It's so hard to come up with an explanation of this that seems intuitively correct, so here's my best stab at it. The important thing is that the host knows where the goats are and is careful to remove only a goat -- it's not a random choice on his part, he made sure to leave you the car (if you didn't pick it already). That's why the odds change. This is easier to see if you imagine 99 goats and 1 car: the odds are overwhelmingly strong that you will initially pick wrong, but if the host then removes 98 goats, the odds are really really good that the one option he left is the car.
Start by assuming the host dosnt know where the car is and also has to make a random choice. If you unknowingly chose the car and he makes a random choice he is 100 percent likely to pick a goat. If you unknowingly chose a goat and he makes a random choice he is 50 percent likely to pick a goat. So the import question becomes what are the odds we both randomly picked a goat and the smart player stays put. Next if we assume the host knows where the car is and is only interested in prolonging the game as much as possible he was always going to turn over 1 of the goats. If you unknowingly pick the car he has to leave a goat, if you unknowingly pick a goat he has to leave the car. Look at it this way, if either one of his cards is the winner he has to leave it on the table for you to pick. You are effectively swapping your 1 card for his 2 cards. So by picking the 1 the host left your odds of getting the car are 66.6 to 33.3 in your favor so the smart player swaps his card. Depending on what assumptions you make as to what the answer is. In the show it was never made clear if the host had to show one of the cards and give a second chance or if it was his decision to make after the player makes his first choice. Then the odds swing back and forth depending on the hosts goals. Yes it could have been made a little clearer but it was a good show and the math is sound.
It is, and it made me finally understand the problem. I've heard the mathematical explanation of the problem too, and then I didn't understand why you had a better chance to win if you switched door only that it was better. So this is a good and simple explanation. Which I love!
I totally get it now! I was so confused; I was going to look this up, until you explained it. They should have used your explanation in the show -- Thanks!
Tv Announcer: We now return to “Numb3rs” starring David Krumholtz as Charlie Eppes (Peter Griffin pulls out a gun and angrily shoots the TV) (BANG! BANG! BANG! BANG!) Chris: Why do you hate that show much? Peter: Because that show prevented Bernard from appearing in The Santa Clause 3: The Escape Clause
Look at it this way. You have 100 doors with 99 goats and 1 car. Well if you pick one door and you don't know which is the car, then of course you have 99% of failing. However, if Monty opens 98 doors with all of them goats, with only one door closed, would you stay with your 99% of losing, or switch? It's the same thing, you win by switching, and thank you Monty for opening 98 doors of goats.
The show makes it seem a little like you could pick the winning card, but it's really an explanation of probability. With only 3 doors, it's easy to NOT see how your odds aren't changing, and are really staying at 1/3 for your initial pick and 2/3 for choosing the other option (set of remaining doors).
Say you choose door 1. The odds that door 1 has the car are 33.3%. Therefore the odds of the car being behind door 2 OR 3 are 66.6%. When the host opens one of the two remaining doors (he will never open the door you chose) it doesn't change the fact that there is a 66.6% chance that you chose wrong the first time. So since we know that there is a 66.6% chance that the car is either behind door 2 or 3, when the host eliminates a door (let's say door 3) there is still a 66.6% chance that the car is behind either door 2 OR door 3. We know there is a 66.6% chance that we chose wrong the first time so door 2 and door 3 still must add up to 66.6%. But now we know that there is a 0% chance that it is behind door 3 so in order for door 2 and door 3 to add up to 66.6% door 2 must equal 66.6%. Therefore, door 1 still equals 33.3%, door 2 now equals 66.6% and door 3 equals 0%.
If the host makes a random choice and reveals a goat then you have to split the odds between the remaining cards. Look at it this way, if the host had turned over the car on the first try the odds from both of the other cards go to zero and the odds of the card he turned over go to 100 percent. So if he randomly turned over a goat it goes in reverse and both cards get half of 33.3 We have to make 2 assumptions for the math to work the way it is shown. First that the Host knows were the car is and second that he wants the game to go as long as possible. In that instance even if he has the winner he will not turn it over. Because of that the odds from the card he turned over should be added to the one both of you did not chose instead of splitting it between them. It is implied multiple times that he knows were the car is. First why would a math expert setup a math demonstration where the odds are greatly against him. If we assume he dos not know were the car is he has to beat the odds multiple times. First he is counting on you not to pick the car, second he has to get lucky and pick one of the goats then he has to beat the odds one last time and pick the car. In this instance he has to get lucky 3 times in a row. When he asks for a volunteer we see he wants more and yet he still picks one of the people who volunteered when he first asks. Next consider he has 2 red x in his hands implying he might need two for the demonstration to work out. Lastly he gives the final clue when he talks about "in this class". Meaning its a setup.
conobs The host does not make a random choice though. That's the point. The host knows where the car is and will only ever reveal a door with a goat. That is stated in the Monty Hall Problem. If the host chose randomly and revealed a door with the car the contestant could not possibly win because the contestant is only allowed to either stay with the door he chose or switch to the door that has not been revealed.
Yes i agree completely but several people were missing it so i attempted to cover it in such detail anyone would understand it. Which judging by the number of folks that claimed 50/50 was needed.
