# The Chris Tarrant Problem

This is a puzzle I presented at the MathsJam conference. It’s a problem that gave me a headache for a week or so, and I thought others might enjoy it, too. I do know the answer, but I’m not going to give it away — you can tweet me @icecolbeveridge if you want to discuss your theories! (As Colin Wright says: don’t tell people the answer).

You’ve heard of the Monty Hall Problem, right?

Well, just in case: the traditional presentation is that, at the end of an episode of the gameshow Let’s Make A Deal, the host – Monty Hall – would offer you a choice of three doors. Randomly arranged behind the doors are two goats, which you don’t really want; behind the third is a shiny car (which you do want). One prize per door, naturally.

You pick a door, and – rather than show you what’s behind it – Monty opens one of the other two doors to reveal a goat, before offering you a simple choice: stick with your original choice, or switch to the other closed door?

It’s a famous puzzle: when Marilyn vos Savant wrote about it, she reportedly received hate mail from irate mathematicians disagreeing with her (correct) solution; it’s also said that Erdős didn’t believe the answer until someone showed him a simulation.

The answer is, the first time you hear it, surprising: to maximise your chances of winning the car, you should switch doors. Switching is the correct move if you were originally wrong, and you’re more likely to have been wrong than right, so you should switch.

### I don’t want to give you that!

Imagine you’re playing Who Wants To Be A Millionaire and a question comes up where – as far as you’re concerned – all four answers are equally likely. You chat with Chris about this, suggest a guess… and Chris then reminds you you still have a lifeline. You can go 50-50.

“Computer,” says Chris, “remove two random wrong answers!”

The computer makes a dramatic noise and removes two of the other answers, leaving you with a choice: do you stick with your original answer, or do you switch to the other?

You see: after you’ve made a choice at random, Monty Hall leaves you in a situation where you have one right and one wrong answer to choose from; Chris Tarrant leaves you in the same situation – so it would seem that switching to the other answer should give you a 75% chance of winning.

But, on the other hand, you didn’t go 25-75, you went 50-50 – and it seems absurd that making a guess before using your lifeline should affect the probabilities at all.

So, the question: which interpretation is correct? And how come?

• #### Colin Beveridge

Colin Beveridge is the author of Basic Maths For Dummies. Based in Weymouth, Dorset, he divides his time between persuading people that C4 is fun and trying to get his head around basic game theory.

### 4 Responses to “The Chris Tarrant Problem”

1. Daniel Jones

My guess is that is makes no difference whether or not you switch in this case. If (as the program assures us) the 2 incorrect answers are removed at random then there is as much chance of your initial answer being removed as not. The fact that it has remained suggests that instead of a 1/4 chance of your hunch being right, it is now a 1/2 chance. Had it been eliminated, you now have to choose between 2 options. The difference with the Monty Hall problem is that the original choice is guaranteed to remain after one of the doors is revealed.

I guess other people will be able to present this argument (or others) more rigorously.

2. Chris Taylor

Here’s another question. Suppose you are on the last (£1m) question. You have 2 lifelines left: 50 – 50 and “ask the audience”. The audience is on your side and each individual will give the correct answer if they know it. Otherwise they will guess at random. (Not, as I’m told, the audience did in the Russian version of the show, where they often tried to mislead the contestant). The question is hard, so you have no idea of the correct answer, and neither do most members of the audience. Should you use your 2 lifelines, and if so, which should you use first?

3. Jesse McKeown

“at random” is an odd phrase for a defined subset… I suppose it means that, even if you think the answer is “C”, when it’s not, the computer will eliminate “C” with probability 2/3?

So pick one of the four at random. 1/4 of the time, you’re right, and “50-50” leaves your guess on the board; 3/4 of the time, you’re wrong, and 2/3 of those times (for 1/2 the time, overall) your guess is taken off the board; the other 1/3 of those times (for 1/4 of the time overall) the guess stays up. So, half the time your guess is not an option, and half the time (evenly split between right and wrong) your guess remains an option.

Monty Hall is weird because there is this communication: your first choice of door constrains which doors Monty can open.