What is the chance for a single bowling pin to remain standing, from
the first throw of each frame only, for all ten frames of a game? 

Note -- this allows
for any of the pins to be eligible to be the sole remaining pin.

as a followup -- what would the chance be if the sole remaining pin
for each frames first throw is a specific pin?

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Request for Question Clarification by lot-ga on 10 Sep 2002 11:42 PDT

Hello,
the answer 'appears' to be quite simple for an expensive question. 
(unless I have it wrong)
What are the deliverables you are expecting in the answer to justify
the $100 price tag?

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Clarification of Question by mdurkee65-ga on 10 Sep 2002 13:00 PDT

Thanks for the response -- yes the answer for the first part seems
relatively simple -- the problem in solving this is in realizing that
there are not just eleven possible outcomes for each first ball
thrown. there are as many outcomes as there are possibilities for pins
to remain standing [not a 1 in 11 chance][examples 10 pins up has 1
possible combination/9 pins up 10 combinations/ 8 pins left up has ??
combinations] it would be the total of all of these combinations
multiplied by the 10 frames first balls that would equal the
percentage of leaving any one random pin up

The followup question is even harder -- this builds upon those pin
possibilities outlined above -- except above where any single pin
counts as good -- now only one pin counts as good.

am i making myself clear -- if you think you can continue forward with
this type of complex calculation i will continue to elaborate.

What i need for the $100 price tag is a detailed analysis of the
possible outcomes from each throw and the statistical chance for
throwing the specific combination outlined in my question.

I'll respectfully disagree with lot-ga; the answer doesn't seem so
simple to me.  (Perhaps the difficulty of the answer depends on one's
perspective; it might be simple, for example, to someone with a degree
in mathematics.)  But the answer can be calculated with some effort.

You are interested in the chance that one pin will be left standing in
all ten frames after the first ball is thrown.  (A game can also last
eleven frames if you convert the spare at the end of the tenth frame;
more about that possibility at the end of the answer.)

I will assume that the probability that any possible set of pins will
be standing after you roll the first ball is equally likely.  In other
words, under this assumption, each grouping of pins has an equal
chance of remaining, from 0 pins to 10 pins to each arrangement of
pins in between.

The first important step of the calculation is to determine whether to
use the formula for permutations or the formula for combinations. 
Permutations, as you may know, are ways of arranging objects into
groups where the order of the objects does matter.  By contrast,
combinations are groups where the order does not matter.  So, for
example, there are two permutations of the most difficult split in
bowling -- you could call it the 1-10 or the 10-1, depending on which
pin you wanted to list first.  But there is only one combination, in
the mathematical sense.  Similarly, there is only one combination for
all ten pins -- again, the order does not matter.  But there are many
permutations of ten pins: 1-2-3-4-5-6-7-8-9-10, 1-3-2-4-5-6-7-8-9-10,
and so on.

It is clear that you are interested in combinations, not permutations.
 If ten pins are left standing, you wouldn't say that
1-2-3-4-5-6-7-8-9-10 is a different group of ten pins than
1-3-2-4-5-6-7-8-9-10.  You might say that they were different if you
were interested in the order that the pins fell; then
1-2-3-4-5-6-7-8-9-10 would not be the same as 1-3-2-4-5-6-7-8-9-10. 
But in your question, ten pins standing is just one possible result.

The formula for calclulating combinations (as well as the formula for
permutations) is explained on various web pages, such as this one:

"Ask Dr. Math: FAQ - Permutations and Combinations"
The Math Forum @ Drexel
http://mathforum.org/dr.math/faq/faq.comb.perm.html

As shown on that page, n_C_k = n! / k!(n-k)!

The probability that one pin will be left standing after the first
ball in each frame is 10C1 divided by the sum of all the possible
combinations (10C0 + 10C1 + 10C2 ... + 10C10) of standing pins.

Using the formula for combinations:

10C0 = 10! / 0!(10-0!) = 10! / 1(10!) = 1
[In other words, there is one result where zero pins remain, just like
there is one result where ten pins remain.]

10C1 = 10! / 1!(10-1)!) = 10! / 1!(9!) = 10*9*8*7*6*5*4*3*2*1 /
1(9*8*7*6*5*4*3*2*1) = 10
[Every number from 9 down to 1 divides by itself in this calcualation,
leaving only the 10.  In the calculations below, I will save a step by
leaving out each number that, in the end, is divided by itself.  I
will also save a step by showing the result of n-k, rather than
displaying the numbers I've plugged into n-k.]

