# On Triumvirate

When last they met, the Baron invited Sir R----- to join him in a wager involving a sequence of coin tosses. At a cost of seven coins Sir R----- would receive one coin for every toss of the coin until a run of three heads or of three tails brought the game to its conclusion.

To evaluate its worth to Sir R----- we begin with his expected winnings after a single toss of the coin.
\begin{align*} \mathrm{E}_T[n] &= \tfrac12 \times \left(1+\mathrm{E}_{TT}[n]\right) + \tfrac12 \times \left(1+\mathrm{E}_{H}[n]\right)\\ \mathrm{E}_H[n] &= \tfrac12 \times \left(1+\mathrm{E}_{T}[n]\right) + \tfrac12 \times \left(1+\mathrm{E}_{HH}[n]\right) \end{align*}
Note that we're exploiting the fact that if the next coin shows a different face to the most recent one then the current coins cannot be part of a run of three and Sir R-----'s further expected takings are identical those that he had at the outset. We can similarly calculate the expected prize after two tosses of the coin with
\begin{align*} \mathrm{E}_{TT}[n] &= \tfrac12 \times 1 + \tfrac12 \times \left(1+\mathrm{E}_H[n]\right)\\ \mathrm{E}_{HH}[n] &= \tfrac12 \times \left(1+\mathrm{E}_T[n]\right) + \tfrac12 \times 1 \end{align*}
Finally, at the start of the game we have
$\mathrm{E}[n] = \tfrac12 \times \left(1+\mathrm{E}_T[n]\right) + \tfrac12 \times \left(1+\mathrm{E}_H[n]\right)\\$
Noting that the probabilities of tossing heads and tails are equal, we can simplify these formulae by exploiting the identities
\begin{align*} \mathrm{E}_H[n] &= \mathrm{E}_T[n]\\ \mathrm{E}_{HH}[n] &= \mathrm{E}_{TT}[n] \end{align*}
yielding by recursion
\begin{align*} \mathrm{E}_{TT}[n] &= \tfrac12 \times 1 + \tfrac12 \times \left(1+\mathrm{E}_T[n]\right)\\ &= 1 + \tfrac12 \times \mathrm{E}_T[n]\\ \mathrm{E}_T[n] &= \tfrac12 \times \left(1+\mathrm{E}_{TT}[n]\right) + \tfrac12 \times \left(1+\mathrm{E}_{T}[n]\right)\\ &= 1 + \tfrac12 \times \left(\mathrm{E}_{TT}[n] + \mathrm{E}_{T}[n]\right)\\ &= 1 + \tfrac12 \times \left(\left(1 + \tfrac12 \times \mathrm{E}_T[n]\right) + \mathrm{E}_T[n]\right)\\ &= \tfrac32 + \tfrac34 \times \mathrm{E}_T[n] \end{align*}
and consequently
\begin{align*} \tfrac14 \times \mathrm{E}_T[n] &= \tfrac32\\ \mathrm{E}_T[n] &= 6 \end{align*}
The expected number of tosses, and Sir R-----'s bounty, is therefore
$\mathrm{E}[n] = 1 + \mathrm{E}_T[n] = 7$
meaning that the game is perfectly fair and I should have suggested to Sir R----- that he could take it or leave it!
$$\Box$$

### Gallimaufry  AKCalc
ECMA  Endarkenment
Turning Sixteen

• Subscribe
• Contact
• Downloads
• GitHub

This site requires HTML5, CSS 2.1 and JavaScript 5 and has been tested with Chrome 26+ Firefox 20+ Internet Explorer 9+