50 player roll-off with favored players

Question

I'm trying to pick $x$ options out of a large number $N$, options by assigning each option a random integer weight in $[1,A]$, and then choosing the options with the $x$ highest values. Some of the options are favored and have an additional $y$ added, such that their available weights are between $[1+y,A+y]$. I am interested, primarily in finding out how the probability to choose a given option (unfavored or favored) changes if I change the weight $y$, but first I need to actually find the base probability for for each option. The best way I can visualize this is by making it into a simple dice game below, where I chose some concrete values for the variables above.

$n$ players $P$ each roll a $300$-sided die. $m$ players $H$ of $n$ have been given a handicap where they add $50$ to their die roll. What is the probability for each player ($H$ or $P$) to roll one of the five highest numbers?

I can brute-force solve a problem with two players $P_1$ and $P_2$. If we only have one winner, then we can look at each number rolled by one player and see how many rolls the other player has that would beat it: $$P(P_1>P_2) = \frac{1}{300}\left(\frac{299}{300}+\frac{298}{300}+\frac{297}{300}+...\right)=\frac{1}{300}\sum_{i=1}^{300}\frac{i-1}{300}=\frac{299}{600}$$ $$P(P_1=P_2) =\frac{1}{300}\sum_{i=1}^{300}\frac{1}{300}= \frac{1}{300}$$ Similarly for two players $P_1$ and $H_1$, except now $H_1$ always beats $P_1$'s lowest 50 rolls: $$P(H_1>P_1) = \frac{1}{300}\left(1+1+...\frac{299}{300}+\frac{298}{300}+...+\frac{50}{300}\right)=\frac{1}{300}\left(50+\sum_{i=51}^{300}\frac{i-1}{300}\right)=\frac{469}{720}$$ $$P(H_1=P_1) =\frac{1}{300}\sum_{i=51}^{300}\frac{1}{300}= \frac{1}{360}$$ $$P(H_1<P_1) = 1-P(H_1>P_1)-P(H_1=P_1)=\frac{249}{720}$$ Next, I need to expand this to more than two players. With three players, I can phrase "roll one of the two highest numbers" as "do not roll the lowest number". To start, if $P_1$ rolls 300, then they can never roll the lowest number, if $P_1$ rolls $298$, then $P_2$ and $P_3$ need to roll some combination of $299$ and $300$, of which there are four combinations. So if we follow that: $$P(P_1<P_2\land P_1<P_3)=\frac{1}{300}\Bigl(0+\bigl(\frac{1}{300}\bigr)^2+\bigl(\frac{2}{300}\bigr)^2+\bigl(\frac{3}{300}\bigr)^2+...\bigl(\frac{299}{300}\bigr)^2\Bigr)$$ $$=\frac{1}{300}\sum_{i=1}^{300}\left(\frac{i-1}{300}\right)^2$$ $$P(P_1=P_2\land P_1=P_3)=\frac{1}{300}\sum_{i=1}^{300}\left(\frac{1}{300}\right)^2$$ Expanding to "don't roll the lowest of $n$ players": $$P(P_1<P_{2..n})=\frac{1}{300}\sum_{i=1}^{300}\left(\frac{i-1}{300}\right)^n$$ Including $m$ players $H$ with $n$ total players: $$P(P_1<P_{2..n-m}\land P_1<H_{1..m})=\frac{1}{300}\left(\sum_{i=1}^{300}\frac{i-1}{300}\right)^{n-m-1}\left(50+\sum_{i=51}^{300}\frac{i-1}{300}\right)^m$$ I'm having difficulty scaling this up from "don't roll the lowest" to "roll one of the $x$ highest" or "do not roll one of the $y$ lowest", as well as finding the probability for $H_m$ to do the same.

Answer 1

Here's something to get you started:

• Use the binomial distribution to figure out how many favored players rolled above $A$. These players cannot be beaten by any non-favored player. If this is greater than or equal to $x$, all $x$ winners will be in this set.
• Otherwise, all of the players above will be among the $x$ winners, and there will be some remaining winner places. Use the binomial distribution to figure out how many non-favored players rolled below $y$. For $x \leq m$ these players cannot win and can be eliminated from consideration.
• Conditional on the above, the remaining players (favored or not) all rolled between $y$ and $A$, and are equally likely to take the remaining winner places.

Note also that if ties are broken uniformly at random, this is equivalent to making the dice continuous distributions rather than discrete, which considerably simplifies the problem.