Expected number of rallies to win a game

[musicgold]

Hi,

I am not sure I am solving this problem the correct way, as my answer seems too low.

Problem: In squash (a two player racquet game), the player who gets 9 points first wins the game. Whoever wins a rally gets the point, irrespective of who served the ball. I am ignoring other rules for the time being. If player A has the 60% chance of winning a rally, how many rallies, on average, will it take her to win a game?

Simplified problem: I assumed that it takes only 3 points, instead of 9, to win a game. Then I drew a decision tree and assumed that on average X number of rallies are required for player A to win a game. Based on that, I got the following four states, where W = won, L= lost.

W W W – Rallies required to win the game= 3

W W L - Rallies required = X + 1 - 2 = X -1
(subtracting 2, to account for 2 points already won, adding 1, to account for the wasted rally)

W L - Rallies required = X + 1 – 1 = X

L - Rallies required = X + 1

Then I got the following equation.

X = 3 * (0.6)^3 + (X-1) * 0.6^2 * 0.4 + X * 0.6 * 0.4 + (X+1) * 0.4

X = 0.784 X + 0.904

X = 4.19 -The Average number of rallies to win 3 points for Player A.

Is this a correct way of looking at this problem?

Thanks,

MG.

 

[tiny-tim]

Hi musicgold!

oooh, you're making this far too complicated …

especially if it was 9 instead of 3 …

Hint: if the game is won on the (n+1)th rally, and needs k+1 rallies to win, then what can you say about the first n rallies?

 

[musicgold]

tiny-tim,

Thanks. I am still not able to figure it out.

 

[tiny-tim]

If the game is won on the (n+1)th rally,

then exactly k of the first n rallies must have been won (and n-k must have been lost),

and the probability of that is … ?

 

[musicgold]

tiny-tim,

Thanks. Are you pointing me to the Binomial formula?

I think there is a problem with this approach. In the table below, the chance of A winning the game on the third rally is 0.216. 'Combin' means the combination term, and 'Term2' is the product term in the binomial formula. The problem starts on the second row. The four comibations in this case also includes the case WWWL, which was accounted for in the first row. Also the third row includes the combinations on the rows 1 and 2.

Rallies	Won-A	lost-A	Combin	Term2	Chance of winning
3       	3    	0      	1    	0.216	  0.216
4       	3    	1      	4    	0.0864	  0.3456
5       	3    	2      	10    	0.03456	  0.3456

Also, we can't add these probabilities, as they represent different events. While 0.216 represents the chance of A wining in 3 rallies, 0.3456 on the second row, represents the chance of winning 3 rallies in four rallies.

 

[tiny-tim]

The four comibations in this case also includes the case WWWL …

I don't really understand what you're doing here …

for 4 rallies, the last one must be a W …

so just count the combinations for the first 3 places (which must be 2 Ws and 1 L) …

as I said before, if the game is won on the (n+1)th rally, and needs k+1 rallies to win, then what can you say about the first n rallies?

 

[musicgold]

if the game is won on the (n+1)th rally, and needs k+1 rallies to win, then what can you say about the first n rallies?

The chance of that occurring = 0.6^(K+1) x 0.4^(n-K)

Rally	Win	Lose	Chance
3	    3  	0		0.216
4	    3  	1		0.086
5	    3  	2		0.035

 

[tiny-tim]

The chance of that occurring = 0.6^(K+1) x 0.4^(n-K)

No, that's the chance of one way of finishing on the (n+1)th rally, isn't it? …

but there are … how many ways?

 

[musicgold]

That means we are using the combination formula to find the other cases. And that is what the first term in the Binomial formula do.

For example, a win in 4 rallies can occur as any of the following.

1. LWWW
2. WLWW
3. WWLW
4. WWWL

However, the fourth case doesn't make any sense, because in that case the game would have been over on the 3rd rally.

 

[tiny-tim]

That means we are using the combination formula to find the other cases. And that is what the first term in the Binomial formula do.

For example, a win in 4 rallies can occur as any of the following.

1. LWWW
2. WLWW
3. WWLW
4. WWWL

However, the fourth case doesn't make any sense, because in that case the game would have been over on the 3rd rally.

No … use the combination formula for the first 3 rallies, not all 4.

 

[musicgold]

Thanks.

I get it now.

3 rallies - possible combinations 2C2 =1, probability = 0.216

4 rallies - 3C2 = 3, proability =0.259

5 rallies - 4C2 = 6 , probability = 0.207


However these probabilities are not from the same event, and so I can't use them as weights to calculate the expected number of rallies to win a game.

 

[tiny-tim]

Just got up … :zzz:

Thanks.

I get it now.

3 rallies - possible combinations 2C2 =1, probability = 0.216

4 rallies - 3C2 = 3, proability =0.259

5 rallies - 4C2 = 6 , probability = 0.207

That's right!

So the general formula for n+1 rallies is … ?

However these probabilities are not from the same event, and so I can't use them as weights to calculate the expected number of rallies to win a game.

What event?

There's no double-counting, so just plug the probabilities into the expectation formula.

 

[musicgold]

tiny-tim

Thanks for working with me on this. I never thought that the binomial formula can be used for situations in which events occur sequentially.

