Let's Make A Deal

[1down5up 7]

Suppose you are on a game show and must choose one of three doors. Only one door hides a prize. You choose Door #1. The host opens Door #2 to reveal the prize is not there. You are now given the option to stay with Door #1 or switch to Door #3. Do you switch or stay?

[bbk 188]

Ha, I learned the actual math behind this classic game show question in a course I'm taking this semester actually.

[Ray Ray 3,566]

what?

[Maksim 1,504]

This was on Mythbusters, you switch, always. You then have a 66% probability.

[bbk 188]

Yea, I didn't want to say it early on, but you always switch.

 

Since you're always shown a door that would garner no prize, it is easy to assume that there is then a 50% probability that either the door you picked or the other one has the prize. However, since the prize is only behind 1 of 3 doors to began with, there is only a 33% probability you picked the door that had the prize to began with. Since they show you said door without the prize, that means by switching, you have a 66% of switching to the door with the correct prize compared to 33% of having picked the correct door originally... and hence you have a double chance of being correct by switching.

 

This whole problem is widely known as the Monty Hall problem:

[papercutninja 24]

I saw Mythbusters too.

[Ray Ray 3,566]

I hate this thread

[NickC 23]

My head hurts.

[hd2000fxdl 422]

This was on Mythbusters, you switch, always. You then have a 66% probability.

 

Didn't you originally have a 33.3% chance of picking the right one, now that you saw that it wasn't the prise, and you were offered (offered another choice is the key here is how I see it) another choice, that would make it a 50-50 shot at picking the right one. Right??

[heaterbob 53]

i gave up 6 minutes of my life for this

[Ray Ray 3,566]

please lock this thread, i'm offended

[bbk 188]

please lock this thread, i'm offended

 

Haha, history of the world. "I don't get it, therefore I shall kill it with fire!"

[Ray Ray 3,566]

Haha, history of the world. "I don't get it, therefore I shall kill it with fire!"

 

No Ben, I do get it. 100% Now lock this thread or I'll report you.

[bbk 188]

No Ben, I do get it. 100% Now lock this thread or I'll report you.

As evident by the fact that I actually took the time to write out something that I thought I had buried into the recesses of my mind, I could use some excitement in the form of a good reporting.

 

ETA:

 

ETA pt. 2: Ray's my hombre. Speaking of, Ray, you haven't texted me in awhile, is your phone broken?

[M4BGRINGO 139]

My Saiga-12 with 12 magazines with 5 rounds each will take care of ALL THE DAMN DOORS!!!!!!!!!!!!!!!! No one gets a prize, I just blew it up!

 

I'm with you Ray, this made my head hurt, make it go-away!

[hd2000fxdl 422]

Mathematically it can work both ways, depends on if your using conditional & marginal or total probabilities.. You can work the numbers to show greater probability one way or another and it should work out that the split when you use each formula it should work out to a 50/50 chance on the results.

[vladtepes 1,060]

was about to call BS.. but then took a breath.. read it again.. and it actually makes perfect sense.... stuff like that is neat..

[vladtepes 1,060]

Didn't you originally have a 33.3% chance of picking the right one, now that you saw that it wasn't the prise, and you were offered (offered another choice is the key here is how I see it) another choice, that would make it a 50-50 shot at picking the right one. Right??

 

that is what I THOUGHT... but you have a better chance of picking a suck prize the first time out.. so it is MORE likely you are holding a suck prize.. so once you are shown the other door that IS a suck prize.. it is more likely that the other door the one remaining.. is a good prize..

[Jon 264]

Saw this in the movie 21. Cool stuff.

[Anselmo 87]

Mathematically it can work both ways, depends on if your using conditional & marginal or total probabilities.. You can work the numbers to show greater probability one way or another and it should work out that the split when you use each formula it should work out to a 50/50 chance on the results.

 

The key is that the host knows which door has the grand prize. That means the door the host opens is not random. Think of this way with 100 doors: You pick one, host opens 98 to show them empty. Now, two doors are left. After you picked one door, you had 1/100 chance to win and 99/100 to lose. The door you picked is still 1/100 and the one door still left is 99/100 to win.

[hd2000fxdl 422]

that is what I THOUGHT... but you have a better chance of picking a suck prize the first time out.. so it is MORE likely you are holding a suck prize.. so once you are shown the other door that IS a suck prize.. it is more likely that the other door the one remaining.. is a good prize..

 

Look up a couple posts or figure it out with: conditional & marginal or total probabilities and you will have a case for both side of the results...

[1down5up 7]

The key is that the host knows which door has the grand prize. That means the door the host opens is not random. Think of this way with 100 doors: You pick one, host opens 98 to show them empty. Now, two doors are left. After you picked one door, you had 1/100 chance to win and 99/100 to lose. The door you picked is still 1/100 and the one door still left is 99/100 to win.

 

Yeah, exactly. Much easier to understand when you add more doors.

[jvheitz 5]

[vladtepes 1,060]

Look up a couple posts or figure it out with: conditional & marginal or total probabilities and you will have a case for both side of the results...

 

I disagree.. I saw it as 50 50 also.. but when it was explained.. and I really thought about it.. your chances DO increase if you swap...

 

you pick one.. at random.. you are now holding good or bad..

 

%33 chance it is good

%66 chance it is bad

 

you are MORE likely to be holding bad..

 

now.. one of the remaining bad is eliminated..

 

since one of the other bad choices are eliminated.. you are now left with a situation...

you can act on the assumption that the one you picked is correct 1 of 3.. or you can assume that you originally picked the bad one.. since that is more likely 2 of 3...

if you assume based on the odds that you picked the wrong one it is more likely that the remaining one is the good one..

[vladtepes 1,060]

I guess this would apply to deal or no deal as well.. to always assume you picked the wrong one..

[Anselmo 87]

I guess this would apply to deal or no deal as well.. to always assume you picked the wrong one..

 

That is different because it's random and the host doesn't know what case holds what.

[vladtepes 1,060]

That is different because it's random and the host doesn't know what case holds what.

 

I mean when you make it to the VERY end.. where they essentially take away all the "wrong ones" and say.. do you want yours? or the one that is left?

[hd2000fxdl 422]

I'll go with it can work out both ways, you have a 50% chance of it being 33% for the win, and 50% chance for it being a 66% chance for the win.

 

Remember: The 50-50-90 rule: Anytime you have a 50-50 chance of getting something right, there's a 90% probability you'll get it wrong

 

And:

 

[Ray Ray 3,566]

ETA pt. 2: Ray's my hombre. Speaking of, Ray, you haven't texted me in awhile, is your phone broken?

 

I sent you some cool pics on Sunday from H-mart. BUT you never answer me back so I stopped texting. I'm a stalker to you.