def how_many_times():
x, y = 0, 1
c = 0
while x != y:
c = c + 1
x, y = roll()
return c, (x, y)
Since I haven't seen it used in answers yet, here's another option using our
new walrus operator
def how_many_times():
roll_count = 1
while (rolls := roll())[0] != rolls[1]:
roll_count += 1
return (roll_count, rolls)
I would go even further, saying there is no need to «roll dices»:
def how_many_times():
nb_times = random.choice([n for n in range(50) for _ in
range(round(10000*(1/6)*(5/6)**(n-1)))])
return nb_times, (random.randint(1, 6),) * 2
If i had more time on my hands, i would do something with bissect to get
nb_times with more precision, as i have (mis)calculated that the
probability of having nb_times = N is N = (1/6) * (5/6) ** (N-1)
Something like this may work:
nb_times = [random.random() < (1/6) * (5/6) ** (N-1) for N in
range(1, 50)].index(True)+1
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