In table tennis & tennis (among other sports?), a player/team must be two points ahead of their opposition to win a game. Thus, a game could technically go on for infinite deuces. However, assuming that the outcome of each rally is an independent event, with a 0.5 chance of either side winning; it is obvious that the expected number of rallies that a game will go for after a deuce is 4. This can be found (among other methods) through the equation:
a = [1/2 * 2] + [1/2 * (2+a)]
a/2 = 2
'a' is the expected number of turns that a game will go for after a deuce. A deuce has a 1/2 chance of finishing after two rallies (if both are won by either player) hence the term [1/2 * 2]. There is also a 1/2 chance of returning to deuce after two rallies, hence the term [1/2 * (2+a)].
The question arises: what would be the expected number of rallies needed to win after a deuce, if a player needed to be 3 points ahead (instead of 2)? This, too can be solved through an analogous method, yielding a result of 9.
The situation is slightly more complex if one needs to be 4 points ahead (denoted as a '4 point deuce'). This is because we must introduce a new variable, 'b', which is the expected number of rallies needed to win the deuce after one player is 2 points ahead.
We now have the equations:
a = 1/2 * (2+b) + 1/2 * (2+a) The trend seems to be that, for an 'n point deuce', the expected number of rallies needed to win it is n^2. The question is, can this be proven? b = 1/2 * (2+a) + 1/2 * 2
This yields a=16, b=12.
It seems that n^2 rallies are to be expected after an n point deuce. The question is, can this be proven?