The Monty Hall problem is a prime example of how, when presented with the simple challenge of selecting one favourable outcome against two unfavourable outcomes, we display a basic inability to correctly weigh up the chances of success.
This is crucial for bettors because in simple terms, if a bettor cannot recognise implied probability and whether a bookmaker’s odds represent ‘value’ they will never make any money in the long term.
The Monty Hall problem
A brand new car is behind one of three doors. Behind the other two is a goat. You must correctly guess which door hides the car in order to win it, but you have no prior knowledge that allows you to distinguish among the doors.
After you choose a door, one of the other doors opens to reveal one of the two goats. You now have another option – do you change door, or do you stick with your original choice?
Named after the host of “Let’s Make a Deal”, a popular US show in the 60s & 70s which formed the basis of the poser, the Monty Hall problem is a seemingly simple mathematical puzzle which effectively demonstrates how people struggle with what appears to be a very straight-forward choice.
With this simple yet cleverly posed teaser, the show demonstrated how the average person can demonstrate counter-intuitive behaviour when faced with probability conundrums – and the same is true of casual bettors. When this question was posed in parade magazine, 10,000 readers complained that the published answer was wrong – including several maths professors.
The Monty Hall solution
The solution to the Monty Hall problem is simple: always switch doors. After the first door is opened, the car is definitely behind one of the two closed doors (although you have no way of knowing which). Most contestants on the show intuitively see no advantage in switching doors, assuming that each door has an equal (1/3) probability.
This is incorrect – in fact, the chances of winning the car double by switching. While it is true that originally each door had a 33.3% chance of hiding the car, after the first goat is revealed, the probability that the car lies behind the remaining door is 66.6%.
It is easiest to calculate these probabilities by imagining that you’re picking between your original door (33.3% probability) and the combined probabilities of the other two doors (33.3% + 33.3%). This is because once you choose your door, the other two are then paired together – there is a 66.6% chance it is behind one of those two doors. When one is then removed, there is still a 66.6% chance that the car lies behind the remaining door.
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