Each of \(m+2\) players pays 1 unit to a kitty in order to play the following
game: A fair coin is to be flipped successively \(n\) times, where \(n\) is an odd
number, and the successive outcomes are noted. Before the \(n\) llips, each
player writes down a prediction of the outcomes. For instance, if \(n=3\) then a
player might write down \((H, H, T),\) which means that he or she predicts that
the first flip will land on heads, the second on heads, and the third on
tails. After the coins are flipped, the players count their total number of
correct predictions. Thus, if the actual outcomes are all heads, then the
player who wrote \((H, H, T)\) would have 2 correct predictions. The total kitty
of \(m+2\) is then evenly split up among those players having the largest number
of correct predictions.
since each of the coin flips is equally likely to land on either heads or
tails, \(m\) of the players have decided to make their predictions in a totally
random fashion. Specifically, they will each flip one of their own fair coins
\(n\) times and then use the result as their prediction. However, the final 2 of
the players have formed a syndicate and will use the following strategy: One
of them will make predictions in the same random fashion as the other \(m\)
players, but the other one will then predict exactly the opposite of the
first. That is, when the randomizing member of the syndicate predicts an \(H,\)
the other member predicts a \(T .\) For instance, if the randomizing member of
the syndicate predicts \((H, H, T),\) then the other one predicts \((T,\) \(T, H)\)
(a) Argue that exactly one of the syndicate members will have more than \(n /
2\) correct predictions. (Remember, \(n\) is odd.)
(b) Let \(X\) denote the number of the \(m\) nonsyndicate players that have more
than \(n / 2\) correct predictions. What is the distribution of \(X ?\)
(c) With \(X\) as defined in part (b), argue that
\(E[\text { payoff to the syndicate }]=(m+2)\)
$$
\times E\left[\frac{1}{X+1}\right]
$$(d) Use part (c) of Problem 59 to conclude that
\(\begin{aligned} E[\text { payoff to the syndicate }]=& \frac{2(m+2)}{m+1} \\\
& \times\left[1-\left(\frac{1}{2}\right)^{m+1}\right] \end{aligned}\)
and explicitly compute this number when \(m=\)
\(1,2,\) and \(3 .\) Because it can be shown that
$$
\frac{2(m+2)}{m+1}\left[1-\left(\frac{1}{2}\right)^{m+1}\right]>2
$$
it follows that the syndicate's strategy always gives it a positive expected
profit.
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