That's really cool! Assume that the game show host will ALWAYS eliminate a goat. He wouldn't want to accidentally reveal the prize before the person has a chance to change their answer. So, if you initially: #1: Picked Goat A. The host reveals the other goat. If you switched now, you'd WIN! #2: Picked Goat B. The host reveals the other goat. If you switched now, you'd WIN! #3: Picked the car. The host reveals a goat. If you switched now, you'd LOSE! 2 out of 3 scenarios WIN by switching. :)
Actually it's quite a good explanation. The only thing he doesn't mention is the fact that its actually the quizmaster that is raising the odds by pointing a goat out. But perhaps that just makes it more confusing :). Actually I KNOW this is right, because I came across this many times during my 6 years of math studies at university. The idea is mainly used for scientific entertainment and math promotion. For instance in introduction classes at university or in popular sience books.
We did this in my cognitive psych class, only it was used as an example of reasoning (sorta like Charlie did). The class had about 300 people in it and I'd say more than half were puzzled when the prof explained it in simple terms, lol.
Ok, I'm not going to say I'm an expert on this. I did believe at first that the odds changed to 1/2 after the reveal of the first goat. But looking at a 'simple solution' spreadsheet, you get to see the overall chances do increase to 2/3 if you switch doors after the first reveal. (Found on wikipedia under 'Monty Hall problem'.) I think what confuses people is that, in effect, the revealed door is still being (and should be) counted towards the percentage. However, after the reveal 'the odds for the two sets (the player's pick and the other two doors) don't change but the odds move to 0 for the open door and 2/3 closed door.' If I'm correct (which I may not be), when people think of the door as revealed, they also think of it as eliminated from the percentage. Along those lines, there are two doors left, and your chances would be 1/2. I think (once again, I might be wrong), it's similar (but not the same) to comparing the overall chance of getting Tails twice in a row on tossing a coin (1/4), as opposed to the chances of getting Tails on each of two tosses (1/2 on the first toss, 1/2 on the second toss). In the case with the goats and car, keeping the third door in the equation is actually more beneficial than eliminating it from the equation entirely. And no, I do not think it was explained well on Numb3rs at all. But there is a 2/3 chance the next episode will explain math concepts better...
No, the odds only become (2/3) when bias (goats only) is applied to revealing a card. If a card is revealed at random and it happens to be a goat, as in this case, then the remaining 2 cards have equal chance of being the car. The REASON for switching DOES change depending on the hosts selection bias. Imagine the problem with a million goats. The host removes 999999 goats on purpose. The card left is likely the car. Removing 999999 cards at random (goats by chance) leaves 2 equal cards.
The religious explanation for this, is you have three choices: Heaven, Hell and Worldly Gains... behind 3 doors. You chose the first, but it was revealed that it leads to Hell; So whatever you choose next, the best that you'll really get during your existence! So, what you'll need someone to reveal what's in the first door, so you'll get things right the first time... an Imam, priest or whatever that have clues about the choice... Or, choose the best on the second try, making the change if needed to gain Heaven on second try; and it's also good to have someone to guide you at this stage... But, if you fail to make the right choice, you can be content by living the full life full of worldly gains... That is, if you don't harm yourself in the first place!
To better explain it, imagine 100 doors. You pick door 1. Door 1 has a 1% chance. The host, (following the rules above) opens all of the doors except yours and door 47 (for example). Your odds for staying with door 1 remain 1%, while switching increases it to 99%, since there was a 99% chance that it was behind one of the other 99 doors to begin with. The fact that he shows you doors that don't have cars behind them doesn't change the odds, which is all you are working with really anyway.
This example was used in 21, and they got it wrong there too. You HAVE to state that the game show host MUST reveal a door, that isn't yours, with a goat behind it for this to be true. If the hosts behavior is random (he can reveal any door, yours with a goat, or the car), then your odds are 50/50. Only when you apply the requirements do your odds go to 66%. Because the hosts behavior is not stated to always follow the pattern it followed this time, the answer is still 50/50.
once you state that the host can't open a door with a car, and can't open your door, then your options go from 4 possible 2nd round random scenarios, with 2 winners (50%), to 6 possible 2nd round random scenarios, with 4 winners (66%).
I'll try to explain this using doors instead of cards. There are two things you need to know to understand why this makes sense. 1) The host KNOWS which one has the car. 2) The host will open ALL of the doors except the one you picked and the one with the car behind it. (In the event you picked the one with the car, he will leave a door closed at random.) (Note that with only 3 doors, the host will always open only 1 door, which is where it probably throws people off.)
It only becomes confusing because there are only three doors. Think of them as sets. You pick door 1 (a set with one item) as it has as just as likely a chance as any other. Given the option to switch over to the set of doors 2 AND 3, (if either has the car, you win) then you of course select that set. When he opens one of the doors, your odds of having initially picked the right door is still 1 in 3. (How could it change to 1/2?)
See, the whole point of the game is to increase your odds of winning, and by switching your answer you'd definitely increase your odds of winning. Think about playing this game with 100 cards (where there is 1 car and 99 goats). You pick any one of the 100 cards. The host then reveals 98 goats, leaving the card you initially chose and one other card left unturned. Would you be wise to switch your answer now? Hell yes! :)
it's mathematically correct IFF the 'host' knows what is behind each card. Otherwise, if you're picking the second card at random, there's 1/3 chance that you win, a 1/3 chance that you lose, and a 1/3 chance that the test is invalid and fails (the 2nd card flipped over is the car)
Behind the other 2 are goatse. You don't want that. In both this and 21 they neglect the most important piece of information: the fact that host has to reveal a goat. That's important because if you pick the car on the first time then the host has 2 doors to choose from and you lose from switching, BUT if you pick one of the goats then the host shows you the other goat and switching gives you the car. So picking a goat the first time always wins, which means you have a 2/3 chance of winning.
