More counterintuitive stats
Another statistical conundrum that will stimulate fierce debate down the pub! Here again we see our inability to interpret statistical data correctly and where our 'common sense' breaks down.
The scenario is based on a 20th century television game show 'Let's Make a Deal' from the USA - hosted by Monty Hall. In the game a contestant is faced with 3 doors. behind 1 door is a Cadillac (expensive American vehicle) and behind the other 2 are goats. The contestent has to choose a door and gets what's behind it. The twist is this: after the choice is made Monty Hall opens one of the other doors to reveal - a goat. So now we have 2 doors (including the contestant's chosen door) left one of which hides the car. Monty now asks the contestant if she would like to change her mind and the question is, should she change or should she stick - or does it make no difference at all?
Most people argue this way: there are 2 doors left so you have a 50:50 chance of getting the car whatever you do - so it really doesn't matter if you change or not. But, rather surprisingly, this is wrong! To maximise your chances of wining you MUST change your selection! In fact this strategy DOUBLES your chance of winning
This result seems quite ridiculous but it's true - and people have run experiments that verify it. The problem here is that people look at the second choice change option as a completely new game (with a 50:50 outcome) but it isn't and also Monty Hall knows which door is which and will always open a goat door.
Here's why it works:
If you are a 'sticker' and don't change your mind then you will win the car if, and only if, your first choice is correct - and obviously that means you have a 1 in 3 chance of winning.
If you are a 'switcher' things are a little more complicated. Lets go through the possible combinations. We have three possible doors goat1, goat2 and car.
If we chose goat 1 initially, Monty will remove goat2 (he knows which door has the car and won't choose that one will he?). at this stage if we swap we win.
If we chose goat 2 then by the same argument if we swap we win.
If we chose the car then by swapping we will lose. Remember that Monty isn't chosing the remaining goat door at random.
So 2 out of 3 times swapping will win.
The conclusion is therefore that the swapping strategy is twice as successful as not swapping!
By-the-way, as I mentioned earlier, this result has been verified in the real world by playing the game countless times using students (probably without the Cadiallac I suspect) - so if you find a contrary proof then you're wrong, unless the proof includes a parallel universe!
The scenario is based on a 20th century television game show 'Let's Make a Deal' from the USA - hosted by Monty Hall. In the game a contestant is faced with 3 doors. behind 1 door is a Cadillac (expensive American vehicle) and behind the other 2 are goats. The contestent has to choose a door and gets what's behind it. The twist is this: after the choice is made Monty Hall opens one of the other doors to reveal - a goat. So now we have 2 doors (including the contestant's chosen door) left one of which hides the car. Monty now asks the contestant if she would like to change her mind and the question is, should she change or should she stick - or does it make no difference at all?
Most people argue this way: there are 2 doors left so you have a 50:50 chance of getting the car whatever you do - so it really doesn't matter if you change or not. But, rather surprisingly, this is wrong! To maximise your chances of wining you MUST change your selection! In fact this strategy DOUBLES your chance of winning
This result seems quite ridiculous but it's true - and people have run experiments that verify it. The problem here is that people look at the second choice change option as a completely new game (with a 50:50 outcome) but it isn't and also Monty Hall knows which door is which and will always open a goat door.
Here's why it works:
If you are a 'sticker' and don't change your mind then you will win the car if, and only if, your first choice is correct - and obviously that means you have a 1 in 3 chance of winning.
If you are a 'switcher' things are a little more complicated. Lets go through the possible combinations. We have three possible doors goat1, goat2 and car.
If we chose goat 1 initially, Monty will remove goat2 (he knows which door has the car and won't choose that one will he?). at this stage if we swap we win.
If we chose goat 2 then by the same argument if we swap we win.
If we chose the car then by swapping we will lose. Remember that Monty isn't chosing the remaining goat door at random.
So 2 out of 3 times swapping will win.
The conclusion is therefore that the swapping strategy is twice as successful as not swapping!
By-the-way, as I mentioned earlier, this result has been verified in the real world by playing the game countless times using students (probably without the Cadiallac I suspect) - so if you find a contrary proof then you're wrong, unless the proof includes a parallel universe!
