[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)
>
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
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
[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 ge
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 in
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 pro
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 w
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.py
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
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 al
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 A
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, whi
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) know
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
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 a
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
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 co
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, wi
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
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 ne
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'
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