What are the odds of winning?

[Halcyon]

So, this has been circling the Internet for years and was once the concept of a gameshow in the U.S.


The question is, what do you pick? What are your chances?

You are a contestant on a game show.

There are three closed doors.

Behind one door is the star prize, which is a brand new car.

The other two doors have a booby prize behind them.

The game show host knows which prize is behind each door.

You are asked to pick one of the doors.

Instead of opening the door you have chosen, the host then opens one of the other doors. And behind this door is a booby prize.

So that leaves two remaining closed doors. Behind one of these doors is the second booby prize. And behind the other door is the car.

You are then asked whether you want to stay with your original choice of door. Or whether you want to switch to the other remaining door.

And the question is, whether you have the better chance of winning the car if you stay with your original choice? - or if you switch to the other door? - or whether each door actually has exactly the same chance of winning the car?

Many people say it is not 50/50 and go into long explanations about it.
In my opinion, once you have been shown the booby prize in one of the doors and you picked the other door, then your chance of winning is 50/50 as there is your door you selected and the other door. So it is in either yours or the other one.






.

https://en.m.wikipedia.org/wiki/Monty_Hall_problem

Most people come to the conclusion that switching does not matter because there are two unopened doors and one car and that it is a 50/50 choice.

This would be true if the host opens a door randomly, but that is not the case; the door opened depends on the player's initial choice, so the assumption of independence does not hold.

Before the host opens a door there is a 1/3 probability that the car is behind each door.

If the car is behind door 1 the host can open either door 2 or door 3, so the probability that the car is behind door 1 and the host opens door 3 is 1/3 x 1/2 = 1/6.

If the car is behind door 2 (and the player has picked door 1) the host must open door 3, so the probability that the car is behind door 2 and the host opens door 3 is 1/3 x 1 = 1 / 3.

These are the only cases where the host opens door 3, so if the player has picked door 1 and the host opens door 3, the car is twice as likely to be behind door 2 as door 1.

The key is that if the car is behind door 2 the host must open door 3, but if the car is behind door 1 the host can open either door.

[spiderplant]

That Wiki explanation is a bit long-winded. Easier to say "the chance of you picking the car to start with is 1/3. So the chance of it remaining behind another door is 2/3. It's not the door the host opens, so the 2/3 chance must be the other door".

Many people say it is not 50/50 and go into long explanations about it.

If you don't believe it, try it. You'll soon find they are correct.

---------- Post added at 14:19 ---------- Previous post was at 14:00 ----------

ps. If you want some software to help you, here's some we prepared earlier:
https://www.cableforum.uk/board/show...9#post34854319

[nomadking]

The 1st round is immaterial, as you still don't know whether your 1st choice was right or wrong. There is a 50/50 chance of selecting the right one. Doesn't matter whether you stick with or swap your original choice.

There is no win or lose event for the 1st round, so you can't multiply the two probabilities together. Doesn't matter which one of the 3 you pick in the first round, as you are starting again with just 2 choices.

If it was 2 rounds with 2 different people separately and independently, making the selection and both had to be right, then yes you can multiply the probabilities together to get a 1 in 6 chance of both people being right.

At the point of your final choice, there are two doors, and you dont know which one has the car, so your odds are 50/50. How can they possibly be anything else ?

[spiderplant]

Originally Posted by Paul

How can they possibly be anything else ?

Because it's more likely to be behind one door than the other.

If you don't believe it, make 3 cards (2 goats and a car), ask a friend to be the host, and play a few games. Or write yourself a computer simulation.

[nomadking]

Originally Posted by spiderplant

Because it's more likely to be behind one door than the other.

If you don't believe it, make 3 cards (2 goats and a car), ask a friend to be the host, and play a few games. Or write yourself a computer simulation.

You could have a million doors in the 1st round. Doesn't matter whether you pick the correct one or one of the 999,999 incorrect ones. The result is the same, you then get to pick from just 2 doors. Any of the 1m 1st round choices is "correct".

But you don’t…

[nomadking]

Originally Posted by Hugh

But you don’t…

You could have a million doors in the 1st round

Originally Posted by Hugh

But you don’t…

Originally Posted by nomadking

The whole point is that there are three doors initially, and it’s about the mathematical odds…

[nomadking]

Originally Posted by Hugh

The whole point is that there are three doors initially, and it’s about the mathematical odds…

Doesn't change the odds. If you choose the right one in the 1st round, you don't win. You don't lose either.

If it was the situation that if you guessed correctly in the 1st round, you would win, then EVERYBODY would win. As if you lost in the 1st round, you would know that it definitely was the one of the remaining 2, you didn't pick 1st time around.

[spiderplant]

Originally Posted by nomadking

You could have a million doors in the 1st round. Doesn't matter whether you pick the correct one or one of the 999,999 incorrect ones. The result is the same, you then get to pick from just 2 doors. Any of the 1m 1st round choices is "correct".

Ok, let's play with a million boxes. That's 999,999 goats and a car.

You pick a box. The chance that you have picked the car is 1 in a million.

So the chance that the car is in one of the other boxes is 999,999 in a million.

The host now opens 999,998 boxes that he knows contain a goat. So the chance that the car is in the one remaining box is 999,999 in a million.

So do you want to keep your original pick (1 in a million), or do you want to swap (999,999 in a million)? Do you like goats?

Originally Posted by spiderplant

Because it's more likely to be behind one door than the other.

I dont think so.

You have two doors, and no idea which is the correct one, your chances are equal, how can one possibly be more likely then the other ?

The whole point is that there are three doors initially

How is that relevant ?
Again, you are left with two doors, and you dont know which one is right, how are your odds at that point anything other than 50/50 ?

---------- Post added at 21:16 ---------- Previous post was at 21:10 ----------

The key is that if the car is behind door 2 the host must open door 3, but if the car is behind door 1 the host can open either door.

Again, how is this "the key" or relevant ?
Only the host knows where the car is, you have no clue at all, as far as you are ever concerned, its behind one door other the other, you have no further information at all to help you decide, so your odds of being right when asked to choose again are 50/50.

[nomadking]

Originally Posted by spiderplant

Ok, let's play with a million boxes. That's 999,999 goats and a car.

You pick a box. The chance that you have picked the car is 1 in a million.

So the chance that the car is in one of the other boxes is 999,999 in a million.

The host now opens 999,998 boxes that he knows contain a goat. So the chance that the car is in the one remaining box is 999,999 in a million.

So do you want to keep your original pick (1 in a million), or do you want to swap (999,999 in a million)? Do you like goats?

Doesn't matter if you get your 1st round choice correct, you don't win at that stage.

You are then asked whether you want to stay with your original choice of door. Or whether you want to switch to the other remaining door.

There is always just the one other door.

[daveeb]

Originally Posted by nomadking

Doesn't matter if you get your 1st round choice correct, you don't win at that stage.
There is always just the one other door.

I told my sister this one years ago and she gave exactly your reply. You're both wrong. Years ago a maths professor in the US explained it and she was ridiculed by many of her peers until she proved it mathematically. Mathematical odds aren't always intuitive.