Parity of Heads

You flip \(10\) independent coins: \(4\) coins land heads with probability \(\tfrac{9}{10}\) and \(6\) coins land heads with probability \(\tfrac{1}{10}\). What is the probability that the total number of heads is odd?

Answer

Let \(N\) be the total number of heads and define

\[ (-1)^N = \begin{cases} 1, & N \text{ even},\\ -1, & N \text{ odd}. \end{cases} \]

We have three insights:

  1. \[ \mathbb{E}[(-1)^N]=\mathbb{P}(N\text{ even})-\mathbb{P}(N\text{ odd}). \]

  2. Write \(N=\sum_{i=1}^{10} I_i\), where \(I_i=\mathbf{1}\{\text{coin }i\text{ is heads}\}\). Then

    \[ (-1)^N=\prod_{i=1}^{10}(-1)^{I_i}, \]

    so by independence,

    \[ \mathbb{E}[(-1)^N]=\prod_{i=1}^{10}\mathbb{E}[(-1)^{I_i}]. \]

    For coin \(i\) with head probability \(p_i\),

    \[ \mathbb{E}[(-1)^{I_i}]=(1-p_i)\cdot 1+p_i\cdot(-1)=1-2p_i, \]

    hence

    \[ \mathbb{E}[(-1)^N]=\prod_{i=1}^{10}(1-2p_i). \]
  3. \[ \mathbb{P}(N\text{ even})+\mathbb{P}(N\text{ odd})=1. \]

Combining all three we have,

\[ \mathbb{P}(N\text{ odd})=\frac{1-\mathbb{E}[(-1)^N]}{2} =\frac{1-\prod_{i=1}^{10}(1-2p_i)}{2}. \]

Here \(p_i=\tfrac{9}{10}\) for \(4\) coins and \(p_i=\tfrac{1}{10}\) for \(6\) coins, so

\[ \prod_{i=1}^{10}(1-2p_i) =\left(1-2\cdot \frac{9}{10}\right)^4\left(1-2\cdot \frac{1}{10}\right)^6 =\left(-\frac{4}{5}\right)^4\left(\frac{4}{5}\right)^6 =\left(\frac{4}{5}\right)^{10}. \]

Therefore,

\[ \boxed{\mathbb{P}(N\text{ odd})=\frac{1-\left(\frac{4}{5}\right)^{10}}{2}\approx 0.446.} \]
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