By the way, that was 321 characters. I was thinking that a lot of words, different explanations and different ways of expressing things can get in the way. It was a lot of work for 321 characters but it is often a kind of effort that is well spent.
Considering mathematical probability and real objects:
"From the first choosing, there is a 2 in 3 probability that the 'right' object was not chosen. The 'right' object is still present at the second choosing. Where it was not chosen before, it is now chosen by switching. So for the whole exercise there is a 2 in 3 probability of obtaining the 'right' object by switching."
Can anyone make a satisfactory, shorter explanation from this or otherwise?
What if there were TWO contestants and each of these two contestants were ardent mathematicians, both with an innate understanding of the principle behind the Hall paradox. After each choose a door, and the third unchosen door was opened to reveal a zonk, would each of the two players, if given the chance, leap at the opportunity to accept the other fellow's door? (And in this scenario the 2 scholars are rivals who would not agree to sell the car and split the proceeds) Hmm?
@ballwizz23 Think of it this way. When you pick a door, you have a 66% chance of having picked a goat. When the host reveals a goat, you STILL have a 66% chance of having already picked a goat. Therefore, if you switch, you double your chances of picking the car.
This is by far the best explanation of this puzze ive ever seen. The visual is what convinced me. I was very much in the "it is still 50-50" boat until i saw that part. Great job man!!! I see how it is smarter to switch now.
in round one, the percentage is a picking percentage, not winning, and in round 2 is a winning percentage, so how can you add both different types of percentages and say you have 66%? it doesnt make any sense, its like saying
x + y = 2x, where x is 33%, it doesnt make sense.
It does become 50/50. You are given 3 choices originally. However the very second he asks "do you want to switch your choice or keep it the same" after revealing the first door makes it a whole new option. You are now picking between 2 doors. By giving you that choice of picking door 1 or 2, One is right the other is wrong. Truthfully the question prior to now (With 3 doors) Is void. The current question is 1/2 chance of winning. You either keep your choice or switch it. In the end, its 50/50
If the host didnt know where the car was, then he might open up the car-door when "giving you a better try" with two doors left.
The probability that the participant chose the correct door to begin with is 1 / 3, and in that case he will lose if he switches door. The probability that he did not choose the correct door initially is 2 / 3, and in that case he will win if he switches door. It pays for the participant to switch, since he has 2 / 3 chance of winning the car.
@Rockwarrior2004 That is the conditional chance given the previous state of the game. So you're right in the 50/50, but wrong in thinking it's independent. Due to dependencies, the odds in the second state are 2/3. If the game were to start out with the second question (skipping the first bit thus losing the conditional part) the odds would truly be 50/50.
A very easy way to see the solution is to to think each door has a likely percentage of having the car behind it. (Let's say the actual order is Goat, Car, Goat).
So Door 1=33%, Door 2=33%, Door 3=33%. We can re-write this as Door 1= 33% and Doors 2+3=66%. If door 3 is revealed to be a goat, its percentage of being a car is now 0%. So Door 1=33%, and Doors 2+3 now = (66% + 0%).
There is no mystery whatsoever with The Monty Hall Problem ! People have used Bayes Method, Probability theories, etc. to try and explain this puzzle..there is no need for this, the explanation is absurdly simple : If you pick a losing cup and SWITCH - you will win 2/3rds of the time. The PROBABILITY of picking a losing cup is of course 2/3..so if you pick a losing cup[and SWITCH] you will win 2/3 of the time. That's it ! It's no more complicated than that. Next puzzle please !
There are four playing cards: Ace of Spades, Ace Of Hearts, Two of Clubs and Jack of Diamonds. I pick two randomly, and announce that one of them is an Ace. What is the probability that the other card is also an Ace?
I repeat, again annoncing that one of them is an Ace, but this time I specify that it is the Ace of Spades. What is the probability that the other card is also an Ace?
Surely giving the suit of the Ace makes no difference; it didn't matter which one it was. Or does it.....?
I don't think it works. You work like the possible scenarios are 3, and in 2 of these 3 scenarios you would win switching. But there are actually only 2 scenarios. In one, you have already chosen the winning door, and the game host would open one random of the remaining doors. In the other scenario, you have chosen one "goat" door and the host would open the OTHER goat door, leaving your door and the winning one. And it's 50% chance. Isn't it?
@ion010101 yes, there is one winning door, one goat door that will stay for the final check, and one goat door that will be opened before the final. Whichever you choose, one goat door will always be opened. It's sure. The scenarios shown as 2 and 3 in this video are actually the same.
@palmomki The probability isn't divided up between "scenarios". It's divided up between the car being in one of three places. There are three total possible events that can happen. The car is between 1, 2, or 3. This means that there is a 1/3 chance the door you pick has the car. And there is a 2/3 chance the door you pick has the goat. If you have a chance to switch, then you have a 2/3 chance of getting a car (since 2/3 of the time you pick a goat door).
@palmomki There would be a total of 2 possible events if the goat was eliminated before game play. The goat door was opened after you chose. Since you had a 2/3 chance of picking a goat, and you now know which other door has a goat, you can switch and have a 2/3 chance of winning a car.
@ion010101 1) you choose a goat, then the other goat is eliminated. 2) you choose a goat, then the other goat is eliminated. Find the differences. ._. The probability changes when the goat is eliminated, you now have 50% probability to have the pack.
@palmomki Suppose a dice had 4 printed on it twice (on two sides). You would now have a 2/6 chance of getting a four on a toss of the dice. But your chance of getting a 3, or any other number, IS STILL 1/6. The two fours don't count as one side. There are still SIX TOTAL possible events, even if "4" was printed on all 6 sides. Two goats don't count as one event. It just means you're more likely to get a goat than you are to get a car, cause there's two goats and only one car.
additional: suppose someone was going to herd 6 goats and a dog past my window. What's the chance I'll see the dog go by, first. It's not 50/50 simply b/c when you've seen one goat, you've seen them all. A 50/50 chance would be the answer to the question, "What kind of animal will go by first, a Cat or a goat?" Then it's between "cat" or "goat". But if the question is, "What will go by first, a cat or a goat?" then, don't even try to do the math. There are way more goats than cats, right!
