Nathan tosses a coin 20 times
Nathan tossed a coin 20 times and tallied the results:
heads 9
tails 11
How do these experimental results differ from the theoretical probability? Here is the answer given in Stephen Hake and John Saxon (1997). Math 76: An Incremental Development, Third Edition, Norman, OK: Saxon Publishing:
"Tails occurred slightly more often than heads. Theoretically, both would have occurred ten times in 20 tosses."
Here are the theoretical probabilities that Nathan would have gotten k = 0 through 20 heads in 20 tosses if the coin were fair. I computed these with a TI-34 calculator. The formula is
(20 choose k)/2^20.
| no. of heads | no. of tails | probability |
| 0 | 20 | .000000954 |
| 1 | 19 | .000019073 |
| 2 | 18 | .000181198 |
| 3 | 17 | .001087189 |
| 4 | 16 | .004620552 |
| 5 | 15 | .014785767 |
| 6 | 14 | .03696... |
| 7 | 13 | .0739... |
| 8 | 12 | .12013... |
| 9 | 11 | .160179... |
| 10 | 10 | .17619... |
| 11 | 9 | .160179... |
| 12 | 8 | .12013... |
| 13 | 7 | .0739 |
| 14 | 6 | .03696... |
| 15 | 5 | .01478... |
| 16 | 4 | .00462... |
| 17 | 3 | .001087... |
| 18 | 2 | .0001811... |
| 19 | 1 | .0000190... |
| 20 | 0 | .000000954 |
The sum of these probabilities is one. Nathan can expect 10 heads and 10 tails about 17.6% of the time. But he can expect 9 heads and 11 tails, or 11 heads and 9 tails, more often, about 32% of the time.