To see the Brain Teaser, click here The answer may surprise you:
If there are at least 23 people on a bus, there is a better than 50:50 chance that 2 of them will share the same birthday.
Here is the maths if you are interested:
Pick a passenger at random (or the driver for that matter, or even the conductor!) Their birthday is BD1. Now pick another person. The chance that their birthday is also BD1 is 1 in 365 (remember, no leap years). This means that the chance that their birthday (chanceBD2) is NOT BD1 is 364 /365. Now pick a third person and repeat. The chance that they have yet another, different birthday (chanceBD3) is 363 /365, and so on to chance BD4, chanceBD5 .. etc. If you can think back to probability lessons at school, to calculate the chance of getting Result 1 AND Result 2 (assuming they are mutually exclusive) you multiply them together. So we multiply chanceBD2 x chanceBD3 x chanceBD4 etc. OR 364/365 x 363/365 x 362 /365…. etc. If we carry on up to and including chanceBD23 the sum calculates to 0.4927 or 49.27% In other words, there is a 49.27% chance that all 23 people have different birthdays OR, to put it another way, a 50.73% (better than 50:50) chance that they don’t all have different birthdays (i.e. at least 2 of them have the same birthday. So yet again, Barak has won the day.