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# Beginners Query - Simple counter problem

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
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)
Sep 6 '07 #1
8 1352
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
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
Sep 6 '07 #2
On Thu, 2007-09-06 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
Sep 6 '07 #3
Carsten Haese wrote:
On Thu, 2007-09-06 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

Sep 6 '07 #4
On Thu, 2007-09-06 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 six-sided 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
Sep 6 '07 #5
Carsten Haese wrote:
On Thu, 2007-09-06 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 six-sided 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

Sep 6 '07 #6
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
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, bell-curve (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.
Sep 6 '07 #7
On Sep 6, 10:29 am, David Barr <david.barr...@btinternet.comwrote:
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)

Sep 6 '07 #8
On Sep 6, 1:44 pm, Carsten Haese <cars...@uniqsys.comwrote:
On Thu, 2007-09-06 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 six-sided 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.

Sep 6 '07 #9

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