I am a school teacher with a love for writing short stories, usually with a humorous twist.
The ‘shell game’ is a gambling activity of dubious legitimacy that involves manipulation of three ‘shells’ -commonly walnuts, coconuts or plastic cups- and a small object, such as a pea or stone. A pea is placed under one shell, the shells dexterously shuffled and the player chooses which shell is covering the pea. At best, the chance of success would be one in three if the host did not use sleight of hand or chicanery to tilt the odds in their favour.
Let's make a deal!
Less biased probabilistic enterprises flourish in quiz shows, where contestants choose a ‘box’, ‘panel’ or ‘screen’ to reveal a prize.
One popular quiz show was Let's Make A Deal, which began on American television in 1963 and was hosted for many years by Monty Hall.
The game segment in the show became so popular -and controversial- that it came to be known as the Monty Hall Problem.
This is how the game is played.
Behind one door is a coveted prize such as a car, whilst a booby prize -represented by a goat- awaits behind the two other doors.
Now here’s the twist. Monty Hall, who knows what is behind each door, reveals one of the doors hiding a goat and asks the contestant if they want to keep the door they have chosen or to swap it for the other door.
What will be discussed is whether the player should keep their chosen door or swap it for the other door.
What are the odds?
At this stage, before a door is revealed, the player has a one in three chance of correctly choosing the door with the car.
Will swapping make a difference? This is the issue that entrapped mathematicians for a long time before the debate was put to rest.
At the time, many people thought that swapping makes no difference, reasoning that since there are three doors to choose from, the chance of choosing the right door will be 1/3, regardless of whether a door with a goat is opened.
If the player chooses to swap, there is also a 1/3 chance that the door chosen contains the car. Hence, there would be no need to swap.
Another forum believed that when one door is revealed, the odds become more favourable because for the remaining two doors, the chance that behind one of them is the car is now ½.
Hence, again, there is no need to swap.
In 1990, Marilyn vos Savant, a well-known American columnist, mathematician and scientist, replied to a reader who had asked her opinion of the problem. Miss Savant’s reply created much controversy, especially since it is counter intuitive.
In essence, her advice is ‘always swap!’ because it guarantees that the chance of winning will increase by 1/3.
The rationale for swapping might be better understood if we imagine that there are 100 doors, behind which there is one car and 99 goats.
This means initially that the chance of choosing the door with the car is 1/100, or 1%.
Monty Hall reveals 98 doors, each displaying a goat. The probability that the car is behind the 100th door is now (1 + 98)/100 = 99/100, or 99%.
So, following this line of reasoning, if there are three doors, the chance that the car is behind the door the contestant can swap with jumps from 1/3 to (1+1)/3 or 2/3, which is approximately a 33% increase.
A good way to demonstrate the advantage in swapping is to perform a computer simulation of the game. This experimental approach was the undoing of the famous mathematician Paul Erdos, who was adament that ‘swapping’ was mathematically erroneous until confronted with simulations data that convinced him that he was wrong.
Shown below are computer simulations for 100, 1000 and 10,000 trials of the game.
Notice that the ratio swap : no swap is about 67% : 33%, which is very close to the predicted ratio of 2/3 : 1/3 or 2 : 1.
Today, there is widespread acceptance amongst mathematicians and others of the counter-intuitive fact that it is better to swap doors. This begs the question:
Did Monty Hall know about this mathematical property when the game was first played on his show?