Choosing a Game Show Door

Off Topic
101

[quote]LIFTICVSMAXIMVS wrote:

  1. win: initially pick the car and DO NOT switch
  2. win: initially pick a goat and DO switch
  3. lose: initially pick the car and DO switch
  4. lose: initially pick a goat and DO NOT switch.

There are precisely always only 2 ways to win and 2 ways to lose. It is 50:50.[/quote]

That 4-scenario table you just presented actually shows EXACTLY my point. Scenario 1 and 2 are not equally likely to happen. You are twice as likely to initially pick the goat as you are to initially pick the car.

Follow this line of logic:

1.) What is the best strategy to do when you initially pick the goat? It is to switch, as you noted in the table above. In fact, when you initially pick the goat, you CANNOT lose if you switch and you will ALWAYS lose if stay.

2.) What is the best strategy to do when you initially pick the car? It is to stay, as you noted above. In fact, when you initially pick the car, you CANNOT lose if you stay and you will ALWAYS lose if you switch.

3.) Is it more likely that you initially picked the goat or the car? Well it is obvious, there are three doors and two of them contain goats and one contains a car. Thus, you are more likely to initially pick a goat.

4.) If you are more likely to pick a goat and the best strategy to do when you pick a goat is to switch, then what should your strategy be? Obviously it is to switch.

The second choice is not a choice at all. It is inconsequential and only gives you the illusion of choice. If you have decided that you will stay, you will ONLY WIN when you initially pick the car (1/3) and ONLY LOSE when you initially pick the goat (2/3). If you have decided that you will switch, you will ONLY WIN when you initially pick the goat (2/3) and ONLY LOSE when you initially pick the car (1/3). There are no other options.

Furthermore, you can use an “undecided guy” (we will call him Random Roger) to prove that switching is the right strategy. Say that this guy has no idea whether he will switch or stay. Essentially, he will “choose” between the two doors after the first goat has already been shown. There are four ways that this can play out:

Scenario One:
He initially picks the door that contains a car (1/3). When a goat is shown, he decided to switch (1/2). Damnit he loses!
(1/3)*(1/2) = 1/6

Scenario Two:
He initially picks the door that contains a car (1/3). When a goat is shown, he decides to stay (1/2). Yay he wins!
(1/3)*(1/2) = 1/6

Scenario Three:
He initially picks the door that contains a goat (2/3). When a goat is shown, he decides to switch (1/2). Yay he wins!
(2/3)*(1/2) = 2/6

Scenario Four:
He initially picks the door that contains a goat (2/3). When a goat is shown, he decides to stay (1/2). Damnit he loses!
(2/3)*(1/2) = 2/6

So it is pretty easy to see that Scenarios Three and Four will happen twice as often as One and Two, simply because the goat is twice as likely to be behind the initial door as the car is.

So Random Roger will win half the time and lose the half because he can’t decide whether to switch or stay. However, his two cousins, Sureminded Sally and Flip-Flop Fred are also going to play.

Sureminded Sally always sticks with the door she initially picked, so you can throw out scenarios One and Three and switch the (1/2) in the “decides to stay” portion to (1), because she always stays. We are left with Scenario Two (where she initially picks the car and wins) and Scenario Four (where she initially picks the goat and loses). So using the above chart you can see that she will win (1/3) of the time and lose (2/3) of the time.

Flip-Flop Fred always switches his pick, so you can throw out Scenarios Two and Four. You can see that now he is twice as likely to win as he is to lose.

102

I posted how you could simulate this problem with one standard six-sided die, a pencil, and a piece of paper back on the first page of this thread.

Lifty, why don’t you take the time to perform about thirty to fifty trials before you come back and argue.

103

[quote]tGunslinger wrote:
This isn’t voodoo.[/quote]

Well it is voodoo. But not for the reason you think.

104

Lift makes a stronger argument. plus he used the word analogous.

why wouldnt the probability factor-in that one door is automatically eliminated 100% of the time which simplifies to there being only 2 doors every time?

105

[quote]jtrinsey wrote:
LIFTICVSMAXIMVS wrote:

  1. win: initially pick the car and DO NOT switch
  2. win: initially pick a goat and DO switch
  3. lose: initially pick the car and DO switch
  4. lose: initially pick a goat and DO NOT switch.

There are precisely always only 2 ways to win and 2 ways to lose. It is 50:50.

That 4-scenario table you just presented actually shows EXACTLY my point. Scenario 1 and 2 are not equally likely to happen. You are twice as likely to initially pick the goat as you are to initially pick the car.
[/quote]

By being given an other chance it’s as if I never made the first pick. In fact, by removing a goat he guarantees that I only have a 50:50 shot at guessing right.

Picking a different curtain is just an illusion. In the end, which is all that matters, I am going to be presented with two options. That is the part no one gets. The first choice doesn’t matter at all. It always plays out the same way over and over. I get a second chance whether I pick correctly the first time or not.

106

I don’t think you’re reading everything a couple of us have written.

Yes, if you are picking randomly, you have a 50% chance of guessing right. That doesn’t change the fact that 1/3 times the door you originally picked is right and 2/3 times the other door is. In other words, you can disregard the original pick and give yourself even odds, or you can understand that the door you originally picked is less likely to be the winner.

