P: n/a

I am brand new to Python (this is my second day), and the only
experience I have with programming was with VBA. Anyway, I'm posting
this to see if anyone would be kind enough to help me with this (I
suspect, very easy to solve) query.
The following code is in a file which I am running through the
interpreter with the execfile command, yet it yeilds no results. I
appreciate I am obviously doing something really stupid here, but I
can't find it. Any help appreciated.
def d6(i):
roll = 0
count = 0
while count <= i:
roll = roll + random.randint(1,6)
count += 1
return roll
print d6(3)  
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David Barr wrote:
I am brand new to Python (this is my second day), and the only
experience I have with programming was with VBA. Anyway, I'm posting
this to see if anyone would be kind enough to help me with this (I
suspect, very easy to solve) query.
The following code is in a file which I am running through the
interpreter with the execfile command, yet it yeilds no results. I
appreciate I am obviously doing something really stupid here, but I
can't find it. Any help appreciated.
def d6(i):
roll = 0
count = 0
while count <= i:
roll = roll + random.randint(1,6)
count += 1
return roll
print d6(3)
A) your direct answer: by using <=, you are rolling 4 dice, not 3.
B) Much more pythonic:
import random
def d6(count):
result = 0
for die in range(count):
result += random.randint(1, 6)
return result
Scott David Daniels Sc***********@Acm.Org  
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On Thu, 20070906 at 11:00 0700, Scott David Daniels wrote:
def d6(count):
result = 0
for die in range(count):
result += random.randint(1, 6)
return result
This, of course, can be further improved into:
def d6(count):
return sum(random.randint(1, 6) for die in range(count))

Carsten Haese http://informixdb.sourceforge.net  
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Carsten Haese wrote:
On Thu, 20070906 at 11:00 0700, Scott David Daniels wrote:
>def d6(count): result = 0 for die in range(count): result += random.randint(1, 6) return result
This, of course, can be further improved into:
def d6(count):
return sum(random.randint(1, 6) for die in range(count))
My stab at it:
>>def roll(times=1, sides=6):
... return random.randint(times, times*sides)
Ian  
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On Thu, 20070906 at 11:24 0700, Ian Clark wrote:
Carsten Haese wrote:
def d6(count):
return sum(random.randint(1, 6) for die in range(count))
My stab at it:
>>def roll(times=1, sides=6):
... return random.randint(times, times*sides)
That produces an entirely different probability distribution if times>1.
Consider times=2, sides=6. Your example will produce every number
between 2 and 12 uniformly with the same probability, 1 in 11. When
rolling two sixsided dice, the results are not evenly distributed. E.g.
the probability of getting a 2 is only 1 in 36, but the probability of
getting a 7 is 1 in 6.

Carsten Haese http://informixdb.sourceforge.net  
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Carsten Haese wrote:
On Thu, 20070906 at 11:24 0700, Ian Clark wrote:
>Carsten Haese wrote:
>>def d6(count): return sum(random.randint(1, 6) for die in range(count))
My stab at it:
> >>def roll(times=1, sides=6):
... return random.randint(times, times*sides)
That produces an entirely different probability distribution if times>1.
Consider times=2, sides=6. Your example will produce every number
between 2 and 12 uniformly with the same probability, 1 in 11. When
rolling two sixsided dice, the results are not evenly distributed. E.g.
the probability of getting a 2 is only 1 in 36, but the probability of
getting a 7 is 1 in 6.
Doh. I stand corrected. Probability was never a fun subject for me. :)
Ian  
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Scott David Daniels wrote:
David Barr wrote:
>I am brand new to Python (this is my second day), and the only experience I have with programming was with VBA. Anyway, I'm posting this to see if anyone would be kind enough to help me with this (I suspect, very easy to solve) query.
The following code is in a file which I am running through the interpreter with the execfile command, yet it yeilds no results. I appreciate I am obviously doing something really stupid here, but I can't find it. Any help appreciated.
def d6(i): roll = 0 count = 0 while count <= i: roll = roll + random.randint(1,6) count += 1
return roll
print d6(3)
A) your direct answer: by using <=, you are rolling 4 dice, not 3.
B) Much more pythonic:
import random
def d6(count):
result = 0
for die in range(count):
result += random.randint(1, 6)
return result
Scott David Daniels Sc***********@Acm.Org
I was surprised by the speed and number of posts. Thanks for the
solutions provided!
>>def roll(times=1, sides=6):
.... return random.randint(times, times*sides)
Although this would probably be quicker than the other approaches, I'm
not using the dice to generate numbers per say, I actually want to
emulate the rolling of dice, bellcurve (normal distribution) as well as
the range.
Thanks again, I already like what (very) little I can do in Python and
it seems to have a great community too.
Cheers,
Dave.  
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On Sep 6, 10:29 am, David Barr <david.barr...@btinternet.comwrote:
yields no results.
Since every response so far has answered everything __Except The
Question You Asked__, your code runs fine on my Linux machine and
prints 15. The error may be before this bit of code so it isn't
getting called. Add some print statements and try again
def d6(i):
print "start of d6()"
roll = 0
count = 0
while count <= i:
print "d6 count =", count, "of", i
roll = roll + random.randint(1,6)
count += 1
print "returning roll =", roll
return roll
print d6(3)  
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On Sep 6, 1:44 pm, Carsten Haese <cars...@uniqsys.comwrote:
On Thu, 20070906 at 11:24 0700, Ian Clark wrote:
Carsten Haese wrote:
def d6(count):
return sum(random.randint(1, 6) for die in range(count))
My stab at it:
>>def roll(times=1, sides=6):
... return random.randint(times, times*sides)
That produces an entirely different probability distribution if times>1.
Consider times=2, sides=6. Your example will produce every number
between 2 and 12 uniformly with the same probability, 1 in 11. When
rolling two sixsided dice, the results are not evenly distributed. E.g.
the probability of getting a 2 is only 1 in 36, but the probability of
getting a 7 is 1 in 6.

Carsten Haesehttp://informixdb.sourceforge.net
Why settle for a normal distribution?
import random
def devildice(dice):
return sum([random.choice(die) for die in dice])
hist = {}
for n in xrange(10000):
the_key = devildice([[1,2,3,10,11,12],[4,5,6,7,8,9]])
if the_key in hist:
hist[the_key] += 1
else:
hist[the_key] = 1
hkey = hist.keys()
m = max(hkey)
n = min(hkey)
histogram = [(i,hist.get(i,0)) for i in xrange(n,m+1)]
for h in histogram:
print '%3d %s' % (h[0],'*'*(h[1]/100))
## 5 **
## 6 *****
## 7 ********
## 8 ********
## 9 ********
## 10 *******
## 11 *****
## 12 **
## 13
## 14 **
## 15 ******
## 16 ********
## 17 ********
## 18 ********
## 19 ********
## 20 *****
## 21 **
They're called Devil Dice because the mean is 13 even
though you cannot roll a 13.   This discussion thread is closed Replies have been disabled for this discussion.   Question stats  viewed: 1156
 replies: 8
 date asked: Sep 6 '07
