Choosing a Game Show Door

Off Topic
121

[quote]Spry wrote:
Affliction wrote:
Listen, I’ll play the 1000 door variation of this game with you all day. You stick to your initial choice, I’ll switch my choice after 998 of the doors have been eliminated. I WILL WIN 999/1000, you WILL LOSE 999/1000. You can’t argue with this. You are wrong.

NO. If you are both still in the game after 998 doors have been opened BUT (improbably) the prize has not been found and the door the other guy and you both selected has not been opened then you now changing to the other door still only results in a 50/50 chance for both of you.

There is no difference.

You assume that after every door you picked another door which never resulted in a prize.

This is incorrect logic.
[/quote]

Wrong.

That is what people possibly are not getting about the problem. The two choices are not independent, because the host knows what is behind each door. So in the 1000-door variation, the host has to open 998 other doors that he knows are not the prize. Thus, as a “switcher,” you only lose when you initially pick the prize.

Everybody say it with me, because this involves no math:

THE SWITCHING STRATEGY ONLY LOSES WHEN YOU INITIALLY PICK THE PRIZE. THAT STAYING STRATEGY ONLY WINS WHEN YOU INITIALLY PICK THE PRIZE

You cannot debate that statement. It is necessarily true. So it is simple common sense (that has already been proved 20 times mathematically this thread) that, if you are more likely to initially pick the prize, you should stay. If you are less likely to initially pick the prize, you should switch.

Again, the switching strategy can only lose when you initially pick the prize. The more doors there are, the more lopsided the game gets, because your chance of winning with the switch strategy is (1-P), where P is the initial probability of picking the prize.

122

johnnytang has provided a source code. pookie has provided a source code. I put together a formula in Excel and will share it with anyone who wants it.

You can also do this on paper if you think the code is garbage or you think we were sneaky and tweaked it.

Whatever you choose, just stop arguing and see for yourself.

Switching doors will win 67% of the time. See for yourself that it’s true, and then spend your energy trying to figure out why it’s true.

123

[quote]Affliction wrote:
Look man, if we initially each picked the same door from the 1000, call it door #362, our odds of initially picking the correct door were 1/1000, agree?[/quote]

I agree.

I agree. But door #568 also has a 999/1000 chance of not being the correct door as well. FAIL.

[quote]Now imagine that someone was kind enough to open all other doors except door #568, which all contained nothing of value. The odds of door #568 containing the prize are now 999/1000, because of the initial odds of the prize not being behind door #362, the door we each initially had selected.

I am now going to change my selection to door #568, because of variable change, and my counterpart will stick with his initial selection of door #362. I will win 999 times out of 1000, BECAUSE I CHANGED MY SELECTION.[/quote]

See above dumbass!

The initial odds of door #568 are also 1/1000.

But that is irrelevant because if there are only 2 doors lefts they both have 50/50 chance.

I FUCKING WIN! Muahahahahhaa!

124

[quote]Spry wrote:
3 Doors. You have 1 in 3 chance.
2 Doors. You have 1 in 2 chance.

The doors don’t talk to each other dumbass. Look up the word MEMORYLESS.

If you didn’t win on guess number 1 now you have 50/50 odds to pick again.

I pick again and I pick the same door I picked the first time.

I STILL JUST PICKED AGAIN. I STILL HAVE THE SAME CHANCE OF WINNING.
[/quote]

What? I don’t think you are understanding the problem.

If I choose one door out of three, my odds are 1/3, right? So how do the odds magically change? That first door always has odds of 1/3.

EDIT: After reading this I realize you aren’t serious.

[quote]Spry wrote:
I FUCKING WIN! Muahahahahhaa!
[/quote]

125

JTrinsey went into more depth and detail that I have time for. Please go read his post above. Consider his evidence/explanation, which is applicable to the 3-door scenario, and then extrapolate that to the 1000-door scenario. Your argument should appear ludicrous at this point.

Someone raised the question earlier in the post that if you honestly believe that you can defeat someone in the 1000-door variation by not changing your selection, then let’s gamble:

If you win, I will give you $100.
If I win, you give me $1.

I will be make more money, every time.

Please do this experiment yourself, with cards or dice, both possibilities have been detailed earlier in the thread as well.

