A question for the Wizard about the Monty Hall paradox.
February 16th, 2010 at 6:31:27 AM
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I used the Monty Hall paradox to liven up the discussion on another message board, and during the discussion something occurred to me and I was wondering about the official answer.
For those who are not familiar with this puzzle, check it out at the Wizard of Odds website.
I completely understand how a contestant's probabilities will change from 1/3 to 2/3 after a losing door is revealed, but using the Let's Make a Deal game show might not be the best model for this puzzle, and here's why.
On the show, there were always two contestants who were allowed to pick doors. Let's call them Jim and Helen. So Jim picks a door and his chances are 1 out of 3 that he has picked the car. But when it's Helen's chance to pick, she only has two doors left to choose between because the third door has been taken out of play by Jim, which makes Helen's chances 1 out of 2.
If Helen is the one who is then given the option to switch later on, there's no way her chances can have been improved by opening another door because she only had a 50/50 chance from the get-go.
Is my logic wrong? I have a feeling it must be because this puzzle has been explained by the best of the best, but always using a single-contestant scenario, which is not the way the actual game was played.
I thank you in advance for your (hopefully simple) explanation.
For those who are not familiar with this puzzle, check it out at the Wizard of Odds website.
I completely understand how a contestant's probabilities will change from 1/3 to 2/3 after a losing door is revealed, but using the Let's Make a Deal game show might not be the best model for this puzzle, and here's why.
On the show, there were always two contestants who were allowed to pick doors. Let's call them Jim and Helen. So Jim picks a door and his chances are 1 out of 3 that he has picked the car. But when it's Helen's chance to pick, she only has two doors left to choose between because the third door has been taken out of play by Jim, which makes Helen's chances 1 out of 2.
If Helen is the one who is then given the option to switch later on, there's no way her chances can have been improved by opening another door because she only had a 50/50 chance from the get-go.
Is my logic wrong? I have a feeling it must be because this puzzle has been explained by the best of the best, but always using a single-contestant scenario, which is not the way the actual game was played.
I thank you in advance for your (hopefully simple) explanation.
February 16th, 2010 at 6:35:20 AM
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You fail to include in the probability that the first contestant already picked the winning door, which would mean Helen's chances are 0/2. I would guess that is why Helen still has a 1/3 chance.
1/3 of the time (Contestant A picks winning door)= 0% chance
2/3 of the time = 50% chance
(1/3*0) + (2/3*.5) = 1/3
1/3 of the time (Contestant A picks winning door)= 0% chance
2/3 of the time = 50% chance
(1/3*0) + (2/3*.5) = 1/3
February 16th, 2010 at 7:11:50 AM
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Your link doesn't work. Here's link right to the problem:
https://wizardofodds.com/ask-the-wizard/122
Your example of Jim & Helen is flawed.
But that's OK. The text of the original question is flawed too.
On Let's Make A Deal, Doors are only used in the final deal with two contestants, and Monty NEVER offers either contestant the chance to switch.
During the game, Monty works with one contestant at a time, and often shows three curtains and gives the contestant the chance to switch after showing one.
----
NOTE: The entire paradox is flawed!
The scenario never occurs except in two specific conditions:
1 - The contestant has picked the top prize or the goat prize. The exposed curtain is always the nice consolation prize.
2 - There is no goat (BOTH of the 'loser' curtains contain nice consolation prizes, giving Monty free will to show either if the contestant picked the top prize, or to show the other, if the contestant picked one of the consolation prizes).
Note: Either Monty knows what's behind each curtain, or a stage hand tells him which to expose, after the contestant makes a selection. I.E. Monty's choice is NOT random.
While these 'specific' conditions may be frequesnt, they are far from guaranteed. If the contestant picks the consolation prize, and the other curtains contain the car and the goat, no curtain is revealed.
If you've watched the show enough, you know that, time permitting, Monty showed the other curtain when the expose/choice thing doesn't happen. Quite often, there was no goat (or other zonk item) behind any curtain.
https://wizardofodds.com/ask-the-wizard/122
Your example of Jim & Helen is flawed.
