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For each $k$, $\mathcal{C}_k$ is biased so that, when tossed, it has probability $\tfrac{1}{(2k+1)}$ of falling heads. If the $n$ coins are tossed, what is the probability that the number of heads is odd? Express the answer as a rational function $n$.

My solution: Let $p_k = 1/(2k+1)$, then the probability of exactly:

1 head is $p_1 + p_2 + ... p_n$

3 heads is $\sum p_i p_j p_k$

5 heads is $\sum p_i p_j p_k p_m p_n$

etc... and the total probability is obtained by summing the above.

Now the difference $(1+p_1)(1+p_2)...(1+p_n) - (1-p_1)(1-p_2)...(1-p_n)$ gives exactly $2$ times the desired probability but when I calculate this difference, it does not give me the answer of $n/(2n+1)$, where did I go wrong?

Ray
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    I'm not sure your "probability of exactly $m$ heads" expressions are correct. For example, $p_1$ is the probability that the first coin is heads, but some of the other coins might be heads too. – angryavian Oct 01 '22 at 19:42
  • Note that when $n=7$, $p_1 + p_2 + p_3 + p_4 + p_5 + p_6 + p_7 = \frac{46027}{45045} \approx 1.0218$, so it cannot be the probability of anything. – Misha Lavrov Oct 01 '22 at 19:51
  • @angryavian yes... I see now. it should be (1-p_1 + p_1)(1-p_2 + p_2 )... - (1-p_1 -p_1)(1-p_2-p_2) which gives the correct answer – Ray Oct 01 '22 at 19:54

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