@ion010101 you wanted to make my mail inbox crash down, didn't you :D anyway, the examples of the dice and the window don't work, it is a different situation. However: if I ever manage to demonstrate it's 50/50, which I fear will never happen, I'll show it to the world u.ù So far, I think the only way to demonstrate it would be testing it a million times. Statistics always work on big numbers.
@ion010101 It's not of vital importance. It's just that seems to me that this solution is "fashioned". It's like so many people hears it and says "oooh, smart and complex, it must be right" and it annoys me.
@palmomki Did you mean "fashionable"? Try this: instead of thinking about your chances of winning a car, think about your chances of winning a goat. There's 2 goats and 1 car, so you've got a better chance of winning a goat, a 2/3 chance. What if every time you won a goat, you could exchange it for the car? The only stipulation is that if you win the car, you have to exchange it for a goat. Since you'll win a goat 2/3 of the time, then you're going home with a car 2/3 of the time!
If you're still stuck on 50/50 as an answer, take a look at my video response "Proving The Monty Hall Problem." It'll give you a way to prove to *yourself* just what the correct answer is.
In a probability distribution, the sum of the probabilities of each possible outcome must add up to 1, so the P(staying w/ initial choice wins the car) + P(switching doors wins the car) + P(the door the host opens wins the car) = 1.
The P(staying) is 1/3 because you have 1 chance of choosing right out of 3 doors. The P(host wins) is 0/3 because he will always choose the door with the goat. The P(switching) must be 2/3 because 1/3 +2/3 + 0/3=3/3 or 1.
If you switch, you will always end up with the opposite of what you initially choose -- thus, 1 time out of 3, a goat; 2 times out of three, a car. Likewise, if you do not switch, then, after you choose, no matter what else happens, you will just get what you chose -- only 1 time out of 3, a car; while 2 times out of 3, a goat.
Interesting, thanks. I was wondering how else it might be put mathematically. --- This is all presuming, of course, that one would prefer to win the car. When living in the Andes, for instance, one might like to have a goat.
@shotta0121 Statistically speaking it might not come out to be exactly right if you only tried it 100 times. But, the great thing about statistics is that if you tried it say 1000 times or 1000000 times eventually the percentages would work out.
I absolutely LOVE this problem--and suffered with it for DAYS (by trying to force my intuition into line) with no success. Finally, I saw it in the way you suggested--I imagined the case where there were a million doors, realized that he could ALWAYS bring it down to 2 doors, yet, there's no way my choice could ever be better than one in a million. Note: Of course I mean that I had seen the problem long before ever seeing this video.
Well, I believe you're wrong, and what's more: I believe I can prove you're wrong.
In that case, even the whole world would be wrong.
I'll tell you what: I believe giving the advice to switch is unnecessary, because you have a 1/2 chance anyway. And I believe I can prove that to you as well.
But because I am still believing it, I will not reveal my logic unless I feel sure about it.
Well, I believe you're wrong, and what's more: I believe I can prove you're wrong.
In that case, even the whole world would be wrong.
I'll tell you what: I believe giving the advice to switch is unnecessary, because you have a 1/2 chance anyway. And I believe I can prove that to you as well.
But because I am still believing it, I will not reveal my logic unless I feel sure about it.
@shmontyhall your wrong and here is your proof. here is a different form of this problem with 1,000,000 doors.
I'm thinking of a number between 1 and 1 million. now you quess a number. ok... now I'm going to eliminate 999,998 wrong numbers and leave yours alone. the remaining number is 178,354 and your number which do you think is right your number or 178,354?
I know about all those possible explanations that lead to 2/3 as an answer, but there's no point in proving me wrong while I haven't explained my view yet.
There are more goats than cars so it's more likely we'll pick a goat..
Now we remove a goat.
Second choice: CAR GOAT
We were more likely to pick a goat at the start. So now the other goat has been removed the remaining door must be the car. The only way switching would lose is if we picked the car from the beginning, which we showed is a 1/3 chance.
This only works this way if the game host KNOWS which doors are the goats. If the host RANDOMLY opens a door, it does not matter if you switch (50/50 chance either way).
@singingbanana does it? I was just pondering this very question. I need the help of you and your outstanding logic. say for example the host does not know which door has the goats but just happens to open the door that has the goat. Would this effect the propability? as far as I can tell the answer is no.
@gtq838 the store clerk just got a new box of 100 lottery tickets there is a jackpot ticket in this box. You buy one ticket and the guy behind you buys the rest. he scratchs off 98 of them and they are all losers what are the odds that you have the winning ticket? 1/00 right? doesn't his last ticket have a 99% chance. the chart still holds true if you graph it out. its even more counter intuitive but forknowledge doesn't make a difference unless the winning ticket is scratched
@gtq838 never mind I made a program and tested the idea I'm wrong and I figured out why. I failed to take into account that the gentleman must pick the same door/ticket in each row of our chart and thus there are only two equal posibilities left.
@singingbanana Games of chance are a form of advantage taking, as there is no skill involved. Unlike tennis for example which requires that skill be honed.
In my opinion, you still have twice as likely chance to pick a goat at the start, and if you're lucky, the game show host will reveal another goat, and you can still switch and have the same odds. It really doesn't matter whether the host knows or not, but it is still more likely you chose a goat at the start.
@Anon909 That you just said, as also our "Host" said, is easily bypassed as it is only a detail, but a crucial one for this problem. The reason so many mathematicians had trouble realizing this is becuase they ignored the fact that it's not a pure mathematical problem, but actually has the influence of a (tricky) human mind.
