Puzzle
- Thread starter Dan26
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That would only be the case if the car was randomly repositioned following the opening of the first door.
- Mar 25, 2003
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Assuming it's a game show, the host will always have to open a door with a goat behind it. If he opens one of the other two doors at random and it has a car behind it then it kinda ruins the show.
I think that covers it
I don't get how that works.
Aren't the two choices completely independent of each other?
If you have zero indication of which it is, how can it be anything other than 50/50?
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- #30
You have a 1 in 3 chance of picking the car.
This means that the two doors you DIDN'T pick have a combined 66.66% chance of having the car behind them. Look at these two doors as one "set"
Of those two doors that you didn't select, the host then opens one of them, leaving two doors. This doesn't change your initial probability of 1 in 3. You knew the host was going to do this no matter what.
The 66.66% chance that was initially split amongst two doors that you didn't select, has just been reduced to one door. Bonus! So, you have a 66.66% chance of getting the car if you change doors.
No, the choices are not independent. The key to this is that the host is not picking at random, he knows where the car is.
So really, when you are offered the chance to swap, you are in effect being given BOTH of the other doors. That fact that he opens one of them isn't really relevant.
That's the rub of it - cheers
- Aug 13, 2013
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Except that if one door is eliminated, then the chance your original selection was correct goes up to 50%
Put it another way, is A, B and C are in a race, and each are of equal ability. So person A has a 33.3% chance of winning, and a 66.6% chance of not winning. Now, if person C does a hammy and is out, are the chances of person A winning still 33.3% and not winning 66.6%? or 50/50?
- Oct 30, 2010
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no it doesn't.
No, that's wrong.
Feel free to test it out if you want...
http://math.ucsd.edu/~crypto/Monty/monty.html
- Thread starter
- #36
Wrong.
The host KNOWS where the car is. He will eliminate one door no matter what. You know that going in to the game.
Think of the other two doors you didn't select as being one "set." This "set" of two doors has a combined 66.66% chance of the car being behind them. The host then gets rid of one of them for you. So this "set" of two doors with a 66.66% chance is now just one door with a 66.66% chance.
- Aug 13, 2013
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Dan,
thanks for that. I was just about to post. I get it now, the key is, as you said the host knows where the car is. I assumed the host didn't know, which in hindsight was obviously wrong.
Which is why my racing example does not apply, the person doing the hammy is totally at random whereas the host's selection of the door is not.
No, it's still 33% because when you made the choice you had a 1/3 chance of getting it correct. On face value it appears to be 50/50 but in reality that is incorrect. The host opens an INCORRECT door next every time. This is the key. You are now given the choice to switch doors or stay with the same door. The odds of you picking the correct door are 33% or 1/3 and being wrong 66% or 2/3 (seeing as those were the odds when you made your selection). Keeping this in mind - say you decided beforehand that no matter what you're going to switch doors when the offer comes. The ONLY way you can be wrong by switching (given that the host always opens an incorrect door first) is if your initial selection was correct, as you'd be switching from the correct door to an incorrect door. However, as discussed, the odds of you picking correct initially is 33% or 1/3 and therefore the odds of you switching and losing are also 33% or 1/3 (as that requires you to have picked correctly initially). As such, the odds of you switching and winning are elevated to 66% rather than 33% or the 50% on face value.
The problem with your example is that the result isn't predetermined like it is with the goat/million dollars example. The result of the latter never changes, the money is always going to be set to be behind either door A, B or C, it doesn't change at all. In a race, the winner and million dollars can always go to any of the three (unless it was a fixed race of course
), so by one person dropping out the odds of each person winning and you betting on the correct one improve to 50/50.
- Thread starter
- #39
There was a game show in the States where this puzzle was available every week. Nearly every contestant elected to not switch. Their logic being that it is 50-50 either way, I might as well just stay with my original choice. So, they won only 33.33% of the time.
It's a good example of emotion overriding logic and mathematics.
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- #40
How about this one.
There are two envelopes. One has twice as much money in it as the other.
