Naturally I would accept a computer program as evidence. Would be odd to say no. However we will have to agree on the algorithm and this whole thing gave me a terrible headache. I'll check macgyver's later.
over and out.
Naturally I would accept a computer program as evidence. Would be odd to say no. However we will have to agree on the algorithm and this whole thing gave me a terrible headache. I'll check macgyver's later.
over and out.
Originally Posted by brewbuck:
Reimplementing a large system in another language to get a 25% performance boost is nonsense. It would be cheaper to just get a computer which is 25% faster.
No matter what happens, the host will remove one of the donkeys. So then it's 50/50.
Does anybody care to try to disprove this?
Ok, my last attempt to explain it:
There is a 2/3 probablility of winning by switching from the outset. Your odds don't change once you select the first door. This is because, at the start there will allways be a 2/3 probability that one of the doors you do not chose to begin with will have the money in it.
First off these are all possibilities of what can be behind the doors:
It needs to be understood that the host has to remove a donkey from the following possibilities:Code:Host | You donkey donkey | money money donkey | donkey donkey money | donkey
This means that there are three possibilities as to what the host can have behind the final door:Code:donkey, donkey money, donkey donkey, money
The door you picked initially also has three possibilities to the outcome, but they are the same as from the start:Code:donkey money money
Now compare the possibilities of what the host can have against what your original door can have:Code:money donkey donkey
This is because the donkey that the host removes only affects the probability that one of his doors have money behind it, it has no effect on the door you picked. Therefore there is a predetermined 2/3 probability that by switching you will win the money.Code:Host | You donkey| money money | donkey money | donkey
Last edited by mike_g; 03-08-2008 at 05:24 PM.
>> Does anybody care to try to disprove this?
I don't know about "proof", but unfortunately it is wrong.
As I mentioned earlier, the host's choice of what to open depends on your original choice of your door, so you cannot start over with a new probability just because a door has been removed.
If the host opened a door at random and it ended up being a donkey, then the chance would be 50-50. But the host doesn't open a door at random. The host opens the door based on your choice and his knowledge of what is behind the doors.
You can also think about this alternative scenario. There are 50 doors. You pick 1. The host opens 48 doors that he knows all have donkeys. Should you switch?
Think I messed up my original program.
Try this one:
My output for 4 iterations:Code:#include <stdio.h> #include <stdlib.h> #include <time.h> int doRound(void); int getRandomDoor(void); int doRound(void) { int iRightDoor, iGuess; iRightDoor = getRandomDoor(); iGuess = getRandomDoor(); return (iRightDoor!=iGuess); } int getRandomDoor(void) { return rand() % 3; } int main(int argc, char *argv[]) { int i, times = 100, iSwitch = 0, iStay = 0; srand(time(NULL)); if(argc > 1) { times = (int)strtol(argv[1], NULL, 10); } for(i=0;i<times;i++) { if(doRound()) { iSwitch++; } else iStay++; } printf("Switching was appropriate %d times\n", iSwitch); printf("Staying was appropriate %d times\n", iStay); return 0; }
Provided there are no bugs, I think this could be submitted as evidence that Daved, CornedBee et al are correct, especially since the difference is about 2:1... ie. 66% chance in getting it right to make a switch.Code:Switching was appropriate 69 times Staying was appropriate 31 times Switching was appropriate 70 times Staying was appropriate 30 times Switching was appropriate 65 times Staying was appropriate 35 times Switching was appropriate 65 times Staying was appropriate 35 times
I need to dissect this because I believe herein lies the crux of the problem.
> If the host opened a door at random and it ended up being a donkey, then the chance would be 50-50.
Take notice of this statement: If it happened to be a donkey, the chance would be 50-50.
> The host opens the door based on your choice and his knowledge of what is behind the doors.
So he knows what door he can open to show the contestant a donkey. He does this all the time, no matter my choice. He can always open a door to a donkey.
So, what you think will be the odds?... 50:50, because that's exactly what the host does: If it happened to be a donkey, the chance would be 50-50.
Originally Posted by brewbuck:
Reimplementing a large system in another language to get a 25% performance boost is nonsense. It would be cheaper to just get a computer which is 25% faster.
Here's another simulation that runs when your browser loads. It also has an informal explanation of why 2/3 for a switch is correct.
http://edp.org/monty.htm
dammit.
Perhaps you would like to read:
Ask Dr. Math: FAQ: The Monty Hall Problem
Wolfram MathWorld: Monty Hall Problem
For a walkthrough simulation:
Let's Make a Deal
Look up a C++ Reference and learn How To Ask Questions The Smart WayOriginally Posted by Bjarne Stroustrup (2000-10-14)
If the donkeys and the money were placed behind the doors after the host made his/her choice, I would have been right.
>> because that's exactly what the host does: If it happened to be a donkey, the chance would be 50-50.
There's a difference. If the host chooses a door at random, then there is a 1/3 chance he opens the door with the money. But if the host chooses a door on purpose, then that 1/3 chance disappears.
Let's say this happens 30 times. We'll label the doors as such:
Door A: money
Door B: donkey
Door C: donkey
So out of 30 times, 10 times the contestant chooses Door A, 10 times the contestant chooses Door B and 10 times the contestant chooses Door C.
Contestant chooses Door A, host opens Door B five times and Door C five times, all ten times a switch is bad.
Contestant chooses Door B, host opens Door C all ten times (because he can't open door A), switch is good all ten times.
Contestant chooses Door C, host opens Door B all ten times (because he can't open door A), switch is good all ten times.
So the switch is good 20 out of 30 times.
Let's do the same thing when the host doesn't know where the money is.
Contestant chooses Door A, host opens Door B five times and Door C five times, all ten times a switch is bad.
Contestant chooses Door B, host opens Door A five times and Door C five times, switch is good five times, and five times we ignore (because the original problem assumes the host opened a door with a donkey).
Contestant chooses Door C, host opens Door A five times and Door B five times, switch is good five times, and five times we ignore (because the original problem assumes the host opened a door with a donkey).
That's 10 times switch is bad, 10 times switch is good, and 10 times that are ignored because the problem assumes the host opens a door with a donkey.
See the difference?
Last edited by Daved; 03-08-2008 at 07:16 PM.
Yup. I can see it now. It's an amazing paradox. Very invigorating. Also took a look at the links and on wikipedia.
It's a pleasure to see how wrong I am. This one is definitely going to my notebook.
Thanks a lot for the patience through the ordeal of trying to make see the light
Originally Posted by brewbuck:
Reimplementing a large system in another language to get a 25% performance boost is nonsense. It would be cheaper to just get a computer which is 25% faster.