Btw @singingbanana I loved your 100 door example, it made me understand this issue alot better as I've tried to understand this problem before :)
Oh, now I see. The odds you chose the car first is only 1 in 3 therefore the odds that it was another door are 2 in 3 . . . . . . . . . I'm only 11 don't blame me . . . .
I ran a test with 100,000 runs and ran it multiple times and it always was around 2/3 of the time, you win when switching. And it really makes sense, if you think about it. It's only 50% if you aren't given the initial choice.
ok, since no one explained me anything, I proved myself wrong :D If you make a really big chart in witch you enter EVERY single case and the chanse of it happening, you see it clearly: 1/3 chanse vs 2/3 chanse or 33% vs 66%
I decided to make a second counting (sorry if your tired of reading it all)
1. I choose Door 1, he opens Door 2: a) the cars are in Order #1 b) the cars are in Order #3
You can make two cases for you choosing Door 1, him opening Door 3; you choosing Door 2, him opening Door 1 etc. etc. I'm still interested in this theory, so if anyone has something to back it up, you can write me a letter.
In bridge, this is called the Law of Restricted Choice (or principle of RC). It works the same way. If you need to play a suit that contains AT32 opposite your K9876 without losing a trick, you play the Ace and say the Jack appears as the last card, its 2/1 that the player who played last didn't have a choice.
It's probably the biggest application of the Monty Hall problem I can think of.
The difference between the Monty Hall problem and games such as Deal or No Deal is that the game show host in the Monty Hall problem knows which door the prize is behind and thus his choice to reveal a door with a booby prize is not random. In Deal or No Deal the choice is completely random as nobody knows where the 250,000 is. For these reasons it is correct that you should always swap in a Monty Hall scenario as you stand a 2/3 chance, however in DoND, the choice of swap or no swap is 50:50
This is a complete joke that this problem has gotten so much attention..Its sooo simply. You change door everytime and then i win 2 out of 3 you only lose if you pick the winner first...simple huh?
this is very wrong, it has nothing to do with intuiton, it's simple science and math. The fault is in the premises of the 3rd explanation. Once you releave one of the options, the equation changes completely and you cannot add the variables from the 1st equation into the second equation. This is garbabge and nonsense. It's still 50/50 not 2/3.
at first u have 67% chance of losing. however if u switch the chance of losing will become the chance of winning and vice versa. (pick a goat and switch= car... pick a car and switch= goat). 67% chance of picking goat... 67% of getting car if u switch. see the problem beginning to end. not just the ending options.
The reason why it is counter-intuitive is because there is an unspoken rule; Monte will never pre-reveal a door that has a car behind it. If you take away that rule then it all because intuitive.
If you change the rules so that Monte can't know which one has the car then everything changes. This would also lead to the possibility of Monte opening a door you didn't pick to reveal the car, thus showing that you lost before you even open the door you picked.
Further elaborating, if you pick a number between 1 and a 100 and I then eliminate 98 numbers (assuming I have no knowledge of the winning number), leaving your number and one other number, 98% of the time I would eliminate the winning number and you would lose. 1% of the time you would have picked the winning number, and 1% of the time I would have given you the option of switching to the winning number.
What bearing does the random removal of boxes have in this regard? If it's just your box and one other box left and 250,000 and 1p left on the board, aren't you being given the same gaurantee, should you take the swap, as in the Monty Hall Problem when two boxes remain? Say your sole intention was to win 250,000. The box you chose initially had a 1/22 chance of containing it, a 21/22 chance of not containing it. If the swap is offered doesn't the boosted probability (21/22) apply as in Monty H?
Well,think of it this way. What are the chances that you picked 1p? 1/22,same odds of you picking 250,000. So the chances don't increase,it'll be 50/50 if you have 1p and 250,000.
thats just either way if you take out 1, of those 3, that is wrong. you'll end up at 1 right and 1 wrong so thats 50/50. but if your talking about "chances", thats the different side of the story. you got 33.33% from the beginning due to 1/3 is your chance, by removing one you get 66.66%. but logically speaking, its still 50/50 at the end, atleast thats what i think
Someone (In America) wrote into a maths equilvalant agony aunt and asked about this. He asked should he stay with the door he has or always switch to another.
the response was he should always switch. most of americas mathematicians got angry at this and the 'agony aunt' was sent lots of hate mail saying things like. thats wrong this is why the rest of the world think that americans are dumb and that you should always stay with the door you picked.
Indeed, surprising mathematical curiosities are called mathmagics (mathemagics), by mathmagicians (mathemagicians). Have a look for Arthur Benjamin. Those things are not necessarily Monty Hall though.
Mythbusters did this one too.
genstian 4 days ago
Amazing! I subbed so hard.
SteDeRaver 5 days ago
At 4;25, The name of the gameshow was Let's Make a Deal, not Deal Or No Deal. Deal Or No Deal is a different show.
swcomer 5 days ago
Best explanation on the interwebs !
BUFFALOEOX 1 week ago
Mythbuster did it!!! it works! ALWAYS SWITCH
robbie4128 1 week ago
By the way, that was 321 characters. I was thinking that a lot of words, different explanations and different ways of expressing things can get in the way. It was a lot of work for 321 characters but it is often a kind of effort that is well spent.
andrewburbidge 1 week ago
This explanation might help some to see it.
Considering mathematical probability and real objects:
"From the first choosing, there is a 2 in 3 probability that the 'right' object was not chosen. The 'right' object is still present at the second choosing. Where it was not chosen before, it is now chosen by switching. So for the whole exercise there is a 2 in 3 probability of obtaining the 'right' object by switching."
Can anyone make a satisfactory, shorter explanation from this or otherwise?
andrewburbidge 1 week ago
This is by far the best explanation I've seen, thank you for this. I'm convinced.
zebase 1 month ago
Aww so adorable =D
Tranquilance 1 month ago
What if there were TWO contestants and each of these two contestants were ardent mathematicians, both with an innate understanding of the principle behind the Hall paradox. After each choose a door, and the third unchosen door was opened to reveal a zonk, would each of the two players, if given the chance, leap at the opportunity to accept the other fellow's door? (And in this scenario the 2 scholars are rivals who would not agree to sell the car and split the proceeds) Hmm?
