[Guest]

Two tennis players, A and B, are playing a game. In that game, A has a 60% probability of winning any given point (with B obviously having a 40% probability). If the game gets to deuce, what is the probability that A will win the game?

I'm interested to see the reasoning behind your answer.

For anyone not familiar with tennis scoring, deuce works as follows: if the score is deuce, then whichever player wins the next point will get the advantage. If the player with the advantage wins the point after that they win the game, if they lose it the score goes back to deuce.

[Guest]

Player_A = .6 to win any point

Player_B = .4 to win any point

®2 possible outcomes from duece

---------------------------------

Player_A has initial advantage = .6

Now 2 more possible outcomes

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Player_A wins set = .6 (WIN_A) .6 * .6 = .36

Player_B forces duece = .4 (Return to ®) .6 * .4 = .24

Player_B has initial advantage .4

Now 2 more possible outcomes

----------------------------

Player_B wins set = .4 (WIN_B) .4 * .4 = .16

Player_A makes it duece = .6 (Return to ®) .4 * .6 = .24

(WIN_A) .36 = 36%

(WIN_B) .16 = 16%

® .24 = 24%

Therefore (WIN_A + R) + (WIN_B + R) = 1

.36 + .24 + .16 + .24 = 1

.6 + .4 = 1

1 = 1

No matter if R is the first time or the fiftyth time if we start at duece (which we do) Player_A has 36% chance of winning(which they do). Rinse and repeat. If we start at duece no matter how many times we get there the odds of winning do not change nor does the odds of Player A beating Player B. Player A has a 2.25 times greater of a chance to beat Player B: .36/.16 = 2.25 therefore roughly 69.2303/30.7697 = 2.24995 according to my Python program I wrote. Out of 1,000,000,000 games played where 923,044,191 dueces occurred. The logic's right my math is shady. Though Player A has only a 36% chance of winning, Player A will win 2.25 times more then Player B. Therefore out of 100% of the games won Player A will win 2.25 times more then player B

Therefore:

B(2.25) + B = 1

B(2.25 + 1) = 1

B(3.25) = 1

B = 1/3.25 = .30769230769230769230 = 4/13 = 30.7692%

therefore A = (1/3.25)2.25 = .69230769230769230769 = 9/13 = 69.2308%

Keep in mind Player A still only has a 36.00% chance of winning any given match; but, Player A will win 69.2308% of the time.

[Glycereine]

I think from a "duece" situation you are still left with 60/40. If we were to say that A had 100% chance of winning the next point, then you would say that they will have 100% chance of being the first to win two consecutive points. If (as with the coins) A has 50% chance of winning the next point then he will have 50% chance of being the first to win two consecutive points. Likewise if A has a 0% chance of winning the next point then he has 0% chance of being the first to win two consecutive points. Why does this logic not prevail for all other combinations of probabilities?

Because this isn't how probabilities work unfortunately. Your case with 100% and 0% are correct but only because they are 100 and 0. no other number would work with that reasoning.

To get something with X probability to happen twice, the probability is X*X.

So if A and B have a 50% chance of winning, A has a .5*.5 chance of winning the game (from deuce, 2 points needed)

B has a .5*.5 chance of winning from deuce.

and there is a .5*.5 + .5*.5 (A wins, B wins then B wins, A wins (wins a point not the game)) chance to return to deuce.

so you have 25% chance of A winning in 2, 25% chance of B winning in 2 and a 50% chance of the game going back to deuce.

As we've already proven above, from deuce A and B have the same probability of winning in the next 2 shots (the only other option is a return to deuce which is repeating).

So the probability of A winning and B winning (in 2 points) are both 25%. .25/(.25+.25) is A's probability of winning overall (eventually no matter how many times it returns to deuce) and .25(.25+.25) is B's probability of winning overall.

these are both 50%. However for no case OTHER than 0% 50% and 100% is the chance of winning the same as the chance of getting the next point.

For the case stated, P(A1) = Probability of A getting a point = .6

P(B1) = Probability of B getting a point = .4

The probability that A will get 2 points in a row is P(A2) = P(A1)*P(A1) = .6*.6 = .36

The probability that B will get 2 points in a row is P(B2) = P(B1)*P(B1) = .4*.4 = .16

The probability of a return to deuce is P(A1B1)+ P(B1A1) = P(A1)*P(B1) + P(B1)*P(A1) = .6*.4 + .4*.6 = .48

Return to deuce is the starting condition and represents the same probabilities again, therefore the probability of A winning (eventually)

P(A) = P(A2)/(P(A2)+P(B2)) = .36/(.36+.16) = .36/.52 = 9/13 or 69.23%

I may not have explained it as thoroughly in my first answer but that doesn't mean I wasn't confident with it.