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# Whatchya Gonna Do (2 goats and a car)

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Posted · Report post

This is the monty hall paradox. always switch. if you work out the probability of getting the jag in each staying or switching, switching wins. most people think its 50/50

which isn't the case.

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Posted (edited) · Report post

No. Door#1 earlier had 1/3, now that it is open, it is out of equation. Only 2 doors are in race now and we have to re-assign prob. of 1/2, 1/2 for #2,#3. Both the cases you mentioned are equally likely (hence 1/2, 1/2) to happen.

Few may think that changing to #3 is better option but there is no place for finding prior prob. in this case (as #1 is removed out of equation instead of just having its prob changed).

You can read my older comment on why i still wish to stick with #2, even though it has only 1/2 of winning ..

You cannot reassign the probabilities like that.

Let me tell you in the way that makes the most sense to me.

Let's say we have two people playing this game, Switcher and Stayer.

Switcher has firmly decided to always switch doors, while Stayer has decided to never switch doors.

At the outset of the game Stayer has a 1/3 probability of picking the door with the car, and 2/3 probability of picking a door with a goat.

If he picks right, which happens about 1/3 of all the times he's taken part in this silly competition, then when Monty shows the other door, Stayer will win, since he initially picked right and stayed with it.

If Stayer picks wrong initially, which happens 2/3 of the time, he will lose after Monty opens a door, because he was initially wrong and stuck with it.

We can therefore say that Stayer will win 1/3 of the time and lose 2/3 of the time.

Now let's look at Switcher.

Just like Stayer, at the outset of the game, Switcher has a 1/3 probability of picking the door with the car, and 2/3 probability of picking a door with a goat.

If he picks right, which happens about 1/3 of the time, then after Monty opens a goat door, switcher will switch doors to the other goat door and lose, since he switched from the correct door to the particular goat door that Monty didn't open.

If Switcher picks wrong initially, which happens 2/3 of the time then he will be pointing at one of the goat doors. Monty, knowing where everything is, will open the other goat door, and Switcher will switch doors to the car door and win. (Remember this happens 2/3 of the time)

We can therefore say that Switcher will win 2/3 of the time and lose 1/3 of the time.

Let me recap, based on the strategies of the two players and a logical analysis of all possibilities,

Stayer will win with probability 1/3 and lose with probability 2/3

Switcher will win with probability 2/3 and lose with probability 1/3.

Who would you rather be?

Think of it this way, Switcher is betting his initial guess is wrong, while Stayer is betting his initial guess is right. Wrong and right have probabilities 2/3 and 1/3 each, so which would you bet happened?

You bet you were wrong! Sometimes that's the best way to be right.

I think looking at the problem like this, causes it to agree with intuition rather than disagreeing with it.

Edited by mmiguel1
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Posted · Report post

mmiguell is right. and he just proved it.

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Posted · Report post

mmiguell is right. and he just proved it.

Thanks matt, (my name is matt too, last name is miguel).

Looking at the length of this thread, I would bet that it's been proven before somewhere in here at least 10 times over.

I don't know if they stated it the same way as me already, but I wouldn't be terribly surprised if it was already stated like this. (Bonanova likes to convey the logic in multiple perspectives)

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Posted · Report post

Thanks matt, (my name is matt too, last name is miguel).

Looking at the length of this thread, I would bet that it's been proven before somewhere in here at least 10 times over.

I don't know if they stated it the same way as me already, but I wouldn't be terribly surprised if it was already stated like this. (Bonanova likes to convey the logic in multiple perspectives)

Yeah it probably has ha ha

well done on my sequence puzzle. theres another one if you fancy a challenge here.

I'm guessing you have a quite high iq?

