Introduction
Once a week, we are playing head or tail in my favorite bar.
There are $N$ people in the room and each person is guessing whether head or tail will come up to the next toss. If you lose you sit down. If you win you remain standing. The last person to be standing brings the victory to its team.
The are $n$ different teams and the number of members in each team is $[x_1, x_2, ...,x_{n-1}, x_n]$ such that $\sum_i^n x_i = N$. Let's call my team "team 1". There are therefore $x_1$ members in my team.
Strategy
A friend of mine came with the strategy that at each toss, half of the team is playing head, while the other half is playing tail.
Question
If everybody is playing randomly, our probability of winning is $\frac{x}{N}$.
- Does my friend's strategy affect this probability?
- What would be the optimal strategy?
- Is the best strategy different depending on what strategy the other teams are playing (and depending the number of individuals in each team)?
If you split your team of $x$ into $t$ tails and $h$ heads, you'll get the average of the game values with $t$ and $h$ people left. If the game value is concave with respect to the number of people on your team, this average is optimal for an even split $t=\lfloor\frac x2\rfloor$, $h=\lceil\frac x2\rceil$. My guess would be that this will always be the case, independent of the other teams' strategies, but the details of proving that seem messy.
Let's look at the game with two teams to get a feel for things. We can derive the game values $v(x_1,x_2)$ starting from the initial values $v(x_1,0)=1$ for $x_1\gt0$ and $v(0,x_2)=0$ for $x_2\gt0$. You didn't specify $v(0,0)$; I'll take it to be $\frac12$ (a split prize).
Here's a table of game values if both teams always split, for $x_1,x_2\le4$ (with $x_1$ downward and $x_2$ to the right):
\begin{array}{c|cc} &0&1&2&3&4\\\hline 0&\frac12&0&0&0&0\\ 1&1&\frac12&\frac14&\frac3{16}&\frac18\\ 2&1&\frac34&\frac12&\frac38&\frac14\\ 3&1&\frac{13}{16}&\frac58&\frac12&\frac38\\ 4&1&\frac78&\frac34&\frac58&\frac12\\ \end{array}
The value is concave in $x_1$ (and accordingly convex in $x_2$), so splitting is indeed optimal. We can also derive the game values if your team splits while the other plays randomly:
\begin{array}{c|cc} &0&1&2&3&4\\\hline 0&\frac12&0&0&0&0\\ 1&1&\frac12&\frac5{14}&\frac{19}{70}&\frac{471}{2170}\\ 2&1&\frac34&\frac{33}{56}&\frac{269}{560}&\frac{13987}{34720}\\ 3&1&\frac{13}{16}&\frac{305}{448}&\frac{5231}{8960}&\frac{566131}{1111040}\\ 4&1&\frac78&\frac{173}{224}&\frac{3079}{4480}&\frac{342339}{555520}\\ \end{array}
Here, too the value is concave in your team count, so your split play is optimal also against random play. Here's a direct comparison with approximate decimal values (split play first):
\begin{array}{c|cc} &0&1&2&3&4\\\hline 0&0.500&0&0&0&0\\ 1&1&0.500&0.250/0.357&0.188/0.271&0.125/0.217\\ 2&1&0.750&0.500/0.589&0.375/0.480&0.250/0.403\\ 3&1&0.813&0.625/0.681&0.500/0.584&0.375/0.510\\ 4&1&0.875&0.750/0.772&0.625/0.687&0.500/0.616\\ \end{array}
For completeness, here are the game values for random play by both teams:
\begin{array}{c|cc} &0&1&2&3&4\\\hline 0&0.500&0&0&0&0\\ 1&1&0.500&0.357&0.271&0.217\\ 2&1&0.643&0.500&0.407&0.343\\ 3&1&0.729&0.593&0.500&0.433\\ 4&1&0.783&0.657&0.567&0.500\\ \end{array}
The values are much closer to $\frac12$ than for optimal play by both teams, since random play is worse at taking advantage of superior numbers.
Here's the code I used to compute the game values.