Suppose you toss two coins in the air at the same time. To find the probability that they both land heads up, we must know how many different possible outcomes there are. Many people would say that there are three possibilities: two heads, two tails, or one of each. This would mean that both coins land heads up a third of the time. Do you agree with this?

Now imagine writing a $1$ on one of the coins, and a $2$ on the other with a marker. Now there are four possible outcomes: both heads (hh), both tails (tt), coin $1$ heads and coin $2$ tails (ht), or coin $1$ tails and coin $2$ heads (th). This would mean that both coins land heads up a quarter of the time. Do you agree with this?

Are these two scenarios really different? Is it possible for the probability of two heads to be $1/3$ in the first scenario, and $1/4$ in the second? How do we know if the possible outcomes are equally likely? Show answer

In the first scenario, we have no reason to believe that the three outcomes stated are all equally likely. Why should two heads be as common as one of each? We must be careful not to assume that just because we can list three outcomes that they’re three equally likely outcomes.

In the second scenario, we can reason that hh must be just as likely as ht, because once coin $1$ has landed heads up, coin $2$ is just as likely to land heads up as tails up. Similarly, ht must be just as likely as tt because given that coin $2$ lands tails up, coin $1$ is just as likely to land heads up as tails up. In this way we can show that all four outcomes are equally likely.

Scenario one is actually no different from scenario two. It doesn’t matter if we’ve actually marked the coins as different; they are different coins regardless. So getting ‘one of each’ must be twice as likely as getting ‘both heads’, because it’s actually two outcomes: ht and th (even if there’s no difference between the two coins that we can see and we can’t actually tell which of these two outcomes has happened).