How to hedge a sports bet? Suppose I've got a $200 ticket on the Golden State Warriors to win the NBA Finals at 5 : 1. The finals start next week, with the Cavs listed at 2 : 1 to beat the Warriors and the Warriors 4 : 9 to beat the Cavs. What are my hypothetical hedging strategies? 
So far I've thought of three, some/all of which might be flawed. In all of these equations, I'm assuming the odds above are accurate (i.e. the Warriors are truly 67.5% to win, the Cavs 32.5%) and making x the amount I'd bet on the Cavs. 


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*Maximize my EV and let the bet ride; do not hedge. I'm getting this from maximizing .675 * (1000 - x) + .325 * (2x - 200) for 0 ≤ x ≤ 1000, which gives a maximum at x = 0. 

*Hedge using a ratio of the likelihood of each outcome. Since the Warriors are (.675 / .325) as likely to win, I want my return on them to be (.675 / .325) of my return on the Cavs. This gives me (1000 - x) = (.675 / .325) * (2x - 200), or x ≈ $275. 

*Hedge using log utility and the Kelly Criterion. This is where I feel like I must be making a sloppy mistake. When I try to maximize .675 * ln(1 + (1000 - x)) + .325 * ln(1 + (2x - 200)) for 0 ≤ x ≤ 1000, I find a maximum at x ≈ $392.50, but that seems too high. 
Would love to hear any corrections or alternative approaches.
 A: Rather than troll, let's answer the question. 
Your third case is close, but you need to consider the wagers as a fraction of your bankroll, or simply provide the bankroll. (The 1 in the log is fine if you have the results expressed as a fraction of the roll, but with the result expressed only in dollars you're treating it as $1, not 100% of your bankroll.) 
For instance, if your bankroll is $10,000, Kelly gives
$E(\log(X)) = 0.625\log(10000 + (1000 - x)) + 0.325\log(10000 + (2x - 200))$
You would then maximize the function for x ≥ 0 by taking the derivative at 0, then considering only the positive result (or x = 0 if there is no positive result). You can likely see how to generalize this to any case. 
A: There is the minimax/maximin solution too:


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*if you bet $x$ on the Cavaliers and they win, then you get $2x-200$, an increasing function of $x$ 

*if you bet $x$ on the Cavaliers and they lose, then you get $1000-x$, a decreasing function of $x$ 

*so the minimum of these two outcomes is maximised when they are equal, when $x=400$


You will be fully hedged if you now bet $400$ on the Cavaliers at $2:1$, in which case you are guaranteed a net outcome of $+600$.  No other betting strategy guarantees this, though it might remove some of the excitement of watching the Finals
