Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probabilities from movie 21

  1. Dec 4, 2012 #1
    This is the problem that a teacher (Kevi Spacey actually) presented the class on the movie 21:
    You are on a TV show and the announcer asks you to choose 1 from 3 doors.
    Behind 2 doors there are a goat and behind 1 there is a new car.
    The guy coose door 1.
    The announcer go and opens door 3, in wich there is a goat, and asks the guy if he want to change.
    He said yes and open door 2 and win the car.
    Now the teacher asks why the guy changed the door? He answer, at the begining he had 33.3% of chance of being right, but when the announcer (that knew where was the car) asked if he would like to change, he thought that changing would increase his chance of winning to 66.6%.

    My question is if this calculation is right? After 1 door was opened I think that probabilities would change to 50%.
    What do you guys tell me?
     
  2. jcsd
  3. Dec 4, 2012 #2
    The calculation is correct; your chance remains 1/3 if you keep the door you selected first, and so the other door promises a 2/3 chance of winning the car. This problem (known as the "Monty Hall problem", which you can google) continues to cause discussions, since many people insist that two remaining doors should both have equal chances of hiding the car. The doors, however, are not equal, since you chose one (without knowing what was behind it), and the TV show host chose the other (knowing quite well what was behind).

    Here's a similar scenario that might convince you that you should swap when the person running the game offers you the chance:

    Imagine a lottery with 1 million tickets. One of these tickets is marked "$1 million", all others are worthless. First you choose one ticket at random. Your chances of having won the big sum are 1/(1 million), okay?

    Now the person in charge looks at all remaining 999,999 tickets and throws 999,998 of them away. He's now left with one single ticket, which he offers you in exchange for the one you have. Will you swap?

    If you now say that your chances are 50% of winning the money and you shouldn't swap, how do you explain the ticket that just had a chance of winning of 1/(1 million) having suddenly become more valuable, although you did absolutely nothing to increase its winning chances?
     
  4. Dec 5, 2012 #3
    Some implied rules about what doors the host can open before offering the choice are often left out of the example for this puzzle. They are:
    1. The host cannot open your door.
    2. The host cannot open the door with the prize.

    The rules seem obvious in the context of a game show, but when mathematically figuring the probability people suddenly forget them.

    If the host cannot open the door that reveals the prize, and 2/3 of the time you pick the wrong door initially, then 2/3 of the time the host has no choice about which door to open -- he must reveal a loser, because his other two choices are invalid -- your losing door, and the winning door.

    If the host were allowed to open your door, then assuming he did so and revealed a goat, your chance of winning after switching would be 50%.
     
  5. Dec 5, 2012 #4
    There's nothing implied here. The host doesn't open my door, and he doesn't open the door with the prize either. What he could doesn't enter the matter.

    If you don't bother about what has actually happened in this story, but you're going to study all possible scenarios, then you ought to consider the case in which the host opens a door and reveals the prize behind it. Now that's the kind of show I'd want to participate in. :smile:
     
  6. Dec 5, 2012 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It is a fairly easy computation to show that if the encee knows which door the prize is behind, and intentionally does NOT open that door, then, after a door is opened, the contestant can, by changing doors, make use of the emcee's knowledge to change the probability of winning from 1/3 to 1/2.

    It is interesting that if we assume the emcee does NOT know which door the prize is behind but, by chance opens a door that does not have the prize, then switching doors does NOT change the probability.
     
  7. Dec 5, 2012 #6
    The change is from 1/3 to 2/3, not 1/3 to 1/2, as 2/3 of the time you picked the wrong door initially and the host had no choice about which losing door to open, as there was only one possible choice for him. This leaves the prize behind the unpicked and unopened door.

    In this scenario can he pick the same door as you, or are his choices limited to picking one of the other two at random?
     
  8. Dec 5, 2012 #7
    I would say the odds change from 1/3 to 2/3.

    ? Are you certain? I don't see how it makes any difference what the host knows in this case. The number of choices is reduced, so the probability changes.
     
  9. Dec 6, 2012 #8

    pwsnafu

    User Avatar
    Science Advisor

    Yes we are certain. See the Ignorant Monty section.

    Edit: Oh wait. You mean it changed from 1/3 to 1/2. Then you're right.
     
  10. Dec 7, 2012 #9
    After the host reveal a door, a second contestant joins in to pick a door from remaining 2 doors shown.

    (1) If he has seen the host revealing the door, is his probability higher than 1/2 if he always pick differently from the first contestant first choice?

    (2) If he has not observed what happened earlier, does his probability change from 1/2 to 2/3 the moment he happens to pick differently from the first contestant first choice?
     
    Last edited: Dec 7, 2012
  11. Dec 7, 2012 #10
    If the 2nd contestant knows which door the 1st picked, and which one the host revealed the goat behind, then he has the same 2/3 chance of you if he picks the remaining door.

    If the 2nd contestant knows which door the 1st picked, but not which door the host revealed a goat behind, his chance is 1/3. He only has two options:
    1) Intentionally pick the same door as the 1st, and assume the same 1/3 chance.
    2) Randomly pick one of the other two doors, a 1/2 shot at the remaining 2/3 -- i.e., a 1/3 chance.

