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That formula works for n=2. But the game can go on more than 2 rolls. This one works for n <= 6:(\frac{\sum_{i=1}^n i}{n^2})
That formula works for n=2. But the game can go on more than 2 rolls. This one works for n <= 6:
(\frac{79n^4+10n^3+5n^2+50n-24}{120n^4})
but I don't see a pattern for the general solution.
The player who fails to improve upon the previous roll loses.
The way I read this, there can only be one round - a tie by the second player results in a loss as does every roll that is lower. The only alternative is when player 2 outright wins - binary event. Unless I'm misinterpreting?
The number of increasing sequences of elements of (\{1, ..., n\}) of length (k) and ending in (m) is (\binom{m-1}{k-1}) (pick the (k-1) elements less than (m) from (\{1, ..., m-1\}) and there's a unique way to order them). There are (m) ways in which the following roll won't improve on the last one. The probability of this scenario is then (\frac{1}{n^{k+1}}\binom{m-1}{k-1}m=\frac{k}{n^{k+1}}\binom{m}{k}).
The desired probability is then (\sum_{m=1}^n m\sum_{k \text{ odd}} \frac{1}{n^{k+1}}\binom{m-1}{k-1}).
To compute the inner sum, we write it as (\frac{1}{n^2}\sum_{k \text{ odd}} \frac{1}{n^{k-1}}\binom{m-1}{k-1}=\frac{1}{n^2}\left[\large(1+\frac{1}{n}\right)^{m-1}-\large(1-\frac{1}{n}\right)^{m-1}\right]).
The probability then becomes
(\frac{1}{n^2}\sum_{m=1}^n\left[m\large(1+\frac{1}{n}\right)^{m-1}-m\large(1-\frac{1}{n}\right)^{m-1}\right]).
To compute the latter two sums, note that they are both of the form (\sum_{m=1}^n mx^{m-1}), which is the derivative of (\sum_{m=1}^n x^m = \frac{x^{n+1}-x}{x-1}).