What fraction of the fund should one bet?

Question

Say we have a gambler who makes money through sports betting. My aim is to develop a model to help our gambler maximise his winnings and minimize losses.

In my model, rather than betting a fixed amount of money, the gambler bets a certain fraction $0 < r < 1$ of his current betting fund. He continues betting that fraction as his betting fund increases or decreases until he cashes out after a certain number of sessions $n$.

The gambler's initial fund shall be $F_0$. His fund after $i$ sessions shall be $F_i$.

His probability of making a correct prediction shall be $0 < p < 1$. If our gambler had a $p$ of $0$ or $1$, then the entire model would be useless.

The average odds with which our gambler deals with is $a > 1$.

The gambler's minimum desired profit upon cash out is $T$.

$$T \le F_n - F_0 \tag{1}$$

If we expressed everything as a multiple of $F_0$, $(1)$ can be rewritten as:

$$T \le F_n - 1 \tag{1.1}$$

It follows that the following are known: $T$, $a$, $F_0$, $p$.

Should our gambler lose a particular session say $i+1$,

$$F_{i+1} = (1-r)F_i \tag{2.1}$$

Should he win that particular session

$$F_{i+1} = F_i(1-r + ra) \tag{2.2}$$

Given that the gambler plays $n$ sessioms before cashing out.

His expected number of wins = $p*n$ $(3.1)$

His expected number of losses = $(1-p)*n$ $(3.2)$

Now there are many different ways to distribute the gambler's losses and wins{$n \Bbb P pn$} and while calculating all scenarios and finding average $F_n$ may be ideal, it is computationally very expensive. So I decided to model the problem assuming the losses take place in the worst way possible( back to back at the very beginning of the match).

The gambler's revenue after $n$ matches is given by the formula:

$F_n = (1-r)^{(1-p)n}\{(1-r)+ra\}^{pn}$ $(4)$

Now we know that our gambler wants to make a minimum profit of $T$ so we transform $(4)$ into an inequality using $(1.1)$

We get:

$(1-r)^{(1-p)n}\{(1-r)+ra\}^{pn}$ $ \ge T + 1$ $(4.1)$

Taking the Natural logarithm of both sides, I get:

$ln(1-r)*(1-p)(n) + ln(1-r + ra)*pn \ge ln(T+1)$ $(4.2)$

$n\{ln(1-r)(1-p) + ln(r(a-1)+1)(p) \} \ge ln(T+1)$ $(4.3)$

Giving the constraints on the variables and constants, I want to determine the minimum value of $n$ and maximum value of $r$ that satisfies $(4.1) / (4.3)$ (whichever is easier to solve) for any given $T$, $a$, $p$.

MAJOR EDIT

Thanks to @Rodrigo de Azevedo, I discovered Kelly's Criterion. I was sold on it, and decided to implement it into my gambling method.

For the purposes of my method Kelly's criterion is given by:

$r_i = p - $ ${1 - p}\over{a_i - 1}$ $(5)$

Where:

$r_i$ is the ratio at session $i$

$a_i$ is the odds at session $i$

Now $r: 0 \lt r \lt 1$ $(5.1)$

Applying $(5.1)$ to $(5)$ we get:

${p(a - 1) - (1 -p)}\over{a - 1}$ $ \gt \frac{0}{1}$

Cross multiply.

$p(a-1) - (1 - p) \gt 0(a-1)$

$pa - p - 1 + p \gt 0$

$pa - 1 > 0$

$pa > 1$

$p > 1/a$ $(5.2)$

Now that that's out of the way, we still have the problem of determining minimum $n$ such that we make a profit $ \ge T$.

In order to do this, we'll assume a "mean" value for $a$ then find the minimum value for $n$ that satisfies $(4.1)$

Due to the fact, that you do not know the odds for the matches in advance, your mean odds at $i$ say $a_{\mu i}$ may not be the mean odds at $n$ $a_{\mu n}$. In order to protect against this(and because I'm not a very big risk taker), I'll assume a value for $a_{\mu}$, that is less than $a_{\mu}$ called $a_{det}$.

$a_{det} = a_{\mu} - k\sigma$

Where $a_{\mu}$ is the Geometric Mean as opposed to the arithmetic mean of the odds and $\sigma$ is associated S.D

Using Chebyshev's Inequality, at least $k^{2} - 1 \over k^2$ of the distribution of the odds lie above $a_{det}$.

Picking a $k$ of $2.5$

$2.5^{2}-1\over 2.5^{2}$

$0.84$

So our $a_{det}$ is lower than at least $84$% of the distribution of the odds. This is safe enough for me.

$a_{det} = a_{\mu} - 2.5\sigma$

Using $a_{det}$, we'll calculate the minimum $n$ that satisfies $(4.1)$

Subbing $5$ and $a_{det}$ into $(4.1)$ we get:

$\left(1-\left(p - \frac{1-p}{a_{det}-1} \right) \right)^{n - np} \cdot \left(\left(p - \frac{1-p}{a_{det}-1} \right)\cdot(a_{det} - 1)\right)^{np}$ $ \ge T + 1$ $(6.0)$

This can be simplified further to: $\left({a_{det}-1-(pa_{det}-1)}\over{a_{det}-1}\right)^{n(1-p)}\cdot\left(pa_{det}-1+1\right)^{np}$

$\left({a_{det}-pa_{det}}\over{a_{det}-1}\right)^{n(1-p)}\cdot\left(pa_{det}\right)^{np}$

$\left(\left(\frac{a_{det}*(1-p)}{a_{det}-1}\right)^{n(1-p)}\cdot\left(pa_{det}\right)^{np}\right)$ $(6.1)$

P.S due to my particularly low $a_{det}$ we'll likely make much more profit than $T$, but that's loads better than choosing a higher $a_{det}$ and making less.

