Jump to content
Sign in to follow this  
Superbowlbuc

Probability Problem

Which is it?  

16 members have voted

  1. 1. Probability Problem

    • Don't switch. The car is definitely behind door 2.
      0
    • Don't switch. The car is more likely to be behind door 2.
      0
    • No need to switch. The probability of either door is 1/2.
      2
    • Switch doors. The car is guaranteed to be behind door 1.
      2
    • Switch doors. The car is more likely to be behind door 1.
      8
    • Fuck Math
      4


Recommended Posts

I'm a gameshow host, you're the contestent.

 

There's three doors, one has a car, the other two have nothing. I am fully aware of which doors contain what.

 

You pick a door (for this example, say you pick door two). I then open door three and show you there is nothing behind door three. I now give you the option to change your door. Which choice above is correct? Should you switch to door 1?

Edited by Superbowlbuc

Share this post


Link to post
Share on other sites

I'm a gameshow host, you're the contestent.

 

There's three doors, one has a car, the other two have nothing. I am fully aware of which doors contain what.

 

You pick a door (for this example, say you pick door two). I then open door three and show you there is nothing behind door three. I now give you the option to change your door. Which choice above is correct? Should you switch to door 1?

 

its behind door 1 :troll:

Share this post


Link to post
Share on other sites

Yes you switch to door 1. The first door you chose (door 2) had a 2/3 chance of being wrong, but since he showed you door 3 as empty, door 1 now only has a 1/3 chance of being wrong and hence a 2/3 chance of being right.

  • Upvote 2

Share this post


Link to post
Share on other sites

Yes you switch to door 1. The first door you chose (door 2) had a 2/3 chance of being wrong, but since he showed you door 3 as empty, door 1 now only has a 1/3 chance of being wrong and hence a 2/3 chance of being right.

 

:clap:

Share this post


Link to post
Share on other sites

Wasn't this in a movie or something?

Share this post


Link to post
Share on other sites

Yes you switch to door 1. The first door you chose (door 2) had a 2/3 chance of being wrong, but since he showed you door 3 as empty, door 1 now only has a 1/3 chance of being wrong and hence a 2/3 chance of being right.

 

Correct.

 

Have you heard this one before? I've shown it to a lot of people. You're the only one to get it right with the correct reasoning.

Share this post


Link to post
Share on other sites

I've heard it before and I still don't understand it.

 

It seems to me that opening door 3 and being shown there is nothing behind it changes the odds entirely. Now your current door has a 1 in 2 chance of being right, and the other door that is not opened also has a 1 in 2 chance of being right.

 

I don't see how the initial situation affects the probability of the second situation.

Share this post


Link to post
Share on other sites

Wasn't this in a movie or something?

yeah, it was in 21:

 

I've heard it before and I still don't understand it.

 

It seems to me that opening door 3 and being shown there is nothing behind it changes the odds entirely. Now your current door has a 1 in 2 chance of being right, and the other door that is not opened also has a 1 in 2 chance of being right.

 

I don't see how the initial situation affects the probability of the second situation.

Here's the explanation, if you want the short version just read the bold part at the end

 

The key is that the host knows what door the car is behind and he doesn't want to open the door the car is behind because that kills the game, keep that in mind through my explanation. With your initial pick, you have a 1/3 chance of being right, which means a 2/3 chance of being wrong, hopefully that much is intuitive enough. Once you have picked, the host reveals what's behind one of the two remaining doors. For ease of explanation let's assume that the car is behind door 2. If you pick door 1 initially, the host is going to open door 3, leaving door 2 there with the car, so switching would give you the car; if you pick door 3 then the host is going to open door 1, again leaving door 2 there with the car; if you chose door 2 then the host could open either door, but either way switching will give you nothing. So, of the three initial options, only one would leave you with nothing after the switch, and that's if you pick correctly the first time, in other words by switching doors you're betting on your first pick being wrong.

Edited by oochymp

Share this post


Link to post
Share on other sites

The other two have nothing? That's lame. Make the other two have tigers behind them or something.

  • Upvote 2

Share this post


Link to post
Share on other sites

In the original problem, the other two have goats. I wanted to change the wording so people wouldn't google it.

 

Here's a more mathematical explanation if you don't get it.

 

We're trying to find the probability of winning if you switch doors.

 

The probability of picking the correct door initially is 1 in 3. The probability of winning if you pick the correct door and switch to a different door is zero.

