Probability of one or more events occurring all with different probabilities [closed]

Solution 1:

There is a possible point of ambiguity. To a mathematician, $2.2/1$ (for example) translates into a win factor of $\displaystyle \frac{1}{3.2}$.

This corresponds to a payout of $2.2$ to $1$. However, it is not uncommon for a casino (for example) to use the syntax $2.2/1$ to represent a payout of $2.2$ for $1$, which implies that if you win, you get back $2.2$, instead of $(2.2 + 1 = 3.2)$.

In the response below, I am going to assume that $2.2/1$ signifies a payout of $2.2$ to $1$, rather than $2.2$ for $1$.

Easiest to use your example for illustration, for an exact answer.

First, you have to convert each odds into a probability $p$ such that $0 < p < 1.$

In horse racing parlance, $X/1$ translates into $\displaystyle p = \frac{1}{X + 1}.$

So, letting $p_k$ denote the probability of the horse winning race $k$, you have that

$p_1, p_2, p_3, p_4, p_5~~$ equal

$\displaystyle \frac{1}{3.2}, \frac{1}{5}, \frac{1}{4.3}, \frac{1}{11}, \frac{1}{6},~~$ respectively.

Set $q_1, q_2, q_3, q_4, q_5~~$ equal to

$(1 - p_1), (1 - p_2), (1 - p_3), (1 - p_4), (1 - p_5)~~$ respectively.

Per your example, I am assuming that you want the probability of exactly $3$ wins for the horse, rather than (for example) at least $3$ wins.

Unfortunately, there isn't much of a shortcut. First, I will show the brute force method, and then discuss possible shortcuts.

There are $\displaystyle \binom{5}{3}$ different possibilities for which of the $3$ races that the horse wins. The brute force method is to add together the following $10$ terms:

$p_1 \times p_2 \times p_3 \times q_4 \times q_5$.
$p_1 \times p_2 \times q_3 \times p_4 \times q_5$.
$p_1 \times p_2 \times q_3 \times q_4 \times p_5$.
$p_1 \times q_2 \times p_3 \times p_4 \times q_5$.
$p_1 \times q_2 \times p_3 \times q_4 \times p_5$.

$p_1 \times q_2 \times q_3 \times p_4 \times p_5$.
$q_1 \times q_2 \times p_3 \times p_4 \times p_5$.
$q_1 \times p_2 \times q_3 \times p_4 \times p_5$.
$q_1 \times p_2 \times p_3 \times q_4 \times p_5$.
$q_1 \times p_2 \times p_3 \times p_4 \times q_5$.

A shortcut of sorts is to notice that the first $6$ terms all have the horse winning the 1st race. This corresponds to $\displaystyle \binom{4}{2} = 6$, which represents that there are $6$ ways of choosing which of the other $2$ races the horse wins.

So, you could ignore the $p_1$ factor, brute force compute the sum of the 1st 6 terms, using only $4$ factors each (i.e. where the $p_1$ factor is excluded in each of the $6$ computations), and then multiply the sum of these $6$ terms by $p_1$. This would get you the actual sum of the first $6$ terms.

Similarly, you could notice that the last $4$ terms all have a factor of $q_1$. So, you could brute force compute these last $4$ terms, excluding the $q_1$ factor in each term, and then collectively apply the $q_1$ factor.

A (quick) estimate is to notice that the probability of the horse winning the $4th$ race is very small $\left( \text{i.e.} ~ \displaystyle \frac{1}{11}\right)$. So, if you assume that the horse loses that race, then instead of having to sum $\displaystyle \binom{5}{3}$ terms, you only have to sum $\displaystyle \binom{4}{3}$ terms.

There are certainly other approximation games that can be played. For example, instead of simply ignoring the $6$ possibilities that correspond to the horse winning the $4$th race, you could reason as follows:

In exactly $3$ of the $10$ races, the horse wins race $3$ and loses race $4$. Call this group A. In exactly $3$ of the races, the horse (instead) wins race $4$ and loses race $3$. Call this group B.

