Win Another Car!! (Or a goat... or possibly a deuce).

DeVore

Firstly poker is about probability (at least a good chunk of it is ) so such probability based questions are good for helping us all think in those terms. And God knows there are some out there that need all the help they can get

So here's the same problem tweaked:
You have a choice of three doors, behind two you can hear the ominous sound of, yes you guessed it, goats! Behind the third is a car. You are told that one of the goats is black and one white. You pick a door and host then chooses randomly from the two remaining doors. If he picks the car, its all over (unless you like goats in which case... \o/ ).
If he picks a door with a goat behind it you get to either switch or stay.

You can either:
A. Always Switch

B. Always Stay

C. Always switch if the host opens the black goat's door but always stay if he opens the white goat's door.

D. Suggest an alternate action with reasoning.


DeV.

ocallagh

another good one...

http://mathforum.org/library/drmath/view/52223.html

NickyOD

If you can hear the goats then you should know which door the car is behind.

Dub13

NickyOD wrote:

If you can hear the goats then you should know which door the car is behind.

If only life was that simple....

Lafortezza

My first impression is that it doesn't matter if you switch or stay. If the host picks (at random) the door with the car then you lose. If he picks a door with goat behind then you're left with 2 doors, 1 car, 1 goat. 50/50 chance since you've no further information to go on.

How does the colour of the goats affect anything?

ocallagh - that envelope question is weird. pm me the answer?

Crumbs

Agree with previous poster. There's no difference in choices A, B and C. You have a 1/3 chance in each and I don't see any other way that gives you a better chance than that.

sendic

you should still switch but your edge is worse.

1/3 of the time the host will open the door to the car and you lose.

2/3 of the time you will be given the CHOICE to switch, in which case its the same decision as the other post.

if you switch 2/3s of the time you will be right and 1/3 of the time you will be wrong. 2/3 * 2/3 = 4/9 (or 44.4%) vs (1/3 * 2/3) + 1/3 = 5/9 (55.5%)

if you stick you will be correct 1/3 of the time and incorrect 2/3 of the time.
1/3 * 2/3 = 2/9 (or 22.2%) vs (2/3 * 2/3) + 1/3 = 7/9 (77.7%)

i cant think of an instance when it would be benificial to allow the colour of the goat to influence your decision. unless im missing something. Dev????

sendic

C gives you a 66.6% lose rate and a 33% win rate.

sendic

C gives you a 66.6% lose rate and a 33.3% win rate.

Tourneque

sendic wrote:

C gives you a 66.6% lose rate and a 33.3% win rate.

You're right that C gives you a 33% win rate but option C is just a cleverly written way of picking A half the time and B the other half.

Both A & B also have a 33% chance of winning.

The colour of the goats isn't relevant.

sendic

Tourneque wrote:

You're right that C gives you a 33% win rate but option C is just a cleverly written way of picking A half the time and B the other half.

Both A & B also have a 33% chance of winning.

The colour of the goats isn't relevant.

I realise C is A half the time and B half the time. By your method A and B are both 33% chances so C = (33+33)/2 = 33%
By my method A is 44% and B is 22% so C = (22+44)/2 = 33%

Why do you think A and B are both 33%? Isint it the same problem as the other thread but with a situation where you can lose before you have the choice to switch?


edit: i had my percentages for A and B the wrong way round, its fixed now

Tourneque

In the other thread, the Host always showed one of the empty(or goat) doors. He had more information that the contestant and by showing one of the empty doors he gave us some of that info.

This case is different in that the Host does not have more info and he chooses randomly.

Lets take a look at the wording of this example.
In it Dev says that when the Host chooses, if he guesses correctly the game is over and you don't get the option to change doors.
This is actually irrelevent because it means we've already picked the wrong door and even if we swap we'll still be picking the wrong door.
So whether we can swap or not after the host doesnt affect the outcome.

Now that we know that, lets look at the options and the outcomes.

(a) We have a 1/3 chance of guessing the right door
(b) The Host has 1/3 (1/2 x 2/3) chance of guessing the right door
(c) The rest of the time, 1/3 , neither of us guesses correctly.

So basically, it doesn't matter whether we change our decision after the host guesses or not. We still only have a 1/3 chance.