 

The probability of picking an incorrect door initially is 2 in 3. The probability of winning if you initially pick the wrong door and switch is 100 percent. The gameshow host is never going to show you the car door or the door you picked. Therefore, if you picked the wrong one, switch, and don't switch to the wrong one the host showed you, the only door left to switch to is the one with the car.

 

So... probability of winning if you switch =

 

P(Picking Correctly initially) * P(Picking correctly initially, switching, and winning) + P(Picking Incorrectly initially) * P(Picking incorrectly initially, switching, and winning)

 

= (1/3)*0 + (2/3)*100

 

= 2/3 chance of winning if you switch

Share this post


Link to post
Share on other sites

I'll remember that if I'm ever on a game show.

Share this post


Link to post
Share on other sites

Correct.

 

Have you heard this one before? I've shown it to a lot of people. You're the only one to get it right with the correct reasoning.

 

Ummm I remembered a similar question from my high school finals ages ago lol. But the rest I just sorta worked out. It's a basic probability sequence when you think about it.

Share this post


Link to post
Share on other sites

It has less to do with winning the car than beating the odds then. After one door has been opened, you have a better chance of beating the odds again by switching to the other unopened door.

 

Now what if there were more than three doors?

Share this post


Link to post
Share on other sites

It has less to do with winning the car than beating the odds then. After one door has been opened, you have a better chance of beating the odds again by switching to the other unopened door.

 

Now what if there were more than three doors?

 

If I understand this correctly, it would not matter how many doors there were, as long as this same scenario plays itself out, you will always have a better chance of winning if you swap doors, than if you stay with the same door.

Share this post


Link to post
Share on other sites

It has less to do with winning the car than beating the odds then. After one door has been opened, you have a better chance of beating the odds again by switching to the other unopened door.

 

Now what if there were more than three doors?

These scenarios are all about beating the odds, in probability there is no such thing as a sure thing, you just want to get the odds as far in your favor as possible

 

For the more than 3 doors scenario, your strategy depends on how many doors he opens, but overall you get the most benefit by switching.

 

If he only opens one then you're still better off switching but less so if there are more doors. I'll see if I can explain it, but it's late and I haven't taken a math class in a while, so bear with me. If there are x doors then your probability of picking the right door the first time is 1/x so there is a 1/x chance of winning if you don't change doors. Once the host has opened a door, there are now x-1 doors, assuming you picked the wrong door initially you have a 1/(x-2) chance of picking the right door (x-2 because of the original x doors there are 2 that you cannot choose: the door you chose initially since you're switching and you can't choose the door the host opened). This is where the math gets a little complicated, so again, bear with me. You had a 1/x chance of picking correctly the first time, which would give 0 probability of winning if you switch and a (x-1)/x probability of picking incorrectly, which would give you a 1/(x-2) probability of winning. So, the probability of winning if you switch is:

 

(1/x)*0+((x-1)/x)*(1/(x-2))=(x-1)/(x*(x-2))

 

At this point, the question is whether it is true that (x-1)/(x*(x-2))>1/x, which I will break down below:

 

(x-1)/(x*(x-2))>1/x

x/(x*(x-2))-1/(x*(x-2))>1/x

x/(x*(x-2))-1/(x*(x-2))+1/(x*(x-2))>1/x+1/(x*(x-2))

x/(x*(x-2))>(1/x)*((x-2)/(x-2))+1/(x*(x-2))

x/(x*(x-2))>((x-2)/(x*(x-2)))+1/(x*(x-2))

x/(x*(x-2))>((x-2)+1)/(x*(x-2))

[x/(x*(x-2))>(x-1)/(x*(x-2))]*(x*(x-2))

x>x-1

 

Note that this works only if we assume that x>2, but since x is the number of doors we established that in the initial qualifications of more than 3 doors, also if you plug x=3 into the equation (from the original problem) you come up with (3-1)/(3*(3-2))=2/(3*1)=2/3.

 

Now, if you have more than 3 doors but the host keeps opening doors until you're down to two doors, it's to your benefit to stick with the same door until you're down to two, then switch, basically betting that you were wrong initially and forcing the host to remove all of the other wrong options. If it's somewhere between opening one and opening all, then you're best off waiting until the last opportunity to switch and switching then on the same principle.

Edited by oochymp
  • Upvote 1

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
Sign in to follow this  

  • Chatbox

    TGP has moved to Discord (sorta) - https://discord.gg/JkWAfU3Phm

    Load More
    You don't have permission to chat.
×