Say I have a highly biased coin that lands heads with $p_h=0.01$ and tails with $p_t=0.99$, and I flip it $98$ times.

The probability of zero heads is ${p_t}^{98} \approx 0.373$.

The probability of one head is $98 \times {p_t}^{97} \times p_h \approx 0.370$ as any of the 98 coin flips could have given H.

The probability decreases for larger numbers of heads.

The expected number of heads is $\Sigma xp_{xH} = 0.98$ where $p_{xh}$ is the probability of getting $x$ heads (this is also of course $p_h \times 98$).

But the expected number of heads appears to be different to the most likely number of heads. How do we account for this?

Is the answer that if I had to bet on how many heads would come up in a single 98-flip experiment, I should place my bet on zero, but if I had to bet on the long run average of many 98-flip experiments I should bet on 0.98?

• In the last sentence, shouldn't it be 0.01 instead of 0.98? Or more precisely, 0.01 times the number of flips? – Richard Hardy Mar 27 '15 at 15:55

The mode corresponding to a very large number $N$ of flips will be approximately $N \cdot 0.01$ (rounding will play a very small role for very large $N$), so you would bet on that.
• @ssdecontrol Understanding one of "$\hat y$" and "$y$" to be the bet and the other to be the outcome, and once you adjust that to account for the cost to play and the fact the payoff is not $1$, you would be in agreement with what I wrote :-). Richard, thank you very much for the clarifying edits (+1). – whuber Mar 27 '15 at 18:10