The Matrix explanation: I am the Architect. I created this Problem. I've been waiting for you. You have many questions, and although the process has altered your consciousness, you remain irrevocably human. Ergo, some of my answers you will understand, and some of them you will not. Concordantly, while your first question may be the most pertinent, you may or may not realize it is also the most irrelevant. Neo: Where is the car? A: Interesting. That was quicker than the others.
You have a deck of cards. Your host knows where the ace of clubs is. You pick a card. Your host will necessarily reveal 50 cards from their deck that aren't ace of clubs. His remaining card will 51 out of 52 times be an ace of clubs. This goat/car thing simply uses a much smaller deck of 3 cards: two goats and a car.
It relies on just probability only, not actually getting it right. If it were 100 choices, you would rather pick 99 of the choices rather than just 1 as your odds would be 99% of getting it right. In this case your being allow to go from 1 choice to 2 choices. (You already know that one of the two will not have a car behind it, so seeing that fact doesn't change the odds any)
@Selahrs1 what difference does it make if it is a conscious choice on the host's part though? (sry if the answer is obvious but maths has never been my forte)
There are only 2 options -- stay or swap. There are not three options -- stay, swap, or pick the open door. It's a 1 out of 2 shot. Suppose you're allowed to select two doors, and the host opens a random bad door. Regardless of whether one of your doors was the prize, you still only have the chance to pick between 2 doors.
we will ALWAYS be presented with the choice to switch or not switch regardless if we initially pick the car or not, so think of these as "strategies" to play this game, and analyze each one. if our strategy is to not switch, this strategy is equivalent (exactly the same) to initially trying to pick the door with the car, which we have 1/3 chance of doing. if our strategy is to switch, this is equivalent to trying to pick a door that does NOT have the car which we have 2/3 chance of doing.
For those who don't believe it, try it yourself with some playing cards. Put 3 rows of cards down and lay it out in a grid where you have a "winning" card in each row in a different column. Then, choose a column as your "pick" and eliminate one of the cards that isn't a winner from each row. You see that by the original row you selected has a 1/3 to win still, and the other column (the one you are offered a switch to) has a 2/3 chance of having the winning card.
Assuming it was 2/3 goats, you have a 66% chance of picking a goat if you were to choose from 3 cards. WIth that in mind, if the host reveals one goat and leaves you with the chance to switch, statistically speaking, you were more likely to pick a goat from the beginning, so holding this original assumption, it's likely that you still have a goat now, thus, it is better to switch. Of course, this isn't 100% thus, sometimes you may be wrong, but in practice, switching will give you the win.
No, afterwards he says "Now knowing what you know, do you want to switch your choice". This negates the need to state in advance that he knows where the car is since the reason for switching remain the same.
it was more probable that he would pick the car, and it better illustrates his point, of course he's right, if you add in the fact that the host knows what is behind all 3 doors it can serve as a test, when presented to a person who is very smart, to determine whether they would react to a problem emotionally or using their math skills, as it was used in 21
If he shows you A MILLION doors. You pick door #18. The host opens 999,998 doors, leaving you with door #18 and door #134,523. Do you want to switch now? Obviously, yes. The chances you picked the right door out of a million was 1/1,000,000. Switching now will win you the car 999,999 times out of a million. It's the exact same problem, just re-stated a different way to make it more obvious that switching is the correct play.
Numb3rs was wrong. 21 was right. "I'm going to reveal to you one of the cards you did not choose" is NOT the same as a game show host's saying "I'm going to give you a little help and eliminate one of the wrong answers." That makes all the difference.
Agreed. If the choice is slated to be random, then it's a matter of pure equivalent statistics, meaning your choice ultimately does not matter. However, if the choice is already known, then it becomes a logical deduction puzzle. The very premise of the puzzle depends on the host's knowledge and intentions. If the host knows, then it's a logical deduction puzzle. If the host doesn't know, then it's just basic statistics which necessitate a change in statistics to accommodate a change in the equation
As made clear by Marilyn Vos Savant, the columnist who popularized the "The Monty Hall Dilemma" - it is ESSENTIAL that the game show host KNOWS which card conceals the valuable prize, and that he consciously chooses to reveal a goat card. If the host just randomly happens to reveal a goat behind a card, the contestant will gain no advantage by switching away from her original choice. The "Numbers" tech advisors got it wrong again by failing to make this aspect of the Monty Hall dilemma explicit.
So basically to win the game, you need to select a goat in your first pick, that way when the host opens the other door with the goat, you can switch to win the car. And the chances of you picking a goat in your first pick is 2/3 hence its better to switch.
@jacobins3000 Try to understand this way: Suppose you are in a game show. There are 100 cards you can choose, one of them has a prize of 1 million dollars. You choose one of them. The host keeps on intentionally removing the cards, KNOWING THAT THEY AREN'T THE PRIZED ONES. Then it comes down to 2 cards. Yours and one random. It isn't 1/2. You chose one of 100. He KNOWS what cards he is taking away. So the odds dont change. It is like he didnt take any away. It doesnt make any difference.
Think about this: there are a 1000 choices (999 goats and 1 car), you pick 1 "door" then the host reveals 998 goats, we're now left with your choice and a blank. There's a 99.9% chance that your choice is a goat since there were 999 goats out of a 1000 choices, so therefore you would easily switch and win the car almost every time.
ooo strike one, you get another chance! actually i've never seen a full episode of Numb3rs. you can try and: (1) guess which top 30 university i attend, (2) which top 5 semiconductor company i intern at, (3) which master's program in electrical engineering i got into [this is easy though, it's the same as #1]. care to take another shot at my "looser jobless life"? honestly, i'm just passing time while i wait for tests to run...