But that's OK. The text of the original question is flawed too.
On Let's Make A Deal, Doors are only used in the final deal with two contestants, and Monty NEVER offers either contestant the chance to switch.
During the game, Monty works with one contestant at a time, and often shows three curtains and gives the contestant the chance to switch after showing one.
----
NOTE: The entire paradox is flawed!
The scenario never occurs except in two specific conditions:
1 - The contestant has picked the top prize or the goat prize. The exposed curtain is always the nice consolation prize.
2 - There is no goat (BOTH of the 'loser' curtains contain nice consolation prizes, giving Monty free will to show either if the contestant picked the top prize, or to show the other, if the contestant picked one of the consolation prizes).
Note: Either Monty knows what's behind each curtain, or a stage hand tells him which to expose, after the contestant makes a selection. I.E. Monty's choice is NOT random.
While these 'specific' conditions may be frequesnt, they are far from guaranteed. If the contestant picks the consolation prize, and the other curtains contain the car and the goat, no curtain is revealed.
If you've watched the show enough, you know that, time permitting, Monty showed the other curtain when the expose/choice thing doesn't happen. Quite often, there was no goat (or other zonk item) behind any curtain.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 16th, 2010 at 7:27:41 AM
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I also have a problem with the paradox as presented. Consider the last lines from the page I linked:
What's to prevent him from switching to door number 2? Doesn't a 2/3 chance imply two equal choices?
Doesn't this explaination assume that the player has switched to door number 1?Quote:Player picks door 3 --> Host reveals goat behind door 2 --> player switches to door 1 --> player wins.
So by not switching the player has 1/3 chance of winning. By switching the player has a 2/3 chance of winning. So the player should definitely switch.
What's to prevent him from switching to door number 2? Doesn't a 2/3 chance imply two equal choices?
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 16th, 2010 at 7:29:34 AM
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Sorry about that link error, thanks for fixing it.
You're right about the original puzzle being incorrect in some of the details according to the way the real show was played, but mathematically it's still valid.
So since my question was flawed because I was using the wrong scenario, why not make my question a new scenario and work out the probabilities using the facts as they stood? It is not an intermediate deal but the final deal, there are two contestants and three doors, and the second contestant, Helen, is given the option to switch doors after the first contestant's (Jim) choice has already been revealed as not being the car.
Do both contestants have a 1/3 chance of picking the car at the beginning or are Helen's chances different than Jim's? If she were given the option to change her door once she knew that Jim did not win the grand prize, could she improve her chances of getting the car?
You're right about the original puzzle being incorrect in some of the details according to the way the real show was played, but mathematically it's still valid.
So since my question was flawed because I was using the wrong scenario, why not make my question a new scenario and work out the probabilities using the facts as they stood? It is not an intermediate deal but the final deal, there are two contestants and three doors, and the second contestant, Helen, is given the option to switch doors after the first contestant's (Jim) choice has already been revealed as not being the car.
Do both contestants have a 1/3 chance of picking the car at the beginning or are Helen's chances different than Jim's? If she were given the option to change her door once she knew that Jim did not win the grand prize, could she improve her chances of getting the car?
February 16th, 2010 at 8:56:40 AM
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Quote: tsmith
Do both contestants have a 1/3 chance of picking the car at the beginning or are Helen's chances different than Jim's? If she were given the option to change her door once she knew that Jim did not win the grand prize, could she improve her chances of getting the car?
Again, then my above math would work. She has a 1/2 chance of picking it only if Jim didn't, so it's 1/3. Her odds would NOT change by switching if Jim's door was revealed to be a failure. Her odds would now be 1/2. Remember, the variable that changes the odds in the original problem is Monty will ALWAYS open a door that does not have the car. That variable is not present in your problem.
I used this problem every semester when I was teaching, even though my class was not really a math class. It got the students minds' going.
February 16th, 2010 at 8:58:44 AM
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Quote: DJTeddyBearDoesn't this explaination assume that the player has switched to door number 1?
What's to prevent him from switching to door number 2? Doesn't a 2/3 chance imply two equal choices?
