The Monty Hall Problem: Help!

INT21

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A hypothetical . (I love hypothetical)

I made an assumption with all the previous discussions, that Monty changes the door between games.

But let us suppose that the car is always behind the same door; say door A.

There are 100 contestants away in a room somewhere and they are called out one by one to play. Then they leave without being able to contact any other contestants.

So, now the only variable avaliable is the 1 in 3 choice the contestant makes. As both other doors have goats every time.

Monty knows this, but obvioulsy the contestants don't.

So, will the effect be the same.

Should anyone like to write a program to run this I'm sure we would be duly appreciative.
 

INT21

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Anyway message is, if you do do a test again take out the 50:50 bit.
Having seen the exchange since I started writing this, this may not help your understanding. But it helps mine.

O
The two ball choice on the second run represents my random choice of stick or swap.
 

oxo66

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What is interesting to me is that apparently in real life Game Show scenarios, so few Contestants switch when given the choice.
Yes you are right. The psychology is the really interesting bit. People generally value more highly something they already have over something they haven't got yet. I think psychologists have a term for it.

By swapping
Your new decision is effectively betting that you were initially wrong.
and people find it hard to do that.

in simple terms - people feel that having the correct answer and throwing it away is worse than correcting an initial wrong answer is 'good'
 

blessmycottonsocks

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Think it was Int21 who, many posts ago asked what if a new player is given the choice of the remaining two doors, after the goat door has been revealed.

If the new player is not aware of the original player's choice, the odds would be 50/50 wouldn't they?
 

Ringo

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Think it was Int21 who, many posts ago asked what if a new player is given the choice of the remaining two doors, after the goat door has been revealed.

If the new player is not aware of the original player's choice, the odds would be 50/50 wouldn't they?
Yes they would. If the new player was just asked to pick one of the two remaining doors, without knowing he original choice, then their odds would be 50/50. Likewise, the original players odds would be reduced to 50/50 if the host had the option of shuffling the things around behind the two remaining closed doors.

But as it's the same person playing, and they have previous experience, they have an advantage.
 

captain_bats

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The Wikipedia page on the problem has a conditional probablity table, to show the odds of switching versus sticking in 300 cases.

It is quite the most lucid way to present the problem. In the two-out-of-three cases where you are initially wrong, the host has no choice which door he can open. If you switch in these cases, you must now win. In the one-out-of-three cases where you are were right, any switch will now cause you to lose. So the odds are 2/3 in favour of the switch.

I would still hate to have to explain this in a maths class! :frust:
This post cleared this up for me.

If anyone's still struggling to get their head around this, maybe try thinking of it from Monty's perspective as if he's trying to win (by not giving the contestant the car).

In the scenario where the contestant switches, Monty is relying on the contestant picking the car door first time, as picking a goat door will mean that Monty has to reveal the other remaining goat door, leaving the car door as the one the contestant would be switching to.

Essentially, if you switch choices, you're always going to get what wasn't behind your original door and, since there are twice as many goat doors as car doors, you're twice as likely to end up on the car door by switching.
 

INT21

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Essentially, if you switch choices, you're always going to get what wasn't behind your original door and, since there are twice as many goat doors as car doors, you're twice as likely to end up on the car door by switching.
Well, yes..

I do believe that I have worked out what the problem was. And in a way it was like the three meals problem. And it is that not all the facts are taken into consideration.

Firstly, I agree that at the beginning you have one out of three chances of being correct. Or to put it another way, 33% chance of choosing correctly, 66% chance of picking the wrong one.

But what threw me, and is the reason I still would not swap, is that once Monty removes one bad door, The odd are NOW 50/50 that if I swap I will win.
I emphasis this because, in my practical test, I picked the correct door 14(I think) times. And thus by swapping I would have lost.

This factor does not appear to be taken into consideration.

Now, If I had more than one try the odds of moving to the correct door improve. And If I had 100 tries then, indeed, I would probably win the majority of the time. But not every time.

And as I have only one chance, I may as well stick with my first choice as, for me, the odds are simply win or lose.
 

JamesWhitehead

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And as I have only one chance, I may as well stick with my first choice as, for me, the odds are simply win or lose.
Any punter is faced with the same 50/50 decision. The larger stick/switch question compares their choice to that of others in their situation.

Two out of three punters will benefit from the switch. Success is not guaranteed. :)
 

blessmycottonsocks

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This puzzle just appeared on Quora and has a vaguely similar vibe:

"The woman in front of you places two envelopes onto the table. One is labeled Envelope A, and one is labeled Envelope B.

“Here’s the deal,” she says. “Each of these envelopes contains a significant sum of cash. One of them contains exactly twice as much money as the other.”

You nod eagerly, excited to see where this is going.

