Dinner time is the perfect time for conversations and a great opportunity to engage with your children. The idea of Dinner Time Maths is to add a small spark of maths to the conversation. Discuss a mathematical idea as a family or solve a problem together.

The Game Show Dilemma

A game show host offers you the choice of three doors and the chance to win an amazing prize which is hidden behind one of the three doors. Behind the other two doors is a booby prize. Once you’ve selected your door the game show host opens one of the other remaining two doors. The host always knows where the prize is hidden and at this stage always opens a door to reveal one of the booby prizes. Now for the dilemma. Before opening the door you’ve selected you are given a choice. Would you like to stick with the door you’ve selected or switch to the other door? Stick or switch?

Is it better to stick? Is it better to switch? Or does it actually make no difference?

Most people would instinctively say it makes no difference. There are two doors and one has the prize so it’s 50/50. While this seems like a logical explanation, it is in fact incorrect and the mathematics behind the truth is very interesting indeed! But before we get to that here is how you can set up the game show at the dinner table.

The Dinner Table

On the dinner table lay out three napkins and under one of the napkins hide a coin (and remember which one). When the kids come to the table tell them there is a prize behind one of the napkins. Ask them to collectively select one napkin. Then out of the other two unselected napkins reveal nothing behind one of them. This is where the dilemma comes into play. There are two napkins left. The one they selected originally and the one which you did not reveal. Give them a final choice, they may stick with their original napkin, or switch to the other. After their decision is made you may disclose where the coin was hidden. No doubt they will want to play again, so without anyone peeking, reset the game and play it over.

Mathematical Discussion

Mathematically there is actually double the chance of winning the grand prize if one chooses to ‘switch’ at the end before the reveal. There is less chance of the prize being behind their originally chosen napkin (or door) than the other remaining napkin.

Please explain?

This short youtube video offers the best and easiest to understand explanation I have come across in a long time. It will help you to explain this probability problem to your children and you never know, it just may help you win the big prize if you’re ever faced with this dilemma on a game show!


If you are after a more in depth explanation then this will definitely satisfy.

Deal or No Deal

Before dinner, do you often have Deal or No Deal on TV? The ultimate goal is of course to win the biggest cash prize. But what happens in the scenario when the largest prize is still available? Have your children decide whether they should keep their case or switch.

In Deal or No Deal, when the situation arises where they get down to the last two cases and there is still the largest cash prize left on the board, if given the choice, the contestant should always switch cases. There is a 1/26 chance that they have the big money and a 25/26 chance that the other case has it! [The assumption we have applied here is that there are two choices in this game, the biggest cash prize or nothing. We know that other prize amounts are still worth something, but we’re assuming contestants are only interested in winning the top prize and thus we’ve simplified the problem into one that is binary]

This is a fun problem to play and to discuss and such brilliant maths involved. This is sure to liven up the dinner table at your place! 🙂

Our inspiration for Dinner Time Maths comes from Laura Bilodeau Overdeck who is the brainchild of Bedtime Math, a brilliant blog dedicated to providing you with ideas to incorporate a maths problem into every day.

Photo credit: More Good Foundation / Foter / Creative Commons Attribution-NonCommercial 2.0 Generic (CC BY-NC 2.0)