Rolling 100-sided "dice" until they are the same number

Question

I've just begun my journey into python (1 week now) and I've written couple of lines incorporating everything I've learned so far. It would be greatly appreciated if anyone could guide me in writing a better formatted (appropriate style) version of this.

I've just written a couple of lines on finding two random rolls that stop when they roll the same number. When the same number is rolled, that number is recorded along with how many rolls it took to get there. I've also made the code to find those values and store them in a list, finding the average of roll number(same number) as well as roll count(How many rolls it took). Here is the code:

import random totalrolls = 0 rollList = [] rollcountlist = [] while totalrolls < 500: totalrolls = totalrolls + 1 rollcount = 0 roll = random.randint(1,100) rolltwo = random.randint(1,100) while roll != rolltwo: roll = random.randint(1,100) rolltwo = random.randint(1,100) #print roll #print rolltwo rollcount = rollcount + 1 if roll == rolltwo: rollList.append(roll) rollcountlist.append(rollcount) print "Random number that matched was: %s, after %s rolls. " % (roll, rollcount) if totalrolls == 500: print "Finished rolling %s times." % totalrolls averageroll = sum(rollList) / len(rollList) averagerollcount = sum(rollcountlist) / len(rollcountlist) print rollList print rollcountlist print averageroll print averagerollcount

200_success · Accepted Answer · 2017-11-02 23:31:05Z

You have an off-by-one error. If you get lucky and get a matching pair immediately, then you would report "Random number that matched was: …, after 0 rolls." I would say that "after 1 roll(s)" would be the right way to report that. (Alternatively, you could phrase it as "after 0 mismatched rolls".)

Naming could be improved slightly. I suggest:

rollcountlist → roll_counts, because it's shorter and more readable. Since englishisnotgerman, you should use an underscore, as recommended by PEP 8.
rollList → roll_values, because "values" makes it clearer what is being stored. Watch your capitalization consistency.
averageroll → average_value and averagerollcount → average_count, for similar reasons.
totalrolls → experiment_count, because totalrolls sounds like it represents a cumulative sum of rollcount, which it isn't. Introducing the term "experiment" makes it clear that you are counting something else.
roll → roll1 and rolltwo → roll2, for consistency. Or, as I've done below, make one two-element tuple called rolls.

The if conditions after the while loops are superfluous. The inner loop can only be exited when the roll values match. The outer loop can only be exited after 500 experiments.

Counting loops can be more elegantly expressed. For a fixed number of iterations, such as for the outer loop, use the idiom for i in xrange(n): …. For an unbounded count, such as for the inner loop, use for i in itertools.count().

In Python 2, the / operator performs integer division (rounding down) when dividing an integer by an integer. For a more accurate mean, convert at least one of the operands to a float first.

When interpolating an integer into a format string, use the %d placeholder instead of %s for clarity.

from itertools import count import random roll_values = [] roll_counts = [] for experiment_count in xrange(1, 501): for roll_count in count(1): rolls = (random.randint(1,100), random.randint(1, 100)) if rolls[0] == rolls[1]: break roll_values.append(rolls[0]) roll_counts.append(roll_count) print "Random number that matched was: %d, after %d rolls." % (rolls[0], roll_count) average_value = float(sum(roll_values)) / len(roll_values) average_count = float(sum(roll_counts)) / len(roll_counts) print "Finished rolling %d times." % experiment_count print roll_values print roll_counts print average_value print average_count

Wow. Thank you so much for your help! It was exactly what I needed to learn python but also to learn the most efficient way and the smarter way to write python. — John Lee
– John Lee, Commented Nov 3, 2017 at 12:36

Ludisposed · Accepted Answer · 2017-11-03 10:42:20Z

@200_succes has some very good advise, but I think I can add some more.

Originally this was supposed to be a comment, but it got to long.

the % has a few issues, better to use .format(). This will work in both python2 and python3
Even better yet, why not use python3 ?

Especially if you're new learning Python, I suggest to learn python3. Support for python2 will be dropped soons since Python2 EOL (End Of Life) is just a few years away.

It seems silly to learn a language that will be dropped soon.

I just learned about .format() and I'll definitely give that a go. — John Lee
– John Lee, Commented Nov 3, 2017 at 12:36
and yes, seems like learning python 3 would be a better idea. — John Lee
– John Lee, Commented Nov 3, 2017 at 12:37
@JohnLee Not a problem, happy to help. Your code will work in python3 if you change your print xxx statements to print (xxx) and use .format() — Ludisposed
– Ludisposed, Commented Nov 3, 2017 at 12:40

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