Non poker probability problem

[LeeG]

The big misunderstanding most people have with the original problem is that the door opened isn't random.

It becomes more obvious as you add doors. If there were 100 doors to start with, and the host opened 98 of them, would you still stay with your original choice?

You have two ways of winning: Choose correct and stay, or choose wrong and switch. Which is more likely?

[Gunslinger]

Its still an active descision, accepting new odds in a new situation.

It's NOT a new situation. It's more information given about the same situation. The host is required to open a door with a goat, regardless of the first door chosen.

If you switch EVERY time you are presented with this problem, then the 1/3 of the time you were initially correct, you would now get a goat, and the 2/3 of the time you were initially wrong, you would now get the car.

2/3 > 1/3

[Kemics]

if you switch EVERY time you are presented with this problem, then the 1/3 of the time you were initially correct

but surely for this to be the case we are ALWAYS initially correct because the host didnt open OUR door. Because it would be a crap riddle if it went, you pick 1 out of 3 doors, you picked wrong. you lose.

Odds dynamically change. If i put 3 stones in a bag. One green, one blue and one red. I then have to try to guess the order in which the colour of the stones are pulled out. I guess green, red then blue. First stone out is green, now am i more or less likely to be wrong about the colour of the second stone. Or is it a 50-50 chance?

Its like saying on the flop, i have 4 to a flush and i know he's got an over pair (hypothetical rounders oreo style tell) So i've 36% chance to win. Turn comes blank. Do you figure you are a 36% chance to win, or has it now dropped to 20%. Do you still calculate your odds from the flop or with the new information presented.

are these senarios the same as the goat?

[Gunslinger]

The difference with your stone/flush draw examples is that no matter what choice or bet you initially make, the order of the stones (or the cards on the turn/river) will not change. They are not affected by your choice.

With the goat problem, your initial choice DOES affect the next piece of information. Here's how: Let's say the car is behind door A, and goats are behind doors B and C. Before you make your choice, the host cannot think to himself, "I'm going to reveal the goat behind door C after he chooses". He can't because what if you choose door C? Then he HAS to reveal door B. If you choose door B, he HAS to reveal door C. In these TWO out of your THREE initial choices, switching doors gets you the car.

If you initially choose door A, the host can reveal EITHER door B or C. In this ONE out of your three initial choices, switching doors gets you a goat.

For a great detailed explanation of all this, see the wikipedia link I posted in the first page of this thread.

[MasterMike]

it makes more sense if you think about it with 100 doors and showing you 98.

Although with 3 doors, it just seems like semantics.

I do understand the logic even though im not thoroughly convinced with the 3 doors.

[Sean_in_NJ]

A simulation for the problem.

[jimmer]

I don't see what all the fuss is about. I'd be happy to win the goat.

[Kemics]

Gunslinger - thanks, i geddit. I didnt realise it was the host who opened a door, i thought it was random. Probably should take more time to read this stuff through

[Godson]

To make this make sense to people who have a hard time figuring out the answer add more doors. Try 50 doors, 49 goats, and 1 car. Where the contestant chooses a door and the host reveals 48 of the doors that aren't it. Obviously the other door is more likely to have the car since the host is not allowed to reveal your door. This is the key point to the problem. THE HOST CAN NOT REVEAL YOUR ORIGINAL SELECTION. Lets break this up into odds now.

1st level:
Choosing the goat initially = (49/50)
Choosing the car initially =(1/50)

2nd level:
Switching:
Choosing the goat initially and then switching= (50/50)
Choosing the car initially and then switching =(0/50)

Staying:
Choosing the goat initially and then staying=(0/50)
Choosing the car initially and then staying =(50/50)

Since you do not know what you picked from the beggining you can only look at the factor of switching versus staying. Shown below.

Switching:
(49/50)(50/50)+(1/50)(0/50)= 49/50
Staying:
(49/50)(0/50)+(1/50)(50/50)= 1/50

The above reasoning shows the actual probabilities. The (50/50)'s and (0/50)'s tends to make things seem more complicated then it needs to be, but it also shows the reasoning from a tree branch method view. In this particular situation you can dismiss the (50/50)'s and (0/50)'s intuitively giving the following below.

Switching:
(49/50)
Staying:
(1/50)

Same results just simpler.

Now taking the original problem into hand you can see the same results by plugging in the other numbers. It will still be in the same form. Thus, giving an answer 2/3 chance of getting the car by switching and 1/3 chance by staying. This means you should switch. I figured by exagerating the problem with more doors would make this make more sense in a concrete way rather than using abstract thinking.

[foxymoron]

A similar problem which I don't fully understand is:

A winner of a quiz show gets to choose from two envelopes, one with twice as much as the other. He picks and opens one envelope and it has $100. He is given the opportunity to change. The other envelope has $50 or $200, but I know that changing does not improve his expectation.

However, if one wins a prize of $100 in a slot machine in Vegas, and is given the opportunity to double or halve the prize, based on a coin toss or similar with 50% chance of success, one should toss the coin.

[HalfSugar]

However, if one wins a prize of $100 in a slot machine in Vegas, and is given the opportunity to double or halve the prize, based on a coin toss or similar with 50% chance of success, one should toss the coin.

That bet has a perpetuity of +25% EV. I'll take it. A billion times please.