“You can keep one envelope, and I’ll keep the other one,” she says. “The decision is yours. But you’re only allowed to look inside one of them before you decide which one you want.”

You take envelope A, open it and count the money inside.

“Now,” she says, “do you want to keep Envelope A, or do you want to switch to Envelope B?”

Here’s the dilemma: No matter which envelope you’re holding, your expected value is always higher if you switch.

Let’s say you looked inside Envelope A, and found $1,000. You know that one envelope contains twice as much money as the other, but you don’t know which envelope has more. That means there’s a 50% chance that Envelope B contains $2,000, and a 50% chance that it only contains $500.

So let’s say you give up Envelope A and take Envelope B instead. If B is the better envelope, then you’ll gain $1,000. And if B is the poorer envelope, then you’ll only lose $500.

That means that if you switch, there’s an equal chance that you’ll gain $1,000 or lose $500. Seems like a good deal, right? Your expected gain is higher than your expected loss, so if your goal is to go home with as much money as possible, it makes sense to switch envelopes.

But once you’re holding Envelope B, the same logic applies. If the amount of money in Envelope B is X dollars, then by switching back for Envelope A, you know you’ll either gain X dollars or lose X/2 dollars. So once again, it makes sense to switch envelopes.

Does that make sense?
 

AlchoPwn

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Seems legit (unsarcastic).
 

Ringo

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Well there's a difference. In the Envelope A or B story, you will always win something whether you switch or not, so switching is not a risk. You'll walk away with more as often as you walk away with less. Meaning that switching doesn't give you an advantage at all. You best bet is to do nothing.
 

Min Bannister

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In this case it doesn't matter what you do since you'll still get some money.
 

blessmycottonsocks

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Well there's a difference. In the Envelope A or B story, you will always win something whether you switch or not, so switching is not a risk. You'll walk away with more as often as you walk away with less. Meaning that switching doesn't give you an advantage at all. You best bet is to do nothing.
No it isn't.
On opening the first envelope, you know you are guaranteed a sum of money. The gamble is whether you win double that amount - a 100% increase, or lose half of it - a 50% decrease. In terms of probability, it's better to go with the 100% chance.
 

INT21

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Well there's a difference. In the Envelope A or B story, you will always win something whether you switch or not, so switching is not a risk. You'll walk away with more as often as you walk away with less. Meaning that switching doesn't give you an advantage at all. You best bet is to do nothing.
Agreed.

The odds are different than they were in the other case.

With this one you gain something ether way.

I'd stick with my first choice.
 

Ringo

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Meant a 100% increase v a 50% decrease.
In terms of walking away with the most money, the chance of 100% increase has to be the better option.
Of course it is the better option but it's just as likely to be the other one. The chances of losing half of the money as just the same as doubling your money. Therefore changing to the other envelope is a gamble but not an advantage. It's 50% chance of losing half and 50% chance of winning double.

So if you want to walk away with the most money you can, doing nothing is the sensible option. Switching is the gamble (based on nothing but luck).
 

blessmycottonsocks

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Of course it is the better option but it's just as likely to be the other one. The chances of losing half of the money as just the same as doubling your money. Therefore changing to the other envelope is a gamble but not an advantage. It's 50% chance of losing half and 50% chance of winning double.

So if you want to walk away with the most money you can, doing nothing is the sensible option. Switching is the gamble (based on nothing but luck).
Obviously the odds of getting each amount are 50/50, but the point is that you stand to win twice as much as you stand to lose.
Surely any gambler has to acknowledge that those are favourable odds?
 

INT21

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Obviously the odds of getting each amount are 50/50, but the point is that you stand to win twice as much as you stand to lose.
Surely any gambler has to acknowledge that those are favourable odds?
The key word there is 'gambler'.

Even a Russian Roulette player has better odds. But one could say it boils down to 50/50 in that case.

Only someone who is ready to throw away a guarenteed amount of money would risk losing it for only half as much,
 

blessmycottonsocks

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Here's another puzzle in a similar vein ie. it seems unintuitive, but is totally logical.

Two prisoners are arrested by a sadistic tyrant and told they will be placed in separate dungeons with no contact between them. Each morning a guard will enter each dungeon and flip a coin. Each prisoner will then be asked to guess the outcome of the coin toss (heads or tails) in the other dungeon. If at least one of them guesses correctly, they are spared to live another day. If neither guesses correctly, they are both executed. Before being taken to their cells they have a brief moment together to confer. What strategy could they devise to guarantee living indefinitely?
 

blessmycottonsocks

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. Prisoner 1 always guesses the same result as the coin toss in their cell and prisoner 2 always guesses the opposite to the coin toss in their cell. As the only possible outcomes are the coin tosses are the same or are different, both bases are covered.
 
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