given the following horses, each with its chance of winning:

Horse 1 -> 0.29
Horse 2 -> 0.34
Horse 3 -> 0.11
Horse 4 -> 0.07
Horse 5 -> 0.14
Horse 6 -> 0.05
Sum -> 1

At the moment, in order to calculate the straight forecast of 1-4, that is the probability that horse 1 wins the race AND horse 2 finishes second, I consider horse 1 as winner and recalculate all the other probabilities excluding horse 1 from recalculation (with proportions). After calculating the probability that horse 4 finishes first on 5 horses total, I multiply the two probabilities. In this way, straight forecast 1-4 and 4-1 are different values. I think this is the correct way to proceed. The same thing I do the for straight tricast, recalculating and multiplying for two horses instead of one.

But I don't know how to calculate:

  • the probability that horse 1 finish first OR second (place bet)
  • the probability that horse 1 finish first OR second OR third (show bet)
  • the probability that result is 1-4 OR 4-1 (reverse forecast)
  • the probability that result is 1, 4 and 3 finish in any order (combination tricast)

I don not simply need a formula, but a full mathematical explanation in order to fully understand. Academic papers on Internet regarding this argument are well accepted.


Firstly, some notation. For $i=1\ldots 6$, define events: \begin{eqnarray*} W_i &=& \mbox{$H_i$ (i.e. Horse $i$) wins} \\ S_i &=& \mbox{$H_i$ comes second} \\ T_i &=& \mbox{$H_i$ comes third}. \end{eqnarray*}


\begin{eqnarray*} P\left(\mbox{$H_i$ is $1^{st}$ or $2^{nd}$}\right) &=& P\left(\mbox{$H_i$ is $1^{st}$}\right) + P\left(\mbox{$H_i$ is $2^{nd}$}\right) \qquad\mbox{(since they are disjoint events)}\\ &=& P\left(W_i\right) + \sum_{\substack{1\leq m \leq 6 \\ m\neq i}}{P\left(\mbox{$W_m$ and $S_i$}\right)} \qquad\mbox{(all the ways that $H_i$ comes $2^{nd}$)} \\ &=& P\left(W_i\right) + \sum_{\substack{1\leq m \leq 6 \\ m\neq i}}{\left(P\left(W_m\right) \dfrac{P\left(W_i\right)}{1 - P\left(W_m\right)}\right)}. \end{eqnarray*}

This last term in brackets is exactly what you have already described: proportioning the given win probability to work out the second place probability for $H_i$ and then multiplying that with the win probability for $H_m$. Here, we want to include the cases when any other horse wins, hence the summation over index $m$. Also, $P\left(W_1\right) = 0.29$, and so on.


Here we use a similar method as in $\left(a\right)$ but it goes one level deeper. We sum over all $m,n$ such that $H_m$ wins, $H_n$ is second, and $H_i$ is third.

\begin{eqnarray*} P\left(\mbox{$H_i$ is $1^{st}$ or $2^{nd}$ or $3^{rd}$}\right) &=& P\left(\mbox{$H_i$ is $1^{st}$ or $H_i$ is $2^{nd}$}\right) + P\left(\mbox{$H_i$ is $3^{rd}$}\right) \qquad\mbox{(disjoint events)}. \end{eqnarray*}

We already have $P\left(\mbox{$H_i$ is $1^{st}$ or $H_i$ is $2^{nd}$}\right)$ from $\left(a\right)$. So now,

\begin{eqnarray*} P\left(\mbox{$H_i$ is $3^{rd}$}\right) &=& \sum_{\substack{1\leq m \leq 6 \\ m\neq i}}{\sum_{\substack{1\leq n \leq 6 \\ n\notin \{i,m\}}}{P\left(\mbox{$W_m$ and $S_n$ and $T_i$}\right)}} \\ &=& \sum_{\substack{1\leq m \leq 6 \\ m\neq i}}{\sum\limits_{\substack{1\leq n \leq 6 \\ n\notin \{i,m\}}}{\left( P\left(W_m\right) \dfrac{P\left(W_n\right)}{1 - P\left(W_m\right)} \dfrac{P\left(W_i\right)}{1 - P\left(W_m\right) - P\left(W_n\right)} \right) }}. \\ \end{eqnarray*}

