The Monty Hall Problem
October 42009
The Monty Hall Problem is a famous (or rather infamous) probability puzzle. Ron Clarke takes you through the puzzle and explains the counter-intuitive answer.
You can read more about this problem, and the controversy, on Marilyn Vos Savant’s website www.marilynvossavant.com
Duration : 0:5:48
[youtube mhlc7peGlGg]
October 4th, 2009 at 3:13 am
I know exactly what …
I know exactly what you’re saying….but i’m trying to explain it in another way, since what you’ve said above has been said in hundreds of posts already!
@ Kibate
Thats exactly what im saying….the more doors you have…..the greater your’e chances of winning if you swap!!!
The 99 other doors ARE one choice…that have a 99% chance of containg the prize! Just because he opens 98…..does not mean the chances of the car being in the last decreases to 50% its still 99%!
October 4th, 2009 at 3:13 am
Oh and just to make …
Oh and just to make this more understandable:
The two goats in one scenario are two different goats (which is obvious in reality) and therefore there are 6 possible scenarios (listed in the other comment) and not 4 (:
or as the video is saying… when you pick a door you are more likely to have picked the wrong door (66%), and you know that the host is going to open the other goat – so the other door is more likely to be the car, as you are likely to have picked the wrong door first round.
October 4th, 2009 at 3:13 am
i wuld rather the …
i wuld rather the goat, my dads got a fkn awsome car lol
October 4th, 2009 at 3:13 am
C [G1] G2 swap, get …
C [G1] G2 swap, get goat
C G1 [G2] swap, get goat
G1 C [G2] swap, get car
G2 C [G1] swap, get car
G1 [G2] C swap, get car
G2 [G1] C swap, get car
You forgot to count the probability of scenario-of-getting-goat twice, as there are two goats.
If there are 2 goats, you need to count the probability of the goats being swapped around, even they are identical, the scenario is counted as two. Hope that helps! (:
October 4th, 2009 at 3:13 am
THanks Mr Clark. …
THanks Mr Clark. well explained!!!
i didnt get from the literature i was studying, but here i got it perfectly
5 stars!
October 4th, 2009 at 3:13 am
meant to add one …
meant to add one thing: 66% of the possible number of total game scenarios leads to a win IF you swap.
October 4th, 2009 at 3:13 am
@Kierz77 It’s …
@Kierz77 It’s tricky, but the numbers if doors is not what we are counting. It’s the number of total events (or in this case, you can call it an individual game) that leads to a winning status. If the hosts knows where the car is and picks a goat, then 66% of the possible number of total game scenarios leads to a win. It’s not based on the number of doors per say. Sorry if that is more confusing.
October 4th, 2009 at 3:13 am
Just 66%. Remember …
Just 66%. Remember 66% represents the total number of events that leads to a win, it’s not the probability between the last 2 doors.
Say we choose the first column below, then the host chooses a goat, and then we swap
CG(G)
C(G)G
That leads to 2 scenarios where we lose.
Now suppose we choose the first column below, then the host MUST choose the only remaning goat, and we swap to the only other option: the car
(G)CG
GC(G)
(G)GC
G(G)C
That is 4 scenarios for a win. 4/6 total = 66%
October 4th, 2009 at 3:13 am
your first door 1) …
your first door 1) and second door 1) are the exact same scenario just backwards. You’re counting it twice when you should only count it once.
You can also write 2) as [G] C G and 3) as [G] G C, but it still won’t affect it.
October 4th, 2009 at 3:13 am
So you’re …
So you’re considering that there are 100 doors?
If so, and as you said “the host opens 98 doors at once with goats”, the chance that you picked a goat to begin with is 99% so then when the host opens the 98 other doors with 98 goats the chance that your initial choice was that of a goat is still 99%.
So when he narrows it down to 2 doors the probability that you have a goat behind the door you picked is still 99% which means when you swap doors there is a 99% chance you’re swapping to the car.
October 4th, 2009 at 3:13 am
Think of it first …
Think of it first with these rules, which are fact:
1. If you choose a car first, when you switch you lose
2. When you choose a goat first, when you switch you win
Considering those, the number of times you choose a goat first is 2/3- 66.6%! And if you always switch, you will always win in this case.
