[superprismatic]

I used to like to go ten pin bowling quite a bit.

I remember that I scored 83 in my first ever game.

It occurred to me that I should compare how I did

in that game to the expected score in a type of

random bowling.

In random bowling, if there are N>0 pins standing,

there is an equal probability for each of the

possible outcomes of your bowl: knocking down

0 pins, 1 pin, ..., N pins. Each of these outcomes

has probability 1/(N+1). The game is scored the

way standard 10-pin bowling is scored.

So, for example, at the beginning of a frame, your

first bowl has 1/11 probability of getting a strike,

a 1/11 probability of knocking nine pins down, and

so on down to a 1/11 probability of knocking no

pins down. If you didn't bowl a strike, there are

some number of pins, say X, still standing. On

your next throw to complete the frame, you have a

1/(X+1) chance of knocking all X down for a spare,

a 1/(X+1) chance of knocking down X-1 pins, and

so on down to a 1/(X+1) chance of knocking none of

them down.

With this scenario, what is your expected score?

[Guest]

Forgive me for not being familiar with the rules, what is the scoring system?

[Guest]

Forgive me for not being familiar with the rules, what is the scoring system?

The basics are as follow:

You receive one point per pin you knock down (i.e. 5 pins = 5 points).

For a spare, you receive ten points for that frame, plus the value of the next ball you bowl (i.e. you spare the second frame, and bowl a 6 on the first ball of the third frame. Your second frame score is 16.)

For a strike, you receive ten points for that frame, plus the value of the next two balls you bowl (i.e. you spare the second frame, and bowl a 9 in the third frame (both balls combined). Your second frame score is 19.) If you spare the next frame, your strike frame is 20 points, and so forth.

Hopefully that covers it for you.

[Guest]

If the first ball is a gutter, the average for the 2nd roll will be 5. If the first is 1 pin, the average 2nd will be 4.5, etc. The total pins for each of these alternatives will be 5, 5,5, 6, etc. for an expected pin count of 7.5. There is a 1/11th chance that you will roll a strike. The chances of a spare depend on the first roll but will be 1/11 if the first is a gutter, 1/10 if the first ball gets 1 pin, etc. If we give 10 points fro a Spare and 20 points for a strike, The expected total point for each 11 possibilies are as follows

First Roll-----Expected Score

0--------------5.909

1--------------6.5

2--------------7.1

3--------------7.75

4--------------8.4

5--------------9.17

6--------------10

7--------------11

8--------------12.33

9--------------14.5

10-------------30

This is an average of 11.15 per frame or 115 per game.

This is an over estimation as it gives full points benefit for each spare or strike. If someone wants to make that correction, I'll be interested in seeing the approach

[superprismatic]

If the first ball is a gutter, the average for the 2nd roll will be 5. If the first is 1 pin, the average 2nd will be 4.5, etc. The total pins for each of these alternatives will be 5, 5,5, 6, etc. for an expected pin count of 7.5. There is a 1/11th chance that you will roll a strike. The chances of a spare depend on the first roll but will be 1/11 if the first is a gutter, 1/10 if the first ball gets 1 pin, etc. If we give 10 points fro a Spare and 20 points for a strike, The expected total point for each 11 possibilies are as follows

First Roll-----Expected Score

0--------------5.909

1--------------6.5

2--------------7.1

3--------------7.75

4--------------8.4

5--------------9.17

6--------------10

7--------------11

8--------------12.33

9--------------14.5

10-------------30

This is an average of 11.15 per frame or 115 per game.

This is an over estimation as it gives full points benefit for each spare or strike. If someone wants to make that correction, I'll be interested in seeing the approach

I didn't check your average of 11.15 per frame, but if it is that, then you get 111.5 per game. A very nice upper bound!

[Guest]

Ok so the guy before isn't entirely correct... he is if you account for two bowls in each round regardless of if you hit ten pins or not. If you don't then he over estimated by ~0.9 . Now, to include spares and strikes... I'm no statistics wiz but when I worked out the numbers I got the average to be 11.62436198 pins per round or... ~116 pins per game. The contribution from strikes is actually quite small, only about 0.15 pins per frame, due to the statistical unlikelihood that you will get them. Again I'm not positive this is correct but I spent about an hour going over the program I wrote to calculate it and it seems like I accounted for everything. Anyway, good luck.

[EventHorizon]

I'm not positive this is completely correct. I analyzed the score for a single frame by itself and multiplied by 10. There may be correlation between frames that throw things off, but here's my work/guess...

