Re: Probability Problem
I think I got it. I noticed my code is essentially the same as Tim Peter's (plus the part of the problem he skipped). I read his code 20 minutes before recreating mine from Alex's hints. Thanks! def main(): ways = ways_to_roll() total_ways = float(101**10) running_total = 0 for i in range(1000-390+1): j = i + 390 running_total += ways[i] * ways[j] print running_total / total_ways**2 print ways[:10] def ways_to_roll(): result = [1] for i in xrange(10): result = combine([1] * 101, result) return result def combine(a, b): results = [0] * (len(a) + len(b) - 1) for i, ele in enumerate(a): for j, ele2 in enumerate(b): results[i+j] += ele * ele2 return results main() # output: 3.21962542309e-05 and # [1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620] # 3.21962542309e-05 is 32 out of a million On Apr 24, 2006, at 9:14 PM, Alex Martelli wrote: Elliot Temple [EMAIL PROTECTED] wrote: On Apr 24, 2006, at 8:24 PM, Alex Martelli wrote: Lawrence D'Oliveiro [EMAIL PROTECTED] wrote: In article [EMAIL PROTECTED], Elliot Temple [EMAIL PROTECTED] wrote: Problem: Randomly generate 10 integers from 0-100 inclusive, and sum them. Do that twice. What is the probability the two sums are 390 apart? I think the sum would come close to a normal distribution. Yes, very close indeed, by the law of large numbers. However, very close (in a math course at least) doesn't get the cigar. You can compute the requested answer exactly with no random number generation whatsoever: compute the probability of each result from 0 to 1000, then sum the probabilities of entries that are exactly 390 apart. That was the plan, but how do I get the probability of any given result? (in a reasonable amount of time) BTW I'm not in a math course, just curious. OK, I'll trust that last assertion (sorry for the hesitation, but it's all too easy to ``help'' somebody with a homework assignment and actually end up damaging them by doing it FOR them!-). I'm still going to present this in a way that stimulates thought, rather than a solved problem -- humor me...!-) You're generating a uniformly distributed random number in 0..100 (101 possibilities), 10 times, and summing the 10 results. How do you get a result of 0? Only 1 way: 0 at each attempt -- probability 1 (out of 1010 possibilities). How do you get a result of 1? 10 ways: 1 at one attempt and 0 at each of the others - probability 10 (again in 1010'ths;-). How do you get a result of 2? 10 ways for '2 at one attempt and 0 at each of the others', plus, 10*9/2 ways for '1 at two attempts and 0 at each of the others' -- probability 55 (ditto). ...and so forth, but you'd rather not work it out... So, suppose you start with a matrix of 101 x 10 entries, each of value 1 since all results are equiprobable (or, 1/1010.0 if you prefer;-). You want to compute the number in which you can combine two rows. How could you combine the first two rows (each of 101 1's) to make a row of 201 numbers corresponding to the probabilities of the sum of two throws? Suppose you combine the first entry of the first row with each entry of the second, then the second entry of the first row with each entry of the second, etc; each time, you get a sum (of two rolls) which gives you an index of a entry (in an accumulator row starting at all zeros) to increment by the product of the entries you're considering... Can you generalize that? Or, do you need more hints? Just ask! Alex -- http://mail.python.org/mailman/listinfo/python-list -- Elliot Temple http://www.curi.us/blog/ -- http://mail.python.org/mailman/listinfo/python-list
Re: Probability Problem
I had a possibly similar problem calculating probs related to premium bond permutation. With 10^12 memory ran out v quickly. In the end I got round it by writing a recursive function and quantising the probability density function. Elliot Temple wrote: Problem: Randomly generate 10 integers from 0-100 inclusive, and sum them. Do that twice. What is the probability the two sums are 390 apart? I have code to do part of it (below), and I know how to write code to do the rest. The part I have calculates the number of ways the dice can come out to a given number. The problem is the main loop has 9 iterations and it takes about 2.5 minutes to begin the 4th one, and each iteration is about 101 times longer than the previous one. So: x = 2.5 * 101**6 x /= (60*24*365.25) x 5045631.5622908585 It'd take 5,000 millennia. (If my computer didn't run out of memory after about 4 minutes, that is.) Any suggestions? Either a way to do the same thing much more efficiently (enough that I can run it) or a different way to solve the problem. Code: li = range(101) li2 = [] range101 = range(101) for x in xrange(9): print x is %s % x li2 = [] for y in li: for z in range101: li2 += [y+z] li = li2 print li.count(800) # prints how many ways the dice can add to 800 This link may help: http://www.math.csusb.edu/faculty/stanton/m262/intro_prob_models/calcprob.html -- Elliot Temple http://www.curi.us/blog/ -- http://mail.python.org/mailman/listinfo/python-list
Re: Probability Problem
[Elliot Temple]
I think I got it. I noticed my code is essentially the same as Tim
Peter's (plus the part of the problem he skipped). I read his code 20
minutes before recreating mine from Alex's hints. Thanks!