10C2 = 10! / 2!(8!) = 10*9 / 2 = 45

10C3 = 10! / 3!(7!) = 10*9*8 / 3*2 = 120

10C4 = 10! / 4!(6!) = 10*9*8*7 / 4*3*2 = 210

10C5 = 10! / 5!(5!) = 10*9*8*7*6 / 5*4*3*2 = 256.

From here on in, the results are 10C6 = 210, 10C7 = 120, 10C8 = 45,
10C9 = 10, and 10C10 = 1.  There is no difference, for instance,
between 10C4 and 10C6.  10C6 = 10! / 6!(4!).  This is equivalent to
10! / 4!(6!), and thus equals 210.  Likewise, 10C7 produces the same
outcome as 10C3, and so on.

Adding them all up, you get 1 + 10 + 45 + 120 + 210 + 256 + 210 + 120
+ 45 + 10 + 1 = 1024.  There are 10 chances that one pin will be left
standing (one chance for each of pin numbers 1 through 10).  Thus, the
probablity of one pin standing after the first ball of a frame is
10/1024, or if you simplify it, 5/512.

Now that's not quite the end of the story.  You have to multiply the
chance that this event will happen in one frame by the possibility it
will happen in every frame.  That would be 5/512 to the 10th power,
which is a remarkably small number -- approximately 3.009 x 10 to the
-26th power (otherwise expressed as 3.009 x 10^-26.  And that's your
answer.

The chance of a specific pin will be standing after the first ball of
a frame is not 10/1024, but 1/1024.  Again, that must be multiplied by
the chance of that this pin will also be standing in the other frames.
 So that is 1/1024 to the 10th power, or approximately 3.081 x 10^-33.

If you want to calculate the probabilities for an eleven-frame match,
you would take 5/512 to the 11th power, or 1/1024 to the 11th power,
respectively.

Of course, in real life, the probabilities are different.  If you are
a decent bowler, you won't throw the ball randomly; moreover, the pins
won't fall at random.  If you hit the pins near the middle, you'll
fairly often have 1 pin standing -- much more than 10 times out of
1024.  Also, you will be more likely to have, say, just the 7 or 10
pin standing than just the 4 or 6.

Thank you for this question.  I used to bowl all the time, but
recently I've been pursuing other kinds of recreation (like answering
questions here!).  Now that I've been inspired by your question, I
think I'll head on over to the Bowl-a-rama!

- justaskscott-ga


Search terms used on Google:

combinations permutations

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Clarification of Answer by justaskscott-ga on 10 Sep 2002 13:55 PDT

I made an error in the final step of my calculations; I plugged the
numbers into my calculator incorrectly the first time:

The actual answers are approximately 7.889 x 10^-21 and 8.078 x
10^-28.

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Clarification of Answer by justaskscott-ga on 10 Sep 2002 14:08 PDT

I've only just now seen your clarification of the question, which was
posted while I was working on the answer.

I believe that I've answered the question as you wanted.  Perhaps I
put in too much detail; you weren't interested in an 11 frame
scenario, and you probably didn't need the information about
"combinations" versus "permutations", since you use the term
"combinations" yourself.

But aside from a couple of mistakes with punctuation marks (and the
original mistake with my calculator!), I don't notice anything that I
would change in my answer and clarification.

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Clarification of Answer by justaskscott-ga on 10 Sep 2002 17:27 PDT

Please see mvguy's comment, and my comment in response, for a correction.

i was very impressed with the speed in which a well thought answer was
provided -- if i am in need of this service again i will use it. one
suggestion -- it would be nice to have some way to have direct
communication with the people who answer the questions - is that in
the works?

Here's how I would calculate it, if I understand the question right:

The chance that in the first frame only the first pin would remain
standing is (1/2)^10 or 1/1024 or .0009765625.

The chance that the same thing would happen in every frame is
(1/1024)^10 or 7.889*(10^-31).

So the chance that only the first pin would be standing every time is
7.889*10^-31.  The chance would be the same that only the second pin
would be standing every time, or the third pin, and so on.