The average score after one throw of the ball if there are 10 pins is 5. So if you get a spare, you can essentially just add 5.

Ignoring the possibility of a strike for the moment, we can use the above to find the average for a frame.

if you get a gutter ball on the first throw, your average score for the frame will be 60/11 (0+1+2+3+4+5+6+7+8+9+15 <-because of +5 for spare)

similarly for 1-9 pins

0:60/11

1:60/10

2:59/9

3:57/8

4:54/7

5:50/6

6:45/5

7:39/4

8:32/3

9:24/2

The sum of the above is 457931/5544.

The average score for a frame will be (average score for strike + (457931/5544))/11.

Now we can analyze what happens if you get a strike. If you get a strike you get an automatic 10 points and the score for the two next rolls. If you get another strike then you get 10 points plus another roll (= average of 5 more points). So if you get another strike the total will be the initial 10 plus an average of 15.

We can do a similar thing to above for combinations for the rest, only it is simpler this time due to the lack of spares.

0:55/11

1:55/10

2:54/9

3:52/8

4:49/7

5:45/6

6:40/5

7:34/4

8:27/3

9:19/2

The sum of the above is 145/2.

The average score for a frame that you initially get a strike is then 10 + (15 + (145/2) )/11

= 10 + (175/2)/11 = 10 + 175/22 = (220+175)/22 = 395/22. Plugging this into the earlier equation gives:

((395/22) + (457931/5544))/11 = (99540 + 457931)/60984 = 557471/60984. So about an average of 9.141267 per frame. For a whole game, this would average 91.41267.

Did I get it? Or is there a mistake here somewhere?

[Guest]

[i didn't work through all of your math but the approach seems reasonable. The frame average with NO strike/spare benefit is 7.5 (75 game). My calculation with MAXIMUM benefit is 11.15 (115 game). The midpoint is 9.33, not far from your 9.14/spoiler]

I'm not positive this is completely correct. I analyzed the score for a single frame by itself and multiplied by 10. There may be correlation between frames that throw things off, but here's my work/guess...

The average score after one throw of the ball if there are 10 pins is 5. So if you get a spare, you can essentially just add 5.

Ignoring the possibility of a strike for the moment, we can use the above to find the average for a frame.

if you get a gutter ball on the first throw, your average score for the frame will be 60/11 (0+1+2+3+4+5+6+7+8+9+15 <-because of +5 for spare)

similarly for 1-9 pins

0:60/11

1:60/10

2:59/9

3:57/8

4:54/7

5:50/6

6:45/5

7:39/4

8:32/3

9:24/2

The sum of the above is 457931/5544.

The average score for a frame will be (average score for strike + (457931/5544))/11.

Now we can analyze what happens if you get a strike. If you get a strike you get an automatic 10 points and the score for the two next rolls. If you get another strike then you get 10 points plus another roll (= average of 5 more points). So if you get another strike the total will be the initial 10 plus an average of 15.

We can do a similar thing to above for combinations for the rest, only it is simpler this time due to the lack of spares.

0:55/11

1:55/10

2:54/9

3:52/8

4:49/7

5:45/6

6:40/5

7:34/4

8:27/3

9:19/2

The sum of the above is 145/2.

The average score for a frame that you initially get a strike is then 10 + (15 + (145/2) )/11

= 10 + (175/2)/11 = 10 + 175/22 = (220+175)/22 = 395/22. Plugging this into the earlier equation gives:

((395/22) + (457931/5544))/11 = (99540 + 457931)/60984 = 557471/60984. So about an average of 9.141267 per frame. For a whole game, this would average 91.41267.

Did I get it? Or is there a mistake here somewhere?

[Guest]

[i didn't work through all of your math but the approach seems reasonable. The frame average with NO strike/spare benefit is 7.5 (75 game). My calculation with MAXIMUM benefit is 11.15 (115 game). The midpoint is 9.33, not far from your 9.14/spoiler]

I'm not positive this is completely correct. I analyzed the score for a single frame by itself and multiplied by 10. There may be correlation between frames that throw things off, but here's my work/guess...

The average score after one throw of the ball if there are 10 pins is 5. So if you get a spare, you can essentially just add 5.

Ignoring the possibility of a strike for the moment, we can use the above to find the average for a frame.

if you get a gutter ball on the first throw, your average score for the frame will be 60/11 (0+1+2+3+4+5+6+7+8+9+15 <-because of +5 for spare)

similarly for 1-9 pins

0:60/11

1:60/10

2:59/9

3:57/8

4:54/7

5:50/6

6:45/5

7:39/4

8:32/3

9:24/2

The sum of the above is 457931/5544.

