1
$\begingroup$

Suppose two people are playing a game with an infinitely generous agent who gives 1 coin at random between the two players.

The agent gives 1 coin to each player at the start. And then the agent proceeds to give coins randomly to each player with a probability proportional to how many coins they have already received.

So, for example: Players A and B start with 1 coin each. The agent assigns a coin to A and B with probability 1/2. Say the coin is given to B. Now B=2, A=1. Now the agent then gives the coin to B with probability 2/3, and A 1/3. And we start this infinite process.

My question is: Is one player 'expected' to run away from the other as the number of iterations approaches infinity? Or are they expected to switch places an infinite amount of times?

I'm 95% sure that they eventually run away from each other, since the expected value of the ratio B/A is still B/A after one iteration. But I'm looking for direction to prove this in a more rigorous way. Not really too sure where to start.

  • 0
    If you define $Z(t)=\frac{\mbox{Num coins $A$ has at time $t$}}{\mbox{Total num coins at time $t$}}$ for $t \in \{0, 1, 2, ...\}$ then you can say something about a bounded martingale. I'm sure you have such a theorem at your disposal. – 2017-02-12

1 Answers 1

0

Let $C_t=|A_t-B_t|$, and see that $\mathbb{E}[C_{t+1}^2|\mathcal{F}_t]=\left(C_t+1\right)^2\frac{t+1+C_t}{2(t+1)}+\left(C_t-1\right)^2\frac{t+1-C_t}{2(t+1)}$. To see that the probabilities are correct, assume $A_t\leq B_t$, so that $B_t-A_t=C_t$, $A_t+B_t=t+1$, so that $B_t=\frac{C_t+t+1}{2}$. This simplifies to $\mathbb{E}[C_{t+1}^2|\mathcal{F}_t]=C_t^2\frac{t+3}{t+1}+1$. We can then prove by induction that $\mathbb{E}[C_t^2]=\sum_{i=3}^{t+1}\frac{(t+2)(t+1)}{(i+1)(i)}=\frac{(t+1)(t+2)}{3}-(t+1)$. Notice that this implies that $\mathbb{E}\left[\frac{C_t}{t+1}\right]\leq\frac{1}{\sqrt{3}}$. You can similarly show that $\mathbb{E}[C_{t+1}]=\frac{t+2}{t+1}\mathbb{E}[C_t]+\mathbb{P}[C_t=0]$, and therefore that $\mathbb{E}\left[\frac{C_t}{t+1}\right]\geq \mathbb{E}[\frac{C_2}{3}]=\frac{1}{3}$. Therefore the absolute difference in proportions can be seen to grow linearly in $t$, with linear constant varying between $\frac{1}{3}$ and $\frac{1}{\sqrt{3}}$. So although $A_t$ and $B_t$ diverge, one doesn't quite dominate in the long term.