def main():
ways = ways_to_roll()
total_ways = float(101**10)
running_total = 0
for i in range(1000-390+1):
j = i + 390
running_total += ways[i] * ways[j]
print running_total / total_ways**2
print ways[:10]
def ways_to_roll():
result = [1]
for i in xrange(10):
result = combine([1] * 101, result)
return result
def combine(a, b):
results = [0] * (len(a) + len(b) - 1)
for i, ele in enumerate(a):
for j, ele2 in enumerate(b):
results[i+j] += ele * ele2
return results
main()
# output: 3.21962542309e-05 and
# [1, 10, 55, 220, 715, 2002, 5005, 11440, 24310, 48620]
# 3.21962542309e-05 is 32 out of a million
You should sanity-check the computation by generalizing it, then
applying it to a case so small you can easily work out the result via
exhaustive enumeration by hand.
For example, suppose you took integers from the much smaller set {0,
1}, and did that only twice. Then the possible sums and their
probabilities are clearly:
0 1/4
1 1/2
2 1/4
If you did this twice, what's the probability that the sums differ by
1? Suitably generalized, your program above would compute 1/4. Is
that actually right? It depends on what exactly the sums differ by
1 means. If it means the second sum is one larger than the first
sum, 1/4 is correct. Ditto if it means the second sum is one smaller
than the first sum. But if it means 1 is the absolute value of the
difference of the sums, the right answer is 1/2. I'm not sure which
meaning you have in mind, but the last one was my guess.
--
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Probability Problem
Problem: Randomly generate 10 integers from 0-100 inclusive, and sum them. Do that twice. What is the probability the two sums are 390 apart? I have code to do part of it (below), and I know how to write code to do the rest. The part I have calculates the number of ways the dice can come out to a given number. The problem is the main loop has 9 iterations and it takes about 2.5 minutes to begin the 4th one, and each iteration is about 101 times longer than the previous one. So: x = 2.5 * 101**6 x /= (60*24*365.25) x 5045631.5622908585 It'd take 5,000 millennia. (If my computer didn't run out of memory after about 4 minutes, that is.) Any suggestions? Either a way to do the same thing much more efficiently (enough that I can run it) or a different way to solve the problem. Code: li = range(101) li2 = [] range101 = range(101) for x in xrange(9): print x is %s % x li2 = [] for y in li: for z in range101: li2 += [y+z] li = li2 print li.count(800) # prints how many ways the dice can add to 800 This link may help: http://www.math.csusb.edu/faculty/stanton/m262/intro_prob_models/ calcprob.html -- Elliot Temple http://www.curi.us/blog/ -- http://mail.python.org/mailman/listinfo/python-list
Re: Probability Problem
In article [EMAIL PROTECTED], Elliot Temple [EMAIL PROTECTED] wrote: Problem: Randomly generate 10 integers from 0-100 inclusive, and sum them. Do that twice. What is the probability the two sums are 390 apart? I think the sum would come close to a normal distribution. -- http://mail.python.org/mailman/listinfo/python-list
Re: Probability Problem
Lawrence D'Oliveiro [EMAIL PROTECTED] wrote: In article [EMAIL PROTECTED], Elliot Temple [EMAIL PROTECTED] wrote: Problem: Randomly generate 10 integers from 0-100 inclusive, and sum them. Do that twice. What is the probability the two sums are 390 apart? I think the sum would come close to a normal distribution. Yes, very close indeed, by the law of large numbers. However, very close (in a math course at least) doesn't get the cigar. You can compute the requested answer exactly with no random number generation whatsoever: compute the probability of each result from 0 to 1000, then sum the probabilities of entries that are exactly 390 apart. Alex -- http://mail.python.org/mailman/listinfo/python-list
Re: Probability Problem
On Apr 24, 2006, at 8:24 PM, Alex Martelli wrote: Lawrence D'Oliveiro [EMAIL PROTECTED] wrote: In article [EMAIL PROTECTED], Elliot Temple [EMAIL PROTECTED] wrote: Problem: Randomly generate 10 integers from 0-100 inclusive, and sum them. Do that twice. What is the probability the two sums are 390 apart? I think the sum would come close to a normal distribution. Yes, very close indeed, by the law of large numbers. However, very close (in a math course at least) doesn't get the cigar. You can compute the requested answer exactly with no random number generation whatsoever: compute the probability of each result from 0 to 1000, then sum the probabilities of entries that are exactly 390 apart. That was the plan, but how do I get the probability of any given result? (in a reasonable amount of time) BTW I'm not in a math course, just curious. -- Elliot Temple http://www.curi.us/blog/ -- http://mail.python.org/mailman/listinfo/python-list