Door Number 2 has been opened by Monty and revealed as not a winner.
DJ, they did indeed do the problem as it's written (not by the original poster, but by the Wizard), though in the early days of the show, with only ONE contestant. Player picks door. (1/3 chance). Monty reveals loser (ALWAYS). Two doors left. Player should switch because he still only had a 1/3 chance he was right the first time, but 2/3 that he was WRONG the first time, so he should change. The way my students used to get it was to add more doors. Let's say there are 10 doors. You pick Door 5 (1/10 chance of being right). Then Monty shows you 8 losers, and only Door 2 is left. So what are the odds you were right the first time? 1/10. Should you switch? Obviously!
February 16th, 2010 at 9:32:06 AM
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That doesn't answer my questions.Quote: cclub79Door Number 2 has been opened by Monty and revealed as not a winner.Quote: DJTeddyBearDoesn't this explaination assume that the player has switched to door number 1?
What's to prevent him from switching to door number 2? Doesn't a 2/3 chance imply two equal choices?
What's to prevent him from picking door number 2?
and
Doesn't a 2/3 chance imply two equal choices?
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 16th, 2010 at 9:35:33 AM
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Quote: DJTeddyBearThat doesn't answer my question.Quote: cclub79Door Number 2 has been opened by Monty and revealed as not a winner.Quote: DJTeddyBearDoesn't this explaination assume that the player has switched to door number 1?
What's to prevent him from switching to door number 2? Doesn't a 2/3 chance imply two equal choices?
What's to prevent him from picking door number 2?
The rules of the game. It would be like choosing a category in Jeopardy that's already done. The door has been revealed to hide nothing behind it. I guess if you wanted to, you could say "I choose the $0 option and to lose the game after all of this effort!" but I don't find it probable enough to change the odds.
February 16th, 2010 at 10:00:26 AM
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The rules of the game prevent me from selecting the exposed door?
What if Monty exsposes the winning door?
Since the 'Rules of the game' specify that Monty will always reveal a loosing option, then the game itself always starts as a 50/50 chance that you will either pick the car or one of the goats.
After you make your selection, Monty reveals a goat, leaving you with two choices. It's 50/50.
---
Or to use your logic, based upon your assertion that I can't select the revealed door:
While it's true that I have a 2/3 chance of selecting a goat, once one door is revealed to contain a goat, one of those chances is removed. So one of my chances to have selected a goat is eliminated. So now I had only a 1/3 chance of selecting a goat. Does that mean I now have a 2/3 chance of having selected the car to begin with?
What if Monty exsposes the winning door?
Since the 'Rules of the game' specify that Monty will always reveal a loosing option, then the game itself always starts as a 50/50 chance that you will either pick the car or one of the goats.
After you make your selection, Monty reveals a goat, leaving you with two choices. It's 50/50.
---
Or to use your logic, based upon your assertion that I can't select the revealed door:
While it's true that I have a 2/3 chance of selecting a goat, once one door is revealed to contain a goat, one of those chances is removed. So one of my chances to have selected a goat is eliminated. So now I had only a 1/3 chance of selecting a goat. Does that mean I now have a 2/3 chance of having selected the car to begin with?
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 16th, 2010 at 10:13:52 AM
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Monty will NEVER reveal the winning door. It's an essential part of the puzzle. It says that either he knows where the car is, or a stagehand tells him which door to reveal.
It's still 1 in 3 when you first choose.
Here are the scenarios:
Let's say the car is behind Door number 3 and you are going to switch no matter what...
You choose Door 1. Monty reveals Door 2 is a loser. You change to Door 3 and win!
You choose Door 2. Monty reveals Door 1 is a loser. You change to Door 3 and win!
You choose Door 3. Monty reveals Door 1 is a loser. You change to Door 2 and lose.
Let's say the car is behind Door number 3 and you are going to stick no matter what...
You choose Door 1. Monty reveals Door 2 is a loser. You stick with Door 1 and lose.
You choose Door 2. Monty reveals Door 1 is a loser. You stick with Door 2 and lose.