~Johnny Radionic ™
Johnnyradionic 1 month ago
haha, he said booby
ezpkns34 1 month ago
I had problem accepting any proof, but the table really did it! I think that the table is the best way to explain it.
RdMrcr 1 month ago
THE BEST EXPLANATION YET!!! IVE NEVER UNDERSTOOD THIS TILL RIGHT NOW! THANK YOU SOO MUCH
K2063 3 months ago
i still don't get it... i mean i understand THAT it works. I just can't wrap my mind around WHY it works.
ballwizz23 3 months ago
@ballwizz23 Think of it this way. When you pick a door, you have a 66% chance of having picked a goat. When the host reveals a goat, you STILL have a 66% chance of having already picked a goat. Therefore, if you switch, you double your chances of picking the car.
killerkerara 3 months ago
@killerkerara Actually that helps a lot. Thanks for simplifying that for me.
ballwizz23 3 months ago
Brillant explanation.
LogicallyCritical 3 months ago
I got here from Wikipedia :D I recognized your name in the referred websites :-)
Also, my win odds are 2/3 from the start, as I love animals and don't have a drivers license. ;-)
This explanation is excellent! Thanks for explaining :-)
FHomeBrew 3 months ago
This is by far the best explanation of this puzze ive ever seen. The visual is what convinced me. I was very much in the "it is still 50-50" boat until i saw that part. Great job man!!! I see how it is smarter to switch now.
spyrofan10 4 months ago
But what if u have the car in the first place and then u switch u lose but if u stay u win
So how would that work
dvdrmsy 4 months ago
@dvdrmsy That's the 1/3 lose chance. You'd lose in that case.
FHomeBrew 3 months ago
This has been flagged as spam show
in round one, the percentage is a picking percentage, not winning, and in round 2 is a winning percentage, so how can you add both different types of percentages and say you have 66%? it doesnt make any sense, its like saying
x + y = 2x, where x is 33%, it doesnt make sense.
ImageGround 4 months ago
It does become 50/50. You are given 3 choices originally. However the very second he asks "do you want to switch your choice or keep it the same" after revealing the first door makes it a whole new option. You are now picking between 2 doors. By giving you that choice of picking door 1 or 2, One is right the other is wrong. Truthfully the question prior to now (With 3 doors) Is void. The current question is 1/2 chance of winning. You either keep your choice or switch it. In the end, its 50/50
Rockwarrior2004 5 months ago
@Rockwarrior2004 To clarify. The question is now
"You have two doors, Would you rather pick door 1 or door 2"
Rockwarrior2004 5 months ago
This has been flagged as spam show
@Rockwarrior2004
If the host didnt know where the car was, then he might open up the car-door when "giving you a better try" with two doors left.
The probability that the participant chose the correct door to begin with is 1 / 3, and in that case he will lose if he switches door. The probability that he did not choose the correct door initially is 2 / 3, and in that case he will win if he switches door. It pays for the participant to switch, since he has 2 / 3 chance of winning the car.
TNBM4Life 3 months ago
@Rockwarrior2004 Remember: Consider the fact that the Game Show Host knows which door contains the car.
GaleStream 4 months ago
@GaleStream what would that have to do with anything? Okay so he knows the answer how does that help your choice?
Rockwarrior2004 4 months ago
Comment removed
TNBM4Life 3 months ago
@Rockwarrior2004 That is the conditional chance given the previous state of the game. So you're right in the 50/50, but wrong in thinking it's independent. Due to dependencies, the odds in the second state are 2/3. If the game were to start out with the second question (skipping the first bit thus losing the conditional part) the odds would truly be 50/50.
FHomeBrew 3 months ago
A very easy way to see the solution is to to think each door has a likely percentage of having the car behind it. (Let's say the actual order is Goat, Car, Goat).
So Door 1=33%, Door 2=33%, Door 3=33%. We can re-write this as Door 1= 33% and Doors 2+3=66%. If door 3 is revealed to be a goat, its percentage of being a car is now 0%. So Door 1=33%, and Doors 2+3 now = (66% + 0%).
Hopefully this way makes it easy to see!
xCelsius451x 5 months ago 2
There is no mystery whatsoever with The Monty Hall Problem ! People have used Bayes Method, Probability theories, etc. to try and explain this puzzle..there is no need for this, the explanation is absurdly simple : If you pick a losing cup and SWITCH - you will win 2/3rds of the time. The PROBABILITY of picking a losing cup is of course 2/3..so if you pick a losing cup[and SWITCH] you will win 2/3 of the time. That's it ! It's no more complicated than that. Next puzzle please !
brindow1 5 months ago
There are four playing cards: Ace of Spades, Ace Of Hearts, Two of Clubs and Jack of Diamonds. I pick two randomly, and announce that one of them is an Ace. What is the probability that the other card is also an Ace?
I repeat, again annoncing that one of them is an Ace, but this time I specify that it is the Ace of Spades. What is the probability that the other card is also an Ace?
Surely giving the suit of the Ace makes no difference; it didn't matter which one it was. Or does it.....?