Re: Probability Problem
Elliot Temple [EMAIL PROTECTED] wrote: On Apr 24, 2006, at 8:24 PM, Alex Martelli wrote: Lawrence D'Oliveiro [EMAIL PROTECTED] wrote: In article [EMAIL PROTECTED], Elliot Temple [EMAIL PROTECTED] wrote: Problem: Randomly generate 10 integers from 0-100 inclusive, and sum them. Do that twice. What is the probability the two sums are 390 apart? I think the sum would come close to a normal distribution. Yes, very close indeed, by the law of large numbers. However, very close (in a math course at least) doesn't get the cigar. You can compute the requested answer exactly with no random number generation whatsoever: compute the probability of each result from 0 to 1000, then sum the probabilities of entries that are exactly 390 apart. That was the plan, but how do I get the probability of any given result? (in a reasonable amount of time) BTW I'm not in a math course, just curious. OK, I'll trust that last assertion (sorry for the hesitation, but it's all too easy to ``help'' somebody with a homework assignment and actually end up damaging them by doing it FOR them!-). I'm still going to present this in a way that stimulates thought, rather than a solved problem -- humor me...!-) You're generating a uniformly distributed random number in 0..100 (101 possibilities), 10 times, and summing the 10 results. How do you get a result of 0? Only 1 way: 0 at each attempt -- probability 1 (out of 1010 possibilities). How do you get a result of 1? 10 ways: 1 at one attempt and 0 at each of the others - probability 10 (again in 1010'ths;-). How do you get a result of 2? 10 ways for '2 at one attempt and 0 at each of the others', plus, 10*9/2 ways for '1 at two attempts and 0 at each of the others' -- probability 55 (ditto). ...and so forth, but you'd rather not work it out... So, suppose you start with a matrix of 101 x 10 entries, each of value 1 since all results are equiprobable (or, 1/1010.0 if you prefer;-). You want to compute the number in which you can combine two rows. How could you combine the first two rows (each of 101 1's) to make a row of 201 numbers corresponding to the probabilities of the sum of two throws? Suppose you combine the first entry of the first row with each entry of the second, then the second entry of the first row with each entry of the second, etc; each time, you get a sum (of two rolls) which gives you an index of a entry (in an accumulator row starting at all zeros) to increment by the product of the entries you're considering... Can you generalize that? Or, do you need more hints? Just ask! Alex -- http://mail.python.org/mailman/listinfo/python-list
Re: Probability Problem
[Alex Martelli] ... You can compute the requested answer exactly with no random number generation whatsoever: compute the probability of each result from 0 to 1000, then sum the probabilities of entries that are exactly 390 apart. [Elliot Temple] That was the plan, but how do I get the probability of any given result? As the link you gave suggests, the number of ways to get a given sum s is the coefficient of x**s in (1 + x + x**2 + ... +x**100)**10. Divide that by 101**10 to compute the probability of getting sum s. (in a reasonable amount of time) BTW I'm not in a math course, just curious. Computing exact results is probably too hard for just curious. This little Python program computes the exact number of ways to get each sum in 0 .. 1000 quickly (fraction of a second on my box): def mul(p1, p2): result = [0] * (len(p1) + len(p2) - 1) for i, x1 in enumerate(p1): for j, x2 in enumerate(p2): result[i+j] += x1 * x2 while result and result[-1] == 0: del result[-1] return result base = [1] * 101 result = [1] for dummy in range(10): result = mul(result, base) assert len(result) == 1001 assert sum(result) == 101**10 Then result[s] is the exact number of ways to get sum s, for each s in range(1001). Figuring out why that works is the probably too hard for 'just curious' part. If you want to pursue that, it's a straightforward implementation of fully multiplying out: (1 + x + x**2 + ... +x**100)**10 A polynomial here is represented by a list L of coefficients, where L[i] is the coefficient of x**i. `mul()` implements polynomial multiplication using this format. For example, mul([1, 1], [-1, 1]) [-1, 0, 1] or, IOW, (x+1)*(x-1) = (x**2 - 1). -- http://mail.python.org/mailman/listinfo/python-list
Re: Dice probability problem
On 2006-04-05, Tomi Lindberg [EMAIL PROTECTED] wrote:
Antoon Pardon wrote:
def __rmul__(self, num):
tp = num * [self]
return reduce(operator.add, tp)
sum3d6 = 3 * D(6)
One basic question: is there any particular reason not to
use __mul__ instead (that would allow me to use both 3 *
D(6) and D(6) * 3, while __rmul__ raises an AttributeError
with the latter)?