You choose Door 3. Monty reveals Door 1 is a loser. You stick with Door 3 and WIN!
You can replace the numbers of the doors, but you'll get the same result every time. You win 2/3 of the time you switch. You'd only win 1/3 if you don't. The only variable not included is whether Monty reveals 1 or 2 when you actually did choose correctly, but it does not change the numbers.
It's still 1 in 3 when you first choose.
Here are the scenarios:
Let's say the car is behind Door number 3 and you are going to switch no matter what...
You choose Door 1. Monty reveals Door 2 is a loser. You change to Door 3 and win!
You choose Door 2. Monty reveals Door 1 is a loser. You change to Door 3 and win!
You choose Door 3. Monty reveals Door 1 is a loser. You change to Door 2 and lose.
Let's say the car is behind Door number 3 and you are going to stick no matter what...
You choose Door 1. Monty reveals Door 2 is a loser. You stick with Door 1 and lose.
You choose Door 2. Monty reveals Door 1 is a loser. You stick with Door 2 and lose.
You choose Door 3. Monty reveals Door 1 is a loser. You stick with Door 3 and WIN!
You can replace the numbers of the doors, but you'll get the same result every time. You win 2/3 of the time you switch. You'd only win 1/3 if you don't. The only variable not included is whether Monty reveals 1 or 2 when you actually did choose correctly, but it does not change the numbers.
February 16th, 2010 at 10:22:17 AM
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Ah, but there are actually SIX scenarios, assuming the car is behind door #3:
You choose Door 1. Monty reveals Door 2 is a loser.
You choose Door 1. Monty reveals Door 3 is a winner.
You choose Door 2. Monty reveals Door 1 is a loser.
You choose Door 2. Monty reveals Door 3 is a winner.
You choose Door 3. Monty reveals Door 1 is a loser.
You choose Door 3. Monty reveals Door 2 is a loser.
You choose Door 1. Monty reveals Door 2 is a loser.
You choose Door 1. Monty reveals Door 3 is a winner.
You choose Door 2. Monty reveals Door 1 is a loser.
You choose Door 2. Monty reveals Door 3 is a winner.
You choose Door 3. Monty reveals Door 1 is a loser.
You choose Door 3. Monty reveals Door 2 is a loser.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 16th, 2010 at 10:34:14 AM
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Quote: cclub79The only variable not included is whether Monty reveals 1 or 2 when you actually did choose correctly, but it does not change the numbers.
yes it does
If the host only gives you the option of changing when you are right, it is not correct to switch
Its correct to switch with 3 assuptions
1. The host knows which door is correct
2. the host will give you the option to switch every time
3. the host will never open the door with the car
February 16th, 2010 at 10:58:12 AM
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Quote: BigsoonerQuote: cclub79The only variable not included is whether Monty reveals 1 or 2 when you actually did choose correctly, but it does not change the numbers.
yes it does
If the host only gives you the option of changing when you are right, it is not correct to switch
Its correct to switch with 3 assuptions
1. The host knows which door is correct
2. the host will give you the option to switch every time
3. the host will never open the door with the car
Correct, but those 3 assumptions are vital parts of the puzzle, which means it's always correct to switch.
February 16th, 2010 at 10:59:13 AM
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Quote: DJTeddyBearAh, but there are actually SIX scenarios, assuming the car is behind door #3:
You choose Door 1. Monty reveals Door 2 is a loser.
You choose Door 1. Monty reveals Door 3 is a winner.
You choose Door 2. Monty reveals Door 1 is a loser.
You choose Door 2. Monty reveals Door 3 is a winner.
You choose Door 3. Monty reveals Door 1 is a loser.
You choose Door 3. Monty reveals Door 2 is a loser.
2 & 4 never occur. It's just part of the puzzle that he doesn't reveal the winner.
February 16th, 2010 at 11:05:55 AM
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If it helps, think of it like this: After choosing a door, you are offered the option to trade the contents of the door you chose for the contents of BOTH other doors. This is, in essence, what your choice boils down to.
