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There are 3 players and one dealer in a casino. The dealer chooses a player randomly($p_1=\frac{1}{3}$). The chosen player tosses a coin($p_2=\frac{1}{2}$).

If the coin lands head, the chosen player will get 3 dollar, the dealer and the other two players will lose 1 dollar each.
If the coin lands tail, the chosen player will lose 3 dollar, the dealer and the other two players will get 1 dollar each.

This game repeats $n$ times. Let $x_{1n},x_{2n},x_{3n}$ be the total net profit(or loss) of players and let $y_n$ be the total net profit(or loss) of the dealer.

Let $g(n)=prob(|y_n|>max(|x_{in}|) )$

For example, $g(1)=0, g(3)=\frac{1}{18}$.

Prove: $g(n)$ is increasing as odd number $n$ goes $n+2$

comments: This question is similar with this question. a coupling probability problem and random walk game Only difference is $|y_n|>max(|x_{in}|)$ condition instead of $y_n>max(x_{in})$

martian03
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  • How far have you gotten with this? – saulspatz Feb 08 '21 at 14:35
  • |y|>|x| doesn't mean "y is the biggest winner or the biggest loser". For example, y=2 and x1=-3. x2=-3, y is biggest winner but |y|<|x| – martian03 Feb 08 '21 at 15:26
  • My bad. Don't have my head on straight yet this morning, I guess. Did you dispose of the "increasing" part of the prior question? – saulspatz Feb 08 '21 at 15:31
  • I think the previous question was solved by user's answers already. I am waiting to confirm the answer because of some reason. You can read the comments of the answers of the prior question. – martian03 Feb 08 '21 at 16:51

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