126

He must be trolling at this point…

127

[quote]Spry wrote:
THE ODDS DO NOT MAGICALLY INCREASE!

The first selection is irrelevant.

Now you have 2 doors.

Picking either is a 50/50 chance of winning.

Not changing doors is the same as picking the different door.

Log off and go back to school.[/quote]

How is the first selection irrelevant? The host can’t open the door you pick.

Again, if you make the second selection randomly your odds are indeed 50/50. At the same time, one door–the door you didn’t pick–is more likely to be the winner.

Let’s say there are only two doors. The game is rigged so that 2/3 of the time the car is behind Door 1 and 1/3 of the time it’s behind Door 2. If you don’t know that and you pick randomly you have a 1/2 shot of getting it right. But if you know how the game works, you’d always pick Door 1 and you’d win 2/3 of the time. Get it?

128

Post the source code! I’m interested!

129

This thread reminds me of any thread discussing god. People that know how the situation works trying to convince people that believe otherwise. The believers rely on a misunderstanding of the situation, and are irritable, and the others are talking to a wall.

130

[quote]Spry wrote:
Believe in magic. See if I care.

3 Doors. You have 1 in 3 chance.
2 Doors. You have 1 in 2 chance.

The doors don’t talk to each other dumbass. Look up the word MEMORYLESS.

If you didn’t win on guess number 1 now you have 50/50 odds to pick again.

I pick again and I pick the same door I picked the first time.

I STILL JUST PICKED AGAIN. I STILL HAVE THE SAME CHANCE OF WINNING.
[/quote]

Now you are just being a condescending bastard and the only thing worse than that is a condescending bastard who is WRONG.

Your comment about going back to school implies you have formal education in mathematics, which I sincerely hope is not the case because you are dead wrong. I am an undergrad and this is basic shit we talked about in our 200-level class.

Perhaps you want to look up the term “memoryless” again because that does NOT apply here. The variables have fundamentally changed because THE HOST KNOWS WHERE THE PRIZE IS. He does not randomly open another door; he always shows a goat.

Since you obviously are incapable of reading yourself, let me hold your hand and walk you through the only 3 scenarios that can happen. This is what we people who have been to school refer to as a “proof by exhaustion.” For the purposes of this proof, I will refer to one goat as Edgar and the other as Thomas.

Scenario One:
The door you initially pick conceals the car. The host will then show you either Thomas or Edgar. Thus, switching is the losing strategy and staying is the correct strategy.

Scenario Two:
The door you initially pick conceals Edgar. The host HAS to show you Thomas. Thus, the other door necessarily conceals the car. Thus, switching is the winning strategy and staying is the correct strategy.

Scenario Three:
The door you initially pick conceals Thomas. The host HAS to show you Edgar. Thus, the other door necessarily conceals the car. Thus, switching is the winning strategy and staying is the correct strategy.

If you assume that you initially have an equal chance of selecting the car, Thomas or Edgar, than it is pretty obvious to see that switching wins when you initially select a goat and staying wins when you initially select the car.

Let me go over that again:

SWITCHING WINS WHEN YOU INITIALLY PICK THE GOAT. STAYING WINS WHEN YOU INITIALLY PICK THE CAR. Thus, your odds of winning with the staying strategy are equal to P, where P is your odds of initially choosing the car. Your odds of winning with the switching strategy are equal to (1-P).

If you would like to disprove that statement, show me a scenario where you initially pick the goat, but staying wins. Alternatively, show me a scenario where you initially pick the car, but switching wins. I will personally buy anybody a Biotest supplement of their choice who is able to do that.

131

Alright, I think at this point we are being trolled.

Damnit.

132

I’m confused as fuck. Many make sense here, but Lifty has some good arguments.

Once you get past addition, subtraction and multiplication with math, I’m lost.

133

[quote]jtrinsey wrote:
Scenario One:
The door you initially pick conceals the car. The host will then show you either Thomas or Edgar. Thus, switching is the losing strategy and staying is the correct strategy.

Scenario Two:
The door you initially pick conceals Edgar. The host HAS to show you Thomas. Thus, the other door necessarily conceals the car. Thus, switching is the winning strategy and staying is the correct strategy.

