Example 1.11 (page 10) in Ross's book(Introduction to Probability Theory)

Question

I do not understand the solution to Example 1.11 on page 10.

There are r players, with player i initially having $n_i$ units,$n_i$>0 and i = 1,...,r. At each stage, two of the players are chosen to play a game, with the winner of the game receiving 1 units from the loser. Any player whose fortune drops to 0 is eliminated, and this continues until a single player has all n = $\sum_{i=1}^r n_i$ units, with that player designated as the victor. Assuming that the results of successive games are independent, and each game is equally likely to be won by either of its two players, find the probability that player i is the victor.

Here is my tuition.

1. It seems there shall be more rounds than n for a player to be the victor.
2. All players have the odds to be the victor. People with less units is more likely to lose, and people with more units is more likely to win.

Solution

To Begin, suppose that there are n players,with each player initially having 1 units. Consider player i. Each stage she plays will be equally likely to result in her either winning or losing 1 unit, with the results from each stage being independent. In addition, she will continue to play stages until her fortune becomes either 0 or N. Because this is the same for all players, it follows that each player has the same chance of being the victor. Consequently, each player has player probability $1/n$ of being the victor.

Now, suppose these n players are divided into r teams, with team i containing $n_i$ players,i = 1,...,r. That is, suppose players 1,...,$n_i$ constitute team 1, players $n_1 + 1$,...,$n_1+n_2$ constitute team 2 and so on. Then the probability that the victor is a member of team i is $n_i/n$. But because team i initially has a total of fortune of $n_i$ units, i = 1,...,r, and each game played by members of different teams results in the fortune of the winner's team increasing by 1 and the of the loser's team decreasing by 1, it is easy to see that the probability that the victor is from team i is exactly the desired probability. Moreover, our argument also shows that the results is true no matter how the choice of the players in each stage are made.

Thanks in advance

Answer 1

Something I did not understand on first reading is that in Ross's solution, when the $n$ players are divided into $r$ teams, each player is given one unit.

Here is a small example using Ross's reasoning which may help. Let's say there are just two players, $A$ and $B$: $A$ has $2$ units and $B$ has $3$ units. We imagine that there are actually $5$ players, each with one unit. We put imaginary players 1 and 2 on "team $A$" and imaginary players 3, 4 and 5 on "team $B$". We let the teams play until there is a final victor. By symmetry, each of the five imaginary players is equally likely to emerge as the final victor. So the probability that the victor is on team $A$ is $2/5$, and the probability that the victor is on team $B$ is $3/5$. But a win by an imaginary player on team $A$ corresponds to a win by the original player $A$, and similarly for $B$. So the probability that $A$ will win is $2/5$, and the probability that $B$ will win is $3/5$.

The part about a win by an imaginary player on team $A$ corresponding to a win by player $A$ is a bit tricky. In the "imaginary teams" game, it is possible that two players on team $A$ could be chosen to play a game. This never happens in the "real game", but since the net effect is zero in this case, it doesn't matter.