A B C
BlackGoat WhiteGoat Car

Con picks A, Host B, Con Sticks and both lose
Con picks A, Host C, Con Sticks and loses (or Con cannot change and still loses)
Con picks B, Host A, Con Sticks and both lose
Con picks B, Host C, Con Sticks and loses (or Con cannot change and still loses)
Con picks C, Host B, Con Sticks and wins
Con picks C, Host A, Con Sticks and wins

Con picks A, Host B, Con Swaps and wins
Con picks A, Host C, Con Swaps and loses (or Con cannot change and still loses)
Con picks B, Host A, Con Swaps and wins
Con picks B, Host C, Con Swaps and loses (or Con cannot change and still loses)
Con picks C, Host B, Con Swaps and both lose
Con picks C, Host A, Con Swaps and both lose

Each outcome has an equal 2/6 chance of happing regardless of sticking, swapping or even if you can/cannot change after the host makes his choice and wins.

(The only problem with your calcs is that you made the assumption that switching has a probablility of 2/3 when it's actually 1/2)

sendic

I thought the difference between the senarios (host knows where the car is/host picks at random) is that we are less likely to win in the senario where he picks at random.

host picks knowing where the car is : 66/33
host picks at random we switch : 44/55
host picks at random we stick : 22/77

when the host picks at random 33% of the time we have no choice to make as he wins and its game over
all probabilities must add up to 100%
if the host picks at random and picks a goat (66% of the time) we now have the option to switch or stick. we have the same information as in the case when the host opens a door knowing its a goat (ie we pick A, host opens B, B is a goat, so car is behind A or C).
so the probability we work out now is a fraction of that 66% and we add 33% to the fraction of the 66% of the time when we lose.
because we lose 22% (1/3 of 66%) of the time when we switch our overall losing is 22+33=55%
because we lose 44% (2/3 of 66%) of the time when we stick our overall losing probability is 44+33= 77%

I’m not 100% sure I’m right though

Tourneque

Once you get your head around the first problem you end up having trouble with the 2nd. It can become counter-intuitive which is what makes the discussion interesting.

The reason the probablity is 2/3 on the first problem is because the Host isn't choosing a door - he's giving information.
The only randomness in his choice is when the Contestant has already picked the Car and he's randomly showing one of the 2 empty doors.

On the 2nd problem.. well.. I'm struggling a little to explain this so I'll tackle it from another way.

You say that A is only 22% ? This can never be so.
You have 3 choices. A B or C. Behind 1 is a car.
When you choose a door, the probably of being correct has to be 1 in 3. Therefore by sticking, your chance cannot fall below 33%. So ProbA = 33

You agree that choosing C is the same as picking A half the time and B the other half right ? Ok. So ProbC = (0.5 x ProbA) + (0.5x ProbB)

And you say that the chance of the host picking the car is 1/3 ? Ok. So ProbH = 33

Since the probabilities must add up, it means that sticking, swapping and the host guessing correctly must be 100%. Ok. So ProbA + ProbB + ProbH = 100

So it's:

(i) ProbA = 33

(ii) ProbC = (0.5 x ProbA) + (0.5x ProbB)

(iii)ProbH = 33

(iv) ProbA + ProbB + ProbH = 100


I'll dump the Prob prefixes and simplify to make things easier.

(i) A = 33

(ii) C = 0.5A + 0.5B ==> 2C = A + B

(iii)H = 33

(iv) A + B + H = 100



I'll substitute A & H into the expressions (ii) and (iv)

(ii) 2C = 33 + B

(iv) 33 + B + 33 = 100 ==> b = 33 (ok, 34 but lets keep the decimals out )


Since B is also 33 it means that C is also 33.

So A, B & C are all 1/3 probability.

Apologies for the math like stuff. I did say I was struggling

DeVore

<Have I Got News For You>
Is the right answer.
</Have I Got News For You>


It doesnt matter a whit what you do here. You can pick, stay, swap and do so on the phases of the moon for all it matters, this is a straight 33% chance regardless of what you do and its the reason why most people think the first question Bohsman posed is 50/50.

My point in posing this one is to isolate the key piece of information that makes them different questions. One where the host chooses randomly and the other where he doesnt. Small difference, but makes a big one to the probabilities!

Oh, the colour thing was a red herring, hoping someone would invoke game theory and the really correct answer is "pick the door you cant hear goats behind and stick to it".
Congrats Nicky, I thought I'd buried that in there sufficently well

DeV.

sendic

Tourneque wrote:

You say that A is only 22% ? This can never be so.
You have 3 choices. A B or C. Behind 1 is a car.
When you choose a door, the probably of being correct has to be 1 in 3. Therefore by sticking, your chance cannot fall below 33%. So ProbA = 33

*light bulb flicks*

you're right. im wrong.

thanks for putting me straight.