# Optimal Fight Purse and Boxing Strategies

The following is all public information available to all the players in this scenario.

The General Setup

In the aftermath of the infamous race between the tortoise and the hare, the salty hare went to trash talk the tortoise in the news. Things come to a head and the two decide to settle their differences in the ring with an $N$ round boxing match. The house manager who owns the venue is wondering how to set three things: venue ticket price $p_t$, the size of the combined guaranteed salaries for both fighters $w$, and the fraction of ticket sales that go to the fight purse, denoted $p_f$. The other animals suspect that pay for the fighters will probably affect the intensity of the fight, and are wondering if the fight will be interesting.

The utility of paying for a fight ticket for each consumer is $$u_j(\mathbb{E}\left(\sum_{i=1}^{N} e_{ti} + e_{hi}\right), v, p_t)$$

That is, the expected total effort exerted by the tortoise and the hare across the $N$ rounds, the number $v \leq J$ of other animals who are in line to buy a ticket, and ticket price $p_t$ are what affect utility. An animal will get in line to buy a ticket if the utility of buying a ticket is $0$ or greater. We will say that $u_j$ is strictly increasing in the first and second arguments, strictly decreasing in the third argument, and that all animals are risk averse (so concave utility). The house manager sets one venue price and does not change it all the way up until the fight gets underway and ticket sales close (it's a cultural thing).

The Fight

Before ticket sales open, and after the house manager has made his decisions, the two fighters must agree on how to split the total fight purse $p_f p_t x$ based on who wins and who loses, where $x$ is the total number of tickets that end up being sold. The winner gets fraction of the fight purse $F$. The combined house salary $w$ is split equally between both fighters. Each fighter has a stock of effort/energy to use in the fight denoted $E_t$ and $E_h$ for the tortoise and hare respectively, where $E_t < E_h$. The total consumption of energy for fighter $f \in \{t, h\}$ is:

$$\sum^N_{i=1} e_{fi} + d(e_{-fi} - e_{fi}) \leq E_f$$

We denote $d(\cdot)$ as a damage function where expending less energy than your opponent in a given round means you lose additional energy. We restrict $d(\cdot) = 0$ if the argument is less than or equal to $0$ and say $d(\cdot)$ is strictly increasing, but negative in the second derivative. For any given round, the fighter who suffers the damage penalty has a chance of being knocked down in the current round $i$ denoted $k_i = \min\{d(\frac{e_{-fi} - e_{fi}}{E_f - \sum^i e_{fi}}), 1\}$. Once knocked down, the chance of getting up is $\mathbb{K} = \frac{E_f - \sum^i e_{fi}}{E_f}$. (Note the way this problem is set-up, only one fighter per round can be knocked down, and only once.)

So the chance of being knocked down is the fraction of extra energy lost from damage divided by total remaining unallocated energy. The chance of getting back up is just the fraction of remaining unallocated energy left relative to the initial stock of energy. Once the fight ends, energy expenditure for all remaining rounds is $0$. If both fighters are still standing at the end of the last round, they get 1 point for each round they spent more energy, and 1 point for every knockdown they got. Most points wins. More knockdowns breaks ties. If that ties, the judge gives the win to the hare, because the hare is seen as more athletic.

For the sake of a consistent set-up, we will say that the fighters set each round's energy expenditure at the start of every round.

The utility of the winning fighter can be expressed as:

$$\overline{u}_f(\frac{w}{2} + F p_f p_t x, \quad v, \quad \frac{\sum_i^N e_{fi}}{E_f})$$

and for the loser:

$$\underline{u}_f(\frac{w}{2} + (1-F) p_f p_t x, \quad v, \quad \frac{\sum_i^N e_{fi}}{E_f})$$

That is, the total pay, the number of animals who show up to watch the fight, and the fraction of energy used through the fight are what affect utility. For the winner, utility is strictly increasing in the first and second arguments, and strictly decreasing in the third argument. For the loser, utility strictly decreases in the second argument instead. Both fighters are risk averse regardless of what outcome they get (concave utility).

We place two last conditions. If the expected utility of the fight for one of the contestants is less than some fixed $S < 0$, they will drop out the fight and take the $\mathscr{L}$ [insert picture of making an L on your forehead]. The other condition is that the two fighters have too much pride to accept a bribe. (At least from each other.)

Just to clarify, we can consider the decision order as:

1. the house manager chooses $w, p_t, p_f$
2. the fighters set $F$
3. the $J$ animals choose queue in line for a ticket
4. the fighters exert $e_h, e_t$ across rounds

Starting Backwards Induction (skip this if you just wanna get to the question)

One thing to note is that if the fighters set $F < \frac{1}{2}$, then there becomes a perverse incentive to throw the fight, which we imagine would cause less expected effort during the fight despite the dis-utility of losing. Knowing this, the consumers would purchase fewer tickets, which would decrease the fight purse, but also decrease the dis-utility of losing. So the change in $F$'s effect on effort is ambiguous in this question's current form.