If you do not switch, you will win the car 1/3 of the time, and a goat 2/3 of the time.
If you do switch, you will win two goats 1/3 of the time, and a goat+car 2/3 of the time.
Assuming the value of a goat is $0 (sorry goats and goat-lovers), switching has the greater expected value.
I recommend the Wikipedia article for more on this.
And if you like things like this, I recommend one of my all-time favorites: the Two envelopes problem.
If you do not switch, you will win the car 1/3 of the time, and a goat 2/3 of the time.
If you do switch, you will win two goats 1/3 of the time, and a goat+car 2/3 of the time.
Assuming the value of a goat is $0 (sorry goats and goat-lovers), switching has the greater expected value.
I recommend the Wikipedia article for more on this.
And if you like things like this, I recommend one of my all-time favorites: the Two envelopes problem.
February 16th, 2010 at 11:09:37 AM
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Thank you for the cover, JB. I understand the difficulty to grasp this problem. It took me awhile and it took many of my students a long time as well.
February 16th, 2010 at 11:20:30 AM
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Quote: cclub79Quote: DJTeddyBearAh, but there are actually SIX scenarios, assuming the car is behind door #3:
You choose Door 1. Monty reveals Door 2 is a loser.
You choose Door 1. Monty reveals Door 3 is a winner.
You choose Door 2. Monty reveals Door 1 is a loser.
You choose Door 2. Monty reveals Door 3 is a winner.
You choose Door 3. Monty reveals Door 1 is a loser.
You choose Door 3. Monty reveals Door 2 is a loser.
2 & 4 never occur. It's just part of the puzzle that he doesn't reveal the winner.
EXACTLY!
So there are FOUR possible scenarios. 2/4 are winners, 2/4 are losers. It's a 50/50 proposition!
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 16th, 2010 at 11:23:50 AM
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You should check out the wikipedia article, there is a section called "Why the probability is not 1/2" But again, it fools a lot of people:
>>>>>>>
When first presented with the Monty Hall problem an overwhelming majority of people assume that each door has an equal probability and conclude that switching does not matter (Mueser and Granberg, 1999). Out of 228 subjects in one study, only 13% chose to switch (Granberg and Brown, 1995:713). In her book The Power of Logical Thinking, vos Savant (1996:15) quotes cognitive psychologist Massimo Piattelli-Palmarini as saying "... no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer."
<<<<<<<<<<<
>>>>>>>
When first presented with the Monty Hall problem an overwhelming majority of people assume that each door has an equal probability and conclude that switching does not matter (Mueser and Granberg, 1999). Out of 228 subjects in one study, only 13% chose to switch (Granberg and Brown, 1995:713). In her book The Power of Logical Thinking, vos Savant (1996:15) quotes cognitive psychologist Massimo Piattelli-Palmarini as saying "... no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer."
<<<<<<<<<<<
February 16th, 2010 at 11:26:59 AM
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Quote: DJTeddyBearQuote: cclub79Quote: DJTeddyBearAh, but there are actually SIX scenarios, assuming the car is behind door #3:
You choose Door 1. Monty reveals Door 2 is a loser.
You choose Door 1. Monty reveals Door 3 is a winner.
You choose Door 2. Monty reveals Door 1 is a loser.
You choose Door 2. Monty reveals Door 3 is a winner.
You choose Door 3. Monty reveals Door 1 is a loser.
You choose Door 3. Monty reveals Door 2 is a loser.
2 & 4 never occur. It's just part of the puzzle that he doesn't reveal the winner.
EXACTLY!
So there are FOUR possible scenarios. 2/4 are winners, 2/4 are losers. It's a 50/50 proposition!
You are using the fact that there are 4 possible scenarios to mean all 4 are equally likely to occur. The first 2 occur 2/3 of the time (when you didn't pick the winner originally), and the second 2 occur 1/3 of the time (when you did pick the winner originally)
February 16th, 2010 at 11:34:05 AM
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I got it.
