# The Answers

## Russian roulette

I have a revolver which holds up to six bullets. There are two bullets in the gun, in adjacent chambers. I am going to play Russian roulette (on my own!), I spin the barrel so that I don't know where the bullets are and then pull the trigger. Assuming that I don't shoot myself with this first attempt, am I now better off pulling the trigger a second time without spinning or spin the barrel first? (Thanks to pusher.)

** Solution**

This is a very typical, simple probability Brainteaser. It doesn't require any sophisticated or lateral thought. Just pure calculation.

Whenever you spin the barrel you clearly have a two in six, or one in three chance of landing on a chamber containing a bullet.

If you spin and pull the trigger on an empty chamber, what are the chances of the next chamber containing a bullet? You are equally likely to be at any one of the four empty chambers but only the last of these is adjacent to a chamber containing a bullet. So there is now a one in four chance of the next pull of the trigger being fatal. Conclusion is that you should *not* spin the barrel. After surviving two pulls of the trigger without spinning the barrel the odds become one in three again, and it doesn't matter whether you spin or not (at least it doesn't matter in a probabilistic sense). After surviving that 'shot' it becomes fifty-fifty and if you are successful four times in a row then the next shot will definitely be fatal.

## Matching birthdays

You are in a room full of people, and you ask them all when their birthday is. How many people must there be for there to be a greater than 50% chance that at least two will share the same birthday? (Thanks to baghead.)

**Solution**

This is a classic, simple probability question that is designed to show how poor is most people s perception of odds.

As with many of these type of questions it is easier to ask what are the chances of two people *not* having the same birthday. So suppose that there are just the two people in the room, what are the chances of them not having the same birthday? There are 364 days out of 365 days that the second person could have, so the probability is 364/365. If there are three people in the room the second must have a birthday on one of 364 out of 365, and the third must have one of the remaining 363 out of 365. So the probability is then 364 *x* 363/3652. And so on. If there are *n* people in the room the probability of no two sharing a birthday is

So the question becomes, what is the smallest *n* for which this is less than one half? And the answer to this is 23.

## Another one about birthdays

At a cinema the manager announces that a free ticket will be given to the first person in the queue whose birthday is the same as someone in line who has already bought a ticket. You have the option of getting in line at any position. Assuming that you don't know anyone else's birthday, and that birthdays are uniformly distributed throughout a 365-day year, what position in line gives you the best chance of being the first duplicate birthday? (Thanks to amit7ul.)

** Solution**

This is solved by an application of Bayes' theorem.

You need to calculate two probabilities, first the probability of having the same birthday as someone ahead of you in the queue given that none of them has a duplicate birthday, and second the probability that none of those ahead of you have duplicate birthdays. If there are *n* people ahead of you then we know from the previous birthday problem that the second

probability is

The first probability is simply n/365. So you want to maximize

This is shown as a function of *n* above. It is maximized when *n =* 19 so you should stand in the 20th place. This maximizes your chances, but they are still small at only 3.23%.