The optimal way of dividing the bet three ways

[Hill]

Homework Statement

In the Olympic finals, only three teams are represented; USA, England, and China. Svetlana believes the USA will win and offers even odds of 1:1. Roberto is supporting England and is offering 2:1 odds. (So, if England wins, Roberto keeps his money. If one of the other teams wins, Roberto pays $2 for each dollar bet.) Finally, Jeff supports China as given by his 5:4 odds. Determine the optimal way of dividing $100 to bet that ensures the largest winning.

Relevant Equations

earnings=gain-loss

If I bet $s$ with Svetlana, $r$ with Roberto, and $j$ with Jeff, my earnings would be,

if USA wins, $-s+2r+ \frac 5 4 j$,

if England wins, $s-r+\frac 5 4 j$,

if China wins, $s+2r-j$,

where $s+r+j=100$.

I think that "the optimal way ... that ensures the largest winning" is to make the earning independent of the result of the final. Then, I get three equations with three unknowns,

$-s+2r+ \frac 5 4 j=s-r+\frac 5 4 j$
$-s+2r+ \frac 5 4 j=s+2r-j$
$s+r+j=100$.

The solution is to bet $17.65 with Svetlana, $35.29 with Roberto, and $47.06 with Jeff.

 

[PeroK]

You can calculate a general solution based on any number of different odds.

England is not an Olympic nation. It's Great Britain.

 

[FactChecker]

Determine the optimal way of dividing $100 to bet that ensures the largest winning.

"the optimal way ... that ensures the largest winning" is to make the earning independent of the result of the final.

IMO, if this is really the intent of the question, a better way of phrasing it would be "with the largest guaranteed win". That would be a "min-max" problem, where you are maximizing the minimum possible winnings. Otherwise, I would interpret problems like this to be talking about maximizing expected winnings.

 

[Hill]

IMO, if this is really the intent of the question, a better way of phrasing it would be "with the largest guaranteed win". That would be a "min-max" problem, where you are maximizing the minimum possible winnings. Otherwise, I would interpret problems like this to be talking about maximizing expected winnings.

I assume that the intent of the question is rather former than latter, because the text starts discussing the expectation values a bit later.

 

[PeroK]

Suppose there are $n$ mutually exclusive possibilities and that (precisely) one must happen. You are offered odds of $a_k:b_k$ for each ($k = 1 \dots n$). The return on a bet of one unit, if event $k$ happens, is:
$$r_k = 1 + \frac{a_k}{b_k} = \frac{a_k+b_k}{b_k}$$You can achieve a guaranteed return (net win or loss) by betting on each event with a proportion $x_k$, where (for all $k, l$):
$$x_kr_k = x_lr_l$$Hence:$$x_l = \frac{r_k}{r_l}x_k$$Now, for all $k$:
$$1 = \sum_{l=1}^{n}x_l = \sum_{l=1}^{n}\frac{r_k}{r_l}x_k = \bigg (\sum_{l=1}^{n}\frac{1}{r_l}\bigg )r_kx_k$$Hence:$$x_k = \frac{1}{r_k(\sum_{l=1}^{n}\frac{1}{r_l})} = \frac{\frac{b_k}{a_k+b_k}}{\sum_{l=1}^{n}\frac{b_l}{a_l+b_l}}$$Note that we can interpret $p_k = \frac{b_k}{a_k+b_k}$ as an estimated likelihood for event $k$. These are not probabilities, simply estimates that may have no relation to how likely the event is. The people offering the odds may be completely wrong. In any case, we have:
$$x_k = \frac{p_k}{\sum_{l=1}^{n}p_l}$$In a different context, where these are probabilities, you can guarantee getting precisely your money back by betting $x_k = p_k$.

Finally, the amount you end up with is:
$$r = x_kr_k = \frac{1}{\sum_{l=1}^{n}p_l}$$This is greater than 1 (net win) iff the sum of all the estimated likelihoods is less than 1. And. it's a net loss if the sum of all the estimated likelihoods is greater than 1.

If you check the odds for any competition with any specific bookmaker, the estimated likelihoods will, indeed, sum to greater than 1. If not, then you can beat the bookie, as it were. Or, you can go round different bookmakers looking for a total estimate for all possibilities of less than 1.

 

[Hill]

These are not probabilities, simply estimates that may have no relation to how likely the event is.

The text calls these, "implied probabilities."