# Can someone help me balance this game (probability question) [closed]

A team of 9 vs a team of 1. Each round each of "the 9" roll a die to "attack" and "the 1" rolls 9 dice to "defend", the nine dice are preassigned to attackers before the roll, "the 1" cannot choose afterward how to assign them. If the attacker die is higher than the corresponding defender die "the 1" loses a life. Ties always go in favour of "the 1".

Then "the 1" rolls 9 dice to attack and "the 9" roll a die each to defend again assigned before rolling. For each die "the 1" rolls that is higher than or equal to "the 9's" rolls that person loses one life if a member of "the 9" loses 3 lives they are out and can no longer roll, "the 1" also rolls one less die. If "the 1" or every member of "the 9" loses their lives the other team wins.

How many lives should "the 1" have to make this a fair game with a 50% chance of each team winning? The 9 always go first.

• Is it obvious that there is a whole number of dice making this game fair? Anyway, the fairest option is between 1 and 27 :-P – chi Apr 26 '16 at 18:48
• I mean fair in the sense that each side has an equal chance of winning. Obviously if the 1 has 1 life they'll lose almost every time and if they have 1000 lives they'll almost never lose. What's the breakeven point? – tomdemaine Apr 26 '16 at 18:51
• do you have your own thoughts on the problem to share? – gt6989b Apr 26 '16 at 19:00
• Only intuition that the team of 1 has the advantage of winning ties so will win slightly more often, it's 27 hits to take out everyone but the team of 9 goes first so I'd estimate something in the range of 30-40 – tomdemaine Apr 26 '16 at 19:08
• I forgot that "the 9" goes first, so it might be more than 27 indeed. However, my point was that it is possible that $n$ lives make "the 9" more likely to win, yet $n+1$ lives make "the 1" more likely to win. If so, there's no point at which the game is perfectly fair. I do understand that what you want in that case is to know $n$. – chi Apr 26 '16 at 21:12