Well 3 * D(6) is similar to the notation used in roleplaying,
while D(6) * 3 would make me think of the distribution
{3:1, 6:1, 9:1, 12:1, 15:1, 18:}
Difference between the two methods is
slightly unclear to me.
I have to look it up myself regularly. But in this case
it was more a matter of some intuition that 3 * D(6)
was not the same as D(6) * 3. You may have a different
intuition.
--
Antoon Pardon
--
http://mail.python.org/mailman/listinfo/python-list
Re: Dice probability problem
Op 2006-04-04, Tomi Lindberg schreef [EMAIL PROTECTED]:
First, thanks to Antoon and Alexander for replying.
Antoon Pardon wrote:
It would be better to construct distributions for one
die and make a function that can 'add' two distributions
together.
As both replies pointed to this direction, I tried to take
that route. Here's the unpolished code I came up with. Does
it look even remotely sane way to accomplish my goal?
-- code begins --
# A die with n faces
D = lambda n: [x+1 for x in range(n)]
# A new die with 6 faces
d6 = D(6)
# Adds another die to results.
def add_dice(sums, die):
# If first die, all values appear once
if not sums:
for face in die:
sums[face] = 1
# Calculating the number of appearances for additional
# dice
else:
new_sums = {}
for k in sums.keys():
for f in die:
if new_sums.has_key(k+f):
new_sums[k+f] += sums[k]
else:
new_sums[k+f] = sums[k]
sums = new_sums
return sums
sums = add_dice({}, d6)
sums = add_dice(sums, d6)
sums = add_dice(sums, d6)
-- code ends --
IMO you are making things too complicated and not general
enough. Here is my proposal.
-
import operator
class Distribution(dict):
'''A distribution is a dictionary where the keys are dice
totals and the values are the number of possible ways
this total can come up '''
def __add__(self, term):
'''Take two distributions and combine them into one.'''
result = Distribution()
for k1, v1 in self.iteritems():
for k2, v2 in term.iteritems():
k3 = k1 + k2
v3 = v1 * v2
try:
result[k3] += v3
except KeyError:
result[k3] = v3
return result
def __rmul__(self, num):
tp = num * [self]
return reduce(operator.add, tp)
def D(n):
''' One die has a distribution where each result has
one possible way of coming up '''
return Distribution((i,1) for i in xrange(1,n+1))
sum3d6 = 3 * D(6)
sum2d6p2d4 = 2 * D(6) + 2 * D(4)
-
--
Antoon Pardon
--
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Re: Dice probability problem
Antoon Pardon wrote: IMO you are making things too complicated and not general enough. I believe that the above is very likely more than just your opinion :) Programming is just an occasional hobby to me, and I lack both experience and deeper (possibly a good chunk of shallow as well) knowledge on the subject. I'll study the code you posted, and make further questions if something remains unclear afterwards. Thanks, Tomi Lindberg -- http://mail.python.org/mailman/listinfo/python-list
Re: Dice probability problem
Tomi Lindberg [EMAIL PROTECTED] writes:
# Adds another die to results.
def add_dice(sums, die):
# If first die, all values appear once
I'd add something like
sums = sums or {}
because otherwise your function will sometimes mutate sums and sometimes
return a fresh object, which usually is a very bad thing as it can easily lead
to quite nasty bugs.
