**Stars
and circles**

In a hat there were three cards. One had a star on both sides. One had a circle on both sides. And the third had a star on one side and a circle on the other. You pulled out one of the cards at random and placed it on a table. The face that's up has a star on it. What's the probability that the face that's down also has a star? (Adapted from problem in Interactive Mathematics Program (IMP), year 1)

**Logistics**

Arguments about this can become very passionate.

You might actually prepare a set of three cards for each group.

**Approaches**

The probability is 1/2, since you know that you got one of two cards and they are equally likely, but only one of the two has a star on the reverse side.

The probability is 2/3, since there are three sides that are equally likely to be facing up, and in two of the three cases there's a star on the reverse side.

A tree diagram like the one to the right shows that, of the six outcomes, our attention is limited to three, two of which come from the card that has a star on both sides.

A calculator simulation depends on whether the person doing it assumes that the two cards or the four sides are equally likely.

Apply Bayes's Rule to get 2/3.

Kyle's Iron Law of Probability: Whenever your intuition tells you the answer is 1/2, the answer is 2/3.

The probability is 0 or 1, because the card is chosen and lying there. You just don't know yet whether the probability is actually 0 or actually 1.

**Your
experience**

Have you used this problem with a class and seen approaches other than(or more specific than) those mentioned above? Or do you have other comments or criticisms or stories? If so, please tell us!

*Last updated 30 November, 2004*