Your odds are only 50/50 if you think of it in sets. 50% of the time the car will be behind set one (door 1) and 50% of the time it will be behind set 2 (doors 2 or 3)
Neo: Is this still about the car? A: There are two doors. The door to your right leads to the Car. The door to your left leads to the Goat. As you adequately put, the problem is choice. But we already know what you are going to do, don't we? Already I can see the chain reaction: the chemical precursors that signal the onset of an emotion, designed specifically to overwhelm logic and reason. An emotion that is already blinding you to the simple and obvious truth: There is no car. Just a goat.
1/3 by switching its now 2/3 it increases the chance of getting the car. but also the sheep too. so its luck. but if you have more than 3 cards, this theory will prove useful it alot of situations of choosing.
I think he was miscommunicating, almost a version of the myth gamblers buy into, where if the roulette wheel comes up red 10 times, it is more probable that the next spin is black. It's a myth, the IMPROBABLE event (10 reds) already happened, it's still 50:50 next spin. She made one of the 2 losing choices. Now that card #3 is revealed, the probability is no longer 1 in 3.
I'm not sure what you are trying to say, vskarate. Are you trying to tack on a fourth scenario (so that there is one for each "wrong" answer the contestant could switch to?). If that's what you're implying, it doesn't quite work that way. It doesn't matter which goat he switches to if he initially picks the car... all that matters is that he would lose. He only has 3 choices initially, and 2 of those three choices will benefit him later if he changes his answer. Does that make more sense?
I really really love Numb3ers, but i think he didn't explain the problem fully. He should have said immediately that the gameshowhost exactly KNOWS where the 2 goats are. So the gameshowhost MUST pick a goat out of the 2 remaining boxes. So if you picked a goat at the beginning (chance 2/3), the host has to eliminate the only other goat. If you switch then u got a 100% chance of getting it. Making it a total chance of 2/3 * 1/1 = 66%
This seems to be an attempt to recreate the famous Monty Hall puzzle. Because he said "reveal to you one of the cards you did not choose", we must assume he chooses to reveal one of the 2 cards at ramdom. If this assumption is correct then he is WRONG. Odds are 50/50. However if he meant to say "I will reveal one of the goats that you did not choose" then he is correct and you should switch your choice to get better odds of winning.
Choose Goat 1, Goat 2 is revealed, switch and you get a car. Choose Goat 2, Goat 1 is revealed, switch and you get a car. Choose Car, Goat 1 or 2 is revealed, switch and you get other goat. You win 2 out of 3 times if you switch.
Try imagining doing this millions of times. You always pick and open door 1 and one of the other two that doesn't have a car is reveled. You'll see that you only win 1/3 of the time, not half of the time.
Of course he has to reveal a goat and not the car you don't even need that in an explanation because if the host reveals the car then you'd of course know where the car is. Right?
Deciding to switch after one card is revealed only raises your chances, it does not guarantee the outcome. The outcome can be good with less chances, or bad with more chances. You can choose to play a game with 1 chance out of 100 to win, and win it. You can choose to play a game with 99 chances out of 100 to win, and lose it. But it's wiser to choose the easiest game.
It LOOKS like 50/50 on the surface, but it is not. That's what makes this equation so fun. Look up "Monty Hall Problem" on the Wikipedia. They have excellent diagrams, and even an in-depth explanation for why it isn't 50/50, despite the fact that the person is choosing between two cards in the end. It's like a magic trick... you have to look deeper than just the surface. BTW, I've only seen 3 episodes of Numb3rs. The fact that this was aired on a TV show has no relevance to me. ;)
In 500 words...I'll have a go: You pick a card. there is a 2in3 chance you have picked a goat and a 1in3 chance that you have picked the car. Of course you don't know which yet. Let's assume that you picked the goat. The host turns over the other goat. Of course you must change your pick to win the car. Assume you picked the car. The host turns over a goat. In this case you must stick to win the car. So in 2 out of 3 situations changing will win the car.
It's so hard to come up with an explanation of this that seems intuitively correct, so here's my best stab at it. The important thing is that the host knows where the goats are and is careful to remove only a goat -- it's not a random choice on his part, he made sure to leave you the car (if you didn't pick it already). That's why the odds change.
This is easier to see if you imagine 99 goats and 1 car: the odds are overwhelmingly strong that you will initially pick wrong, but if the host then removes 98 goats, the odds are really really good that the one option he left is the car.
The real question is who would REALLY choose a Ford Capri over a goat?
Plrododos
The monty hall problem! I love it!
I remember when I first saw this in a statistics textbook and was completly blown away.
BONE!!!!!!!
ahhh finally, this is the first explanation of the monty hall problem i actually understand!!!!
True. It's always interesting getting a reply two years later lol. I was 13 when I posted that.
Very nice show.
Start by assuming the host dosnt know where the car is and also has to make a random choice.
If you unknowingly chose the car and he makes a random choice he is 100 percent likely to pick a goat.
If you unknowingly chose a goat and he makes a random choice he is 50 percent likely to pick a goat.
So the import question becomes what are the odds we both randomly picked a goat and the smart player stays put.
Next if we assume the host knows where the car is and is only interested in prolonging the game as much as possible he was always going to turn over 1 of the goats. If you unknowingly pick the car he has to leave a goat, if you unknowingly pick a goat he has to leave the car.
Look at it this way, if either one of his cards is the winner he has to leave it on the table for you to pick. You are effectively swapping your 1 card for his 2 cards. So by picking the 1 the host left your odds of getting the car are 66.6 to 33.3 in your favor so the smart player swaps his card.
