OK OK OK! We know it’s the weekend. You’re winding down, recharging your (metaphorical) batteries and would probably rather not do things that could result in a headache (hangovers don’t count).
This, however, is pretty great. Just don’t blame us for the head-scratching that may follow.
We know how you feel.
1. The Monty Hall Problem
Say you’re on a game show where there are three doors. Behind two of the doors, there are goats. Yes, goats. Behind one of the doors, there is a brand new car. You want the car, not the goats.
The host says that once you pick a door, he’ll open one of the doors you didn’t pick to reveal a goat. You will then have the option of either staying with your door you had already selected or switching to the last unopened door.
Do you switch or stay?
Answer: Switch
This is actually based on a real game show, and the result has been the source of controversy for years.
Essentially, when you first made the selection, you had a one in three chance of correctly selecting the door that had a car behind it. Switching raised that probability to two in three that you’ll select a car.
Said another way: A player whose strategy is to always switch will only lose when the door they initially selected has a car behind it. A contestant who selects either of the two doors with a goat behind it and then switches will always get the car.
Here’s a final way to look at it, provided the contestant selected door number one.
You run an office that employs 23 people. What is the probability that two of your employees have the same birthday? For the purposes of the problem, ignore 29 February. It’s not that these people don’t deserve cake, it’ll just make things easier.
Answer: 50 per cent
Once the population of an office hits 366 people, it’s a certainty that two people in your office have the same birthday, since there are only 365 possible days of birth.
Still, assuming that each birth date (except 29 February) is equally likely, it turns out that once your office has 57 people in it there is a 99 per cent chance that two of them share a birthday. When there are 23 people, that probability is 50 per cent.
Here’s why. Instead of calculating the probability that two people share a birthday, instead calculate the converse probability that two people don’t share a birthday.
Since these are mutually exclusive scenarios (has to be one or the other), first probability plus the second probability has to equal one.
Here’s how we figure this out, then.
Select two people in the office. The probability of the second person not sharing a birthday with the first is 364/365. The probability of the third person not sharing a birthday with the first or second is 363/365. Going through the office and multiplying these together, we see this:
365/365 x 364/365 x 363/365 x 362/365 x … x 343/365 = 0.4927.
So, the probability that nobody in an office of 23 people share a birthday is 0.4927, or 49.3 per cent. That means that the probability that two people in the office share a birthday is one – 0.4927 = 0.5073, or 50.7 per cent.
That was suitably mind-numbing. Pretend it’s your birthday and eat some cake as a reward.
3. Gambler’s Ruin
A gambler has a certain amount of money (“B”) and is playing a game of chance with some win probability less than one. Every time he wins, he raises his stake to a certain fraction, 1/N, of his bankroll, where N is a positive number. The gambler doesn’t reduce his stake when he loses.
Every time he wins, he’ll raise his stake to €B/N, or his bankroll divided by N. When B= €1000 and N=4, for example, he’ll gamble €250 each time going forward. Should he win, he’ll raise it again. Should he lose, he’ll keep his stake at €250.
If he keeps at it, what are his expected winnings?
Answer: They’ll lose everything
When it comes down to it, if our gambler bets 1/N of his bankroll each time and then maintains the amount as he loses, the gambler is N losing bets in a row away from bankruptcy.
Assuming that the player keeps on playing and there is some chance that the player can lose – we are gambling, after all – then the player remains N losing bets away from a broken bank each time.
If our gambler sounds like something of an idiot, know that this is actually a rather common betting strategy. Casinos also endorse it by ensuring that players are stocked with mostly high denominational chips as they go on winning streaks in order to encourage higher bets.
Boo-urns.
4. Abraham Wald’s Memo
This is excellent.
Abraham was tasked with reviewing damaged planes coming back from sorties over Germany in the Second World War. He had to review the damage of the planes to see which areas must be protected even more.
After inspecting them, he found that the fuselage and fuel system of the returned planes are much more likely to be damaged by bullets or flak than the engines. So what should he have recommend to get extra protection?
Answer: Protect the parts that don’t have damage
Abraham Wald, a member of the Statistical Research Group at the time, saw this problem and made an unconventional suggestion that saved countless lives.
Don’t arm the places that sustained the most damage on planes that came back. By virtue of the fact that these planes came back, these parts of the planes can sustain damage.
If an essential part of the plane comes back consistently undamaged, like the engines in the previous example, that’s probably because all the planes with shot-up engines don’t make it back.
Wald’s memos on this situation – in addition to being a remarkable historical statistical document – shed additional light of the statistics developed during the Second World War that would go on to found the field of Operations Research.
Who’s a clever boy then?
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