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Pick a cup, any cup

Sun Jan 3 12:55:23 GMT 2010

Over the festive period my Dad posed an interesting probability conundrum based on a Shell Game. The conundrum is as follows.

In our shell game the gamemaster places a coin beneath one of three upturned cups on a table. He shuffles the cups such that the player doesn't know which cup conceals the coin. This done, the player now picks a cup and moves it aside. Of the remaining two cups, the gamemaster removes one empty cup and offers the player a choice: either to stick with his original choice of cup, or to switch to the remaining cup. Which option is more likely to win the player the coin ?

Well, this puzzle generated a lot of debate amongst my family members, with a split between two conflicting points of view. The first was that the player's final choice makes no difference, since the coin is equally likely to be under either of the two cups. The second was that the player should always switch cups since the cup he originally chose had a probability of 1 in 3: therefore he is twice as likely to win by switching cups.

What do you think ?

As unintuitive as it may seem, the second point of view is correct. The player should always switch cups. The question is how to explain this to those who hold the first view -- after all, it makes logical sense: there are two cups and one coin, so surely the choice is 50:50 ?

The best way to think about it that I've come up with is that the gamemaster is devious. Once you assume that, it is easier to see how he tricks the player into thinking the choice is 50:50 when in actual fact it is not. Let's just step through the game and look at the probabilities to see how it breaks down.

1. At the start of the game the coin is under a randomly chosen cup, hence the first cup the player chooses has a 1 in 3 chance of containing the coin, and correspondingly there is a 2 in 3 chance that the coin is under one of the remaining two cups.
2. Now, the gamemaster removes an empty cup from the two remaining cups. Just think for a moment what extra information is presented here. Is there anything the player knows that he didn't know before ? Of course not: he always knew that the remaining two cups must necessarily have comprised at least one empty cup. So has anything changed with regards to the probable location of the coin ? Again, of course not: there is a 1 in 3 chance that the player's original cup contains the coin, and a 2 in 3 chance that one of the remaining two cups contain the coin -- whether or not the empty cup remains on the table is immaterial.
3. At this point, the player makes his final choice of cups. The naive player sees two cups with a coin under one, and assumes the final choice makes no difference and he may just as well stick with his original choice. The cunning player sees that by switching his choice of cups he effectively selects a cup with a 2 in 3 chance of containing the coin: the gamemaster's deviousness doesn't trick him !

In order to explore the problem further, and to satisfy myself as to the correctness of this explanation, I wrote a small program which simulates playing the game many times -- and it shows that the player is indeed more likely to win by switching than by sticking: the experimental results break down as 1 in 3 win rate for the player who always sticks versus 2 in 3 for the player who always switches. You can download it here if you'd like to play with it yourself -- it compiles cleanly on an Ubuntu 9.04 system using gcc.

Have fun !