Bingo winning strategy

I will write about a problem by my friend Edgardo Roldán Pensado. He told me it long time ago. If you are a mathematician the solution may be trivial, however, I personally admire the way he faced the situation and came out with this nice problem.

I won’t bore you with a longer introduction, let start the problem.

Suppose you bought two days ago (in amazon.com for example) a Royal Bingo Supplies Wooden Bingo Game (like the one in the picture)

Bingo

  • Wooden Bingo Game Set with instructions.
  • Perfect for old-fashioned fun with a nostalgic twist.
  • Includes with 18 Bingo cards, 150 Bingo chips, a Bingo board, brass cage and 75 wooden balls.
  • Great for parties, barbeques or family game nights.
  • Recommended for ages 3 and up.

Today you received your game and you invite seven other friends to play together. Each player takes a bingo card and you start playing. Five hours later, when everybody is already tired of playing, you decide to count the times each player has won. Everybody is very  surprised how lucky a single player was, who won many more times than anybody else.

Is there a Bingo winning strategy? Or do you have a very lucky friend?

Well…Everybody has a very lucky friend, so no discussion about this. However, there is also a Bingo winning strategy, so, there is a chance the winner was a Bingo´s tactician (with a bit of luck).

How does the Bingo´s tactician play?

This guy chose his bingo card at the end, after analysing the bingo cards of the other player. Each card was (hopefully, I do not know if it is true) created with a uniform random distribution (for each square you choose a number from 1 to 75 with probability 1/75.). He considered a metric on the set of bingo cards, for example, given two bingo cards a=(x_1,\ldots,x_{24}) and b=(y_1,\ldots,y_{24}), d(a,b)=\#\{i:x_i\neq y_i\}, with the convention that \# \emptyset :=0. He was a bit lucky enough to be able to find a bingo card that maximises the d distance with respect to the bingo cards of the other player.

Why is this a winning strategy?

Suppose that the 7 bingo cards chosen by the opponents of the winner were very close with respect to the distance d, moreover, suppose that all the 7 bingo cards were exactly the same. On the other hand, the “lucky one” chose a different bingo card. So there is \frac{1}{2} probability one of the other seven players (and then all) wins, and \frac{1}{2} probability the “lucky one” wins. Now, suppose, the 7 bingo cards chosen by the opponents are very close (with respect to d), but all different. Then the “lucky one” player wins with probability close to \frac{1}{2}, whilst  and the other players with probability close to \frac{1}{14}.

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