Depending on what assumptions you make as to what the answer is.
In the show it was never made clear if the host had to show one of the cards and give a second chance or if it was his decision to make after the player makes his first choice. Then the odds swing back and forth depending on the hosts goals.
Yes it could have been made a little clearer but it was a good show and the math is sound.
It is, and it made me finally understand the problem. I've heard the mathematical explanation of the problem too, and then I didn't understand why you had a better chance to win if you switched door only that it was better. So this is a good and simple explanation. Which I love!
I totally get it now! I was so confused; I was going to look this up, until you explained it. They should have used your explanation in the show -- Thanks!
Tv Announcer: We now return to “Numb3rs” starring David Krumholtz as Charlie Eppes
(Peter Griffin pulls out a gun and angrily shoots the TV)
(BANG! BANG! BANG! BANG!)
Chris: Why do you hate that show much?
Peter: Because that show prevented Bernard from appearing in The Santa Clause 3: The Escape Clause
DAMN IT I WANT A GOAT!.
Look at it this way. You have 100 doors with 99 goats and 1 car. Well if you pick one door and you don't know which is the car, then of course you have 99% of failing. However, if Monty opens 98 doors with all of them goats, with only one door closed, would you stay with your 99% of losing, or switch? It's the same thing, you win by switching, and thank you Monty for opening 98 doors of goats.
My stats teacher explained this too. Except his explanation was really good, and I got totally enlightened.
The show makes it seem a little like you could pick the winning card, but it's really an explanation of probability. With only 3 doors, it's easy to NOT see how your odds aren't changing, and are really staying at 1/3 for your initial pick and 2/3 for choosing the other option (set of remaining doors).
we did this in math class. I can't believe it actually worked. It's hard to grasp the concept at first. I did get it eventually.
Say you choose door 1. The odds that door 1 has the car are 33.3%. Therefore the odds of the car being behind door 2 OR 3 are 66.6%. When the host opens one of the two remaining doors (he will never open the door you chose) it doesn't change the fact that there is a 66.6% chance that you chose wrong the first time. So since we know that there is a 66.6% chance that the car is either behind door 2 or 3, when the host eliminates a door (let's say door 3) there is still a 66.6% chance that the car is behind either door 2 OR door 3. We know there is a 66.6% chance that we chose wrong the first time so door 2 and door 3 still must add up to 66.6%. But now we know that there is a 0% chance that it is behind door 3 so in order for door 2 and door 3 to add up to 66.6% door 2 must equal 66.6%. Therefore, door 1 still equals 33.3%, door 2 now equals 66.6% and door 3 equals 0%.
Well said, and correct.
If the host makes a random choice and reveals a goat then you have to split the odds between the remaining cards.
Look at it this way, if the host had turned over the car on the first try the odds from both of the other cards go to zero and the odds of the card he turned over go to 100 percent. So if he randomly turned over a goat it goes in reverse and both cards get half of 33.3
We have to make 2 assumptions for the math to work the way it is shown. First that the Host knows were the car is and second that he wants the game to go as long as possible. In that instance even if he has the winner he will not turn it over. Because of that the odds from the card he turned over should be added to the one both of you did not chose instead of splitting it between them.
It is implied multiple times that he knows were the car is. First why would a math expert setup a math demonstration where the odds are greatly against him. If we assume he dos not know were the car is he has to beat the odds multiple times. First he is counting on you not to pick the car, second he has to get lucky and pick one of the goats then he has to beat the odds one last time and pick the car. In this instance he has to get lucky 3 times in a row. When he asks for a volunteer we see he wants more and yet he still picks one of the people who volunteered when he first asks. Next consider he has 2 red x in his hands implying he might need two for the demonstration to work out.
Lastly he gives the final clue when he talks about "in this class". Meaning its a setup.
conobs The host does not make a random choice though. That's the point. The host knows where the car is and will only ever reveal a door with a goat. That is stated in the Monty Hall Problem. If the host chose randomly and revealed a door with the car the contestant could not possibly win because the contestant is only allowed to either stay with the door he chose or switch to the door that has not been revealed.
Yes i agree completely but several people were missing it so i attempted to cover it in such detail anyone would understand it. Which judging by the number of folks that claimed 50/50 was needed.
That's really cool!
Assume that the game show host will ALWAYS eliminate a goat. He wouldn't want to accidentally reveal the prize before the person has a chance to change their answer.
So, if you initially:
#1: Picked Goat A. The host reveals the other goat. If you switched now, you'd WIN!
#2: Picked Goat B. The host reveals the other goat. If you switched now, you'd WIN!
#3: Picked the car. The host reveals a goat. If you switched now, you'd LOSE!
2 out of 3 scenarios WIN by switching. :)
this is well explained. personally i didn't get it, but my stats teacher said this is a really good explanation. i get it now after asking my teacher.
Actually it's quite a good explanation. The only thing he doesn't mention is the fact that its actually the quizmaster that is raising the odds by pointing a goat out. But perhaps that just makes it more confusing :).
Actually I KNOW this is right, because I came across this many times during my 6 years of math studies at university.
The idea is mainly used for scientific entertainment and math promotion. For instance in introduction classes at university or in popular sience books.
We did this in my cognitive psych class, only it was used as an example of reasoning (sorta like Charlie did). The class had about 300 people in it and I'd say more than half were puzzled when the prof explained it in simple terms, lol.
Ok, I'm not going to say I'm an expert on this. I did believe at first that the odds changed to 1/2 after the reveal of the first goat.