The average score for a frame will be (average score for strike + (457931/5544))/11.

Now we can analyze what happens if you get a strike. If you get a strike you get an automatic 10 points and the score for the two next rolls. If you get another strike then you get 10 points plus another roll (= average of 5 more points). So if you get another strike the total will be the initial 10 plus an average of 15.

We can do a similar thing to above for combinations for the rest, only it is simpler this time due to the lack of spares.

0:55/11

1:55/10

2:54/9

3:52/8

4:49/7

5:45/6

6:40/5

7:34/4

8:27/3

9:19/2

The sum of the above is 145/2.

The average score for a frame that you initially get a strike is then 10 + (15 + (145/2) )/11

= 10 + (175/2)/11 = 10 + 175/22 = (220+175)/22 = 395/22. Plugging this into the earlier equation gives:

((395/22) + (457931/5544))/11 = (99540 + 457931)/60984 = 557471/60984. So about an average of 9.141267 per frame. For a whole game, this would average 91.41267.

Did I get it? Or is there a mistake here somewhere?

[superprismatic]

I'm not positive this is completely correct. I analyzed the score for a single frame by itself and multiplied by 10. There may be correlation between frames that throw things off, but here's my work/guess...

The average score after one throw of the ball if there are 10 pins is 5. So if you get a spare, you can essentially just add 5.

Ignoring the possibility of a strike for the moment, we can use the above to find the average for a frame.

if you get a gutter ball on the first throw, your average score for the frame will be 60/11 (0+1+2+3+4+5+6+7+8+9+15 <-because of +5 for spare)

similarly for 1-9 pins

0:60/11

1:60/10

2:59/9

3:57/8

4:54/7

5:50/6

6:45/5

7:39/4

8:32/3

9:24/2

The sum of the above is 457931/5544.

The average score for a frame will be (average score for strike + (457931/5544))/11.

Now we can analyze what happens if you get a strike. If you get a strike you get an automatic 10 points and the score for the two next rolls. If you get another strike then you get 10 points plus another roll (= average of 5 more points). So if you get another strike the total will be the initial 10 plus an average of 15.

We can do a similar thing to above for combinations for the rest, only it is simpler this time due to the lack of spares.

0:55/11

1:55/10

2:54/9

3:52/8

4:49/7

5:45/6

6:40/5

7:34/4

8:27/3

9:19/2

The sum of the above is 145/2.

The average score for a frame that you initially get a strike is then 10 + (15 + (145/2) )/11

= 10 + (175/2)/11 = 10 + 175/22 = (220+175)/22 = 395/22. Plugging this into the earlier equation gives:

((395/22) + (457931/5544))/11 = (99540 + 457931)/60984 = 557471/60984. So about an average of 9.141267 per frame. For a whole game, this would average 91.41267.

Did I get it? Or is there a mistake here somewhere?

EventHorizon has it! Just to check things, I wrote a program to simulate random games. After 1,000,000,000 random games, the average score from the simulator was 91.4135. Congrats!

[Guest]

EventHorizon has it! Just to check things, I wrote a program to simulate random games. After 1,000,000,000 random games, the average score from the simulator was 91.4135. Congrats!

I am having problems uploading the simulator I made in excel, but I am getting around 88 pins per game. I know a problem a lot of people have with scoring bowling is paying attention to the differences in the 10th frame. I will try again uploading my simulator so anyone can critique my work.

[Guest]

I am having problems uploading the simulator I made in excel, but I am getting around 88 pins per game. I know a problem a lot of people have with scoring bowling is paying attention to the differences in the 10th frame. I will try again uploading my simulator so anyone can critique my work.

Here is a graph of the distrobution I recieved.

Bowling.bmp

[Guest]

i wrote a bowling sim to solve the problem and got about the same, 91.4278

note i took in things like the 10th frame specialness and stuff like the chance of getting 2 strikes in a row.

[Guest]

Here is a graph of the distrobution I recieved.

I have attached my bowling sim. The first game is setup and can be checked for accuracy. You can drag all the formulas in the thick outline down to simulate however many games you would like. I used 15K and got the results I posted earlier. I welcome any comments. Thanks.

Bowling Sim.xls

[Guest]

all your formulas look correct redstuff, but when i ran the sim i got 91.41.