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Match Equity brainteaser for the mathematically inclined

Posted By: AP
Date: Monday, 9 July 2012, at 2:40 p.m.

I have been pondering an interesting match-equity problem that is not backgammon specific, but still should be interesting to many of the forum readers. The notation I use below is P(-x,-y). This means the probability that Player 1 will beat Player 2 when Player 1 is x points away from victory, and Player 2 is y points away from victory.

Variant 1: Coinflipping tournament

Suppose we are having a coin flipping match where you win a flip if its heads and I win a flip if its tails. What is the generic formula to calculate P(-x,-y)? I know that the general formula for solving this formula is the binomial formula.

But what if we change the rules a bit:

Variant 2: Instead of a game with only 2 outcomes like the coin flip, say there are 4 outcomes:

i) Player A wins 2 points with probability p

ii) Player A wins 1 point with probability q

iii) Player B wins 2 points with probability p

iv) Player B wins 1 point with probability q

Is there a general formula for solving this type of formula?

For those that have an easier time with a concrete example, here goes: Suppose we are playing a game of pure chance where we each win a single game 30% of the time and a gammon win 20% of the time. The score is 40-away, 50-away. What is the probability that Player A, the one who is 40-away, will win?

I am looking for a general formula and just the answer generated by computer code. Thanks!

Messages In This Thread

• Match Equity brainteaser for the mathematically inclined
  AP -- Monday, 9 July 2012, at 2:40 p.m.
  • Match Equity brainteaser for the mathematically inclined
    Bob Koca -- Monday, 9 July 2012, at 3:21 p.m.
    • Match Equity brainteaser for the mathematically inclined
      AP -- Tuesday, 10 July 2012, at 5:02 p.m.
  • Match Equity brainteaser for the mathematically inclined
    Maik Stiebler -- Monday, 9 July 2012, at 3:36 p.m.