But looking at a 'simple solution' spreadsheet, you get to see the overall chances do increase to 2/3 if you switch doors after the first reveal. (Found on wikipedia under 'Monty Hall problem'.)
I think what confuses people is that, in effect, the revealed door is still being (and should be) counted towards the percentage. However, after the reveal 'the odds for the two sets (the player's pick and the other two doors) don't change but the odds move to 0 for the open door and 2/3 closed door.'
If I'm correct (which I may not be), when people think of the door as revealed, they also think of it as eliminated from the percentage. Along those lines, there are two doors left, and your chances would be 1/2.
I think (once again, I might be wrong), it's similar (but not the same) to comparing the overall chance of getting Tails twice in a row on tossing a coin (1/4), as opposed to the chances of getting Tails on each of two tosses (1/2 on the first toss, 1/2 on the second toss). In the case with the goats and car, keeping the third door in the equation is actually more beneficial than eliminating it from the equation entirely.
And no, I do not think it was explained well on Numb3rs at all. But there is a 2/3 chance the next episode will explain math concepts better...
oh the Monty Hall Problem ... so many memories ... statistics and psychology
No, the odds only become (2/3) when bias (goats only) is applied to revealing a card. If a card is revealed at random and it happens to be a goat, as in this case, then the remaining 2 cards have equal chance of being the car. The REASON for switching DOES change depending on the hosts selection bias. Imagine the problem with a million goats. The host removes 999999 goats on purpose. The card left is likely the car. Removing 999999 cards at random (goats by chance) leaves 2 equal cards.
you know from now on im gonna do that from now on
The religious explanation for this, is you have three choices:
Heaven, Hell and Worldly Gains... behind 3 doors.
You chose the first, but it was revealed that it leads to Hell; So whatever you choose next, the best that you'll really get during your existence!
So, what you'll need someone to reveal what's in the first door, so you'll get things right the first time... an Imam, priest or whatever that have clues about the choice...
Or, choose the best on the second try, making the change if needed to gain Heaven on second try; and it's also good to have someone to guide you at this stage...
But, if you fail to make the right choice, you can be content by living the full life full of worldly gains... That is, if you don't harm yourself in the first place!
I love Charlie too much
Glad my explanation helped. :)
To better explain it, imagine 100 doors. You pick door 1. Door 1 has a 1% chance. The host, (following the rules above) opens all of the doors except yours and door 47 (for example). Your odds for staying with door 1 remain 1%, while switching increases it to 99%, since there was a 99% chance that it was behind one of the other 99 doors to begin with. The fact that he shows you doors that don't have cars behind them doesn't change the odds, which is all you are working with really anyway.
This example was used in 21, and they got it wrong there too. You HAVE to state that the game show host MUST reveal a door, that isn't yours, with a goat behind it for this to be true. If the hosts behavior is random (he can reveal any door, yours with a goat, or the car), then your odds are 50/50. Only when you apply the requirements do your odds go to 66%. Because the hosts behavior is not stated to always follow the pattern it followed this time, the answer is still 50/50.
once you state that the host can't open a door with a car, and can't open your door, then your options go from 4 possible 2nd round random scenarios, with 2 winners (50%), to 6 possible 2nd round random scenarios, with 4 winners (66%).
I'll try to explain this using doors instead of cards.
There are two things you need to know to understand why this makes sense.
1) The host KNOWS which one has the car.
2) The host will open ALL of the doors except the one you picked and the one with the car behind it. (In the event you picked the one with the car, he will leave a door closed at random.) (Note that with only 3 doors, the host will always open only 1 door, which is where it probably throws people off.)
I remember that episode :)
but what if i change my choice by picking the same one
It only becomes confusing because there are only three doors. Think of them as sets. You pick door 1 (a set with one item) as it has as just as likely a chance as any other.
Given the option to switch over to the set of doors 2 AND 3, (if either has the car, you win) then you of course select that set.
When he opens one of the doors, your odds of having initially picked the right door is still 1 in 3. (How could it change to 1/2?)
simplicity is the essence of genius. something that isnt simple is impractical.
See, the whole point of the game is to increase your odds of winning, and by switching your answer you'd definitely increase your odds of winning.
Think about playing this game with 100 cards (where there is 1 car and 99 goats). You pick any one of the 100 cards. The host then reveals 98 goats, leaving the card you initially chose and one other card left unturned. Would you be wise to switch your answer now? Hell yes! :)
does anyone know the background song played in this video? Sounds really nice!
it's mathematically correct IFF the 'host' knows what is behind each card. Otherwise, if you're picking the second card at random, there's 1/3 chance that you win, a 1/3 chance that you lose, and a 1/3 chance that the test is invalid and fails (the 2nd card flipped over is the car)
Behind the other 2 are goatse. You don't want that.
In both this and 21 they neglect the most important piece of information: the fact that host has to reveal a goat. That's important because if you pick the car on the first time then the host has 2 doors to choose from and you lose from switching, BUT if you pick one of the goats then the host shows you the other goat and switching gives you the car. So picking a goat the first time always wins, which means you have a 2/3 chance of winning.
Wait a minute, was that continuity error? The car was in the middle and then it was on the left.
The Matrix explanation:
I am the Architect. I created this Problem. I've been waiting for you. You have many questions, and although the process has altered your consciousness, you remain irrevocably human. Ergo, some of my answers you will understand, and some of them you will not. Concordantly, while your first question may be the most pertinent, you may or may not realize it is also the most irrelevant.
Neo: Where is the car?
A: Interesting. That was quicker than the others.
