on writing a while loop for rolling two dice

[lucas]

>> 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