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View Poll Results: Should the contestant stick with his door, or switch?
Definitely stick with his original door
13.73%
Stick with his original door, but it's close
0
0%
It doesn't matter; it's 50-50 either way
54.90%
Switch to the other door, but it's close
1.96%
Definitely switch to the other door
29.41%
Voters: 51. You may not vote on this poll

Interesting logic problem

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Old Oct 18, 2005 | 09:25 AM
  #31  
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Although the answer is not obvious.....it's there. The show would not have been such a sucess from the VERY begining had there been a statistical advantage to either the show or the player.

It was a success because it was a show based on giving a contenstant an actual 50/50 chance of winning a car. That's better odds than ANY casino will give you.
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Old Oct 18, 2005 | 09:26 AM
  #32  
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Originally Posted by WestSideBilly,Oct 18 2005, 09:16 AM
Bill - I was typing when you posted that.
I knew that.

I'm a physic.
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Old Oct 18, 2005 | 09:29 AM
  #33  
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Originally Posted by SIIK2NR,Oct 18 2005, 09:17 AM
. . . Monte is also making choices during this process and he cannot emliminate or choose the door with the car in it.
This is the key to the correct answer.

Make sure that you fully understand the implications of what you (correctly) wrote here.
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Old Oct 18, 2005 | 09:37 AM
  #34  
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You could look at it from another perspective also...

When faced with 3 doors...... pick one of the 3 that you think is a GOAT.

When Monte shows you the other one...... change your door to the final door.

Works either way.... your picking a door you think has a GOAT or your picking a door you think has a CAR....

fun to think about either way...
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Old Oct 18, 2005 | 09:38 AM
  #35  
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Originally Posted by SIIK2NR,Oct 18 2005, 09:22 AM
Did you read my post on the flaw behind your theory?
Billy's theory - analysis, actually - isn't flawed.

If you have the time you should try a simulation:

Imagine that the car is behind door #1.

Roll a fair (six-faced) die.

If it comes up one or two, the contestant chooses door #1.

If it comes up three or four, the contestant chooses door #2.

If it comes up five or six, the contestant chooses door #3.

If the contestant chooses door #1, roll the die again.

If it comes up one, two, or three, Monty opens door #2.

If it comes up four, five, or six, Monty opens door #3.

If the contestant chooses door #2, Monty opens door #3; if the contestant chooses door #3, Monty opens door #2.

Keep a tally of whether the contestant should stick with his door to win the car, or should switch to win the car.

Do this, say, 30 times. That should be enough of a sample to give meaningful statistics.
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Old Oct 18, 2005 | 09:39 AM
  #36  
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Originally Posted by SIIK2NR,Oct 18 2005, 12:25 PM
Although the answer is not obvious.....it's there. The show would not have been such a sucess from the VERY begining had there been a statistical advantage to either the show or the player.

It was a success because it was a show based on giving a contenstant an actual 50/50 chance of winning a car. That's better odds than ANY casino will give you.
Shows that have high winning percentages are much more popular. That's why the "losers" get nice door prizes on most shows. Look at the successful shows - Jeopardy (2nd and 3rd always get something), WoF (same thing), Price is Right (people who get 'up' win far more often than not, and have already won something), etc. Game shows where people go home empty handed don't succeed.
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Old Oct 18, 2005 | 09:48 AM
  #37  
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Originally Posted by magician,Oct 18 2005, 09:38 AM
If the contestant chooses door #2, Monte opens door #3; if the contestant chooses door #3, Monte opens door #2.

Keep a tally of whether the contestant should stick with his door to win the car, or should switch to win the car.

Do this, say, 30 times. That should be enough of a sample to give meaningful statistics.
Once you stopped using the fair six sided dice to choose for Monte....it's no longer relevant...

Monte picks the door he HAS TO.... but to the contestant it is just like Monte rolled the dice to pick it. BUT he doesn't.. because he can't. Hence the reason they didn't use BIG DICE on the show.

The results of your sampling are bogus because the original pick is decided by randomness of the fair dice....but the 2nd to last choice which is Monte's is NOT random. The LAST choice is random and can be made with dice, coin flip or just picking one or the other door remaining....it's 50/50.
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Old Oct 18, 2005 | 09:50 AM
  #38  
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Another way to look at it: Among 3 doors, Monte will never show you what's behind your door, and never show the car. Among the nine combinations of pick/car, 3 are your doors (1 right, 2 wrong) and 2 more are cars (both right). Monte takes one of the other 4 cases. That leaves you with 5 combinations, 3 cases you win and 2 you don't (statistically, you really have a 60% chance of winning the car), and of those 3 winning combos, two involve switching to the other door.

If my brain is working, this gives a 2:1 advantage of switching.
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Old Oct 18, 2005 | 09:52 AM
  #39  
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Originally Posted by WestSideBilly,Oct 18 2005, 09:39 AM
Shows that have high winning percentages are much more popular. That's why the "losers" get nice door prizes on most shows. Look at the successful shows - Jeopardy (2nd and 3rd always get something), WoF (same thing), Price is Right (people who get 'up' win far more often than not, and have already won something), etc. Game shows where people go home empty handed don't succeed.
Yes....but back when that show was popular....it was a big deal just to have to opportunity or chance to win something big...

Today's audiences just want something to show for it regardless of it's monetary value.
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Old Oct 18, 2005 | 09:56 AM
  #40  
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Originally Posted by WestSideBilly,Oct 18 2005, 09:50 AM
Another way to look at it: Among 3 doors, Monte will never show you what's behind your door, and never show the car. Among the nine combinations of pick/car, 3 are your doors (1 right, 2 wrong) and 2 more are cars (both right). Monte takes one of the other 4 cases. That leaves you with 5 combinations, 3 cases you win and 2 you don't (statistically, you really have a 60% chance of winning the car), and of those 3 winning combos, two involve switching to the other door.

If my brain is working, this gives a 2:1 advantage of switching.
wow.... I'm sorry but although that looks nice on paper and seems to make sense to you....it's wrong.

Think of it this way...keep it simple.

Change the show...

3 doors....

Monte picks first.....

Contestant picks from remaining 2 doors... 50/50

Same result....not as exciting... the only reason they swapped it up... more fun but same 50/50 odds in the end.
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