TheLoonwolf 5 months ago
I don't think it works. You work like the possible scenarios are 3, and in 2 of these 3 scenarios you would win switching. But there are actually only 2 scenarios. In one, you have already chosen the winning door, and the game host would open one random of the remaining doors. In the other scenario, you have chosen one "goat" door and the host would open the OTHER goat door, leaving your door and the winning one. And it's 50% chance. Isn't it?
palmomki 5 months ago
@palmomki Remember, there are two goats, not just one.
ion010101 5 months ago
@ion010101 yes, there is one winning door, one goat door that will stay for the final check, and one goat door that will be opened before the final. Whichever you choose, one goat door will always be opened. It's sure. The scenarios shown as 2 and 3 in this video are actually the same.
palmomki 5 months ago
@palmomki The probability isn't divided up between "scenarios". It's divided up between the car being in one of three places. There are three total possible events that can happen. The car is between 1, 2, or 3. This means that there is a 1/3 chance the door you pick has the car. And there is a 2/3 chance the door you pick has the goat. If you have a chance to switch, then you have a 2/3 chance of getting a car (since 2/3 of the time you pick a goat door).
ion010101 5 months ago
@ion010101 but then they show you that one of the other two doors has a goat, and then 1/2 of the times you chosen the winning door
palmomki 5 months ago
@palmomki There would be a total of 2 possible events if the goat was eliminated before game play. The goat door was opened after you chose. Since you had a 2/3 chance of picking a goat, and you now know which other door has a goat, you can switch and have a 2/3 chance of winning a car.
ion010101 5 months ago
@ion010101 1) you choose a goat, then the other goat is eliminated. 2) you choose a goat, then the other goat is eliminated. Find the differences. ._. The probability changes when the goat is eliminated, you now have 50% probability to have the pack.
palmomki 5 months ago
@palmomki Suppose a dice had 4 printed on it twice (on two sides). You would now have a 2/6 chance of getting a four on a toss of the dice. But your chance of getting a 3, or any other number, IS STILL 1/6. The two fours don't count as one side. There are still SIX TOTAL possible events, even if "4" was printed on all 6 sides. Two goats don't count as one event. It just means you're more likely to get a goat than you are to get a car, cause there's two goats and only one car.
ion010101 5 months ago
additional: suppose someone was going to herd 6 goats and a dog past my window. What's the chance I'll see the dog go by, first. It's not 50/50 simply b/c when you've seen one goat, you've seen them all. A 50/50 chance would be the answer to the question, "What kind of animal will go by first, a Cat or a goat?" Then it's between "cat" or "goat". But if the question is, "What will go by first, a cat or a goat?" then, don't even try to do the math. There are way more goats than cats, right!
ion010101 5 months ago
Respond to this video... somehow a Cat substituted the dog... beg yer pardon. BTW, I envy people who can draw.
ion010101 5 months ago
@ion010101 you wanted to make my mail inbox crash down, didn't you :D anyway, the examples of the dice and the window don't work, it is a different situation. However: if I ever manage to demonstrate it's 50/50, which I fear will never happen, I'll show it to the world u.ù So far, I think the only way to demonstrate it would be testing it a million times. Statistics always work on big numbers.
palmomki 5 months ago
@palmomki I don't get what is meant by "different situations".
Playing the game many times (and simulating it on a computer) has been done. Please do this, if nothing else will convince you.
ion010101 5 months ago
@ion010101 It's not of vital importance. It's just that seems to me that this solution is "fashioned". It's like so many people hears it and says "oooh, smart and complex, it must be right" and it annoys me.
palmomki 5 months ago
@palmomki Did you mean "fashionable"? Try this: instead of thinking about your chances of winning a car, think about your chances of winning a goat. There's 2 goats and 1 car, so you've got a better chance of winning a goat, a 2/3 chance. What if every time you won a goat, you could exchange it for the car? The only stipulation is that if you win the car, you have to exchange it for a goat. Since you'll win a goat 2/3 of the time, then you're going home with a car 2/3 of the time!
ion010101 5 months ago
What if Monty can move the goat?
11Agamemnon235 6 months ago
those of you into monty hall stuff, check stayorswitch (dot) com, it's not bad.
mumiscrunk 7 months ago
I love this problem!
Crabbi5 7 months ago
The reason the booby prizes are goats is because Monty Hall's old show Let's Make a Deal actually had goats & donkeys as prizes.
TNO73 7 months ago
The chart was VERY good!
Great way to explain it !
JeppeTrabergS 7 months ago
The chart was a great way of explaining it. thanks.
braydenwis 8 months ago
This has been flagged as spam show
If you're still stuck on 50/50 as an answer, take a look at my video response "Proving The Monty Hall Problem." It'll give you a way to prove to *yourself* just what the correct answer is.
kirkbocek 8 months ago
at 3:50 the last 2 scenerios are exactly the same, there is no need for the duplicate. it's 50/50
pivotman64 8 months ago
sorry, just realised, ignore my comment, you HAVE.
ECproductions991 8 months ago
50:50 Chance... END! Noobs! -.-
Lyeaux 9 months ago
But if the door is opened to the mogwai, then the light from the sparkling jacket will kill it!
looney1023 10 months ago
Comment removed
MaThWa92 10 months ago
Haha.. around 2 days ago somebody told me this in school
Lightn0x 10 months ago
ohhhhhhhhhhhhhh I get it now
mewmewmewm1 11 months ago
@RationalConclusion
This may help?
In a probability distribution, the sum of the probabilities of each possible outcome must add up to 1, so the P(staying w/ initial choice wins the car) + P(switching doors wins the car) + P(the door the host opens wins the car) = 1.
The P(staying) is 1/3 because you have 1 chance of choosing right out of 3 doors. The P(host wins) is 0/3 because he will always choose the door with the goat. The P(switching) must be 2/3 because 1/3 +2/3 + 0/3=3/3 or 1.
ElectronDotts 11 months ago
@ElectronDotts
I'm make no claims at being a statistician, so calling this a probability distribution may not be correct, but I think the conclusion is valid.
ElectronDotts 11 months ago
If you switch, you will always end up with the opposite of what you initially choose -- thus, 1 time out of 3, a goat; 2 times out of three, a car. Likewise, if you do not switch, then, after you choose, no matter what else happens, you will just get what you chose -- only 1 time out of 3, a car; while 2 times out of 3, a goat.
comic4relief 1 year ago
@comic4relief
...or a goose, or a pig, or...
comic4relief 1 year ago
@comic4relief Said a bit more precisely: by switching you are betting that your initial pick was not the prize.
