I recently came across a paradox known as the Monty Hall problem. It goes like this: imagine you're on a game show, and the host, (Monty) asks you two pick one of three doors. Two of the doors have a goat behind them and one door has a car behind it. When you pick your door, Monty will choose one of the remaining doors for himself. However, Monty's a nice guy, and he ALWAYS chooses a goat. Now you have a choice: keep your door, or take the one Monty left?

Surprisingly, the odds of you winning by taking the last door are 2/3. This is counter-intuitive for most people (myself included!), but this guy provides a nice explanation of the math behind it: http://betterexplained.com/articles/understanding-the-monty-hall-problem/#comment-5738

the commenter DerekSmith gave a very good explanation:

DerekSmith on September 12, 2009 at 1:53 pm said:

We humans have a habit of being acquisitive. Once we choose something, we have a tendency to want to hang on to it. The act of choosing a door, makes it ‘our door’ and so there is a natural reluctance to part with it.

An easier way of intuitively understanding the Monty Hall game is to hold off selection until the ‘door’ has been opened, and to show the higher probability of switching choice.

Take a pack of cards – it contains a single Ace of Diamonds – this is the winning card – pick the Ace of diamonds and you win.

Shuffle the pack and take a single card out – lay it on the table. Put the rest of the pack on the table beside the single card.

Now, the choice is pick the single card or pick the rest of the pack – which do you thinks is most likely to give you the Ace of Diamonds. Clearly, most people would choose the pack with 51 chances that they have chosen the Ace.

Now we add in the ‘opening of the doors’

As before, pick a card and lay it on the table, but before putting the pack down, somebody sorts through the pack and if it contains the Ace of Diamonds, they put it on the top of the pack and then put the pack on the table.

Now you have the choice of taking the single card or the top card off the pack – again, automatically folks will take the top card from the pack.

Now we put it all together – ‘Choose’ a card and put it on the table (we now own it). Next the pack is sorted and if the Ace is present it is put on the top of the pack.

The top card is taken off the pack and placed beside your chosen card and the rest of the pack are turned face up to show that they are not the Ace of Diamonds.

Now you are offered the choice of changing the card you chose for the remaining card from the pack. Even though ‘ownership’ is now involved, most people would not have any doubt about changing, and understand that there is not the remotest chance that the first card has a 50:50 chance of being the Ace.

they will understand that they had a one in 52 chance of picking the Ace first time, and almost dead certainty of it being the other card – they would switch…

-Hayden

Surprisingly, the odds of you winning by taking the last door are 2/3. This is counter-intuitive for most people (myself included!), but this guy provides a nice explanation of the math behind it: http://betterexplained.com/articles/understanding-the-monty-hall-problem/#comment-5738

the commenter DerekSmith gave a very good explanation:

DerekSmith on September 12, 2009 at 1:53 pm said:

We humans have a habit of being acquisitive. Once we choose something, we have a tendency to want to hang on to it. The act of choosing a door, makes it ‘our door’ and so there is a natural reluctance to part with it.

An easier way of intuitively understanding the Monty Hall game is to hold off selection until the ‘door’ has been opened, and to show the higher probability of switching choice.

Take a pack of cards – it contains a single Ace of Diamonds – this is the winning card – pick the Ace of diamonds and you win.

Shuffle the pack and take a single card out – lay it on the table. Put the rest of the pack on the table beside the single card.

Now, the choice is pick the single card or pick the rest of the pack – which do you thinks is most likely to give you the Ace of Diamonds. Clearly, most people would choose the pack with 51 chances that they have chosen the Ace.

Now we add in the ‘opening of the doors’

As before, pick a card and lay it on the table, but before putting the pack down, somebody sorts through the pack and if it contains the Ace of Diamonds, they put it on the top of the pack and then put the pack on the table.

Now you have the choice of taking the single card or the top card off the pack – again, automatically folks will take the top card from the pack.

Now we put it all together – ‘Choose’ a card and put it on the table (we now own it). Next the pack is sorted and if the Ace is present it is put on the top of the pack.

The top card is taken off the pack and placed beside your chosen card and the rest of the pack are turned face up to show that they are not the Ace of Diamonds.

Now you are offered the choice of changing the card you chose for the remaining card from the pack. Even though ‘ownership’ is now involved, most people would not have any doubt about changing, and understand that there is not the remotest chance that the first card has a 50:50 chance of being the Ace.

they will understand that they had a one in 52 chance of picking the Ace first time, and almost dead certainty of it being the other card – they would switch…

-Hayden