Thanks to a paragraph in the Wikepedia article JB linked. Here's the key parts:
Thanks to a paragraph in the Wikepedia article JB linked. Here's the key parts:
Sorry, CClub. Your student example of 10 doors wasn't enough to get thru this thick noggin.Quote:Increasing the number of doors
It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three. The player picks a door. The game host then opens 999,998 of the other doors revealing 999,998 goats.
Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 16th, 2010 at 11:41:55 AM
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10 just fit on the blackboard....I assure you I was even more unwilling to see the light when I first found the problem. No apology is necessary. I miss sharing this with my students.
February 16th, 2010 at 12:05:08 PM
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One reason I was stubborn is because I like math. And I was applying Algebra. The 1/3 vs 2/3 remains unchainged argument made no sense. I had all these proofs. The 'duh' moment eluded me.
I was even going to argue that if you were given the choice to switch BEFORE being shown the contents of one door, that it would still resolve to a 50/50, since you KNEW that at least one of those doors had the goat...and that would have held true for your 10 door student example.
Then I read that thing about the million doors.
Did I mention that I had a thick noggin?
I was even going to argue that if you were given the choice to switch BEFORE being shown the contents of one door, that it would still resolve to a 50/50, since you KNEW that at least one of those doors had the goat...and that would have held true for your 10 door student example.
Then I read that thing about the million doors.
Did I mention that I had a thick noggin?
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 16th, 2010 at 12:18:39 PM
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It has been a while since I have written about this problem. However, it seems I'm arriving late, and the question has been resolved. Thanks to those who helped.
I did take notice that the way I framed the question on my site did not match the show. After watching about six old episodes on YouTube, what actually happened is Monty picked two contestants (in early shows I think he only picked one) and they each chose a door. Then Monty opened the doors one at a time, always opening the door with the best prize LAST. However, he did not offer a chance to switch. The way the question should be framed is what should the contestant do if Monty opens one door first (revealing a lesser prize as usual), and then offers a chance to switch. Indeed, at this point, the contestant should switch. As was mentioned already, it is essential to understand he always opens a losing door first. I went back and reworded that question on my site, by the way.
I did take notice that the way I framed the question on my site did not match the show. After watching about six old episodes on YouTube, what actually happened is Monty picked two contestants (in early shows I think he only picked one) and they each chose a door. Then Monty opened the doors one at a time, always opening the door with the best prize LAST. However, he did not offer a chance to switch. The way the question should be framed is what should the contestant do if Monty opens one door first (revealing a lesser prize as usual), and then offers a chance to switch. Indeed, at this point, the contestant should switch. As was mentioned already, it is essential to understand he always opens a losing door first. I went back and reworded that question on my site, by the way.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
February 16th, 2010 at 12:34:22 PM
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But all this has not answered my original question, which is, are the second contestant's odds the same as the first contestant's, i.e., 1/3, if the first contestant has taken a door out of play before the second contestant gets to pick his door?
February 16th, 2010 at 12:40:22 PM
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Quote: tsmithBut all this has not answered my original question, which is, are the second contestant's odds the same as the first contestant's, i.e., 1/3, if the first contestant has taken a door out of play before the second contestant gets to pick his door?
Actually I answered it twice, with the equation:
There are 3 possibile scenarios for the 2nd contestant:
1/3 of the time the 1st contestant will pick the winner, meaning the 2nd contestant has a 0% chance
2/3 of the time the 1st contestant will be wrong, leaving 2 possible choices, of which one is correct --> 50% chance
(1/3*0) + (2/3*.5) = 1/3
The second contestant still has a 1/3 chance of winning.
February 16th, 2010 at 12:55:28 PM
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Sorry, I missed them. Okay, I think I understand, thanks.
February 16th, 2010 at 12:58:31 PM
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Not exactly the same thing, but in our office, we had a similar discussion regarding the Secret Santa drawings.
The rule was, everybody take a name, but NOBODY look until everyone has chosen. Why? Because that way, everyone has an equal chance of getting anyone.
I didn't really understand that until now.
The rule was, everybody take a name, but NOBODY look until everyone has chosen. Why? Because that way, everyone has an equal chance of getting anyone.
