A penny is tossed 3 times and a nickel is tossed 2 times. What is the probability that more heads are tossed using the penny than using the nickle?
Well, we need to determine how many outcomes of flipping would result in a penny "win" vs a penny "tie or loss".
For a penny, there's 2^3 possibilities (number of options to the power of number of flips), and for nickels it'd be 2^2. The chance of getting each result is then "x in 4" or "x in 8".
| Outcome | Penny | Nickel |
| 0 heads | 1 in 8 | 1 in 4 |
| 1 head | 3 in 8 | 2 in 4 |
| 2 heads | 3 in 8 | 1 in 4 |
| 3 heads | 1 in 8 | 0 in 4 |
So add the results of the penny runs that beat the each nickel run together:
If Nickels get zero, Pennies win 7 in 8 runs
If Nickels get one, Pennies win 4 in 8 runs
if Nickels get two, pennies win 1 in 8 runs
Multiply each of these by its weight (i.e. how likely the scenario is to occur):
7 in 8 * 1 in 4 ||\(7/8 * 1/4 = 7/32\)
4 in 8 * 2 in 4 || \(4/8 * 2/4 = 1/4 or 8/32\)
1 in 8 * 1 in 4 || \(1/8 * 1/4 = 1/32\)
So I see 16 in 32 or simply 1 in 2.
Half the time, Penny wins :)