Mathematicable 1 year ago
@Mathematicable
Interesting, thanks. I was wondering how else it might be put mathematically. --- This is all presuming, of course, that one would prefer to win the car. When living in the Andes, for instance, one might like to have a goat.
comic4relief 1 year ago
I saw this in the movie 21. It was fascinating to me then, still is. :)
Quzit999 1 year ago
thumbs up for knowing what a mogwai is
YellowCakeKid 1 year ago
I've watched various videos on this problem but I still don't understand it.
RationalConclusion 1 year ago
So if we tried this say a 100 times would this theory work?
shotta0121 1 year ago 3
@shotta0121 Yes it would.
singingbanana 1 year ago 11
@shotta0121 Statistically speaking it might not come out to be exactly right if you only tried it 100 times. But, the great thing about statistics is that if you tried it say 1000 times or 1000000 times eventually the percentages would work out.
kevin2fla 3 months ago
i litterally thought the title said "montey python"
chandlerj333 1 year ago
dammit i picked one
TheLifesJoker 1 year ago
Best explanation of this I've seen.
Libervurto 1 year ago
The Monty Hall Problem is why do you throw a card that reads "The Monty Hall Problem" across the room?
RandomNinjaOfEvil 1 year ago
there is 1 big problem in this game
its the game master nows everything
and he wont act always the same way
mokachie 1 year ago
@mokachie
He would never reveal the car. That would be giving it away.
Libervurto 1 year ago
so on deal or no deal you should always switch at the end?
TheCheweeRevolutions 1 year ago
@TheCheweeRevolutions
No because that show is random, there is no one removing boxes who knows what's in them.
Libervurto 1 year ago
this was also in the Movie 21. blackjack counting cards movie
maxlee1000 1 year ago
Let's use the notation P(door) = x, which means the chance that the car is behind door 'door' is x.
If you choose door 1, the chance you picked the car is 1/3:
P(1) = 1/3.
Therefore the chance that the car is behind door 2 or 3, is 2/3:
P(2 or 3) = 2/3 .... which is P(2) + P(3)
(Using an 'or' means that we must add the changes together).
Now Monty opens door 3, which has a goat.
Because P(2) + P(3) remains 2/3 and P(3) appears to be 0, P(2) must be 2/3.
uitlegklas 1 year ago
I absolutely LOVE this problem--and suffered with it for DAYS (by trying to force my intuition into line) with no success. Finally, I saw it in the way you suggested--I imagined the case where there were a million doors, realized that he could ALWAYS bring it down to 2 doors, yet, there's no way my choice could ever be better than one in a million. Note: Of course I mean that I had seen the problem long before ever seeing this video.
GetMeThere1 1 year ago
This has been flagged as spam show
Well, I believe you're wrong, and what's more: I believe I can prove you're wrong.
In that case, even the whole world would be wrong.
I'll tell you what: I believe giving the advice to switch is unnecessary, because you have a 1/2 chance anyway. And I believe I can prove that to you as well.
But because I am still believing it, I will not reveal my logic unless I feel sure about it.
Anyway, thanks for the video
shmontyhall 1 year ago
Well, I believe you're wrong, and what's more: I believe I can prove you're wrong.
In that case, even the whole world would be wrong.
I'll tell you what: I believe giving the advice to switch is unnecessary, because you have a 1/2 chance anyway. And I believe I can prove that to you as well.
But because I am still believing it, I will not reveal my logic unless I feel sure about it.
Anyway, thanks for the video
shmontyhall 1 year ago
@shmontyhall : You're wrong. See if my post above makes sense to you.
GetMeThere1 1 year ago
@shmontyhall your wrong and here is your proof. here is a different form of this problem with 1,000,000 doors.
I'm thinking of a number between 1 and 1 million. now you quess a number. ok... now I'm going to eliminate 999,998 wrong numbers and leave yours alone. the remaining number is 178,354 and your number which do you think is right your number or 178,354?
gtq838 1 year ago
@gtq838
I know about all those possible explanations that lead to 2/3 as an answer, but there's no point in proving me wrong while I haven't explained my view yet.
shmontyhall 1 year ago
@shmontyhall
I'll try explaining it slightly differently.
First choice: CAR GOAT GOAT
There are more goats than cars so it's more likely we'll pick a goat..
Now we remove a goat.
Second choice: CAR GOAT
We were more likely to pick a goat at the start. So now the other goat has been removed the remaining door must be the car. The only way switching would lose is if we picked the car from the beginning, which we showed is a 1/3 chance.
Libervurto 1 year ago
This only works this way if the game host KNOWS which doors are the goats. If the host RANDOMLY opens a door, it does not matter if you switch (50/50 chance either way).
Anon909 1 year ago 2
@Anon909 As I said.
singingbanana 1 year ago 7
@singingbanana does it? I was just pondering this very question. I need the help of you and your outstanding logic. say for example the host does not know which door has the goats but just happens to open the door that has the goat. Would this effect the propability? as far as I can tell the answer is no.
continued in next comment
gtq838 1 year ago
@gtq838 the store clerk just got a new box of 100 lottery tickets there is a jackpot ticket in this box. You buy one ticket and the guy behind you buys the rest. he scratchs off 98 of them and they are all losers what are the odds that you have the winning ticket? 1/00 right? doesn't his last ticket have a 99% chance. the chart still holds true if you graph it out. its even more counter intuitive but forknowledge doesn't make a difference unless the winning ticket is scratched
gtq838 1 year ago
@gtq838 never mind I made a program and tested the idea I'm wrong and I figured out why. I failed to take into account that the gentleman must pick the same door/ticket in each row of our chart and thus there are only two equal posibilities left.
gtq838 1 year ago
thats a different set of conditions.
DarkKnightBob1o1 1 year ago
@singingbanana Games of chance are a form of advantage taking, as there is no skill involved. Unlike tennis for example which requires that skill be honed.
TomMarAlem1987 1 year ago
@Anon909 '
Can you explain?