If you commit to always switch, you choose a goat twice as often as the car first, and you win twice as often!
October 4th, 2009 at 3:13 am
sorry, still …
sorry, still doesn’t make any sense to me (i replied to DanceInYourRoom down there too, please look at it too)
but considering we stay with door1(or the host open 98 doors at once with goats) that means we have to put those 99 other doors together as one choice, therefore making it 50/50.
But according to your logic, the more doors there are, the better the chance is we get to win? In your example we would have a 99% of winning by swapping in the last part of the game
October 4th, 2009 at 3:13 am
this is where i see …
this is where i see the problem(and still believe its 50/50)
because it should be:(the [] are the doors the host opens, assuming we always choose the first door first)
1) C [G] G
1) C G [G]
2) G C [G]
3) G [G] C
people forget that the host can choose from 2 options if the car is in door1
or at least this is what bugs me to the end
October 4th, 2009 at 3:13 am
It makes perfect …
It makes perfect sense!!
The chances of you picking the door with the car with ur first pick, is 1/100.
The fact that the host knows where the car is, and proceeds to open all but one of the other doors, means that door has a 99% of hiding the car!!
By swapping, you’re turning the initial chances of picking a goat (99/100) into the chances of you picking a car!
October 4th, 2009 at 3:13 am
LUCK people…LUCK. …
LUCK people…LUCK…:)))))
October 4th, 2009 at 3:13 am
wow this has been …
wow this has been buggin me for a long time. good explaination!
October 4th, 2009 at 3:13 am
haha, best comment …
haha, best comment ever! I understand it but I Completely agree it goes against my intuition.
October 4th, 2009 at 3:13 am
The probability 66% …
The probability 66% does not represent what’s behind the last two doors. It represents the 3 events 1) 2) 3). Of course it is 50%…between those two doors. But we’re looking at the whole picture: which starts with 3 doors and involves the host changing the choices halfway through – which really makes it counter-intuitive.
October 4th, 2009 at 3:13 am
It’s 66%. There are …
It’s 66%. There are 3 options:
1) C G G
2) G C G
3) G G C
If you pick option 1) and swap, you lose. If you pick 2) or 3) and swap, you win. Hence 66%.
What people who say 50/50 don’t understand is that you’re not taking into account that after you choose a curtain, the host is forced to choose a goat. Try it using the little diagram I left and you’ll see it’s 66%.
October 4th, 2009 at 3:13 am
One last try: I …
One last try: I cannot see my new comments even though I can see the first two, but even though it BURNS, BURNS my logic centers to a crisp, I do not believe people would lie about scientific tests.
I am leaving out the link this time just in case that is somehow causing it.
October 4th, 2009 at 3:13 am
Or to look at it …
Or to look at it another way, the basic claim is that after the first goat is revealed the odds that your initial pick was right are still 1/3.
By that logic, if there are only TWO doors, and he shows you what was behind the other, are the odds that your door hides a car still 50/50?
October 4th, 2009 at 3:13 am
Dice have no memory …
Dice have no memory. The “stick/swap” option is in actuality a COMPLETELY NEW CHOICE which (as there are only 2 doors left) each have a 50% chance of having a car.
Otherwise you’re saying that if he were to lead you away from the first set of doors, and show you a NEW set of doors — this time only two — and each one also has a goat and car respectively behind it, then the probability would be somehow different from the old set of doors.
October 4th, 2009 at 3:13 am
ya i understand how …
ya i understand how it works, but would u know if u picked the car or not
October 4th, 2009 at 3:13 am
it would work with …
it would work with only three doors. Besides if you pick a door that has a car hidden inside it, than your odds are totally different, because if you go with ur first choice than u win the car, but if u swap at that situation than u win a goat. This problem only works if u pick a door in the first place that has a goat hidden, and host reveals the another door with a goat instead.
October 4th, 2009 at 3:13 am
wait but how would …
wait but how would u know if u picked the goat or the car, cuz say u did pick the car the first time, but u wouldn’t know that, but in the video it saids u should always swap. wat would happen if theres 4 doors?