What season? what episode?
in which season and episode is this scene?
You have a deck of cards. Your host knows where the ace of clubs is. You pick a card. Your host will necessarily reveal 50 cards from their deck that aren't ace of clubs. His remaining card will 51 out of 52 times be an ace of clubs. This goat/car thing simply uses a much smaller deck of 3 cards: two goats and a car.
what episode is this
21 came two years after this
It relies on just probability only, not actually getting it right. If it were 100 choices, you would rather pick 99 of the choices rather than just 1 as your odds would be 99% of getting it right.
In this case your being allow to go from 1 choice to 2 choices. (You already know that one of the two will not have a car behind it, so seeing that fact doesn't change the odds any)
@Selahrs1 what difference does it make if it is a conscious choice on the host's part though? (sry if the answer is obvious but maths has never been my forte)
Wow he's right, thats awsome!
@777Skeptic Nicely put!
There are only 2 options -- stay or swap. There are not three options -- stay, swap, or pick the open door. It's a 1 out of 2 shot. Suppose you're allowed to select two doors, and the host opens a random bad door. Regardless of whether one of your doors was the prize, you still only have the chance to pick between 2 doors.
I love math! And i love to explain this to people! now i dont have to cuz i can just show them this
@Song4Alex No it isnt. Your not supposed to chose again at the two once, just switch from your first choice, THAT is why it doubles the odds...
we will ALWAYS be presented with the choice to switch or not switch regardless if we initially pick the car or not, so think of these as "strategies" to play this game, and analyze each one. if our strategy is to not switch, this strategy is equivalent (exactly the same) to initially trying to pick the door with the car, which we have 1/3 chance of doing. if our strategy is to switch, this is equivalent to trying to pick a door that does NOT have the car which we have 2/3 chance of doing.
For those who don't believe it, try it yourself with some playing cards. Put 3 rows of cards down and lay it out in a grid where you have a "winning" card in each row in a different column. Then, choose a column as your "pick" and eliminate one of the cards that isn't a winner from each row. You see that by the original row you selected has a 1/3 to win still, and the other column (the one you are offered a switch to) has a 2/3 chance of having the winning card.
Assuming it was 2/3 goats, you have a 66% chance of picking a goat if you were to choose from 3 cards.
WIth that in mind, if the host reveals one goat and leaves you with the chance to switch, statistically speaking, you were more likely to pick a goat from the beginning, so holding this original assumption, it's likely that you still have a goat now, thus, it is better to switch. Of course, this isn't 100% thus, sometimes you may be wrong, but in practice, switching will give you the win.
No, afterwards he says "Now knowing what you know, do you want to switch your choice".
This negates the need to state in advance that he knows where the car is since the reason for switching remain the same.
@vernonzhou Are you suggesting there are stupider things you had seen or read that day?
@baisuh look up 'monty hall theorem' .. its where it began.. and a good youtube clip explains it
it was more probable that he would pick the car, and it better illustrates his point, of course he's right,
if you add in the fact that the host knows what is behind all 3 doors it can serve as a test, when presented to a person who is very smart, to determine whether they would react to a problem emotionally or using their math skills, as it was used in 21
Maths is everything
If he shows you A MILLION doors. You pick door #18. The host opens 999,998 doors, leaving you with door #18 and door #134,523.
Do you want to switch now?
Obviously, yes. The chances you picked the right door out of a million was 1/1,000,000. Switching now will win you the car 999,999 times out of a million.
It's the exact same problem, just re-stated a different way to make it more obvious that switching is the correct play.
Which episode was this anyone know?
Numb3rs was wrong. 21 was right.
"I'm going to reveal to you one of the cards you did not choose" is NOT the same as a game show host's saying "I'm going to give you a little help and eliminate one of the wrong answers." That makes all the difference.
Agreed. If the choice is slated to be random, then it's a matter of pure equivalent statistics, meaning your choice ultimately does not matter. However, if the choice is already known, then it becomes a logical deduction puzzle. The very premise of the puzzle depends on the host's knowledge and intentions. If the host knows, then it's a logical deduction puzzle. If the host doesn't know, then it's just basic statistics which necessitate a change in statistics to accommodate a change in the equation
This is the infamous Monty Hall Problem. A good explanation appears on Wikipedia.
I still don't get it! If she were to pick the left most card, then it wud fail wudn't it?
That's trippy
As made clear by Marilyn Vos Savant, the columnist who popularized the "The Monty Hall Dilemma" - it is ESSENTIAL that the game show host KNOWS which card conceals the valuable prize, and that he consciously chooses to reveal a goat card. If the host just randomly happens to reveal a goat behind a card, the contestant will gain no advantage by switching away from her original choice. The "Numbers" tech advisors got it wrong again by failing to make this aspect of the Monty Hall dilemma explicit.
So basically to win the game, you need to select a goat in your first pick, that way when the host opens the other door with the goat, you can switch to win the car. And the chances of you picking a goat in your first pick is 2/3 hence its better to switch.
@jacobins3000
Try to understand this way: Suppose you are in a game show. There are 100 cards you can choose, one of them has a prize of 1 million dollars. You choose one of them. The host keeps on intentionally removing the cards, KNOWING THAT THEY AREN'T THE PRIZED ONES. Then it comes down to 2 cards. Yours and one random. It isn't 1/2. You chose one of 100. He KNOWS what cards he is taking away. So the odds dont change. It is like he didnt take any away. It doesnt make any difference.
