##### Question 14.1.1

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