# Sunday Times Teaser 2846 – Bingo!

*by Mike Fletcher*

#### Published April 9 2017 (link)

The infant teacher played a bingo game with his class. He had two identical dice, the numbers on the six faces of each being 1, 2, 2, 3, 3 and 3. He told the class that he would throw the pair of dice, add up the two numbers showing, and call that number out in a game of bingo. He then asked each member of the class to make their own bingo card consisting of five numbers of their own choice. He explained that they could repeat a number on their card if they wished (and then delete just one occurrence of the number whenever it was called). Most of the class chose the five possible different totals as their bingo numbers, but one very clever girl chose the best possible selection.

What were her five numbers?

My initial thought was that the card would have the most probable number on it five times. But, remembering Mike Fletcher’s earlier dice puzzles, I suspected that the puzzle was more subtle and this was confirmed by Peter Hayes (in a private communication) soon after the teaser was published.

We can calculate the probabilities of all the different sequences of five dice throws as follows:

It is also possible to analytically compute the average number of Bingo calls for a win when each card is played in isolation. I thought that this would be really hard but John Crabtree explained how to do this for one set of numbers (thanks John!) and I was then able to do this for all 126 cards as follows:

As an alternative, the following program runs a large number of random Bingo trials for each possible card to find the one that wins with the fewest calls on average:

Congratulations to Mike Fletcher on producing a Teaser for which the ‘obvious’ solution turns out to be wrong! For many solvers, myself included, it was easy to conclude that the best card would be one that would have the most probable dice result repeated five times. It hence comes as some surprise to find that this is far from the best. The reason is most easily explained by considering a simplified game with Bingo cards having only two numbers.

A card with two fives on it will have a probability of being called of (1/3) x (1/3), which is 1/9. A card with a five and a four will win when a five is called (1/3) and then a four (5/18), giving a probability of 5/54, which is lower than 1/9 (5/45). But it will also win when a four is called followed by a five, which, with the same probability, gives an overall probability of 5/27, much larger than 1/9 (3/27).

The point here is that if you put all fives on your card, there is only one sequence of numbers that gets you to Bingo; but if you put a mixture of numbers on your card there are then several sequences of calls, all of which will get you there. And although each specific sequence is less likely to win than the all fives sequence, the fact that there are a number of them more than compensates for this.

I had to code this one to verify the counterintuitive result:

Its nice to find that Arthur’s code and my own produce consistent probability and frequency results. For those who prefer it, here is the equivalent of Arthur’s code in Python 3:

It does seem very counter-intuitive, as surely each call is statistically independent. I guess the point is that in early calls there is benefit to having more than one way of ticking a number on my card. So if my card has {2,3,4,5,6} in the more extreme case, I’ll be guaranteed to score on the first call and will reduce my card to 4 spaces.

Yes, we can see how this works by considering a card with just two numbers. If we put two 5’s on it there is only one sequence of calls that will win – (5, 5) . But if we put a 3 and a 4 on the card, we then win with two sequences of calls – (3, 4) and (4,3) – and while each is less likely than (5, 5) the fact that there are two winning sequences more than compensates for this.