Again, this messy looking last term is exactly how you are calculating your straight tricast. Here, we are summing them over all possible $1^{st}$ and $2^{nd}$ places (indexes $m$ and $n$).


\begin{eqnarray*} P\left(\mbox{($H_i$ $1^{st}$ and $H_j$ $2^{nd}$) or ($H_j$ $1^{st}$ and $H_i$ $2^{nd}$)}\right) &=& P\left(\mbox{$W_i$ and $S_j$}\right) + P\left(\mbox{$W_j$ and $S_i$}\right) \\ &&\qquad\mbox{(because they are disjoint events)}. \end{eqnarray*}

You already know how to calculate the two values on the RHS (they are straight forecasts) so we are done.


\begin{eqnarray*} P\left(\mbox{$H_i$ $H_j$ $H_k$ combination tricast}\right) &=& \sum\limits_{a,b,c}{ P\left(\mbox{$W_a$ and $S_b$ and $T_c$}\right)} \end{eqnarray*}

The sum here is over all permutations $(a,b,c)$ of the values $(i,j,k)$. There are $3! = 6$ terms in this sum. Again, you know how to calculate these terms (they are straight tricasts) so we are done.

Straight Forecast

The probability that $H_i$ wins and $H_j$ is second:

\begin{eqnarray*} P\left(\mbox{$W_i$ and $S_j$}\right) &=& P\left(W_i\right) \dfrac{P\left(W_j\right)}{1 - P\left(W_i \right)} \end{eqnarray*}

Straight Tricast

The probability that $H_i$ wins, $H_j$ is second and $H_k$ is third:

\begin{eqnarray*} P\left(\mbox{$W_i$ and $S_j$ and $T_k$}\right) &=& P\left(W_i\right) \dfrac{P\left(W_j\right)}{1 - P\left(W_i \right)} \dfrac{P\left(W_k\right)}{1 - P\left(W_i \right) - P\left(W_j \right)} \end{eqnarray*}

Example (the one in the question)

Straight forecast on $H_5$ and $H_3$:

\begin{eqnarray*} P\left(\mbox{$W_5$ and $S_3$}\right) &=& P\left(W_5\right) \dfrac{P\left(W_3\right)}{1 - P\left(W_5 \right)} \\ && \\ &=& 0.14 \times \dfrac{0.11}{1 - 0.14} \\ && \\ &=& 0.01791 \end{eqnarray*}

Straight tricast on $H_5$, $H_3$ and $H_2$:

\begin{eqnarray*} P\left(\mbox{$W_5$ and $S_3$ and $T_2$}\right) &=& P\left(W_5\right) \dfrac{P\left(W_3\right)}{1 - P\left(W_5 \right)} \dfrac{P\left(W_2\right)}{1 - P\left(W_5 \right) - P\left(W_3 \right)} \\ && \\ &=& 0.14 \times \dfrac{0.11}{1 - 0.14} \times \dfrac{0.34}{1 - 0.14 - 0.11} \\ && \\ &=& 0.008118 \end{eqnarray*}

$H_4$ comes first or second:

\begin{eqnarray*} P\left(\mbox{$H_4$ is $1^{st}$ or $2^{nd}$}\right) &=& P\left(W_4\right) + \sum_{\substack{1\leq m \leq 6 \\ m\neq 4}}{\left(P\left(W_m\right) \dfrac{P\left(W_4\right)}{1 - P\left(W_m\right)}\right)} \\ && \\ &=& 0.07 + P\left(W_1\right) \dfrac{0.07}{1 - P\left(W_1\right)} + P\left(W_2\right) \dfrac{0.07}{1 - P\left(W_2\right)} \\ && \qquad + P\left(W_3\right) \dfrac{0.07}{1 - P\left(W_3\right)} + P\left(W_5\right) \dfrac{0.07}{1 - P\left(W_5\right)} \\ && \qquad + P\left(W_6\right) \dfrac{0.07}{1 - P\left(W_6\right)} \\ && \\ &=& 0.07 + 0.29 \times \dfrac{0.07}{0.71} + 0.34 \times \dfrac{0.07}{0.66} \\ && \qquad + 0.11 \times \dfrac{0.07}{0.89} + 0.14 \times \dfrac{0.07}{0.86} \\ && \qquad + 0.05 \times \dfrac{0.07}{0.95} \\ && \\ &=& 0.1584 \end{eqnarray*}

