One way to think about options is simply to model them as a binary game where there are two players, and only
two outcomes, winning (W) or losing (L):
Or to put it another way, which will be useful shortly, we can write:
\[P(\text{winning}) = 1 - P(\text{losing})\]Now let us consider a series of games, where the outcome is denoted by a W for a win and a L for a loss, and for simplicity’s sake let us consider a fair coin toss as the game, and for now only do 1 trial (i.e. $n = 1$). We can then easily calculate the $2^n = 2^1 = 2$ outcomes and visualize them as strings:
We can easily see there is only 1 outcome where we win out of the $2^n = 2^1 = 2$ total outcomes, and hence the probability of “winning at least one game” is:
\[P(\text{winning at least once}) = \frac{1}{2}\]Now let us repeat the game for 2 trials (i.e $n = 2$), which creates $2^n = 2^2 = 4$ outcomes:
\[\{WW, WL, LW, LL\}\]We can see that, in the scenario where we are only concerned with “winning at least one game” then we have 3 different ways to “win” out of the $2^n = 2^2 = 4$ total outcomes:
\[\{WW, WL, LW\}\]The probability of “winning at least one game” is then:
\[P(\text{winning at least once}) = \frac{3}{4}\]If we repeat the same series of games but use 3 trials (i.e. $n = 3$), we now have $2^n = 2^3 = 8$ outcomes:
\[\{WWW, WWL, WLW, WLL, LWW, LWL, LLW, LLL\}\]We can see that, in the scenario where we are only concerned with “winning at least one game” then we have 7 different ways to “win” out of $2^n = 2^3 = 8$ total outcomes:
\[\{WWW, WWL, WLW, WLL, LWW, LWL, LLW\}\]And the probability of “winning at least one game” is then:
\[P(\text{winning at least once}) = \frac{7}{8}\]So in general, for this fair toss game the ways in which we can “win at least one game” in a series of games (i.e. multiple trials) follows this pattern:
\[P(\text{winning at least once}) = \frac{2^n - 1}{2^n}\]which simplifies further:
\[\begin{align*} P(\text{winning at least once}) &= \frac{2^n - 1}{2^n} \\ &= \frac{2^n}{2^n} - \frac{1}{2^n} \\ &= 1 - \left(\frac{1}{2}\right)^n \end{align*}\]This formula can the be further abstracted for the probability of the fair coin toss game (i.e. $p_l = \frac{1}{2} $) and we are back to our starting formulation:
\[\begin{align*} P(\text{winning at least once}) &= 1 - \left(\frac{1}{2}\right)^n \\ &= 1 - p_l^n \\ &= 1 - P(\text{losing})^n \end{align*}\]and we can now see that the starting formula is just a special case of $n = 1$:
\[\begin{align*} P(\text{winning}) &= 1 - P(\text{losing})^n \\ &= 1 - P(\text{losing})^1 \\ &= 1 - P(\text{losing}) \end{align*}\]But what about games where each game in the series (i.e. trial) has a different probability:
\[p_i \neq p_{i+1} \neq \ldots \neq p_n\]We can generalize the formula to work with variable probabilities as follows:
\[\begin{align*} P(\text{winning}) &= 1 - \prod_{i=1}^{n} P(\text{losing}) _ i \\ &= 1 - \prod_{i=1}^{n} p_i \end{align*}\]Where:
i-th game out of n total gamesAs the number of games n approaches a greater and greater number, the player’s advantage (i.e. the player’s probability of winning at least one game) approaches $1$ or 100% (i.e. they are guaranteed to win at least one game):
What we can see from the above formulas is that the more options your or your oponent possess, the greater their advantage (i.e. the higher probability they will win at least one game). Even if the $p_l$ is high (meaning a high probability of losing), the more options added, the more the $p_w$ (probability of winning at least once) will approach unity (i.e. 100%).