@thomasproparr, then why change the choice ?? If i stick to my previous option, then too the odds are same(as a goat is now revealed)
Think about this: there are a 1000 choices (999 goats and 1 car), you pick 1 "door" then the host reveals 998 goats, we're now left with your choice and a blank.
There's a 99.9% chance that your choice is a goat since there were 999 goats out of a 1000 choices, so therefore you would easily switch and win the car almost every time.
Was this show anything like The Sopranos?
Ahhh the monty hall problem. I love the way they try to explain it in this episode.
What is this episode called
impressive
ooo strike one, you get another chance! actually i've never seen a full episode of Numb3rs. you can try and: (1) guess which top 30 university i attend, (2) which top 5 semiconductor company i intern at, (3) which master's program in electrical engineering i got into [this is easy though, it's the same as #1]. care to take another shot at my "looser jobless life"?
honestly, i'm just passing time while i wait for tests to run...
oh man... now I got it hehehe
thanks ;]
Your odds are only 50/50 if you think of it in sets. 50% of the time the car will be behind set one (door 1) and 50% of the time it will be behind set 2 (doors 2 or 3)
i can`t believe the amount of people that don`t get this... (posting from argentina)
I know.. that could be late my ask but.. this episode is number 13 of the first season
Monty Hall Problem - old trick :) Always love how even the most smartest of people get this one wrong.
Neo: Is this still about the car?
A: There are two doors. The door to your right leads to the Car. The door to your left leads to the Goat. As you adequately put, the problem is choice. But we already know what you are going to do, don't we? Already I can see the chain reaction: the chemical precursors that signal the onset of an emotion, designed specifically to overwhelm logic and reason. An emotion that is already blinding you to the simple and obvious truth: There is no car. Just a goat.
1/3
by switching its now
2/3
it increases the chance of getting the car.
but also the sheep too.
so its luck.
but if you have more than 3 cards, this theory will prove useful it alot of situations of choosing.
It's funny and irritating when people still try to argue that the chances are 50-50.
I think he was miscommunicating, almost a version of the myth gamblers buy into, where if the roulette wheel comes up red 10 times, it is more probable that the next spin is black. It's a myth, the IMPROBABLE event (10 reds) already happened, it's still 50:50 next spin.
She made one of the 2 losing choices. Now that card #3 is revealed, the probability is no longer 1 in 3.
I'm not sure what you are trying to say, vskarate. Are you trying to tack on a fourth scenario (so that there is one for each "wrong" answer the contestant could switch to?).
If that's what you're implying, it doesn't quite work that way. It doesn't matter which goat he switches to if he initially picks the car... all that matters is that he would lose. He only has 3 choices initially, and 2 of those three choices will benefit him later if he changes his answer.
Does that make more sense?
I really really love Numb3ers, but i think he didn't explain the problem fully. He should have said immediately that the gameshowhost exactly KNOWS where the 2 goats are. So the gameshowhost MUST pick a goat out of the 2 remaining boxes. So if you picked a goat at the beginning (chance 2/3), the host has to eliminate the only other goat. If you switch then u got a 100% chance of getting it. Making it a total chance of 2/3 * 1/1 = 66%
This seems to be an attempt to recreate the famous Monty Hall puzzle. Because he said "reveal to you one of the cards you did not choose", we must assume he chooses to reveal one of the 2 cards at ramdom. If this assumption is correct then he is WRONG. Odds are 50/50. However if he meant to say "I will reveal one of the goats that you did not choose" then he is correct and you should switch your choice to get better odds of winning.
The good old Monty Hall Problem
Movie????
Choose Goat 1, Goat 2 is revealed, switch and you get a car.
Choose Goat 2, Goat 1 is revealed, switch and you get a car.
Choose Car, Goat 1 or 2 is revealed, switch and you get other goat.
You win 2 out of 3 times if you switch.
Try imagining doing this millions of times. You always pick and open door 1 and one of the other two that doesn't have a car is reveled. You'll see that you only win 1/3 of the time, not half of the time.
Numb3rs uses the same background actors and locations as burn notice i watch both alot and i see it
this thing works only if you pick the goat first, if not you loose.
Of course he has to reveal a goat and not the car you don't even need that in an explanation because if the host reveals the car then you'd of course know where the car is. Right?
But what if she ended up picking the car in the first place? Would that just be blind luck? Probability?
Deciding to switch after one card is revealed only raises your chances, it does not guarantee the outcome. The outcome can be good with less chances, or bad with more chances. You can choose to play a game with 1 chance out of 100 to win, and win it. You can choose to play a game with 99 chances out of 100 to win, and lose it. But it's wiser to choose the easiest game.
I want to say it's from "Man Hunt" from Season One; Episode Thirteen. I'm not 100% sure, though.
Its from Numb3rs
It LOOKS like 50/50 on the surface, but it is not. That's what makes this equation so fun.
Look up "Monty Hall Problem" on the Wikipedia. They have excellent diagrams, and even an in-depth explanation for why it isn't 50/50, despite the fact that the person is choosing between two cards in the end. It's like a magic trick... you have to look deeper than just the surface.
BTW, I've only seen 3 episodes of Numb3rs. The fact that this was aired on a TV show has no relevance to me. ;)
In 500 words...I'll have a go:
You pick a card. there is a 2in3 chance you have picked a goat and a 1in3 chance that you have picked the car. Of course you don't know which yet.
Let's assume that you picked the goat. The host turns over the other goat. Of course you must change your pick to win the car.
Assume you picked the car. The host turns over a goat. In this case you must stick to win the car.
So in 2 out of 3 situations changing will win the car.
Season 1, last episode
i love this movie
Liviu B not a movie