Hi all,

This is a little maths/ probability challenge I like- the obvious answer is not the right one

You are in a game show. There are three doors. Behind 1 door is a car, behind the other two is nothing.

The game show host knows what is behind each door. He asks you to pick a door. You do. He then looks at the two doors left, and opens one to reveal nothing behind it (he knew this would happen).

There are now two doors remaining. The door you picked (that hasn't been opened) and the other door.

He says to you, "Now, would you like to stick to your original choice, or change to the other unopened door?"

Is your probability of winning greater if you stick with your original choice, or change to the other remaining door?

I've noticed that quite often host Eddie McGuire, of the "Hot Seat" quizz show asks people if they want to change their answer when they have made the wrong choice. However I don't know what the mathematical probability of changing doors being a better choice is.

I'm no mathematician but the odds on the second choice would be 50/50 whether you stick with your first choice or change. The only variable would be if you think the game show host is trying to help you win or trying to stop you from winning.

Incorrect partickt. Consider what has happened before. Anyone else like to have a 'stab'?

This is an old one which I remember from some years/decades ago ...

I copied the information below from the internet, because it is more concise and accurate than anything I would write.

When you initially made the first choice, you had a one in three chance of winning. The other scenario, in which the prize is behind one of the other two doors and so only one of those two doors is a losing choice, has a probability of two-thirds. Let’s focus on that scenario, and assume — since we don’t have any information to the contrary — that the host always opens a door that reveals nothing. In that case, the door he didn’t open and that you didn’t choose always contains the car. So if you switch, there’s a two-thirds chance that you’ll win the car. If you don’t, you’re stuck with a one-third chance.

You flip a coin twenty times. The odds against flipping a coin and getting heads twenty times in a row are astronomical. However, if you flipped a coin nineteen times and gotten a head each time and you are preparing to flip again, the odds for getting another head is fifty-fifty.

In the problem given, the odds of picking the right door on the first try would be one out of three. The odds on picking the right door on the second try would be 50-50. Which door you pick or why you select it would seem to be irrelevant.

the first choice you make has a 1/3 chance of winning.

The probability that the prize is behind the other two doors is 2/3.

the host opens one of the doors you didnt pick, picking a door with out the prize and knowing it.

The proability that the prize is behind the other two doors remains 2/3. and one of them has been opened

Therfore the probability that the prize is behind the unopened door that you didnt pick is 2/3.

for you the probability that the prize is behind your door stays at 1/3.

switching increases your chances to 2/3.

Patrick, you are right about the coin toss as each toss is independent of the time before. However, your 2nd pick in the 3 dor problem is dependent on prio probabilities of the previous decisions, not independent of them.

I agree with Patrick, once one of the three doors has been eliminated the chance of picking the correct door is 50% as there are only two choices.
If two doors are eliminated then the chance of picking the prize is 100%.

For those who think the correct answer is still fifty-fifty and don't understand why it isn't, I encourage you to look up the "Monty Hall" problem on wikipedia so you can better understand how the probability works.

Never mind the theory, once one door has been eliminated it is no longer in contention/the equation.

Therefore if one has two doors and there is a car/prize/bliss behind one of them then the chances of opening the right one is 50/50.