Scenario Three:
The door you initially pick conceals Thomas. The host HAS to show you Edgar. Thus, the other door necessarily conceals the car. Thus, switching is the winning strategy and staying is the correct strategy.
[/quote]

WTF?

When you’re the contestant how do you know you just scored scenario 1 or 2 or 3?

In scenario 1 the host HAS to show you a donkey just the same as 2 and 3.

There is no difference except you capitalised the word ‘HAS’.

If I was the host and you picked the car first time guess what I am going to do (espicially if its you)?

Show you a donkey and get you to swap choices!

Then you would get a donkey you dumbfuck!

Muhahahahhahaha!

134

1001 POSTS!

Woooooooo!

Excellent thread for it!

135

Excel’s Magic Button, in the red circle. Ignore the yellow arrow.

Per Spry’s request, here is the formula I put together in Excel to model this situation.

“D1: Selected Door” means to enter “Selected Door” into cell D1.
“A2:A5002: =rand()*3” means to enter “=rand()*3” into cells A2, A3, …, A5002. Use the Magic Button in the picture above to speed up this process.

To use Magic Button: Click and hold Magic Button, scroll to highlight cells into which you want to copy the formula, and release the mouse button. Voila!

Just cut and paste the formulas into a cell, fill in the rest, and press F9 to run the calculations.

Without further ado:

B1: Winning Door
D1: Selected Door
F1: Eliminated Door
G1: Switched Door
H1: Wins
I1: Winning %
A2:A5001: =rand()*3
B2:B5001: =IF(A2>2,3,IF(A2>1,2,1))
C2:C5001: =rand()*3
D2:D5001: =IF(C2>2,3,IF(C2>1,2,1))
E2:E5001: =rand()*2
F2:F5001: =IF(NOT(B2=D2),6-(B2+D2),IF(D2=3,IF(E2<1,1,2),IF(D2=2,IF(E2<1,1,3),IF(E2<1,2,3))))
G2:G5001: =6-D2-F2
H2:H5001: =IF(G2=B2,1,0)
I2: =SUM(H2:H5001)/5000 (THIS IS WHERE THE RESULTS ARE POSTED)

Winning Door is the randomly selected door that holds the prize.

Selected Door is the randomly selected initial selection.

Eliminated Door is the door that is eliminated by the host. If the player did not initially pick the winning door, then the second losing door is eliminated. If the player DID initially pick the winning door, one of the two losing doors is eliminated randomly.

Switched Door is the door to which the player switches his selection.

Wins produces either a 1 if the player wins, or a 0 if the player loses.

Winning % sums the # of wins, and then divides by 5000.

If you do not want to fill 5000 cells, then fill as many as you wish, and then edit to

I2: =SUM(H2:H n+1)/n, where n is the number of cells you filled.

Thus, if you want to run only 10 iterations, put

I2: =SUM(H2:H11)/10

Press F9 to calculate everything again. You only need to enter the formulas one time.

136

[quote]jehovasfitness wrote:
I’m confused as fuck. Many make sense here, but Lifty has some good arguments.

Once you get past addition, subtraction and multiplication with math, I’m lost.[/quote]

The contents of the door never change. What was a 1/3 chance before the host interacts with the doors, is still a 1/3 chance of being ‘correct’(and a 2/3 chance of being incorrect). You are not ‘picking again’ after the host interacts, you are deciding whether to stay with your 1/3 chance, or switch to another variable, since there were 3 doors total, since the contents of the doors do not change, and since that other variable was 2/3… you are left with 2/3 of winning if you switch.

Does that help?

137

[quote]Spry wrote:
jtrinsey wrote:
Scenario One:
The door you initially pick conceals the car. The host will then show you either Thomas or Edgar. Thus, switching is the losing strategy and staying is the correct strategy.

Scenario Two:
The door you initially pick conceals Edgar. The host HAS to show you Thomas. Thus, the other door necessarily conceals the car. Thus, switching is the winning strategy and staying is the correct strategy.

Scenario Three:
The door you initially pick conceals Thomas. The host HAS to show you Edgar. Thus, the other door necessarily conceals the car. Thus, switching is the winning strategy and staying is the correct strategy.