Now to recalculate for it not mattering which pin stands:

The chance that any one pin would still be standing in the first frame
would be 10*(1/1024) or 10/1024 or 1/102.4 or .009765625.  (That's the
chance that only the first pin stands plus the chance that only the
second stands plus the chance that only the third one stands, etc.)
The chance that one pin (doesn't matter which) is standing in 10
frames would be (10/1024)^10 or 7.889*(10^-21).

I'm pretty confident this is right (if I understand the question).  If
someone could show me where my logic or calculation is wrong, I'll
gladly retract this comment.  I'm not attempting to undermine
Justaskscott-ga's work, but I do think his calculations (not his
logic) are incorrect.

Mvguy has submitted a valuable correction.  We both have the same
answer to the first question: 7.889 x 10^-21.  Mvguy has calculated
the correct answer to the second question: 7.889 x 10^-31.  What I did
was take 1/512 to the 10th power.  This was an error: I had forgotten
that the original fraction was 1/1024, not 1/512.  (I had been using
5/512 for the other fraction, so I was thinking of 512 as the
denominator.)

Thanks, mvguy!  It's a little embarrassing for me to admit that I made
an error, but it would be worse if no one had submitted a correction.

mdurkee65 -

Thank you for your comments on the work that mvguy and I did.

Unfortunately, I think that I'm not supposed to communicate privately
about questions posted on Google Answers.

However, I'll pass along your suggestions to the Google Answers
editors about private communication and splitting money between
Researchers.  These are good ideas.

- justaskscott

No problem -- i realise that this is new and they need to work out a few of the bugs

For me the important part is the bit of the answer which says "Of
course, in real life, the probabilities are different.  If you are a
decent bowler, you won't throw the ball randomly".

In real life this is referring to "skill", statistically, you must
take into account the "independence" between each bowl. For example,
consider the following two questions. The questions sound the same but
the answers are very different:

(1) If a football/soccer team win the coin toss in their first nine
games, what is the probability of them winning the toss in their tenth
game.
(2)  If a football/soccer team win their first nine games, what is the
probability of them winning their tenth game.

The answer to the first question is 0.5 because each throw is
independent.
The answer to the second question is much higher because the team are
obviously playing well and are a good team.

thanks for the comment alan0-ga  -- you are absolutely correct there
is a great deal of difference in the answer when "real life" is taken
into account. doing so would involve a much greater challenge to
determine a statistically correct answer. do you have any ideas on how
this could be done? the reason is asked for a random probability
answer is that this is the upper limit to the chances -- real life
will always have an increased chance for the single pin to be left
standing. i would love to continue this discourse -- here or directly
mdurkee at msi-mfg.com

i am interested in continuing this effort but i do not think that this
forum is the right place to conduct the research

mdurkee65-ga - not sure I agree that "real life will always have an
increased chance for the single pin to be left standing". If the
player is really bad then the chances are very low that any pins will
be knocked down, let alone nine. If the player is pretty good then the
chances are high that one pin will be left. If the player is very good
then the chance gets lower again as the player will probably knock
down all ten (unless the player knows what the aim is!).

Anyway, to get a real life answer we need to leave the world of
permutations and combinations and move in to distributions - what is
the distribution of players success, what is the average number of
pins knocked down, etc.

A Google search on "bowling scores distribution" uncovered:

(1) a web site with some commercial software which you can feed you
own scores in and work out your own distribution. See
http://www.mnsi.net/~edd/index.htm

(2) A link to a page that would not display but looked like it was a
homework answer about distribution of bowling scores. See
http://w.public.iastate.edu/~dnett/S401A/hw7.pdf

(3) Some books:

Cooper, C. N. and Kennedy, R. E. "A Generating Function for the
Distribution of the Scores of All Possible Bowling Games." In The
Lighter Side of Mathematics (Ed. R. K. Guy and R. E. Woodrow).
Washington, DC: Math. Assoc. Amer., 1994.

Cooper, C. N. and Kennedy, R. E. "Is the Mean Bowling Score Awful?" In
The Lighter Side of Mathematics (Ed. R. K. Guy and R. E. Woodrow).
Washington, DC: Math. Assoc. Amer., 1994.

Hope this helps. What is this all for anyway (and why was it worth
$100?)

Thanks for the additional information alan0-ga.