Earlier today I set you the following problem, about a game show where objects are hidden behind three doors. Behind one door is a car. Behind a second door are the car keys. Behind the third door is a goat. The car, the keys and the goat were placed there randomly, meaning that each item has a 1/3 chance of being behind any particular door.
Twins Timmy and Tammy, the contestants, are backstage on the game show. They are told the rules:
1) Timmy will be taken on stage first. He will be asked to open two of the doors, and then shut them. Timmy will then be led off stage to a holding room on his own.
2) Tammy will then be taken on stage. She will be asked to open two of the doors.
If Timmy opens the door with the car, and Tammy opens the door with the keys, then they both get to keep the car. In all other outcomes, they leave with nothing.
The twins are given 10 minutes to think up a door-opening strategy before Timmy goes on stage. What strategy gives them the best chance of winning the car?
Just to be clear: The twins do not know what is behind any of the doors before they ask for a door to be opened. When one door is opened, all they can see is what is behind that door. The car, car keys and goat stay behind the same door for the duration of the programme. When Timmy is on stage opening his two doors, Tammy cannot see or hear what is going on. Thus when Tammy is choosing her two doors, she has no idea what was behind the two doors that Timmy opened. Also (as was pointed out by readers of the earlier article), Timmy does not need to open both his doors at the same time. He can open one, and then based on what he sees, decide which one to open second. As can Tammy.
If the twins had no strategy, that is, if both of them choose two doors at random, the probability Timmy gets the car is 2/3, and the probability Tammy gets the keys is also 2/3. The probability they get to keep the car is thus 2/3 x 2/3 = 4/9 = 44 per cent.
Yet, rather incredibly, there is a strategy that gives them well over 50 per cent chance of keeping the car. What is it?
Solution
Let’s call the doors 1, 2 and 3. The optimal strategy is for Timmy to open 1 first. If it reveals either the car or the keys, he should open door 2 next. But if door 1 reveals the goat, he should open door 3.
Tammy should then open door 2. If it reveals either the car or the keys, she should open door 1 next. But if door 2 reveals the goat, he should open door 3.
Here are the six possible arrangements of car, keys and goat behind doors 1, 2 and 3:
CKG
CGK
KGC
KCG
GCK
GKC
I have marked in bold the ones in which Timmy gets the car, and Tammy the keys, which happens in four out of the six equally likely cases. This the chances of them winning the car is 4/6 = 66 per cent.
If you discovered this solution, please discuss your thought processes below.
Today’s puzzle is taken from Surprises in Probability- Seventeen Short Stories, by Dutch mathematician Henk Tijms
I set a puzzle here every two weeks on a Monday. I’m always on the look-out for great puzzles. If you would like to suggest one, email me.
My latest puzzle book is So You Think You’ve Got Problems.
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