### Coin Tossing Games

You and I play a game involving successive throws of a fair coin. Suppose I pick HH and you pick TH. The coin is thrown repeatedly until we see either two heads in a row (I win) or a tail followed by a head (you win). What is the probability that you win?

### Fixing the Odds

You have two bags, four red balls and four white balls. You must put all the balls in the bags although you are allowed to have one bag empty. How should you distribute the balls between the two bags so as to make the probability of choosing a red ball as small as possible and what will the probability be in that case?

### Penta Colour

In how many different ways can I colour the five edges of a pentagon red, blue and green so that no two adjacent edges are the same colour?

# The Birthday Bet

##### Age 14 to 16 Challenge Level:

This solution comes from Tom Ridley:

Assuming that the problem stated means that you win if AT LEAST two people have the same birthday, then the following method is appropriate. For the purpose of simplifying the problem, people with their birthday on the 29th of February have been excluded.

Instead of looking at the problem in the stated way (i.e. the probability that two or people will have the same birthday) it is simpler to look at it in terms of the probability that no one will have the same birthday. Once this has been calculated, the probability that two or more people will have the same birthday can be calculated by subtracting this value from 1.

There are 365 days in the year. Assume that primarily we only have two people. The probability that they have different birthdays is 365/365 $\times$364/365. This is because if the second person's birthaday is not the same as the first person's birthday then his/her birthday can be on any of the 364 other days of the year. Therefore for ten people, the probability that nobody will have the same birthday is

365/365 $\times$364/365 $\times$363/365 $\times$362/365 $\times$ 361/365 $\times$360/365 $\times$359/365 $\times$358/365 $\times$357/365 $\times$356/365.

This approximately equates to: 0.883.The probability therefore that at least 2 people will have the same birthday is 1 - 0.883 = 0.117. So, the question still remains: Is it worth it? If you're wrong, you lose £1, if you're right, then you gain £20. As the potential loss is 1/20 of the potential gain, I would argue that for the bet to be worth making, the chance of winning should be greater than 1/20. 1/20 = 0.05 and 0.116948177> 0.05 therefore I believe that the bet is worth making. Even if the prize were only £10, probability would still be in favour of you making youre money back as 0.116948177 > 0.1

Several other correct solutions to this tricky puzzle were received -- well done everyone!