if not sums:
for face in die:
sums[face] = 1
# Calculating the number of appearances for additional
# dice
else:
new_sums = {}
for k in sums.keys():
for f in die:
if new_sums.has_key(k+f):
new_sums[k+f] += sums[k]
else:
new_sums[k+f] = sums[k]
sums = new_sums
return sums
'as
--
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Re: Dice probability problem
Antoon Pardon wrote: def __rmul__(self, num): tp = num * [self] return reduce(operator.add, tp) sum3d6 = 3 * D(6) One basic question: is there any particular reason not to use __mul__ instead (that would allow me to use both 3 * D(6) and D(6) * 3, while __rmul__ raises an AttributeError with the latter)? Difference between the two methods is slightly unclear to me. Thanks, Tomi Lindberg -- http://mail.python.org/mailman/listinfo/python-list
Dice probability problem
Hi,
I'm trying to find a way to calculate a distribution of
outcomes with any combination of dice. I have the basics
done, but I'm a bit unsure how to continue. My main concern
is how to make this accept any number of dice, without
having to write a new list comprehension for each case?
Here's a piece of code that shows the way I'm doing things
at the moment.
-- code begins --
# A die with n faces
D = lambda n: [x+1 for x in range(n)]
# A pool of 3 dice with 6 faces each
pool = [D(6)] * 3
# A List of all outcomes with the current 3d6 pool.
results = [x+y+z for x in pool[0] for y in pool[1]
for z in pool[2]]
# A dictionary to hold the distribution
distribution = {}
# If outcome is already a key, adds 1 to its value.
# Otherwise adds outcome to keys and sets its value
# to 1.
def count(x):
if distribution.has_key(x): distribution[x] += 1
else: distribution[x] = 1
# Maps the results with above count function.
map(count, results)
-- code ends --
Thanks,
Tomi Lindberg
--
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Re: Dice probability problem
Tomi Lindberg [EMAIL PROTECTED] writes: I'm trying to find a way to calculate a distribution of outcomes with any combination of dice. I have the basics done, but I'm a bit unsure how to continue. My main concern is how to make this accept any number of dice, without having to write a new list comprehension for each case? You need to think about the right way to break the problem down into some operation that can be repeated one fewer times than there are dice (if you just have a single dice, nothing needs to be done) and then repeat it. An obvious candidate is adding a single dice to the sums computed so far: def addDice(sums, dice): return [x+y for x in dice for y in sums] If you have less than 1 dice the answer is # len(pool) == 1 pool[0] After that, each time you add a dice you need to call addDice on the sum computed for all the previous dice and the new dice: # len(pool) == 2 addDice(resultFor1, pool[1]) addDice(pool[0], pool[1]) then # len(pool) == 3 addDice(resultFor2, pool[2]) addDice(addDice(resultFor1, pool[1]), pool[2]) addDice(addDice(pool[0], pool[1]), pool[2]) finally you get # len(pool) == n addDice(addDice(addDice(..., pool[n-3]), pool[n-2]) pool[n-1]) OK, so how do we get the repetition? Conveniently the pattern f(...f(f(x[0],x[1]),x[2])...,x[n-1]) or equivalently, if we write the infix operator * for f: x[0]*x[1]*...*x[n-1], can just be written as reduce(f, x) in python. So we get: reduce(addDice, pool) == reduce(lambda sums, dice: [x+y for x in dice for y in sums], pool) You should presumably also try writing this out as a single function, without using reduce (but recognizing the (left) reduction pattern is useful, even if you don't use python's reduce). 'as -- http://mail.python.org/mailman/listinfo/python-list
Re: Dice probability problem
Alexander Schmolck [EMAIL PROTECTED] writes: addDice(resultFor1, pool[1]) addDice(pool[0], pool[1]) sorry should have spelled out that successive lines are meant to be equivalent, i.e. addDice(resultFor1, pool[1]) == addDice(pool[0], pool[1]) 'as -- http://mail.python.org/mailman/listinfo/python-list
Re: Dice probability problem
Op 2006-04-04, Tomi Lindberg schreef [EMAIL PROTECTED]: Hi, I'm trying to find a way to calculate a distribution of outcomes with any combination of dice. I have the basics done, but I'm a bit unsure how to continue. My main concern is how to make this accept any number of dice, without having to write a new list comprehension for each case? IMO you are looking at it from the wrong side. It would be better to construct distributions for one die and make a function that can 'add' two distributions together. So for 3D6 you first add the distribution of a D6 to the distribution of a D6 and to this result you add the distribution of a D6 again. If you need more to start, just ask. Here's a piece of code that shows the way I'm doing things at the moment. -- code begins -- # A die with n faces D = lambda n: [x+1 for x in range(n)] # A pool of 3 dice with 6 faces each pool = [D(6)] * 3 # A List of all outcomes with the current 3d6 pool. results = [x+y+z for x in pool[0] for y in pool[1] for z in pool[2]] This is very inefficient. I wouldn't want to calculate the distribution of 10D10 this way. Try to think how you would do this with only D2's. (Triangle of Pascal) and generalize it. -- Antoon Pardon -- http://mail.python.org/mailman/listinfo/python-list
Re: Dice probability problem
First, thanks to Antoon and Alexander for replying.