In my opinion, you still have twice as likely chance to pick a goat at the start, and if you're lucky, the game show host will reveal another goat, and you can still switch and have the same odds. It really doesn't matter whether the host knows or not, but it is still more likely you chose a goat at the start.
BDpartnercoJM 11 months ago
@Anon909 That you just said, as also our "Host" said, is easily bypassed as it is only a detail, but a crucial one for this problem. The reason so many mathematicians had trouble realizing this is becuase they ignored the fact that it's not a pure mathematical problem, but actually has the influence of a (tricky) human mind.
Btw @singingbanana I loved your 100 door example, it made me understand this issue alot better as I've tried to understand this problem before :)
kennethkrist 9 months ago
@Anon909 No. Him knowing would not matter
mstrgrl 8 months ago
Oh, now I see. The odds you chose the car first is only 1 in 3 therefore the odds that it was another door are 2 in 3 . . . . . . . . . I'm only 11 don't blame me . . . .
TheWolzen 1 year ago
I ran a test with 100,000 runs and ran it multiple times and it always was around 2/3 of the time, you win when switching. And it really makes sense, if you think about it. It's only 50% if you aren't given the initial choice.
TheAwesomeBrandon 1 year ago
ok, since no one explained me anything, I proved myself wrong :D If you make a really big chart in witch you enter EVERY single case and the chanse of it happening, you see it clearly: 1/3 chanse vs 2/3 chanse or 33% vs 66%
eneses93 1 year ago
@eneses93 I'm glad you did it, but it might be possible to convince yourself that a chart like mine is all you need :)
singingbanana 1 year ago
I decided to make a second counting (sorry if your tired of reading it all)
1. I choose Door 1, he opens Door 2: a) the cars are in Order #1 b) the cars are in Order #3
You can make two cases for you choosing Door 1, him opening Door 3; you choosing Door 2, him opening Door 1 etc. etc. I'm still interested in this theory, so if anyone has something to back it up, you can write me a letter.
eneses93 1 year ago
I think that your not counting the cases right:
1. I choose Door 1, the cars are in Order #1 he opens Door 2 (don't change)
2. I choose Door 1, the cars are in Order #1 he opens Door 3 (don't change)
3. I choose Door 1, the cars are in Order #2 he opens Door 3 (change)
4. I choose Door 1, the cars are in Order #3 he opens Door 2 (change)
Because he could open Door 2 or 3 if I choose Door 1, and that should be counted as a different case.
eneses93 1 year ago
i just tried to explain this to a friend of mine and he didnt believe me . . .
keggerous 1 year ago
I love the little Mogwai!!!
BigbyWolf20 1 year ago
Question: can this really be classified as a "paradox" ?
Tolstoievsky 1 year ago
@Tolstoievsky No paradox, just counter-intuitive.
singingbanana 1 year ago
I was viewing it wrong the whole time! The chart really helped. I thought I was going mad with thought. Thanks for clearing it up for me.
MasterAdam100 1 year ago
I just watched it on scam school :)
It really does make sense with the 100 doors! I get this fully now
MattTheGreat2008 1 year ago
this is in movie 21
Frosteees 1 year ago
love your board
rotnakleugim 1 year ago
you are a genius, im 14 years old and you have opened my mind like it is not possible
needpants 2 years ago
In bridge, this is called the Law of Restricted Choice (or principle of RC). It works the same way. If you need to play a suit that contains AT32 opposite your K9876 without losing a trick, you play the Ace and say the Jack appears as the last card, its 2/1 that the player who played last didn't have a choice.
It's probably the biggest application of the Monty Hall problem I can think of.
manudude02 2 years ago
That's interesting. I don't play bridge, but that sounds right.
singingbanana 2 years ago
The difference between the Monty Hall problem and games such as Deal or No Deal is that the game show host in the Monty Hall problem knows which door the prize is behind and thus his choice to reveal a door with a booby prize is not random. In Deal or No Deal the choice is completely random as nobody knows where the 250,000 is. For these reasons it is correct that you should always swap in a Monty Hall scenario as you stand a 2/3 chance, however in DoND, the choice of swap or no swap is 50:50
tyler86media 2 years ago
This is a complete joke that this problem has gotten so much attention..Its sooo simply. You change door everytime and then i win 2 out of 3 you only lose if you pick the winner first...simple huh?
donrane 2 years ago
this is very wrong, it has nothing to do with intuiton, it's simple science and math. The fault is in the premises of the 3rd explanation. Once you releave one of the options, the equation changes completely and you cannot add the variables from the 1st equation into the second equation. This is garbabge and nonsense. It's still 50/50 not 2/3.
eastofakron 2 years ago
Comment removed
polywoly15 2 years ago
at first u have 67% chance of losing. however if u switch the chance of losing will become the chance of winning and vice versa. (pick a goat and switch= car... pick a car and switch= goat). 67% chance of picking goat... 67% of getting car if u switch. see the problem beginning to end. not just the ending options.
richardtherichard 2 years ago
@eastofakron You are incorrect. The first equation is still in play.
You had a 1 in 3 chance of picking the right door.
That is a 2 in 3 chance of picking the WRONG door.
(So, 2/3 of the time you pick a WRONG door to start with.)
Now, the host opens the OTHER WRONG door...
(The host knows where the prize is, so it's 100% chance he opens a wrong door.)
2/3 of the time, you BOTH picked WRONG doors...
Should you switch?
Answer: Yes, unless you're feeling REALLY lucky!! =D
QCHitch 1 year ago
Finally I understand the why-part. Watched some other videos which just told how it is. Very good!
123456789Aksu 2 years ago
This is a very good explanation!
The reason why it is counter-intuitive is because there is an unspoken rule; Monte will never pre-reveal a door that has a car behind it. If you take away that rule then it all because intuitive.
If you change the rules so that Monte can't know which one has the car then everything changes. This would also lead to the possibility of Monte opening a door you didn't pick to reveal the car, thus showing that you lost before you even open the door you picked.
