# Conditional expected value of coin flips

Suppose you flip four coins and you're guaranteed that at least 2 are heads. What is the expected value of the number of heads? I have the following calculation.

\begin{align} & \operatorname E\left[ X \mid X \geq 2\right] \\[8pt] = {} & \sum_{n = 2}^4 nP [X=n\mid X \geq 2] \\[8pt] = {} & 2 \cdot \frac{\binom{4}{2}}{11} + 3 \cdot \frac{4}{11} + 4 \cdot \frac{1}{11} \\[8pt] = {} & 2.5454. \end{align}

Is this calculation correct? Specifically is the use of the conditional probability distribution in the first step correct?

• Looks good to me! – Matthew Daly Dec 8 '19 at 21:06

\begin{align} \mathbb E[X\mid X\geqslant 2] &= \frac{\mathbb E[X\mathsf 1_{\{X\geqslant 2\}}]}{\mathbb P(X\geqslant 2)}\\ &= \frac{\sum_{k=2}^4 k\binom 4k 2^{-4}}{\sum_{k=2}^4\binom 4k 2^{-4}}\\ &= \frac{\sum_{k=2}^4 k\binom 4k}{\sum_{k=2}^4\binom 4k}\\ &= \frac{28}{11}. \end{align}