How to win at “Deal or No Deal”

I have never actually watched “Deal or No Deal“, but I am assuming it is the same as the “Let’s Make A Deal” TV gameshow described in Ian McEwan‘s spy romp Sweet Tooth– which I have just read in the last 18 hours of plane flights, mid-air turn arounds (bad weather at the destination rather than anything more serious) and airport lounges.

The novel is voiced by a Cambridge maths graduate (and low grade MI5 spook) who at one point explains to her lover the “paradox of choice” based on “Let’s Make A Deal”:

You have three boxes, one contains a prize, the two others are empty.

You pick one and then the dealer reveals one other to be empty. Do you stick with your original choice or pick the other box?

The naive answer is to stick with your original choice. After all the odds of the prize being in the box you first picked, or any other box –  are one in three – now increased to one in two.

But that is wrong.

In fact you double your chances of winning by picking the other box.

Think of it this way. If you picked an empty box as your first choice – and the odds are that you did as there are two of them and only one with a prize – then the dealer has no choice but to reveal the other empty box and so the third box must be the one with the prize.

So the odds of you picking the prize in this way are $\frac{2}{3} \times 1 = \frac{2}{3}$. The odds that you picked the prize the first time round are just $\frac{1}{3}$.

Of course, this is not a guaranteed way to win. But you would win two times in every three games if you followed this strategy.

Update: Having now read the Wikipedia entry on “Deal or No Deal” I can see it is similar to “Let’s Make a Deal”, but not the same and, in fact, rather more complex. I am not sure the above analysis of the so-called “Monty Hall Problem” (named after the presenter of “Let’s Make A Deal”) would be much of a guide to a contestant!