I didn't really understand that until now.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
February 16th, 2010 at 1:04:36 PM
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Ahhh, but what if, like Kevin on The Office, you get your own name? We used to have numbered cards and we'd each take one, making sure not to take our own.
February 16th, 2010 at 2:21:06 PM
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Quote: cclub79Ahhh, but what if, like Kevin on The Office, you get your own name? We used to have numbered cards and we'd each take one, making sure not to take our own.
That is my second favorite episode, second only to "Diversity Day." I liked that one so much, I bought the domain diversitytomorrow.com, but have not done anything with it yet. My third favorite is Casino Night, perhaps not surprisingly.
A good math problem is what is the probability nobody gets his/her own name. As the number of people gets larger, the probability approaches 1/e.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
February 16th, 2010 at 2:34:00 PM
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With the risk of turning this into an Office thread, I'd say the 2nd and 3rd seasons were the most consistantly good, but my favorite episode was the one that aired after the Super Bowl. But the Secret Santa one has a special place in my heart, because it was the first one I saw when I knew I loved the show.
In "Casino Night", there were a couple of casino errors I noticed, one blatant, the other less so. Anyone who's a fan of the show notice them? It's surprising for a show that has about as much respect for continuity and accuracy as I've seen in some time.
In "Casino Night", there were a couple of casino errors I noticed, one blatant, the other less so. Anyone who's a fan of the show notice them? It's surprising for a show that has about as much respect for continuity and accuracy as I've seen in some time.
February 16th, 2010 at 4:52:45 PM
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Sorry to contribute to this getting off topic, but I love The Office. That and Lost are the only two fictional shows I watch. Any real casino night is bound to have lots of errors anyway. One error I recall is when Toby put Michael all in before the flop. With any cards, Michael could still have won. However, the dealer gave Toby the pot without ever even doing a flop.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
February 16th, 2010 at 5:06:11 PM
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I am in the exact same boat as you. I was just telling someone how Lost and The Office are my only two "must watch" shows at present. I like to have sports on most of the time because you can get other work done while that's on!
Yes, the Michael All-in was sloppy on their part, and that was the big error that always makes me mad when I watch that episode. There was also a scene with Michael at the Craps table where he didn't make his point, but didn't roll a Seven, and everyone was all dejected as if they lost...when in fact, they just didn't resolve the bet yet.
Yes, the Michael All-in was sloppy on their part, and that was the big error that always makes me mad when I watch that episode. There was also a scene with Michael at the Craps table where he didn't make his point, but didn't roll a Seven, and everyone was all dejected as if they lost...when in fact, they just didn't resolve the bet yet.
February 17th, 2010 at 2:27:17 PM
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Quote: cclub79I am in the exact same boat as you. I was just telling someone how Lost and The Office are my only two "must watch" shows at present. I like to have sports on most of the time because you can get other work done while that's on!
Yes, the Michael All-in was sloppy on their part, and that was the big error that always makes me mad when I watch that episode. There was also a scene with Michael at the Craps table where he didn't make his point, but didn't roll a Seven, and everyone was all dejected as if they lost...when in fact, they just didn't resolve the bet yet.
Great minds think alike! I should have added Curb Your Enthusiasm to the list, but I dropped HBO from my cable service, so have lost touch with the show. Eventually the latest season will appear on Netflix.
Good catch on the craps game, I never noticed that. Lost is a great show. I think the seasons 4 and 5 were too rushed, and at times confusing, but still well worth watching. The episode last night I thought was the best so far of season 6. However, if they have a Sopranos type ending, I'm going to be raging mad, and I don't think I'll be the only one.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
February 17th, 2010 at 3:48:24 PM
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Quote: WizardGreat minds think alike!
Well, if we're allowed to drift off topic... my favorite shows ...
#1 Family Guy
#2 The Office
#3 30 Rock
Both Dexter and Curb you Enthusiasm would be on my list if I got those channels. I have "basic basic cable," which means that you can't get fewer channels and still be a cable subscriber.
--Dorothy
"Who would have thought a good little girl like you could destroy my beautiful wickedness!"
