Can one player win more games while scoring fewer points over multiple trials a simple probabilistic game?

Question

The Game

Two players $P_{1}$ and $P_{2}$ will play the following game:

1. Three non-negative numbers $a$, $b$, and $c$ will be selected at random from a distribution unknown to either player. The players know only that the numbers will be non-negative.

2. Player $P_{1}$ will be given a choice of two options, after they make their selection the values of $a$, $b$, and $c$ will be revealed, and $P_1$ and $P_2$ will score points depending on the values of $a$, $b$, and $c$ as described below. ($P_2$ makes no choices.)

3. Player $P_{1}$'s two options are:

   • Option 1: Score points equal to $a+b$.
   • Option 2: Score points equal to the maximum value of $a$, $b$, or $c$.

   Note: Neither player knows the values of $a$, $b$, or $c$ at the time this choice is made.

4. Whichever option $P_{1}$ selects, $P_{2}$ will score the points of the other unselected option.
   For example, if:

   • $P_1$ selects Option 2
   • $a$, $b$, and $c$ turn out to be $1$, $17$, and $\pi$ respectively

   Then:

   • $P_1$ scores $17$ points (e.g. the maximum of $1$, $17$, and $\pi$)
   • $P_2$ scores $18$ points (e.g. $a+b = 1+17 = 18$)

5. The player who scores the most points wins the game. (Ties are also possible.)

The Scenario

Let's assume:

• The game will be played some number $N$ times, with the unknown distribution from which $a$, $b$, and $c$ are selected remaining fixed over these $N$ games.
• $P_1$ must select the same choice of Option 1 or 2 in each of the $N$ games. (e.g. they make only one choice at the beginning of Option 1 or 2, which is then applied in every game.)
• We tabulate both:
  • The number of times each player wins
  • The cumulative points scored by each player over the $N$ games

The Question

1. Does there exist:

   • A distribution of numbers from which $a$, $b$, and $c$ are selected
   • A choice by $P_1$, and
   • Sufficiently large $N$

Such that both:

• $P_1$ is expected to win at least as many games as $P_2$, and
• $P_1$ is also expected to score no more points than $P_2$

? (Thanks to Marcus M for answering this with a clear example!)

2. Does the answer change if all values must be positive, as opposed to non-zero?

My sense is that there is no such distribution possible when $a$, $b$, and $c$ are positive - but I'm not sure how to go about proving that?

It's fairly easy to construct distributions such that Option 1 or Option 2 both lead to $P_1$ expecting to win more games and scoring more points, for example:

• If $a$, $b$, and $c$ are always some permutation of 0, 0, and 1 then as $N$ approaches infinity:
  • Option 2 ties 2 games out of 3 and wins 1 game out of 3
  • Option 2 scores 1 point per game whereas Option 1 expects to score $2/3$ points per game
• If $a$, $b$, and $c$ are selected over a uniform continuous distribution then as $N$ approaches infinity:
  • Option 1 expects to win 5 games out of 6, and score $4/3$rds as many points as Option 2.

However I'm interested if it's possible to have distribution where there is a split outcome of winning more while also scoring less overall.

Thanks for reading!

Answer 1

I believe that it is indeed possible and the following distribution does it. Consider $a, b$ and $c$ are a permutation of $0, 1$ and $10$, with every permutation chosen with equal probability. If $P_1$ always chooses $a + b$, then the possibilities (with equal probability) are

• $c = 0,~a + b = 11$: $P_1$ scores $11$, $P_2$ scores $10$.
• $c = 1,~a + b = 10$: $P_1$ scores $10$, $P_2$ scores $10$.
• $c = 10,~a + b = 1$: $P_1$ scores $1$, $P_2$ scores $10$.

Then $P_1$ wins the first case, ties the second case, and loses in the third case, meaning that $P_1$ wins exactly as often as $P_2$. However, $P_1$'s expected score - $P_2$'s expected score is $\frac{1}{3}\cdot 1 + \frac{1}{3}\cdot 0 + \frac{1}{3}\cdot (-9) = \frac{-8}{3} < 0$. Thus $P_1$ has the same expected number of wins as $P_2$, but has a lower expected score than $P_2$.