WarrenMI48091 2 years ago
Further elaborating, if you pick a number between 1 and a 100 and I then eliminate 98 numbers (assuming I have no knowledge of the winning number), leaving your number and one other number, 98% of the time I would eliminate the winning number and you would lose. 1% of the time you would have picked the winning number, and 1% of the time I would have given you the option of switching to the winning number.
WarrenMI48091 2 years ago
I don't think does work for deal or no deal does it though because the boxes are taken away at random. Well explained vid though.
EricSeanPayn 2 years ago
Which is why I said this doesn't work for Deal or No Deal because the boxes are taken away at random.
singingbanana 2 years ago
My bad, I was reading through the comments towards the end, not listening to you with full attention. :)
Hope you can forgive me.
EricSeanPayn 2 years ago
What bearing does the random removal of boxes have in this regard? If it's just your box and one other box left and 250,000 and 1p left on the board, aren't you being given the same gaurantee, should you take the swap, as in the Monty Hall Problem when two boxes remain? Say your sole intention was to win 250,000. The box you chose initially had a 1/22 chance of containing it, a 21/22 chance of not containing it. If the swap is offered doesn't the boosted probability (21/22) apply as in Monty H?
ExplodingggBoy 2 years ago
Well,think of it this way. What are the chances that you picked 1p? 1/22,same odds of you picking 250,000. So the chances don't increase,it'll be 50/50 if you have 1p and 250,000.
Feno900 2 years ago
I figured it out in 2 minutes at a Marines game when my friend Chris proposed it to me.
I am no special person.
I am a nice guy.
milfballs 2 years ago
I GET IT!! So many people have tried explaining this put I never understood, UNTIL NOW!! Thank you Dr. Grime I'll sleep better now.
BlAiR6996 2 years ago
mogway!
PivotWithBiscuits 2 years ago
thats just either way if you take out 1, of those 3, that is wrong. you'll end up at 1 right and 1 wrong so thats 50/50. but if your talking about "chances", thats the different side of the story. you got 33.33% from the beginning due to 1/3 is your chance, by removing one you get 66.66%. but logically speaking, its still 50/50 at the end, atleast thats what i think
mipangaluth 2 years ago
no the chance of you picking the right door from the start is 1/3 that fact doesent change when he opens another door.
in fact the 1/3 chance that was on the door he opened is transfered to the remaining doors that u didnt pick.
well because the car is either in the door you picked OR in a door that you didnt pick. 2 options but not with the same chances.
Alfaomegabravo 2 years ago
thank you, finally some common sense
eastofakron 2 years ago
Yes, you are right
eastofakron 2 years ago
i was gonna ask you if this wold be like deal or no deal! lol im glad you said it at the end
RickyAustinR 2 years ago
Marilyn vos Savant. Name says it all...
ComanderLock 2 years ago
I really had no idea how it could not be 50/50 when you have 1/3 choice. Whoever thought of that must be damn smart
Moorich 2 years ago
cool
TheAceOfSpades100 2 years ago
that sam thing was in the movie 21
Magicguy2112 2 years ago
yeah but now i understand the reason behind it all
polywoly15 2 years ago
i dont get it but ur really smart
sportzman10 2 years ago
very very smart...
stribor666 2 years ago 9
cool!!
rotnakleugim 2 years ago 5
Comment removed
hyldago 2 years ago
cool
octan488 2 years ago 2
really interesting and you are really cute ^^"
emathed1 2 years ago 2
I am completely lost. But you're cute.
onefridaynight 2 years ago
Singingbanana, you're the man, love your vids!
funtube99999 2 years ago 3
where is ur accent from??? itz kool
ConnBommFM 2 years ago
i just watched 21 and this is how the movie begins
ClickHereifyouwill 2 years ago
I haven't seen the film myself, but this problem has become quite well know, it's a very good problem.
singingbanana 2 years ago
it was under the table actually
jimmyrich1 2 years ago 3
u didnt do any of this is, its an exact copy of what marilyn vos savant did.
jonatan9505 2 years ago
Well done. You've caught me. I didn't invent mathematics. Does that mean you win?
I did reference the Parade article in the video. Yet the problem was old when Marilyn was asked about it in 1990. At least 15 years old, if not older.
It's a modern classic, I said that in the video too didn't I? The proof is my own, but would be similar to any mathematician's proof.
If you prefer, next time I will make a video about my own maths research in combinatorial representation theory.
singingbanana 2 years ago
blame wiki not me P:
jonatan9505 2 years ago
easy
rhinomaster1000 2 years ago
man ur too smart
blizzmonkin 2 years ago
You are supposed to switch cuz it like 1/3 and 2/3 or sumthing
JellyfishPie00 2 years ago
OMFG! I love this problem! This problem was on Numb3rs and Charlie explained this to his students, it was so awesome!
thamiks 2 years ago
this was in a card counting movie but i cant think of the name. it was made in 06 or 07
jimmaa66 2 years ago
The movie was called 21.
cardsfan7012 2 years ago
i knew it was behind #2!!!
yippie29 2 years ago
I Love This.
Someone (In America) wrote into a maths equilvalant agony aunt and asked about this. He asked should he stay with the door he has or always switch to another.
the response was he should always switch. most of americas mathematicians got angry at this and the 'agony aunt' was sent lots of hate mail saying things like. thats wrong this is why the rest of the world think that americans are dumb and that you should always stay with the door you picked.
XDeVaStAtOrZ 2 years ago
and infact they were all wrong.
Gutted!
XDeVaStAtOrZ 2 years ago
Fascinating. Many thanks! Perhaps this has an application in a magician's arsenal? Must put my thinking cap on.
norrisonthespot 2 years ago
Indeed, surprising mathematical curiosities are called mathmagics (mathemagics), by mathmagicians (mathemagicians). Have a look for Arthur Benjamin. Those things are not necessarily Monty Hall though.
singingbanana 2 years ago
how long did it take you to get your phD in math?
Barthayn 2 years ago