Suppose $F > \frac{1}{2}$. Say that the fight is already on and the wages for the winner and loser and the number of animals at the venue are fixed. The only change in utility happens with effort expended.

For $N = 1$, even here it is difficult to think of a pure Nash equilibrium. If the tortoise exhausts effort and the hare matches (thus winning), the tortoise will want to deviate by lowering effort. If the tortoise exerts no effort, the hare would want to match, but that would cause the tortoise to want to exert effort, which would give the hare the opportunity to increase effort. And so on.

The only way a pure Nash equilibrium exists is if at effort level $0$, the tortoise would find the marginal cost of exerting more effort and getting the winning payoff to be greater than the marginal benefit of getting that winning payoff. $e_t = 0, e_h = 0$ would then be Nash (then there is no incentive to lower effort, because the tortoise can't). On the flip side, maybe the hare would be in this position while the tortoise wasn't, so $e_t = 0 + \epsilon, e_h = 0$ would be some (ill-defined) Nash.

What about randomized strategies? Consider a discrete case:

\begin{array} {|r|r|r|r|} & e_h = 0 \;& e_h = 1 \;& e_h = 2 \;\\ \hline e_t = 0 & \underline{u}(e = 0), \overline{u}(e = 0)& \underline{u}(e = 0), \overline{u}(e = 1) & \underline{u}(e = 0), \overline{u}(e = 2)\\ \hline e_t = 1 & \overline{u}(e = 1), \underline{u}(e = 0) & \underline{u}(e = 1), \overline{u}(e = 1) & \underline{u}(e = 1), \overline{u}(e = 2)\\ \hline \end{array}

$e_h = 2$ is strictly dominated by $e_h = 1$, so we just look at cases where $e_h = 0, 1$ and similarly so where $e_t = 0, 1$ for the tortoise.

$u_t(\sigma_h, e_t = 0) \geq u_t(\sigma_h, e_t = 1)$ if

$p\underline{u}(e = 0) + (1-p)\underline{u}(e = 0) \geq p\overline{u}(e = 1) + (1-p)\underline{u}(e = 1)$

$\implies \frac{\underline{u}(e = 0) - \underline{u}(e = 1)}{\overline{u}(e = 1) - \underline{u}(e = 1)} \geq p$

where $p$ is the probability the hare picks $e_h = 0$. Similarly, with $q$ the probability the tortoise picks effort $e_t = 0$, our condition here is

$q \geq \frac{\overline{u}(e = 1) - \underline{u}(e = 0)}{\overline{u}(e = 0) - \underline{u}(e = 0)}$

with these conditions, as long as $p, q \in [0, 1]$, we can get conditions for randomized Nash. For the continuous game, things are quite a bit trickier. Depending on what the fixed values are for the first and second arguments in fighter utility are as well, the existence of a randomized strategy might change. From here, we'd have to move back and see what each previous decision leading up to the fight would change. Then we could look at a two period fight, where the chance of getting knocked down would then come into play.

Specified Functional Forms

Let's say

• $u_j = \alpha_j \ln(\mathbb{E}\left(\sum_{i=1}^{N} e_{ti} + e_{hi}\right) + \beta_j \ln(v) + \gamma_j \ln(C - p_t)$ (so utility is also additive, but note $v$ and $\mathbb{E}(\cdot)$ may be endogenous) for $\alpha, \beta, \gamma, C > 0$
• $d = (e_{-fi} - e_{fi})^{\frac{1}{2}}$ with the same constraints on the domain
• $\overline{u} = \lambda \ln\left(\frac{w}{2} + F p_f p_t x \right) + \delta \ln(v) + \theta \ln\left(D - \frac{\sum_i^N e_{fi}}{E_f} \right)$
• $\underline{u} = \lambda \ln\left(\frac{w}{2} + (1-F) p_f p_t x \right) + \delta \ln(E - v) + \theta \ln\left(D - \frac{\sum_i^N e_{fi}}{E_f} \right)$

with $\lambda, \delta, \theta > 0; \quad D > 1; \quad E > J$

• $N = 3$

And here's a big one, let's say consumers are identical. So $\alpha_j = \alpha$ and so on. Once one animal gets in line, all the others will too (at least, I take that for granted given the incentives for the fighters).

The Question

• What is the probability of the tortoise winning the fight once the house sets their optimal prices and we go through the whole game?

You may express the values generally or plug in your own numbers for the fixed constants, as long as the values plugged in don't give trivial results like, "all prices equal zero and the fight is called off" or "the tortoise always loses". Basically, what's an interesting equilibrium?

I'll also upvote answers that partially solve the game. So for example you could assume the number of viewers is fixed, as well as the size of the purse, but then show what the optimal $F$ is to be set along with the optimal fighting strategies, from where the chances of each player winning can be calculated.