On Pennies From Heaven

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Recall that the Baron and Sir R-----'s most recent wager first had the Baron place three coins upon the table, choosing either heads or tails for each in turn, after which Sir R----- would follow suit. They then set to tossing coins until a run of three matched the Baron's or Sir R-----'s coins from left to right, with the Baron having three coins from Sir R----- if his made a match and Sir R----- having two from the Baron if his did.

When the Baron described the manner of play to me I immediately pointed out to him that it was Penney-Ante[1], which I recognised because one of my fellow students had recently employed it to enjoy a night at the tavern entirely at the expense of the rest of us! He was able to do so because it's an example of an intransitive wager in which the second player can always contrive to make a choice that will best the first player's.
To understand why, we need simply enumerate the Baron's possible choices and figure the probabilities of his winning should Sir R----- have chosen his coins wisely. I explained as much to the Baron but I fear that he may not have fully comprehended my meaning.

Now, it is quite evident that Sir R----- should not have chosen the same coins as the Baron since that would inevitably have resulted in their both matching the tossed coins at the same time and a loss of one of his coins.
With his choosing different coins we need only consider the outcomes of games in which the Baron's first coin shows heads since, by symmetry, we may exchange tails for heads throughout our reckoning to figure the outcomes were it to show tails.
Representing heads with an \(H\) and tails with a \(T\) let us first imagine that the Baron chose \(HHH\) and Sir R----- countered with \(THH\). If three heads were tossed in a row, with a chance of just one in eight, then the Baron would clearly have had his prize. If but one of those coins showed tails, however, then Sir R----- must surely have prevailed since there could thereafter never be a trio of heads before there were a tail followed by a pair of them.
Let us imagine instead that the Baron had chosen \(HHT\) and that Sir R----- had again responded with \(THH\). Clearly the Baron should have won the game had the first two tosses shown heads, at a chance of one in four, since there could have been no way by which there might have subsequently been a tail in the leftmost place before there were one in the rightmost. Contrariwise, if either of the first two tosses had been tails then Sir R----- should have triumphed as the two heads that the Baron required to precede a tail would have themselves first been preceded by one.
Next, let us suppose that the Baron had elected for \(HTH\) and that Sir R----- had consequently settled upon \(HHT\). If the first toss were tails then we should have had to wait until a head had been tossed and shifted to the leftmost place before either of them stood a chance to win and so we need only figure on it having shown heads. There is a one in four chance that it would have been followed by a tail and then a head to give the Baron the game. If it were followed by another head, however, then Sir R----- would have been guaranteed the win since there could not have been a tail in the second place before there was one in the third. If the second and third tosses had come up tails, at a chance of one in four, then we should have again had to wait for a head to occupy the first place at which point we can figure the probabilities in the same fashion so that the Baron's chance of victory is given by
\[ \begin{align*} p &= \frac14 + \frac14 \times p\\ \frac34 \times p &= \frac14\\ p &= \frac13 \end{align*} \]
Finally, if the Baron's choice had been \(HTT\) and Sir R-----'s had been \(HHT\), then we should again have to wait for a leading head after which the Baron would have had one chance in four of an immediate win and Sir R----- would have had one chance in two of a guaranteed win should it have been followed by another head. In the case of it being followed by a tail and a head, again at a chance of one in four, then neither could possibly take a prize until that last head had advanced to the front. By precisely the same argument as above, this means that the Baron should have had one chance in three of victory.

In conclusion, Sir R-----'s worst chance of winning the wager was two in three and so his expected winnings were no less than
\[ \frac23 \times 2 - \frac13 \times 3 = \frac{4-3}{3} = \frac13 \]
and I should have most emphatically suggested that he rise to the Baron's challenge!
\(\Box\)

References

[1] Penney, W., Problem 95. Penney-Ante., Journal of Recreational Mathematics, Vol. 2, No. 4, 1969.

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