Probability of Dodgeball


New member
Mar 24, 2017

I have a Probability question that I'm struggling to answer.

Let's say you have a game of Dodgeball. Every team has 6 players. The winning team is the team who eliminates the other team first.

Let's assume that at any stage of the game the next player to be removed from the game has an even chance of being from either team. What I am looking for is a formula that can work out what the probability a team will win the game at any particular position.

So for example, lets say there is 1 player remaining on Team A and 2 players remaining on Team B. So team A needs to remove the next 2 players. Team B just needs to remove the 1 remaining Team A player any time in the next 2 eliminations. So to step out the possibilities:
* Team A Wins if both eliminations are Team B players = 50% * 50% = 25%
* Team B Wins if first elimination is Team A or 2nd Elimination is Team A = 50% + 25% = 75%

However, this gets very complicated when there is more players in the game. I am sure there is a formula that can be used to calculate the probability given how many players still remain on both sides.

And one more request, how would this formula change if eliminations weren't always 50/50. For example, how would the formula handle it if Team A always had a 60% of getting the next Elimination and Team B only had a 40% chance.

Any help is much appreciated.