    Where does the 1/2 keep coming from? The scenarios that lead to a 1/2 chance are very convoluted.
     
  12. Dec 7, 2012 #11
    How can it be 1/3 if it is randomly picked? A new scenario with only 2 doors shown to a new player who has not observed anything should at least be 1/2 chance.
     
  13. Dec 7, 2012 #12
    Both of the doors he has to choose from could hold goats. 1/3 of the time, they both will. Therefore his chance of winning is 1/2 * 2/3, not 1/2 * 3/3.
     
  14. Dec 7, 2012 #13
    For this new player who has not observed anything, there is no longer a 3rd door in the choice equation (it can be hidden behind a curtain).
    There are just 2 doors for him to pick from when he arrives.
    He can randomly pick one of the two.i.e. Prob=1/2
     
    Last edited: Dec 7, 2012
  15. Dec 7, 2012 #14
    He can *pick* one with a probability 1/2, but the chance that the car is behind one of those two doors is only 2/3. The 1st door still exists, and still may contain the car. Even if you hide it behind a curtain. ;)
     
  16. Dec 7, 2012 #15
    Its already stated that the hidden door is no longer relevant since the host had shown it has no car. As far as the new player who has not observed anything is concerned, he is just presented with a NEW game where there are only 2 doors, one with the car. However, rather than guessing the 2 doors, the monty hall problem is saying the new player just need to bet that the first player is a loser.
     
    Last edited: Dec 8, 2012
  17. Dec 8, 2012 #16
    For some reason I thought it was the 1st contestants door you were 'hiding', not the one the host opened, sorry! It wasn't clear to me what you meant until now.
     
  18. Dec 8, 2012 #17
    So it is the info of which door first player picks rather than knowing which door host reveal that gives the edge from 1/2 to 2/3. The new player may not be aware but from the audience standpoint, is there some kind of a hidden transient change in the probability the moment he happens to pick the opposite as the first player by himself before the 2 doors are open?

    Given a fixed time period,say 1 week, the stock market can go up,down or stay sideways.
    A hedge fund can long, short or use rangebound(options strategy) to bet for that week closing level.

    Hedge fund manager A opened long position the market on monday.
    On tuesday, there is some major news/economic data that can be interpreted either way so the market will breakout of the trading range to move in one unknown direction(3rd door is out).

    Hedge fund manager B comes along after the news. Should he
    (1) randomly pick one direction?
    (2) instead of analysing market direction, find out fund manager A's position and bet the opposite?
     
    Last edited: Dec 8, 2012
  19. Dec 8, 2012 #18
    What gives the advantage is a combination of two factors.
    1. The contestants initial chance of picking the right door is 1/3. Nobody will argue this.
    2. The host then opens one of the other two doors, reducing that doors chance of holding the prize from 1/3 to 0/3, and shifting that probability to the remaining unpicked door -- as combined they had a 2/3 chance of holding the winner.

    If a 2nd player comes in who knows only which door the host opened, he has a 1/2 chance of picking the right door from the remaining two. If he knows only what door the 1st contestant picked, his odds of picking the right door are also 1/2. If he knows both things, then he can pick the door that the 1st contestant should switch to, and get the same 2/3 odds.

    I don't know what you mean here, but I described all possible scenarios above.

    In this scenario, sure, short everything. However stock markets are not nearly so unpredictable, and news that comes out on monday is going to affect the market on monday as soon as the news breaks -- not the next day.
     
  20. Dec 12, 2012 #19
    Wrong.Only applies to the first contestant.2nd player is shown 2 doors only.

    Correct, the critical question is does it matter whether the host had opened the SAME empty door knowingly or randomly(independently guess) from the other two doors?
    If the host knows everything, it is guarranteed that the game will proceed to the 3rd stage(to switch or not) since he will always pick an empty door.
    If the host just guess one of the other 2 doors, the game may stop at 2nd stage(host reveal door).
    The question is whether the overall situation is affected if the host guess the same empty door and the game still proceeds to stage 3 as per normal? Mathematicians seem to think it is...but why in layman terms?
    http://math.stackexchange.com/questions/41807/variation-on-the-monty-hall-problem?lq=1
    2nd player probability depends on whether he knows how many other doors existed before he is shown the 2 doors game he is playing and 1st contestant's choice. If he does not know anything, from his perspective, it is always 50-50.
     
    Last edited: Dec 12, 2012
  21. Dec 12, 2012 #20
    Yes, "from his perspective." If door A is the 1st contestants choice, B is the hidden door with the goat, and door C is the remaining door, there is still a 2/3 chance that door C holds the prize.

    The fact that the 2nd contestant doesn't know which door (A or C) the 1st contestant picked does not change the odds of a door holding the prize -- which is what the puzzle is all about -- it just changes the odds of the 2nd contestant picking the door with better odds.

    Do you see the difference?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Probabilities from movie 21
Loading...