Answer 1

To find the minimum $n$ that satisfies $(6.1)$.

$\left(\left(\frac{a_{det}∗(1−p)}{a_{det}−1}\right)^{n(1−p)}\cdot\left(pa_{det}\right)^{np}\right) \ge T+1$ $(6.1)$

On the LHS, $n$ is a common exponent. Take the natural logarithm of both sides

$ln\left(\frac{a_{det}∗(1−p)}{a_{det}−1}\right)\cdot{n(1−p)} + ln\left(pa_{det}\right)\cdot{np} \ge ln(T+1)$

Factorise LHS with $n$.

$n\left(ln\left(\frac{a_{det}∗(1−p)}{a_{det}−1}\right)\cdot(1−p) + ln\left(pa_{det}\right)\cdot p\right) \ge ln(T+1)$ $(6.2)$

Rewriting $(6.2)$

$n \ge \left({ln(T+1)}\over{ln\left(\frac{a_{det}∗(1−p)}{a_{det}−1}\right)\cdot(1−p) + ln\left(pa_{det}\right)\cdot p}\right)$ $(6.3)$

Presto.

I realised this worked because Kelly's criterion aimed to maximise expected logarithmic growth.

Answer 2

Given

• odds $\omega_1, \omega_2, \dots, \omega_n > 1$.
• probabilities of winning $p_1, p_2, \dots, p_n \in [0,1]$.

let

• $X_0, X_1, \dots, X_n$ be random variables that denote the fund's size at step $k$.
• $u_k \in [0,1]$ be the fraction of the fund to be put at stake at step $k$. Let $u_k$ depend solely on $\omega_k$ and $p_k$, and not on $X_{k-1}$. Hence, $u_k$ is not a random variable.

Hence, we have the discrete-time stochastic process

$$X_k = \begin{cases} (1 + (\omega_k - 1) \, u_k) \, X_{k-1} & \text{with probability } p_k\\\\ (1 - u_k) \, X_{k-1} & \text{with probability } 1-p_k\end{cases}$$

---

Maximizing the expected return

The return at step $k$ is, thus,

$$R_k := \frac{X_k - X_{k-1}}{X_{k-1}} = \frac{X_k}{X_{k-1}} - 1 = \begin{cases} (\omega_k - 1) \, u_k & \text{with probability } p_k\\\\ - u_k & \text{with probability } 1 - p_k\end{cases}$$

Taking the expected value of the return, we obtain

$$\mathbb E (R_k) = (\omega_k - 1) \, u_k \, p_k - u_k \, (1 - p_k) = (\omega_k \, p_k - 1) \, u_k$$

Maximizing the expected value of the return,

$$\bar{u}_k := \arg \max_{u_k \in [0,1]} \mathbb E \left( R_k \right) = \arg \max_{u_k \in [0,1]} (\omega_k \, p_k - 1) \, u_k = \begin{cases} 1 & \text{if } \omega_k \, p_k - 1 > 0\\ 0 & \text{if } \omega_k \, p_k - 1 \leq 0\end{cases}$$

where $\omega_k \, p_k - 1$ is the expected profit per unit bet at step $k$. Thus, the optimal betting policy, $\bar{u}_k = \pi (\omega_k, p_k)$, is

$$\boxed{\pi (\omega, p) := \begin{cases} 1 & \text{if } \omega \, p > 1\\ 0 & \text{if } \omega \, p \leq 1\end{cases}}$$

In words,

• when the expected profit is non-positive, we bet nothing.
• when the expected profit is positive, we go all in.

Needless to say, this is an extremely aggressive betting policy. It would be wise to maximize another objective function.

---

Maximizing the expected logarithmic growth

Taking the expected value of the logarithm of the growth at step $k$,

$$\mathbb E \left[ \log \left( \frac{X_k}{X_{k-1}} \right) \right] = \mathbb E \left[\log (1 + R_k)\right] = p_k \log \left( 1 + (\omega_k - 1) \, u_k \right) + (1 - p_k) \log \left( 1 - u_k \right)$$

Using SymPy to find where the derivative with respect to $u_k$ vanishes,

>>> from sympy import *
>>> p, u, w = symbols('p u w')
>>> f = p * log(1 + (w-1) * u) + (1 - p) * log(1 - u) 
>>> diff(f,u)
  p*(w - 1)     -p + 1
------------- - ------
u*(w - 1) + 1   -u + 1
>>> solve(diff(f,u),u)
 p*w - 1 
[-------]
  w - 1  

Hence,

$$\bar{u}_k := \arg \max_{u_k \in [0,1]} \mathbb E \left[ \log \left( \frac{X_k}{X_{k-1}} \right) \right] = \begin{cases} \dfrac{\omega_k \, p_k - 1}{\omega_k - 1} & \text{if } \omega_k \, p_k - 1 > 0\\\\ 0 & \text{if } \omega_k \, p_k - 1 \leq 0\end{cases}$$

where $\omega_k \, p_k - 1$ is the expected profit per unit bet at step $k$. This is the Kelly betting policy [0]

$$\boxed{\pi (\omega, p) := \begin{cases} \dfrac{\omega \, p - 1}{\omega - 1} & \text{if } \omega \, p > 1\\\\ 0 & \text{if } \omega \, p \leq 1\end{cases}}$$

We again bet nothing when the expected profit is non-positive, but we no longer go all in when the expected profit is positive. Note that

$$\dfrac{\omega \, p - 1}{\omega - 1} = 1 - \left(\frac{\omega}{\omega - 1}\right) (1 - p) \leq 1$$

---

Reference

[0] Edward O. Thorp, The Kelly criterion in blackjack, sports betting, and the stock market, The 10th International Conference on Gambling and Risk Taking, Montreal, June 1997.