## Posts Tagged ‘probability’

### The Monty Hall Problem

Thursday, March 19th, 2009

If you’ve never heard of the Monty Hall Problem before, here is a brief summary.

You are in a game show where you are faced with three closed doors. Behind one of the doors is a brand new car and the other doors hide a goat each. You then choose a door in an attempt to win the car but before the door is opened the host (Monty Hall) opens one of the remaining doors he knows contains a goat. You are then given the opportunity to choose whether to stay with your original selection or switch to the other, remaining closed door.

Intuitively most of us tend to think that it doesn’t really matter because we think there is a 50/50 chance either way but the reality is that statistically you have a greater chance of winning the car if you switch. In fact you stand a 2/3 chance of winning with the switch strategy as opposed to a 1/3 chance with the stay strategy.

If you remain unconvinced feel free to try it out for yourself with a friend or a die and a piece of paper.

Anyway, the point of this post is to say that in the two years or so that I’ve known about this I’ve never heard a concise and easy-to-understand explanation for why this is so. But now I have one…

You have a 2/3 chance of selecting a goat instead of a car. This means that two thirds of the time Monty Hall will reveal the last remaining goat which means that two thirds of the time if you use a switch strategy you will win.

To highlight how this works, imagine the same scenario but with 1000 doors and only 1 car. You have a 1/1000 chance of selecting the car or a 999/1000 chance of selecting a goat. If Monty Hall opens 998 other doors with goats behind them it is extremely likely (999/1000) that the last remaining door will have the car. (Thanks A3!)

You have two strategies: switch or stay. Using the stay strategy, if you accidentally choose a car first off (1/3 chance) you will win but if you choose a goat (2/3) you’ll lose. Using the switch strategy you’ll reverse these odds giving you statistically a 66.6% chance of winning every time.

Simple huh?

### The Chances of Sharing a Birthday

Tuesday, May 27th, 2008

In any given group of people, how many people do you think there need to be in order for there to be a 50% chance of at least two people sharing the same birthday?

365 / 2 = ~183 people?

Nope. Think again. This is not a trick question, just plain old mathematical probability.

The answer, which I found very counter-intuitive at first, is 23. It’s called the Birthday Paradox. The mistake I was making (and that most people would probably make) is that I was picking a single birthday and thinking of the probability of any given birthday matching it rather than moving on and testing for every other birthday possibility in the group.

I still struggle with it however, when I think back to school days where there should have been a 100% chance of two kids sharing a birthday in any two classes. I can’t remember anyone sharing a birthday at all.

(Or have I done the math wrong here by assuming that two classes of 23 students will give a 100% chance? Perhaps this equates to 75% instead or remains at 50%? Gaaah! I knew I should have listened in school!).

Jack, if you are reading this perhaps you could test this to see if it really works in a class situation?