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a) Let x and y be the odds of the first cash bet and the subsequent free bet, respectively, and w be the amount of the wager.
There is a (0.975/x) probability you’ll win the first bet, which returns w*x.
There is a (1 – 0.975/x)*(0.975/y) probability that you’ll lose the first bet and then win the free bet, which returns w*(y-1). Otherwise you lose both bets, which returns 0.
Your expected profit is then (0.975/x)*(wx) + (1 – 0.975/x)*(0.975/y)*w*(y-1) - w.
Some algebra makes this (0.950625/xy – 0.950625/x – 0.975/y + 0.95) * w.
This approaches 0.95w as x and y go to infinity. So you want to place the largest possible bet ($1,000) at the longest possible odds, and if it loses do the same with the free bet - your expected profit by doing so approaches $950. (By contrast, if you place the bets at the shortest possible odds, i.e. x = y = 1.0, your expected loss is -$25, as implied by the margin.)
b) Of course, the expected value in (a) is based on a small chance of winning a lot of money, offset by a very high probability of simply losing your $1,000. If we were risk-averse and wanted to guarantee a profit, let’s work backwards.
Assuming we have a $1,000 free bet, we could place it on a selection and then wager an amount of cash on the opposite selection such that either outcome returns the same amount (we’re looking for the maximum guaranteed amount, so if they didn’t return the same amount there would be a nonzero chance we’d get the smaller amount).
Given the odds of a selection, we can compute the corresponding odds of their opponent by adding back the 2.5% margin, computing the implied probability, subtracting from 1 and then applying 2.5% margin back to the result. So if the odds on a selection are x, the odds on their opponent would be 0.975 / (1 – 0.975/x).
With a bit more algebra we find that for a selection with odds of x, we can guarantee ourselves a profit P = 1025.64 – 25.64x – 1000/x. If we take the derivative and set it equal to 0 we find that P is maximum when x ~= 6.25.
We could therefore place a $1,000 free bet at odds of 6.25, and wager $4,544.62 on their opponent at odds of 1.155. Either outcome returns $5,250 for a net profit of $705.38.
Stepping back, we want to ensure the same return if we don’t get the free bet, so we would make our first wager of $1,000 cash on a selection at odds of 5.25, and place a subsequent wager of $3,795.51 on their opponent at odds of 1.197. If the first wager wins we get back $5,250; if it loses we get back $4,544.62 and receive a free bet convertible into an additional $705.38.
Either way we’ve wagered a total of $4,795.51 in cash and are guaranteed to receive $5,250 back, for a net profit of $454.49.
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