  • $\begingroup$ Wow! Thank you so much, beautiful explanation... I have a lot to learn. About straight forecast and straight tricast, it was my idea but I don't know how to transform that concept in a formula, also because I'm not so familiar with this kind of "renderer" from text to formula. Maybe you can help as I see you're a master. It's all clear, except for place and show (ehehe). Start from place bet... does the sum of all that parameters exceed 1 (that is the maximum)? If I want to translate that formula into an algorithm, they would be two nested for loop? $\endgroup$
    – Snake
    Jul 13 '14 at 17:04
  • $\begingroup$ I forgot a question. What do you mean with ampersand? (&) $\endgroup$
    – Snake
    Jul 13 '14 at 17:46
  • $\begingroup$ Hi @SnakeX. "&" means "and" but I've replaced it now anyway. I've added formulas for forecast and tricast. I've worked through some of the formulas using your example. I also simplified the formulas for $(a)$ and $(b)$ a little. They give the same value as before but are simpler. To answer your questions: (a) None of the sums should exceed $1$. The algorithm for (a) (i.e. place bet) will now be one loop, the algorithm for (b) (i.e. show bet) will be two nested loops. $\endgroup$
    – Mick A
    Jul 14 '14 at 13:24
  • $\begingroup$ Ok, I've implemented your explanation as an algorithm. A strange thing it's happening... odds generated from probabilities are correct, but the probability sum for each type of bet exceed 1. The result I have is: sum 1 for winners, sum 2 for place bet, sum 3 for show bet, sum 1 for straight forecast, sum 2 for reverse forecast, sum 1 for straight tricast, sum 6 for combination tricast. I already double-checked the algorithm... $\endgroup$
    – Snake
    Jul 14 '14 at 19:32
  • $\begingroup$ You would expect that to happen. (Sorry, I think I misunderstood your earlier question about the sums.) E.g. two horses will place in the race so those probabilities won't add to $1$. P($H_1$ places) really is counting all possible str. forecasts with $H_1$ involved: 12, 13, 14, 15, 16, 21, 31, 41, 51, 61. If you do that for all 6 horses, every str. forecast will appear exactly twice. (13 for both $H_1$ and $H_3$). So it should sum to $2$. Similarly, for show bets: it counts all str tricasts with the given horse involved and overall each tricast is counted 3 times (254 for horses 2,4,5). $\endgroup$
    – Mick A
    Jul 14 '14 at 21:53

This is fine if you are dealing with dice or a specific cards out of a deck. Unfortunately, when handicapping horses, The odds of winning do not necessarily reflect the odds of finishing second or third. I love calculating odds, but there are reasons this will not work with horses.

Imagine a 10 horse race where the odds of winning on numbers 1 and 2 are both 5-2. Horses 3 and 4 are both 7 to 1.

This would make one think that the odds of the exact of being 13, 14, 23, and 24 would be equal.

However, this is horse racing. What if the numbers 1 and 3 are speed horses, and numbers 2 and 4 are very late running closers.

1 and 3 want a slow pace. 2 and 4 need a fast pace.

13 has a better chance than 14, and 24 better than 23.

With horses, the pace can make things great for a horse or knock him out of contention, and often, if you tell me a certain horse is going to win, I can say with greater accuracy which horse is going to be second. If you told me that the 2 horse had won this race, and I knew nothing else aside from the 1&3 horses being speed horses and the 2&4 being closers, odds are the 4 had a better chance for a second then the 3.

This is where handicapping comes into play as opposed to using pure mathematics in casino games.


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