Suppose the envelope in your hand has $20. That means the other envelope has either $10 or $40, right?
Should you switch envelopes?
The answer is yes, because on average, the amount in the envelope you switch to will be $25 (50% chance of it being $40, 50% chance of it being $10 so on average this is $25)
So you switch.
The problem is, now you encounter EXACTLY the same problem with your new envelope. It is in your best interests to switch. So, you switch again, and again, and again for infinity.
How does this paradox get resolved?
There are two envelopes. One has twice as much money in it as the other.
Suppose the envelope in your hand has $20. That means the other envelope has either $10 or $40, right?
Should you switch envelopes?
The answer is yes, because on average, the amount in the envelope you switch to will be $25 (50% chance of it being $40, 50% chance of it being $10 so on average this is $25)
So you switch.
The problem is, now you encounter EXACTLY the same problem with your new envelope. It is in your best interests to switch. So, you switch again, and again, and again for infinity.
How does this paradox get resolved?
Quivorir
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* off Dan, making a confusing thread even more confusing.
I find it easiest to explain the original problem like this:
If you pick a loser door, and switch, you win.
You have a 66% chance of picking a loser door.
Therefore switching has a 66% chance of winning.
If you pick a loser door, and switch, you win.
You have a 66% chance of picking a loser door.
Therefore switching has a 66% chance of winning.
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- #43
Say there are 23 people at a dinner party.
What are the chances that any two of them share the same birthday (not the YEAR, just the date.) Not counting leap years or anything.
To the nearest 5%(i.e 5% chance 10%, 15%, 20%, 25% etc) what are your guesses?
Closest to the pin gets an accolade, personally by Dan26
What are the chances that any two of them share the same birthday (not the YEAR, just the date.) Not counting leap years or anything.
To the nearest 5%(i.e 5% chance 10%, 15%, 20%, 25% etc) what are your guesses?
Closest to the pin gets an accolade, personally by Dan26
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akkaps
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- #46
Mythbusters did an episode on this problem. Here's the link to the episode: http://gorillavid.in/g4us0womplvk
JuddsABlue
Norm Smith Medallist
are you saying we get a new set on envelopes?
because if its the same set, as soon as you get the $40 then its over, the other two are $10 and $20....
If the money increases, so when you get $40 the other two become $20 and $80, then you keep going until your a billionair I guess, then just sacrifice the equity and cash out when you think you've got enough
its not really a paradox
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- #48
But you don't know how much is in the envelopes. You just know one has twice as much as the other.
So if the envelope in your hand has, say, X, then you know that the other envelope has EITHER 2X or 1/2X. On average this is 1.25X. So you are better off switching.
But as soon as you switch, you have exactly the same problem. And it becomes worthwhile to switch again. And again. And again.
How does this get resolved?
JuddsABlue
Norm Smith Medallist
Then that's the second scenario, where the envelopes increase in size or decrease depending on which one you get
In that case, that's who it's solved. You just keep picking forever, there nothing wrong with that
Whoever is filling the envelope with the money might run out at some point
But like I said, it's not really a paradox, the system is set up so it's always wise to pick again. So just keep picking, it works
Reminds me of this joke
A barber is cutting a mans hair when a young boy walks in. The barber says to his client, "watch this! This kid the dumbest one around"
The barber puts out his hands with $1 in one hand and 50c in the other and says to the boy "pick whichever one you want"
The kid thinks for a moment, grabs the 50c and runs out
"You see" says the barber, "he takes the 50c every time!"
The man leaves the barber shop and sees the kid outside. Curious, he asks the kid why he took the 50c instead of the dollar
"Because" the kid replies, "when I take the dollar the game stops"
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- #50
I think the envelope problem is solved like this:
"X" in my previous post, stands for different things. If there is a $40 and a $20, X could be either. This, apparently is illegitimate and the reason for the paradox.
If the equation is written like this:
X + 1/2X =1.5X then the problem is essentially solved. X means only one thing, not two different things. 1.5X, is the total value of both envelopes.
Because 1.5X is less that "twice X" you are better off keeping your original envelope.