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Posted · Report post

Consider the host gave me another chance to pick a door, without opening the first one. Because i could be more wrong in first attempt (2/3), would switching improve my second chance? I doubt, bcoz it was not yet proved that i was wrong. "I could be more wrong initially" was your statement if i am correct. That was correct, when 3 doors were still locked. However then, it was proved that "i was not wrong" on first round. And the second round starts with "I could be equally right or equally wrong" whereas you think "i still could be more wrong".

first round after door#1 opened:

i was wrong (chance: 1/3) - car inside

i was not wrong yet (chance: 2/3) - goat inside

[your statement "i could be still wrong" means you wish to carry forward the odds to next round but you shouldnt do it. Let me explain]

There is a biased coin that throws "GOAT" 2/3 times and "CAR" 1/3 times, that always end up with a car in each 3 throws. You state that you will get the "C" on second time.

G1 G2 C

G1 C G2

G2 G1 C

G2 C G1

C G1 G2 - shdnt carry forward

C G2 G1 - shdnt carry forward

after round-1 we know that one of these have occurred (door1 has G)

G1 G2 C

G1 C G2

G2 G1 C

G2 C G1

Now, if you see, even if i stick with throw#2 or switch to throw#3, i have EQUAL chance of winning (or losing) the car.

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Posted · Report post

Consider the host gave me another chance to pick a door, without opening the first one. Because i could be more wrong in first attempt (2/3), would switching improve my second chance? I doubt, bcoz it was not yet proved that i was wrong. "I could be more wrong initially" was your statement if i am correct. That was correct, when 3 doors were still locked. However then, it was proved that "i was not wrong" on first round. And the second round starts with "I could be equally right or equally wrong" whereas you think "i still could be more wrong".

first round after door#1 opened:

i was wrong (chance: 1/3) - car inside

i was not wrong yet (chance: 2/3) - goat inside

[your statement "i could be still wrong" means you wish to carry forward the odds to next round but you shouldnt do it. Let me explain]

There is a biased coin that throws "GOAT" 2/3 times and "CAR" 1/3 times, that always end up with a car in each 3 throws. You state that you will get the "C" on second time.

G1 G2 C

G1 C G2

G2 G1 C

G2 C G1

C G1 G2 - shdnt carry forward

C G2 G1 - shdnt carry forward

after round-1 we know that one of these have occurred (door1 has G)

G1 G2 C

G1 C G2

G2 G1 C

G2 C G1

Now, if you see, even if i stick with throw#2 or switch to throw#3, i have EQUAL chance of winning (or losing) the car.

Consider the host gave me another chance to pick a door, without opening the first one. Because i could be more wrong in first attempt (2/3), would switching improve my second chance? I doubt, bcoz it was not yet proved that i was wrong. "I could be more wrong initially" was your statement if i am correct. That was correct, when 3 doors were still locked. However then, it was proved that "i was not wrong" on first round. And the second round starts with "I could be equally right or equally wrong" whereas you think "i still could be more wrong".

What justification do you have for changing these probabilities?

If I am pointing at a car before Monty opens a goat door, then I am still pointing at a car after Monty opens a goat door.

Monty opening some other door has no effect on what is lying behind my door.

If I stay with this door and win, it is only because the car was always behind the door and I got lucky when I picked this door out of the 3 choices. If I definitely choose to stay, then whether or not I win or lose is decided only at the time when I make the initial pick of the 3 choices. Although I may not know if I have won or lost yet, it is decided at this point. And the probability of me winning at this point is 1/3. As long as I stay with this door, Monty can do whatever he wants (within reason), and it will not affect my odds of winning. Thus, by staying, I win ONLY if my initial guess was right and I lose if my initial guess was wrong. My initial guess is right with probability 1/3. Therefore by staying, I win ONLY if a certain event happens whose probability is 1/3, and I lose otherwise. This certain event being that my initial choice of 3 doors was correct.

If I decide to switch, that is a different story. If my initial guess was right and I switch, then no matter what Monty does, I will lose. If my initial guess was wrong, then I point at one goat, Monty shows the other goat, and the only possible door left I can switch to is the one holding the car. It HAS to be the case that I win if I was initially wrong and I switch. Monty's only purpose here is to make sure that it HAS to be the case that I win if I was initially wrong and I switch.

Therefore, if I do in fact switch, I will win if and only if I was wrong in my initial guess of 3 doors. Winning becomes an equivalent outcome to being initially wrong as long as I decide to switch. There is no randomness to that idea.

