Consider a pack of $10$ cards marked $1,2,3,3,4,4,5,6,7,8$ ($3$ and $4$ are repeated). There are two players that chose $3$ cards each (a total of $6$ cards without replacement are drawn). The player with the lowest total wins and the loser pays the winner an amount equal to the smallest card the winner is holding. If the totals are equal there is no winner. So if player $A$ has drawn $2,3$ and 5 and player $B 1,4$ and $7, A$ is the winner and B pays her an amount of 2. Let $Y$ be the amount that $A$ wins or loses by after each turn. Find the distribution of $Y$ .
Continuing from the previous question, the rules are modified a bit. Player $A$ has the right to choose one of the three cards in advance and the other $5$ cards are drawn from the remaining pack. Choosing $1$ is tempting because it maximises the chances of winning but then every time she wins she will only win by 1. She is thinking of choosing card 2 as the chances of winning are still good and the winning amount will be 2 more often than 1. Also the possible losing payments will be a bit lower. Can she do even better? The game is neither fair nor symmetric any more.
I am pretty new to probability and stochastic processes. I came across this problem and am not sure on how to go about it. Any thoughts and hints will be appreciated. Thanks.
Edit: I am trying to write a Matlab code for the question but I am stuck on how to go ahead with the code. From what I can understand, there will be around $4200$ cases to check which cannot be done mathematically.