Introduction
Following the treatment in Dorhauer (2019), suppose we have a game where
- a team has \(p\) probability of winning a round of the game (aka hit-rate)
- \(q=1-p\) is the probability of losing a round of the game
- \(n\) is the number of independent rounds of the game
- \(r \in [0,\ldots,n]\) denotes the length of a streak
- An \(r\)-game losing streak means you lose \(r\) consecutive games.
How many \(r\) game windows are possible within \(n\) rounds?
To make this concrete, let \(r=3\) and \(n=10\). There are \(8\) different \(3\)-game windows where a streak could occur within \(10\) game rounds (games 1-3, 2-3,…,8-10). So there are \(n-r\) different \(r\) game windows possible within \(n\) rounds.
How many opportunities are there to start a winning streak?
In \(n\) rounds, any \(r\)-game winning streak will have to start by game \(n-r+1\) at the latest. Thus, in \(10\) rounds, any \(3\)-game winning streak will have to start by game \(8\) at the latest. Any loss in the first \(n-r\) games can potentially be followed by a \(r\)-game winning streak. For a \(p\) hit-rate team, we expect \((n-r)q\) losses over the first \(n-r\) games, plus the first game (which can always start a winning streak as there is no loss from a previous game to consider). That means a \(p\) hit-rate team will have, on average, about \(1+(n-r)q\) opportunities to begin a winning streak of at least \(r\) games.
Similarly, in \(n\) rounds, any \(r\)-game losing streak will have to start by game \(n-r+1\) at the latest. Any win in the first \(n-r\) games can potentially be followed by a \(r\)-game losing streak. For a \(p\) hit-rate team, we expect \((n-r)p\) wins over the first \(n-r\) games, plus the first game (which can always start a losing streak as there is no win from a previous game to consider). That means a \(p\) hit-rate team will have, on average, about \(1+(n-r)p\) opportunities to begin a losing streak of at least \(r\) games.
The expected number of losing streaks
The expected number of losing streaks is the product of the expected number of opportunities to start a losing streak times the probability of \(r\) consecutive losses, so \((1+(n−r)p)q^{r}\)
The probability of at least one losing streak
The probability of at least one losing streak is 1 minus the probability of \(0\) losing streaks. We first determine the probability of not completing a losing streak at every one of the opportunities we have for starting a losing streak. So, in other words, we will find the probability of having a winning streak at every one of these opportunities.
Now \(q^{r}\) is the probability of having a losing streak at any opportunity. Hence, \((1-q^{r})\) is the probability of having a winning streak at any opportunity. So, \((1-q^{r})^{1+(n−r)p}\) is the probability of having a winning streak at each and every one of the \(1+(n−r)p\) opportunities for starting a losing streak.
Finally, we have that \(1 - (1-q^{r})^{1+(n−r)p}\) is the probability of at least one losing streak.
The longest expected losing streak
From SoccerWidow (2016), the length of the longest expected losing streak is given by \(\frac{|ln(n)|}{|ln(q)|}\).
Similarly, the length of the longest expected winning streak is \(\frac{|ln(n)|}{|ln(p)|}\).
The number of rounds played to encounter a losing streak of a particular length
From SoccerWidow (2016), for a game where the probability of losing at any given round is \(q\), we expect to have encountered a losing streak of length \(r\) somewhere on or before \(\frac{1}{q^{r}}\) rounds.
Starting Bankroll
For strategies with a hit rate between 45% to 55% the starting bank roll should be (Length of maximum losing streak X planned stake per bet X 5)