The way the game works is you are given natural numbers from 1 - 10, and one of these numbers is selected. You have to then guess whether the next randomly selected number will be higher or lower than the previous one, hence the name Hi-Lo.
For example, first number is 6. Choose "Lower". Next number is 3. Here, I am correct [because 3<6]. I choose "higher". Next number is 2. I am wrong. And so on.
My question is, given that I will always pick the more probable solution [aka if number is 8, I obviously will pick a number lower than 8 because my chance of being correct is higher], what will my success/fail ratio look like?
As an extension, what would success/fail ratio look like for numbers selected between 1 - n, where n is a natural number?
Note: I am a year 12 high school student so if possible please try to keep explanation simple.
Notice that we have independence here so the "success to fail ratio", if we carry on this experiment in infinitum, is the same as in an individual trial. Assuming that we lose in the case of a tie, the chances of winning given that the first number is $1$ is $\frac{9}{10}$, $2$ is $\frac{8}{10}$, $3$ is $\frac{7}{10}$, $4$ is $\frac{6}{10}$, $5$ is $\frac{5}{10}$, $6$ is $\frac{5}{10}$, $7$ is $\frac{6}{10}$, $8$ is $\frac{7}{10}$, $9$ is $\frac{8}{10}$, and $10$ is $\frac{9}{10}$.
Each of these numbers are equally likely to be selected initially so we just average these probabilities:
$$\frac{\frac{9}{10}+\frac{8}{10}+\frac{7}{10}+\frac{6}{10}+\frac{5}{10}+\frac{5}{10}+\frac{6}{10}+\frac{7}{10}+\frac{8}{10}+\frac{9}{10}}{10}=0.7$$
Or in terms of "success to fail ratio", we have $7$ successes for every $3$ losses giving $\frac{7}{3}$.
In general, for even $n$ the probability is
$$\frac{2\cdot\displaystyle\sum_{i=\frac{n}{2}}^{n-1}i}{n^2}$$
In general, for odd $n$ the probability is
$$\frac{\left\lfloor{\frac{n}{2}}\right\rfloor+2\cdot\displaystyle\sum_{i=\left\lceil\frac{n}{2}\right\rceil}^{n-1}i}{n^2}$$
or more simply
$$\frac{\frac{n-1}{2}+2\cdot\displaystyle\sum_{i=\frac{n+1}{2}}^{n-1}i}{n^2}$$
The success to fail ratio would just be $\frac{p}{1-p}$