Linear Algebra question The betting odds in a three-horse race are as follows: H
ID: 3010348 • Letter: L
Question
Linear Algebra question
The betting odds in a three-horse race are as follows:
Horse A: 1/1 Horse B: 4/1 Horse C: 8/1
So, for example, if you bet $1 on Horse C, you win $8 (You get back your original $1 bet, plus an additional $8, for a profit of $8).
Suppose you bet $4 on A, $2 on B, and $1 on C. Then if A wins, you win
$4 on A and lose $2 on B and $1 on C, for a profit of $1. Similarly, if B wins, your profit is $2. And if C wins, your profit is $2. So, you can’t lose. We call such a betting strategy, where you win money regardless of the outcome of the race, a winning strategy. A losing strategy is one where it is possible for you to lose money. A neutral strategy is one where the worst you can do is to break even.
1.Describe a losing strategy for the above race.
2. Is there a neutral strategy for the above race? If so, describe it. If not, show why. You are required to bet.
3. Let d1, d2, and d3 be positive real numbers. Show that in the three-horse race described above, there exists a betting strategy that returns d1 dollars if A wins, d2 dollars if B wins, and d3 dollars if C wins.
4. Keeping the odds on Horses A and B unchanged, let rC denote the odds on Horse C.
a) For what values of rC can you find a winning strategy?
b) For what values of rC is every strategy a losing one?
c) For what values of rC is there a neutral strategy? Describe this strategy.
5. Let’s go back to the original race, with odds 1/1, 4/1, and
8/1. Suppose you have a fixed amount of money to bet. Your goal is to find a winning strategy that maximizes the least amount of money you can possibly win. How would you would apportion your bets among the three horses? Justify your answer.
Explanation / Answer
Suppose we have a series of 12 races with
4 horses, given the following betting odds therefore for simplicity, we'll only
consider the odds of winning each race):
Horse Odds* Probability (from odds)
----- ---- -----------------------
Horse A 1:1 1/(1+1) = 1/2 = 6/12
Horse B 1:2 1/(1+2) = 1/3 = 4/12
Horse C 1:3 1/(1+3) = 1/4 = 3/12
Horse D 1:5 1/(1+5) = 1/6 = 2/12
-----
Total Probability = 15/12 > 1
* Odds are listed here as (chances for):(chances against), which
is consistent with the notation used in the discussion. Historically,
in horse racing odds are expressed as (chances against):(chances for).
So what we have shown as odds of 1:3 would be shown on a racing sheet
as 3:1.
Now suppose I were to bet $1 on each horse for each race. In order
for me to break even on each horse, horse A would have to win 6 of the
12 races - then I'd win +$6 on the races A won and lose -$6 on the
races A lost. Similarly, horse B would have to win 4 of the 12 races
for me to break even -- I'd win 2 * $4 = +$8 (because of the 2:1 odds)
on the wins but lose -$8 on the losses. Horse C would have to win 3 of
the 12 races (3*$3 = +$9 on the wins, -$9 on the losses), and horse D
would have to win 2 of the 12 races (2*$5 = +$10 on the wins, -$10 on
the losses). Of course, this means that all together, they have to
have 15 wins in 12 races, so somewhere they're going to fall 3 short
of my "break even" requirement.
If, for example, horse A only wins 5 races and horse C only wins 2
races, then I've lost -$2 on horse A (+$5, -$7) and -$6 on horse C
(2*$2 = +$4, -$10). The house has just collected $8 from my pocket.
As long as no horse wins more often than its "probability" (based on
odds), the house wins. Of course, it is possible that horses D and B
will win 4 races each, horse B will win 3 races, and horse A will only
win 1 race. In this case, I will lose -$10 (+$1, -$11) on horse A,
break even on horses B and C, and win +$12 (4*$5 = +$20, -$8) on horse
D for a net winning of $2 - But you can "bet" that that won't happen
too often.
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