odds improve from 1/n to (n-1)/n by switching.
If the host randomly revealed the goats, there is no advantage in switching. That's the Deal or No Deal problem. At the end of the game, with 2 doors left, your odds of winning are 1/2 no matter what you do.
Posted on 8/12/17 at 10:01 pm to Spock's Eyebrow
It's insanely simple when you view it this way:
Call the doors ABC
you always pick A
B+C=2/3
If B=0 then C=2/3
If C=0 then B=2/3
You don't have to be Mayor of NOLA to understand this.
Call the doors ABC
you always pick A
B+C=2/3
If B=0 then C=2/3
If C=0 then B=2/3
You don't have to be Mayor of NOLA to understand this.
Posted on 8/13/17 at 4:27 pm to pensacola
The MHP is misunderstood because it is almost always - including the few answers I've read here, even the ones that get the right answer - explained incorrectly.
Let's say you always pick door #3. At this point, there are three possibilities that I hope everybody can agree with:
(A) There is a 1/3 chance that the car is behind door #1.
(B) There is a 1/3 chance that the car is behind door #2.
(C) There is a 1/3 chance that the car is behind door #3.
What happens next is what everybody seems to get confused about. Monty Hall has to open a door other than door #1, and it has to have a goat. How he does this is different in one of the three cases, than the other two:
(A) There is a 1/3 chance that the car is behind door #1, and he must open door #2.
(B) There is a 1/3 chance that the car is behind door #2, and he must open door #1.
There is a 1/3 chance that the car is behind door #3, BUT NOW HE HAS A CHOICE. So case (C) must get split into two:
(C1) There is a 1/6 chance that the car is behind door #3 and he opens door #1.
(C2) There is a 1/6 chance that the car is behind door #3 and he opens door #2.
What everybody seems to miss, is that WE SEE WHICH DOOR HE OPENED. If it is #1, we know we are in either case (B) or case (C1). If it is #2, we know we are in either case (A) or case (C2).
Either way, the probability of arriving at the point where WE SEE WHAT DOOR IS OPEN and are asked if we want to switch is 1/3 if you picked a goat, and 1/6 if you picked the car. This is what makes it twice as likely that you picked a goat than the car, not the fact that you originally had a 2/3chance to pick a goat.
Most of the answers that get the right answer combine cases (A) and (B) into a single case, as well as cases (C1) and (C2). WITHOUT RECOGNIZING THAT THEY DO SO. As long as the split of (C) into (C1) and (C2) is even, the answer has to be the same, but that doesn't make it correct to combine them this way.
Let's say you always pick door #3. At this point, there are three possibilities that I hope everybody can agree with:
(A) There is a 1/3 chance that the car is behind door #1.
(B) There is a 1/3 chance that the car is behind door #2.
(C) There is a 1/3 chance that the car is behind door #3.
What happens next is what everybody seems to get confused about. Monty Hall has to open a door other than door #1, and it has to have a goat. How he does this is different in one of the three cases, than the other two:
(A) There is a 1/3 chance that the car is behind door #1, and he must open door #2.
(B) There is a 1/3 chance that the car is behind door #2, and he must open door #1.
There is a 1/3 chance that the car is behind door #3, BUT NOW HE HAS A CHOICE. So case (C) must get split into two:
(C1) There is a 1/6 chance that the car is behind door #3 and he opens door #1.
(C2) There is a 1/6 chance that the car is behind door #3 and he opens door #2.
What everybody seems to miss, is that WE SEE WHICH DOOR HE OPENED. If it is #1, we know we are in either case (B) or case (C1). If it is #2, we know we are in either case (A) or case (C2).
Either way, the probability of arriving at the point where WE SEE WHAT DOOR IS OPEN and are asked if we want to switch is 1/3 if you picked a goat, and 1/6 if you picked the car. This is what makes it twice as likely that you picked a goat than the car, not the fact that you originally had a 2/3chance to pick a goat.
Most of the answers that get the right answer combine cases (A) and (B) into a single case, as well as cases (C1) and (C2). WITHOUT RECOGNIZING THAT THEY DO SO. As long as the split of (C) into (C1) and (C2) is even, the answer has to be the same, but that doesn't make it correct to combine them this way.
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