Consider $\displaystyle p_4 \times q_3 = \frac{1}{11} \times \frac{3.3}{4.3}$.
Denote this as $R$.

Consider $\displaystyle q_4 \times p_3 = \frac{10}{11} \times \frac{1}{4.3}$.
Denote this as $S$.

As a shortcut to computing the collective sum of the $3$ races in group B, you can take the collective sum of the $3$ races in group A, and multiply it by $\displaystyle \frac{R}{S} = \frac{3.3}{10}.$

Solution 2:

First of all, you need the probabilities that the horse will win each particular race. Those are not necessarily the reciprocals of the odds, in fact if they were, the betting house would go bankrupt!

If the odds are $A/B$, this means with probability $p$ of winning you get $A/B$ dollars (pounds etc.) per each invested dollar, and with $1-p$ you lose your invested dollar, so the betting house gets on average $(1-p)-p\frac{A}{B}$, and for it to break at least even they need:

$$(1-p)-p\frac{A}{B}\ge 0$$

$$1\ge p\left(1+\frac{A}{B}\right)=p\frac{A+B}{B}$$

$$p\le \frac{B}{A+B}$$

So, if the odds are, say, $4:1$, this means that the probability of winning is at most $1/5=20\%$ (as per the betting house's judgement). Of course, you are betting that they are wrong, which they may be, so $p=\frac{B}{A+B}$ is probably a reasonable estimate of the probability of winning.

Now, let your probabilities be $p_1, p_2, p_3, p_4, p_5$. All you need to calculate to check the probability of winning in exactly three out of five races is:

$$p_1 p_2 p_3(1-p_4)(1-p_5)+p_1 p_2(1-p_3)p_4(1-p_5)+\ldots+(1-p_1)(1-p_2)p_3 p_4 p_5$$

(there are $10$ terms above - one per each choice of three out of five races). There may be more compact ways to write this down (using $\sum$ and $\prod$ notation) - but this won't help much, so I am not going to do it.

You may tidy up the above expression a bit: it is equal to:

$$S(5,3)-4S(5,4)+10S(5,5)$$

where:

$$S(5,3)=p_1p_2p_3+p_1p_2p_4+p_1p_2p_5+p_1p_3p_4+p_1p_3p_5+p_1p_4p_5+p_2p_3p_4+p_2p_3p_5+p_2p_4p_5+p_3p_4p_5$$ $$S(5,4)=p_1p_2p_3p_4+p_1p_2p_3p_5+p_1p_2p_4p_5+p_1p_3p_4p_5+p_2p_3p_4p_5$$ $$S(5,5)=p_1p_2p_3p_4p_5$$

are the symmetrical polynomials in five variables of degree $3, 4, 5$, respectively. One way of getting those is expanding the polynomial:

$$(X-p_1)(X-p_2)(X-p_3)(X-p_4)(X-p_5)=X^5-S(5,1)X^4+S(5,2)X^3-S(5,3)X^2+S(5,4)X-S(5,5)$$

i.e. finding the coefficients multiplying $X^2$, $X$ and the constant term - as per Vieta's formulas.

Exercise: In your case, we have

$$p_1=\frac{1}{1+2.2}=0.3125$$ $$p_2=\frac{1}{1+4}=0.2$$ $$p_3=\frac{1}{1+3.3}\approx 0.23256$$ $$p_4=\frac{1}{1+10}\approx 0.09091$$ $$p_5=\frac{1}{1+5}\approx 0.16667$$

Now try expanding the following polynomial (I've used Wolfram Alpha for that):

$$(X-0.3125)(X-0.2)(X-0.23256)(X-0.09091)(X-0.16667)$$

to get

$$X^5 - 1.00263 X^4 + 0.388746 X^3 - 0.0726216 X^2 + 0.00649665 X - 0.000220226$$

i.e.

$$S(5,3)\approx 0.0726216, S(5,4)\approx 0.00649665 , S(5.5)\approx 0.000220226$$

and, finally

$$S(5,3)-4S(5,4)+10S(5,5)\approx 0.04883726\approx 4.9\%$$