Non poker probability problem

[LeeG]

This is a classic problem for those who haven't heard it. The Monty Hall problem.

Lets Make A Deal has a game that works like this:
There are 3 doors. Behind 1 is a car. Behind the other 2 are goats. The contestant chooses a door (but does not open it). The host then opens a door that has a goat behind it (he knows which doors have what).

The contestant can then choose to stay with his original choice, or switch to the other closed door.

Question:
Does have a better chance to win if:
A: He stays
B: He switches
C: Doesn't matter - it's the same chance

[xDiamond_CutteRx]

Better if he switches, because there is a greater chance he was intially wrong. It's counterintuitive, but I am fairly certain that is the correct answer.

[HalfSugar]

Better if he switches, because there is a greater chance he was intially wrong. It's counterintuitive, but I am fairly certain that is the correct answer.

True.

Chance of picking the car first off is 1/3. Chance of picking a goat is 2/3.

Assuming he has picked a goat, the host has to open the door to the other goat as his other option would be to reveal the car which he would not do.

It is better to switch because there is a 2/3 chance that the car is behind the other door since there is only a 1/3 chance that he picked the car initially. By opening a known goat, the host removes it from the problem but the odds do not drop to 50:50, they change from 1/3, 1/3, 1/3 to 1/3, 2/3 in favour of the switch.

Of course, if you picked the car in the first place, you were unlucky because you would switch away from it. However, 2/3 times, it is right to switch!

[Gunslinger]

http://en.wikipedia.org/wiki/Monte_Hall_Problem

[suitedaces84]

Since everyone knows this problems already, here are a few variants:

1) 1 car; 2 goats. You and another contestent each choose a door. At least one of you will choose a goat. If only one player chooses a goat the host will open that player's door. If both players choose the goat the host will randomly select one player and open his door.

The host opens the other player's door, revealing a goat. Stay or switch?

2) 1 car; 4 goats. You get to pick 2 doors. After you pick the host will open 2 of the 3 doors that you did not pick revealing goats.

You will then have the option to trade your 2 doors for the 1 unopened door that you did not select. Should you trade your 2 doors for the 1 door?

[Garoen]

The original problem seems stupid to me, and I think it's overcomplicated.

You are given three choices. One choice is correct, two are wrong.

When you make your choice, a 3rd party will reveal the outcome of one of your unchosen selections. This outcome will always be bad.

This action in and of itself does not mean that we have made an incorrect selection, ultimately it does not mean a single thing at all.

Because no matter what, after we've made our selection our 3rd party will reveal an incorrect choice. And whether we've chosen a goat or a car, we will be shown the unchosen goat. Now how the hell is it advantageous for us to change our selection? When we pick the car, WE ARE SHOWN A GOAT. When we pick a goat, WE ARE SHOWN A GOAT.

the core facts of this whole thing shout same odds to me.

[kingetje]

^^^yea but if you choose a goat in the first place, YOU WILL ALWAYS WIN THE CAR WHEN YOU SWITCH. and seeing as there is the biggest chance of picking a goat initially, that works out

[supafrey]

Since everyone knows this problems already, here are a few variants:

1) 1 car; 2 goats. You and another contestent each choose a door. At least one of you will choose a goat. If only one player chooses a goat the host will open that player's door. If both players choose the goat the host will randomly select one player and open his door.

The host opens the other player's door, revealing a goat. Stay or switch?

2) 1 car; 4 goats. You get to pick 2 doors. After you pick the host will open 2 of the 3 doors that you did not pick revealing goats.

You will then have the option to trade your 2 doors for the 1 unopened door that you did not select. Should you trade your 2 doors for the 1 door?

It's been a while since I've used pen and paper on THP.

1. With 3 choices and 2 players, there's 6 different ways this first option can turn out, two of which involve us immediately picking the car. This gives us immediate odds of 1/3. After the other player is revealed to have a goat, this effectively eliminates two of the 6 scenarios (scenarios where you had a goat, but your opp had a car). Of the remaining 4, two of them involve you winning a car. Your odds are now 1/2 of having a car. Seems like you have to switch.