The probability of winning a game on n+1^th rally and k^th point

= ⁿ C _k-1 * 0.6 ^k * 0.4^n+1-k

One needs to get probailities for rallies between k and 2k-1.

Note that these probailities can't be directly used with the rally numbers to calculate the expected value. They need to be weighted based on the probability of player A winning the game.


As a result, the expected number of rallies to for A to win a 9 point game is 13.80 or 14.

 

[tiny-tim]

The probability of winning a game on n+1^th rally and k^th point

= ⁿ C _k-1 * 0.6 ^k * 0.4^n+1-k

One needs to get probailities for rallies between k and 2k-1.

Yes.

Note that you can simplify it to 0.4^{n n} C _k-1 (0.6/0.4)^k-1

Note that these probailities can't be directly used with the rally numbers to calculate the expected value. They need to be weighted based on the probability of player A winning the game.

Not following you …

the only weight needed is (n+1)

As a result, the expected number of rallies to for A to win a 9 point game is 13.80 or 14.

that was sudden!

where did that come from?

 

[musicgold]

Thanks.

the only weight needed is (n+1)

This is what I understand. In the three point game, if you simply multiply the # of rallies with the chance of winning and add those terms you get 2.719 as the answer. That is not the correct answer.
You need to convert these chances as a % of 0.682. Then use those percentages to weight the rallies and you get the answer : 3.99 or 4.

Rallies	Chance of winning	Wt. Avg
3	      0.216	                 0.648
4	      0.259                    1.036
5	      0.207	                 1.035

Total      0.682	                 2.719

where did that come from?

I used the formula.

 

[tiny-tim]

You need to convert these chances as a % of 0.682. Then use those percentages to weight the rallies and you get the answer : 3.99 or 4.

oh of course!

i'd forgotten how the question was worded!

 

[musicgold]

tiny-tim,

I am trying to make the situation more realistic. In some countries, squash rules are a bit different, in that the server gets the point. For example, if Player A serves the ball and wins the rally, he gets the point. If Player A loses the rally, nobody gets a point, and Player B gets to serve, and so on. Again, the player who gets 9 points first wins the game.
Player A has a 60% chance of winning a rally, and we want to know the expected number of rallies to win a game for Player A.

Is it possible to find a solution to this problem the way we did it earlier or it gets too complicated?

Thanks,

MG.

 

[CRGreathouse]

Is it possible to find a solution to this problem the way we did it earlier or it gets too complicated?

It's possible, even easy. Just manipulate the probabilities until they match what you had before.

Before: probability p of winning a point, probability 1-p of the opponent winning a point.

Now: probability a of winning a point and 1-a of turning over the serve, at which point the probability is 1-b of the opponent winning a point and b of getting the serve back.

So p = a + (1-a)*b*a + (1-a)^2*b^2*a + ...

Let c = b(1 - a). Then p = a(1 + c + c^2 + ...) = a/(1-c) = a/(1 - b(1 - a)) = a/(1 + ab - b).

Now just solve the original problem with this value of p.

 

[musicgold]

tiny-tim,

I am kinda stuck with multiple questions and not able to think through this problem.

p = a/(1 + ab - b)

if a= 0.6, b =0.4 , then p= 0.714

Question 1 p=0.714. I understand the calculation part, but I find it a bit counter-intuitive that the chance of winning a point has increased even though the receiver has to struggle more to get a point, i.e he has to now win two consecutive rallies to earn a point.

If we use p and 1-p and calculate probabilities of winning a game as a certain combination of points, we the following table.

A's points	B's points	Total points     Combin	Factor 2	Probability

9	          0	          9	          1	         0.048400	 5%
9	          1	          10	         9	         0.013829	12%
9	          2	          11	         45	         0.003951	18%
9	          3	          12	         165	         0.001129	19%
9	          4	          13	         495	         0.000323	16%
9	          5	          14	         1287         0.000092	12%
9	          6	          15	         3003         0.000026	8%
9	          7	          16	         6435         0.000008	5%
9	          8	          17	         12870       0.000002	3%

					                                    Total          97%

Thus almost 50% of the time, A will win the game on between 11 and 13 total points (not rallies).

Question 2. I do not understand how the probability of A winning has increased to 97%.



Calculating the number of rallies to win a game.

Winning a point could involve playing multiple rallies, sometimes as the server and sometimes as the receiver.

As we are interested in the average number of rallies A needs to play to win a 9 point game. I used the decision tree method to find average number of rallies (x) to win a point.

x = 0.6 + ( 0.4 * 0.6 *( x+2 ) )
x = 1.08 / 0.76 = 1.42

Question 3. This answer (x=1.42) seems counter-intutive as it is lower than the 1.67 we got in the case of the game where either server or receiver could win a rally ( one rally per won or lost point).

Now if we assume that x=1.42 rallies per point is the correct answer, how do I incorporate to find the average number of rallies to win a game?

Question 4. Should I find the expected number of rallies to win a point for B as well, and use it along with X=1.42 to calculate the average number of rallies in each case in the table above? I don’t know if this is making any sense to you.

Thanks,

MG.

 

[musicgold]

CRGreathouse,

I am sorry. I missed that you had replied to my last question on Feb 2. I mistakenly assumed that tiny-tim had replied it and I addressed the message to him.

Do you have any pointers for the questions in my last message on Feb 4?

Thanks,

MG.