# Beginners Query - Simple counter problem

Discussion in 'Python' started by David Barr, Sep 6, 2007.

1. ### David BarrGuest

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)

David Barr, Sep 6, 2007

2. ### Scott David DanielsGuest

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

Scott David Daniels, Sep 6, 2007

3. ### Carsten HaeseGuest

This, of course, can be further improved into:

def d6(count):
return sum(random.randint(1, 6) for die in range(count))

Carsten Haese, Sep 6, 2007
4. ### Ian ClarkGuest

My stab at it:
... return random.randint(times, times*sides)

Ian

Ian Clark, Sep 6, 2007
5. ### Carsten HaeseGuest

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, Sep 6, 2007
6. ### Ian ClarkGuest

Doh. I stand corrected. Probability was never a fun subject for me.

Ian

Ian Clark, Sep 6, 2007
7. ### David BarrGuest

I was surprised by the speed and number of posts. Thanks for the
solutions provided!
.... 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.

David Barr, Sep 6, 2007

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)

9. ### mensanatorGuest

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.

mensanator, Sep 7, 2007
10. ### Dennis Lee BieberGuest

Many responses on coding the d6 function... But I have to ask: WHY
are you using execfile!
--
Wulfraed Dennis Lee Bieber KD6MOG

HTTP://wlfraed.home.netcom.com/
(Bestiaria Support Staff: )
HTTP://www.bestiaria.com/

Dennis Lee Bieber, Sep 7, 2007