The trouble is you’re talking about apples when the original question was about oranges. The question was not, “what are the odds of picking correctly?” It was, “should you stay or switch?”

107

TO ANYBODY WHO DOESN"T UNDERSTAND THIS:

You have a 2/3 chance of winning ONLY if you ALWAYS switch. You HAVE to switch for this to work.

SO…:

If you ALWAYS switch, you can pick EITHER goat on the first try and win (2/3). You will lose if you pick the car on the first guess (1/3).

108

[quote]LIFTICVSMAXIMVS wrote:
By being given an other chance it’s as if I never made the first pick. In fact, by removing a goat he guarantees that I only have a 50:50 shot at guessing right.

Picking a different curtain is just an illusion. In the end, which is all that matters, I am going to be presented with two options. That is the part no one gets. The first choice doesn’t matter at all. It always plays out the same way over and over. I get a second chance whether I pick correctly the first time or not.[/quote]

If you are guessing you will only have a 50% chance of winning. 2/31/2 + 1/31/2=.5

The beauty of this problem is that you have prior knowledge that you can use to increase your odds to 2/3 by switching every time.

This has now been discussed and explained to exhaustion. The only way to further explain it is for you to actually sit down with a deck of cards or something and see for yourself. 2 jacks, 1 king. Pick one card and hold it. Look at the the other two cards and discard 1 jack. Now record how many times you are holding the king and how many times it is still on the table. You should figure this out pretty quick.

109

[quote]LIFTICVSMAXIMVS wrote:
Picking a different curtain is just an illusion. In the end, which is all that matters, I am going to be presented with two options. That is the part no one gets. The first choice doesn’t matter at all. It always plays out the same way over and over. I get a second chance whether I pick correctly the first time or not.[/quote]

Well, let’s play the 100 doors version for money then, shall we?

You stick with your first choice every time (since it’s 50/50, it doesn’t matter) and every time you win I pay you 100$ and you pay me 100$ every time you lose (gonna keep those amounts equal, 50/50 odds don’t call for unequal payoff, I’m sure you agree).

So either put your money where your mouth is, or shut the fuck up.

That md5 hash back there is still good, give me your door number.

110

Well, you could flip a coin to decide whether you switch or not.

This will get your chance to 50/50 (and this is what Lifticus essentially says)

But I’d rather switch and enjoy the nice 66% chance to claim the prize.

111

You gamblers think you are sneaky.

I get it.

112

[quote]LIFTICVSMAXIMVS wrote:
You gamblers think you are sneaky.

I get it.[/quote]

Not enough balls to put your money where your mouth is?

No brains and no balls. Must suck to be you.

113

[quote]LIFTICVSMAXIMVS wrote:
jtrinsey wrote:
LIFTICVSMAXIMVS wrote:

  1. win: initially pick the car and DO NOT switch
  2. win: initially pick a goat and DO switch
  3. lose: initially pick the car and DO switch
  4. lose: initially pick a goat and DO NOT switch.

There are precisely always only 2 ways to win and 2 ways to lose. It is 50:50.

That 4-scenario table you just presented actually shows EXACTLY my point. Scenario 1 and 2 are not equally likely to happen. You are twice as likely to initially pick the goat as you are to initially pick the car.

By being given an other chance it’s as if I never made the first pick. In fact, by removing a goat he guarantees that I only have a 50:50 shot at guessing right.

Picking a different curtain is just an illusion. In the end, which is all that matters, I am going to be presented with two options. That is the part no one gets. The first choice doesn’t matter at all. It always plays out the same way over and over. I get a second chance whether I pick correctly the first time or not.[/quote]

Your first pick affects the host’s actions. He can pick randomly among the two goats if you picked the prize, but he is forced to choose the other goat if you picked a goat. If he showed you a goat chosen randomly from the two goats, before you made your initial selection, then you would know nothing except one remaining door has a goat and one has the prize. But because your first choice might have limited his actions, you now know that it is more likely the uninvolved door has the prize.

I’m starting to think I would rather have a goat anyway.

114

[quote]pookie wrote:
Well, let’s play the 100 doors version for money then, shall we?
[/quote]

More doors is the joke.

115

Lifticus,
you’ve spend so much time arguing, why don’t you just try it out on paper?

116

[quote]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.[/quote]

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.

117

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.

118

[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]

No, you are incorrect. I’m not talking about the probability of either of us having the car “if we are both still in the game”. I’m talking about me switching my choice, and him not. You need to re-read the thread, apparently.

119

[quote]Affliction wrote:
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.

No, you are incorrect. I’m not talking about the probability of either of us having the car “if we are both still in the game”. I’m talking about me switching my choice, and him not. You need to re-read the thread, apparently.

[/quote]

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.

120

[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]

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?

The odds of the winning door not being door #362 are 999/1000, can we agree on that as well?

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.

He will win (with door #362) 1 time out of 1000, because the initial odds have not changed. The initial probability of the prize being behind door #362 were 1/1000.

This is fact, my friend. Your attempts do be derogatory, i.e. “log off and go back to school” are both childish and ironic.