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# Whatchya Gonna Do (2 goats and a car)

### #181

Posted 20 October 2010 - 07:53 AM

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.

### #182

Posted 20 October 2010 - 08:18 AM

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.

### #183

Posted 20 October 2010 - 08:38 AM

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

Pls visit http://en.wikipedia..../Bayes'_theorem

### #184

Posted 20 October 2010 - 08:38 AM

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?

*Vidi vici veni.*

### #185

Posted 20 October 2010 - 08:55 AM

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

Based on that, i will stick.

**Edited by aaronbcj, 20 October 2010 - 08:58 AM.**

### #186

Posted 20 October 2010 - 08:56 AM

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

Pls visit http://en.wikipedia..../Bayes'_theorem

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, 20 October 2010 - 08:58 AM.**

### #187

Posted 20 October 2010 - 09:21 AM

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.

### #188

Posted 20 October 2010 - 12:48 PM

thats about as simple as i can put it.

### #189

Posted 20 October 2010 - 07:25 PM

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.

### #190

Posted 20 October 2010 - 07:51 PM

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