Given that I switch:

I was wrong initially = I Win

I was right initially = I Lose

Therefore the probability that I win is equal to the probability that I was initially wrong BECAUSE those two ideas are equivalent, thanks to Monty. No one debates that the probability that I was initially wrong is 2/3. Because of Monty, given that I switch, Winning is the same outcome and must have the same probability of 2/3.

I hope that was clear.

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Posted · Report post

Hi mmiguel

The point i am stating here is

1. Round 1: Random probability

2. Round 2: Conditional probability (prior probability)

We cannt still apply random prob that we once applied without any knowledge. Prior prob of getting car was 1/3. After the occurrence of 1st event (door#1 having goat), getting car has a new probability now (1/2). You said, switch is the better option earlier and i said it doesnt matter bcoz winning or losing is now down to 1/2,1/2

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Posted · Report post

Now, if you see, even if i stick with throw#2 or switch to throw#3, i have EQUAL chance of winning (or losing) the car.

aaronbcj, suppose after picking a door, Monty said, you can stay with that choice OR pick BOTH of the other doors.

Would you switch?

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Posted (edited) · Report post

Let me take it back and give me some time.. that link i gave coincidentally addresses this problem and im going thru it to see any assumptions going there.

Meanwhile pls validate my logic that i used with coin example.

Based on that, i will stick.

Edited by aaronbcj
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Posted (edited) · Report post

Hi mmiguel

The point i am stating here is

1. Round 1: Random probability

2. Round 2: Conditional probability (prior probability)

We cannt still apply random prob that we once applied without any knowledge. Prior prob of getting car was 1/3. After the occurrence of 1st event (door#1 having goat), getting car has a new probability now (1/2). You said, switch is the better option earlier and i said it doesnt matter bcoz winning or losing is now down to 1/2,1/2

Ahh yes, I love the math. I just usually find that people find conceptual sentences more convincing.

P[initially right] = 1/3

P[initially wrong] = 2/3

Total Probability Theorem:

P[Win|stay] = P[Win|stay and initially wrong] P[initially wrong] + P[Win|stay and initially right] P[initially right]

P[Win|stay and initially wrong] = 0 if you are wrong and you stay on the wrong door, you cannot win

P[Win|stay and initially right] = 1 you had the right door and stayed with it, you win!

P[Win|stay] = 0*P[initially wrong] + 1*P[initially right]

P[Win|stay] = P[initially right] = 1/3

P[Win|stay] = 1/3

Similarly,

P[Win|switch] = P[Win|switch and initially wrong] P[initially wrong] + P[Win|switch and initially right] P[initially right]

P[Win|switch and initially wrong] = 1, the only door you can switch to thanks to Monty is the car door

P[Win|switch and initially right] = 0, you had the right door, but you left it for a goat door

P[Win|switch] = 1*P[initially wrong] + 0*P[initially right] = P[initially wrong]

P[Win|switch] = P[initially wrong]

P[Win|switch] = 2/3

Result:

P[Win|switch] = 2/3

P[Win|stay] = 1/3

The probability of winning given that you switch is 2/3.

The probability of winning given that you stay is 1/3.

You can't really apply Baye's rule, because you cannot really calculate P[switch] or P[stay], as those are factors that you can set, not random things you only observe.

Edited by mmiguel1
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Posted · Report post

Thanks all, i got.

We need to take into account WHY Monte opened door#1.

optionA

if C in #2, he has equal prob. of opening #1 and #3 , so opening #1 is 50% and opening #3 is 50%

optionB

if C in #1, (#2 selected by me), he has to select #3 , so opening #3 is 100%

optionC

if C in #3, (#2 selected by me), he has to select #1 , so opening #1 is 100%

considering he opened #1, we never know because of option A or B, but his action of opening #1 is more bcoz #3 had prize, rather #2 having it.

So switching is better. Even in coin ex, i missed out the point "why" he opened #1 instead of #3.

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Posted · Report post

At the start there is higher chance of picking wrong. after the host has revealed the first wrong. there is still that higher chance that you picked wrong in the first place. so switching is all ways better.

thats about as simple as i can put it.