Antoon Pardon wrote:
It would be better to construct distributions for one
die and make a function that can 'add' two distributions
together.
As both replies pointed to this direction, I tried to take
that route. Here's the unpolished code I came up with. Does
it look even remotely sane way to accomplish my goal?
-- code begins --
# A die with n faces
D = lambda n: [x+1 for x in range(n)]
# A new die with 6 faces
d6 = D(6)
# Adds another die to results.
def add_dice(sums, die):
# If first die, all values appear once
if not sums:
for face in die:
sums[face] = 1
# Calculating the number of appearances for additional
# dice
else:
new_sums = {}
for k in sums.keys():
for f in die:
if new_sums.has_key(k+f):
new_sums[k+f] += sums[k]
else:
new_sums[k+f] = sums[k]
sums = new_sums
return sums
sums = add_dice({}, d6)
sums = add_dice(sums, d6)
sums = add_dice(sums, d6)
-- code ends --
Thanks,
Tomi Lindberg
--
http://mail.python.org/mailman/listinfo/python-list
Re: Dice probability problem
Tomi Lindberg wrote: # A die with n faces D = lambda n: [x+1 for x in range(n)] That can be written: D = lambda n : range(1,n+1) Gerard -- http://mail.python.org/mailman/listinfo/python-list
Re: Dice probability problem
That looks reasonable. The operation you are implementing is known as 'convolution' and is equivalent to multiplying polynomials. It would be a little more general if you had the input 'die' be a sequence of the count for each outcome, so d6 would be [1]*6 (or [0]+[1]*6 if you prefer). That would allow allow you to represent unfair dice and also to add not just a die to a distribution, but to add any two distributions, so you can play tricks like computing 16d6 as (d6)*2*2*2*2. (The code above is a convolution that restricts the second distribution 'die' to have only 0 and 1 coefficients.) The general convolution can be implemented much like what you have, except that you need another multiplication (to account for the fact that the coefficient is not always 0 or 1). My not particularly efficient implementation: def vsum(seq1, seq2): return [ a + b for a, b in zip(seq1, seq2) ] def vmul(s, seq): return [ s * a for a in seq ] # Convolve 2 sequences # equivalent to adding 2 probabililty distributions def conv(seq1, seq2): n = (len(seq1) + len(seq2) -1) ans = [ 0 ] * n for i, v in enumerate(seq2): vec = [ 0 ] * i + vmul(v, seq1) + [ 0 ] * (n - i - len(seq1)) ans = vsum(ans, vec) return ans # Convolve a sequence n times with itself # equivalent to multiplying distribution by n def nconv(n, seq): ans = seq for i in range(n-1): ans = conv(ans, seq) return ans print nconv(3, [ 1 ] * 6) print nconv(3, [ 1.0/6 ] * 6) print nconv(2, [ .5, .3, .2 ]) [1, 3, 6, 10, 15, 21, 25, 27, 27, 25, 21, 15, 10, 6, 3, 1] [0.0046296296296296294, 0.013888, 0.027776, 0.046296296296296294, 0.069448, 0.097238, 0.11574074074074074, 0.125, 0.125, 0.11574074074074073, 0.097224, 0.069448, 0.046296296296296294, 0.027776, 0.013888, 0.0046296296296296294] [0.25, 0.2, 0.29004, 0.12, 0.040008] -- http://mail.python.org/mailman/listinfo/python-list