2. There are 10 different variations, of which 4 give us the car (2/5 odds). Obviously, if we have the car, we don't want to switch away from them. However, if we haven't originally picked the car, then Monte is effectively pointing out which of the other 3 spots has the car in it! Meaning, if we haven't picked the car, he's going to pick it for us, and we must switch. The remaining 6 out of 10 variations all fall into this. Considering there's a 60% chance our original pick was wrong, we must switch.

Huge miscalculations?

[supafrey]

The original problem seems stupid to me, and I think it's overcomplicated.

You are given three choices. One choice is correct, two are wrong.

When you make your choice, a 3rd party will reveal the outcome of one of your unchosen selections. This outcome will always be bad.

This action in and of itself does not mean that we have made an incorrect selection, ultimately it does not mean a single thing at all.

Because no matter what, after we've made our selection our 3rd party will reveal an incorrect choice. And whether we've chosen a goat or a car, we will be shown the unchosen goat. Now how the hell is it advantageous for us to change our selection? When we pick the car, WE ARE SHOWN A GOAT. When we pick a goat, WE ARE SHOWN A GOAT.

the core facts of this whole thing shout same odds to me.

They're supposed to shout same odds, but they're not. That's the point of being a counterintuitive problem =). It only seems overcomplicated because you don't get it.

The point is that when you're originally choosing one of three doors, you have a 1/3 chance of picking right. When Monte tells you which of the other picks aren't it, all he's doing is effectively now making it a 1/2 chance of your door having the car. 1/2 is better than 1/3, so we have to in some way "choose" from this new set of parameters. "Switching" makes us basically "choose again" from only the two cases.

[HalfSugar]

When we pick the car, WE ARE SHOWN A GOAT. When we pick a goat, WE ARE SHOWN A GOAT.

the core facts of this whole thing shout same odds to me.

This is why it seems counter-intuitive to a lot of people but don't forget that initially, there is a 2/3 chance that we picked a goat. It is therefore true that we want to move away from this choice 2/3 of the time which makes switching advantageous because if we did pick wrong and we were then shown a goat, we know for sure we are getting the car. Of course 1/3 of the time we will move away from the car because we lucked out and picked it in the first place but statistically speaking, switching is twice as good as staying put.

[supafrey]

1/3 of the time you'll be wrong in your coinflip decision whether to switch! Clear as mud!

[HalfSugar]

The switching decision is not a coin flip unless your coin is three sided and two of them say change.

[suitedaces84]

1. With 3 choices and 2 players, there's 6 different ways this first option can turn out, two of which involve us immediately picking the car. This gives us immediate odds of 1/3. After the other player is revealed to have a goat, this effectively eliminates two of the 6 scenarios (scenarios where you had a goat, but your opp had a car). Of the remaining 4, two of them involve you winning a car. Your odds are now 1/2 of having a car. Seems like you have to switch.

You're missing something here. Think about what will happen if you both choose a goat.

2. There are 10 different variations, of which 4 give us the car (2/5 odds). Obviously, if we have the car, we don't want to switch away from them. However, if we haven't originally picked the car, then Monte is effectively pointing out which of the other 3 spots has the car in it! Meaning, if we haven't picked the car, he's going to pick it for us, and we must switch. The remaining 6 out of 10 variations all fall into this. Considering there's a 60% chance our original pick was wrong, we must switch.

Yep.

[suitedaces84]

1) 1 car; 2 goats. You and another contestent each choose a door. At least one of you will choose a goat. If only one player chooses a goat the host will open that player's door. If both players choose the goat the host will randomly select one player and open his door.

There are 6 possible outcomes. In outcomes 1 and 2 you pick the goat and your opponent picks the car; in outcome 3 you both pick goats and your opponent's door is opened; in outcome 4 you both pick goats and your door is opened; in outcomes 5 and 6 you pick the car and your opponent picks the goat.

Since your opponent's door was opened we can eliminate outcomes 1, 2 and 4.

2 of the 3 remaining outcomes occur when you picked the car. So you should stay and will have picked the car 2/3 of the time.

[Kemics]

Can someone explain the gap in my logic here

I understand that it is correct to change in theory because 2:1 is better than 3:1. But my changing we're arn't adding another door, by choosing to stay are we not also accepting 2:1 odds on staying? Its still an active descision, accepting new odds in a new situation.