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Posted · Report post

This is always a fun puzzle to debate. I haven't read through all 17 pages posted here so this may have been brought up before. I first encountered this puzzle while reading Marilyn Vos Savant's column in the paper. Someone posed the puzzle and asked if you should switch or not. Marilyn gave the correct answer that you should switch.

Then she received a ton of hate mail saying how stupid she was along with lots of "logical" proofs showing why staying was always better. She was honestly shocked at how many people honestly believed switching was bad and devoted a 2nd column to explaining the solution. Again she received more hate mail blasting her for not admitting when she was wrong and trying to defend her position. I think she wrote a 3rd column but finally gave up trying to convince the readers. She included a lot of this in one of the books she wrote about her column.

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Posted · Report post

I believe that

Door #2 has the car.

The reason for that is that MB said (the show's voice man or whoever Joe is)

to why tje WAX was worth.I suppose that my leting the woman listen something

about what she earns at door 3# but not telling anything about the restaurant

the producers of the show try to make her feel more comfortable at the idea of choosing

door #3, and becouse they don't want to loose any money, they have placed the goat behind

3#

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Posted · Report post

Again, i got this 1 nagging.

Hpw many times would host open a door, had i selected wrongly at first go.

If opening a door is simply optional for host, i will STICK.

if opening a door is part of the game(always), i will SWITCH.

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Posted · Report post

you change your choice due to variable change, when you first picked you had a one in three chance of getting the car, now you have a one in two chance, but you must change your opinion for the odds to go up

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Posted · Report post

you change your choice due to variable change, when you first picked you had a one in three chance of getting the car, now you have a one in two chance, but you must change your opinion for the odds to go up

No, I disagree with your odds. The puzzle is assuming that the host knows where the car is and also assumes that the host will open one of the losing doors everytime no matter what you chose.

When you first select a door, you have a 1/3 chance of picking the car. (Or, a 2/3 chance of picking wrong). If you switch, you now have a 2/3 chance of winning the car. The host helps you win by always eliminating a wrong choice.

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She should stick to her original choice. the game host would only want her to change doors, if he knew she had won the Jag, so he tempts her to change.

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Say you are given a heap of 50 coins, in which only one coin is real, while all other coins are fake. you can not recognize the fake or real coin by seeing or touching them.

When you grab a coin from the heap of 50 coins, there was only one out of 50 chances that you grabed right coin.

Suppose the host now removes 47 fake coins. Only two coins are left in the heap and one in your hand. Now what you will do?

Stick to the coin which you selected When there were 50 coins and there was only a very remote chance that the coin selected was real...?

OR

You will leave that and grabe another coin....?

Definitely when you do not know which coin is real, you shall like to leave the coin in hand which had more chances to have been fake, and grab a new coin which has now better chances to be real.

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Posted · Report post

Swap.

33.3 % before for the girl, 66.6% for the host.

The odds haven't changed, but the host is offering her his 66.6%probability.

Take it.

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Posted · Report post

No, I disagree with your odds. The puzzle is assuming that the host knows where the car is and also assumes that the host will open one of the losing doors everytime no matter what you chose.

When you first select a door, you have a 1/3 chance of picking the car. (Or, a 2/3 chance of picking wrong). If you switch, you now have a 2/3 chance of winning the car. The host helps you win by always eliminating a wrong choice.

No, I disagree with your odds. The puzzle is assuming that the host knows where the car is and also assumes that the host will open one of the losing doors everytime no matter what you chose.

When you first select a door, you have a 1/3 chance of picking the car. (Or, a 2/3 chance of picking wrong). If you switch, you now have a 2/3 chance of winning the car. The host helps you win by always eliminating a wrong choice.

No, I disagree with your odds. The puzzle is assuming that the host knows where the car is and also assumes that the host will open one of the losing doors everytime no matter what you chose.

When you first select a door, you have a 1/3 chance of picking the car. (Or, a 2/3 chance of picking wrong). If you switch, you now have a 2/3 chance of winning the car. The host helps you win by always eliminating a wrong choice.

Correct.

Variable change.

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