WTF?

When you’re the contestant how do you know you just scored scenario 1 or 2 or 3?

In scenario 1 the host HAS to show you a donkey just the same as 2 and 3.

There is no difference except you capitalised the word ‘HAS’.

If I was the host and you picked the car first time guess what I am going to do (espicially if its you)?

Show you a donkey and get you to swap choices!

Then you would get a donkey you dumbfuck!

Muhahahahhahaha!

[/quote]

Yes, in scenario 1, you lose. In scenario 2 and 3 however, you win. I will take winning two times out of three over winning one time out of three.

Here is one final way to look at this problem. If you stay with your initial choice every time, meaning, you pick you door and then just close your eyes and don’t even pay attention when the host opens a door and shows you a goat, you will win one out of every three times. That can’t be argued. There are two goats and one car. Stay with your initial choice and you will win one out of every three times. However, the various doubters have “proved” that if you make a random choice after the host has already shown you the goat, you have a 50/50 shot. Well… how do you think your odds improve from 1/3 to 1/2? I will show you how:

**Half of the time you choose “staying”, which has a 1/3 chance of winning. [ (1/2)(1/3) ] = 1/6
**Half of the time you choose “switching”, which has a 2/3 chance of winning. [ (1/2)(2/3) ] = 2/6
(1/6)+(2/6) = 1/2

I’m pretty sure you are just trolling, but for some reason I feel compelled to respond.

138

[quote]jehovasfitness wrote:
I’m confused as fuck. Many make sense here, but Lifty has some good arguments.

Once you get past addition, subtraction and multiplication with math, I’m lost.[/quote]

No he really doesn’t. This problem is not even about math, it’s about common sense.

The “switching” strategy only loses when you initially pick the car. You know what I mean? If you decide to switch door when the host asks you, the only way you lose is if the door you initially pick contains the car.

The “staying” strategy only wins when you initially pick the car. If you decide to stay when the host asks you, the only way you can win is if the door you initially pick contains the car.

So the only question to ask yourself is how likely you were to pick the car in the first place.

139

For anybody still not getting this, let us play another game. This game is called Obvious Game.

Host: “Hello everybody and welcome to Obvious Game!”

Audience: “Hello host!”

Host: “Alright now ladies and gentlemen, we have a treat for you tonight. It’s the most exciting game show on the planet, the one where we give out a new car every night… or some unlucky loser gets stuck with a new pet goat!”

Audience: (laughs)

Host: “First let me introduce our first contestant; his name is Brad from South Carolina. Brad, what do you do for a living?”

Brad: “Umm-”

Host: “It doesn’t even matter, you only need half a functioning game to play Obvious Game. Now let’s get started we all know the rules so let’s go over to the playing area.”

(they turn to the three doors)

Host: “Now behind one of these doors is a car and behind each of the other two is a goat. Brad, I’m going to ask you to think carefully and choose a door.”

Brad: “Door number one please.”

Host: “Okay Brad, you’ve chosen door number one, excellent choice my boy. Now I’m going to ask you one more question. You can stick with that door, or you can decide to choose to pick BOTH of the other doors. If the car is in EITHER of those other two doors, you win.”

Brad: “Okay, let me just make sure I have this straight: if I think the car is behind door number one, I stick with that door; if I think the car is behind either of the other two doors, I can choose BOTH of them?”

Host: “That’s corr-”

Brad: “Okay, I switch.”

Host: “Are you sure?”

Brad: “Yeah. I had fourteen margaritas backstage but I’m still pretty sure there’s a better chance that the car is behind either of the other two doors than it is behind the first door I picked.”

Audience: (why are we still watching this)

If any of you were playing Obvious Game, would you stick with the original door?

140

If you play this game only one time and do not switch you have exactly 50:50 chance.

If you keep playing it the odds go to the house’s favor, 2/3 – thus the numerous iterations it takes to prove that the odds are 2/3.

It is a gamblers trick to lure you into keep playing with the same strategy of not switching. But, as was pointed out already, you need to ALWAYS switch to get these odds.

Since most of us would only play this game one time it doesn’t matter whether you switch the first time or not.

Does anyone know the series that proves this?