John makes a bet with Donald that he’s able to flip heads on a coin before Donald throws 2 in a die. What’s the probability of Donald winning, considering John begins the game and they play alternately?
Let the probability of tossing an head be \(p=~^1/_2\) and tail \(q=~^1/_2\), and the probability of tossing any die face other than a 2 be \(r=1-~^1/_6=~^5/_6\). If we consider the outcome of John winning, then:
| outcome | probability |
|---|---|
| \(H\) | \(p\) |
| \(T⚅H\) | \(qrp\) |
| \(T⚅T⚅H\) | \(qrqrp\) |
| … | \((qr)^np\) |
Therefore, the converse probability is given by:
\[1 - \sum_{n~=~0}^{\infty}{^1/_2~\left(^5/_6\cdot~^1/_2\right)^n}\]Since this is a geometric series with \(\lvert~r~\rvert =~^5/_6\cdot~^1/_2 < 1\), then \(\sum_{n~=~0}^{\infty}a_i=\frac{a}{1-r}\), so:
\[\begin{align} 1 - \sum_{n~=~0}^{\infty}{^1/_2~\left(^5/_6\cdot~^1/_2\right)^n}&=1-\frac{^1/_ 2}{1-~^5/_{12}}\\ &=~^1/_7 \end{align}\]