MATHS WEEK STARTED on Saturday and, as is our annual tradition, we’re setting our readers some puzzles. Give them a go!

Day five -  An unfinished game

Yesterday we introduced the Chevalier de Méré and his question that got the brilliant Blaise Pascal working on probability.

De Méré (whose real name was Antoine Gombaud) was troubled by another question about gambling, the problem of points. Pascal in turn reached out to a more established mathematician, Pierre de Fermat (1607 – 1665). A very fruitful correspondence resulted.

The problem of points (also called the division problem) is about an unfinished game. If a game of chance is interrupted before the expected end, how should the prize money be divided?

Suppose two players put up a stake of €5 each and agree that whoever gets the best out of five tosses of a coin wins the stake. Áine with heads wins the first two and then Ben wins the third with tails. On the fourth toss, the coin rolls down a drain. Unfortunately, they have no other coin and must abandon the game. How should they divide the stake?

There are many different approaches. Perhaps they should each take back their own stake. As Áine is leading 2:1, perhaps she should get two thirds and Bill one third.

Alternatively, what about dividing in proportion of the likelihoods of either of them winning?

Fermat had the idea that the likelihood of winning could be determined by assessing all the future possibilities. In other words, if we consider all the different ways the final two tosses could have gone, this will give the probability for each player winning.

How would Fermat suggest dividing the stake?

Bonus questions:

With a fair coin you have equal chance of getting heads or tails on any toss. What are the chances of getting at least two tails from three tosses of a coin?

What are the chances of getting two heads in a row from three tosses of a coin?

Ciara tosses two coins and covers them. She uncovers one and you see it is heads. What are the chances that the other one is heads? 

Tuesday’s puzzle: The answer

• There are six equal possible results of which only one is a six. Therefore, the probability is one in six or 1/6.

• For each of the six possible results on the first die there are six possible results on the second die. That is 36 possibilities. However, there is only one outcome that gives a double six. Therefore, the probability is 1/36.

• There are 36 possibilities and six of these will add up to seven (1+6; 2+5, 3+4, 4+3, 5+2, 6+1). Therefore, the probability is 6 /36 or 1/6.

• There are 36 possibilities overall. For you to win there are four possibilities on the first die (1-4) and for each of these four possibilities on the second die (also 1-4) so that is 16 winning possibilities. That leaves 20 winning outcomes for the gambler. The lesson is that good gamblers, bookmakers and casinos understand probability a lot better than the rest of us.

Come back tomorrow at 7.30pm for the answers to today’s puzzle. 

The puzzles this week have been compiled by Eoin Gill and Colm Mulcahy of Maths Week Ireland / South-East Technological University (SETU).