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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 $n$flips, 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, 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 $\frac{n}{2}$correct predictions. (Remember, $n$is odd.)

(b) Let $X$denote the number of the $m$non-syndicate players who have more than $\frac{n}{2}$correct predictions. What is the distribution of $X$?

(c) With $X$as defined in part (b), argue that

role="math" localid="1647513114441" $E[payofftothesyndicate]=(m+2)\begin{array}{rl}=& (m+2)×E\left[\frac{1}{X+1}\right]\end{array}$

d) Use part (c) of Problem 7.59 to conclude that
role="math" localid="1647513007326" $E[payofftothesyndicate]=\frac{2(m+2)}{m+1}×\left[1-{\left(\frac{1}{2}\right)}^{m+1}\right]$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.

Step by step solution

01

Given Information (Part a)

Total Players $=m+2$

First prediction $=(H,H,T)$

Second Prediction $=(T,T,H)$

Argue one of the syndicate members will have more than$\frac{n}{2}$correct predictions.

02

Explanation (Part a) 

Remember that their predictions are complementary. If both players would have less than $\frac{n}{2}$correct predictions, that would lead that there was less than $n$flips, which is a contradiction. If both players would have more that $\frac{n}{2}$ correct predictions, that would yield that there was more than $n$flips in total, which is also a contradictions.$\frac{n}{2}$ correct predictions.

03

Final Answer (Part a) 

So, there has to exist one and only one player that has more than$\frac{n}{2}$correct predictions.

04

Given Information (Part b)

Total Players$=m+2$

First prediction$=(H,H,T)$

Second Prediction $=(T,T,H)$

Number of$m$non-syndicate players who have more than$\frac{n}{2}=X$

05

Explanation (Part b) 

The probability that a single non-syndicate member have more than $\frac{n}{2}$correct predictions is equal to $\frac{1}{2}$ since $n$ is odd number and probabilities for Head and Tail in a single flip are equal.

06

Final Answer (Part b) 

So, if we know that non-syndicate members make their predictions independently, we have that$X~\mathrm{Binom}\left(m,\frac{1}{2}\right).$

07

Given Information (Part c)

Total Players$=m+2$

First prediction$=(H,H,T)$

Second Prediction $=(T,T,H)$

Argue that$\text{E[payoff to the syndicate}]=(m+2)=(m+2)×E\left[\frac{1}{X+1}\right]$

08

Explanation (Part c) 

We have showed in part (a) that there is always one and only one player winner from the syndicate. Also, we have that there are $X$ non-syndicate players winners. Hence, there are $X+1$ of winners and they split $(m+2)$ of money.

09

Final Answer (Part c) 

Hence, the expected winnings of the syndicate is equal to$E\left(\frac{m+2}{X+1}\right)=(m+2)E\left(\frac{1}{X+1}\right).$

10

Given Information (Part d)

Total Players$=m+2$

First prediction$=(H,H,T)$

Second Prediction $=(T,T,H)$

Show that$m=3⇒\frac{2(3+2)}{3+2}\left(1−{\left(\frac{1}{2}\right)}^{3+1}\right)≈2.34$

11

Explanation (Part d) 

In problem 59, part (c) we had that if $B(n,p)$, then

$E\left(\frac{1}{B+1}\right)=\frac{1-(1-p{)}^{n+1}}{(n+1)p}$

Here we have specified parameters $n=m$and $p=\frac{1}{2}$since $X~Binom\left(m,\frac{1}{2}\right)$.

Therefore $E\left(\frac{1}{X+1}\right)=2·\frac{1-{\left(\frac{1}{2}\right)}^{m+1}}{(m+1)}$

Which implies that (just plug into the result from part (c) of this exercise)

$E\left[\text{payoff to the syndicate}\right]=\frac{2(m+2)}{m+2}\left(1−{\left(\frac{1}{2}\right)}^{m+1}\right)$

12

Explanation (Part d) 

Calculate directly that,

$m=2⇒\frac{2(2+2)}{2+2}\left(1−{\left(\frac{1}{2}\right)}^{2+1}\right)=2.33$

$m=3⇒\frac{2(3+2)}{3+2}\left(1−{\left(\frac{1}{2}\right)}^{3+1}\right)≈2.34$

13

Final Answer (Part d)

So, we see that this strategy yields positive expected winnings for $m=1,2,3$since these players have given 1 unit of money each (which is 2 in total) and they can expect more than 2 units of earnings.

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