Dice Roll (Find out how many rolls until specified streak)

[DustinKlent]

I am trying to figure out how many rolls of dice it would take to get a specified number of snake eye rolls. For instance if I want to get snake eyes 5 times in a row and roll until I get it, how many rolls would it take when it's random?

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import random   streaks = 0 a = random.randint(1, 6) b = random.randint(1, 6) rollnum = 1 while a + b != 2:     a = random.randint(1, 6)     b = random.randint(1, 6)     rollnum +=1     if a + b == 2:         streaks += 1         print(f"number of rolls: {rollnum}")

So far my code is very basic as I'm a beginner, but what I'd like to do is have a variable that the user can specify (x) and then the dice will roll again and again until it hits snake eyes x times in a row. Then it tells you how many times it had to roll before it hit snake eyes x times in a row.

[Gribouillis]

You could learn generators for this task. For example

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>>> from random import randint >>> from itertools import islice >>> def rolls(): ...     while True: ...         yield 1 if (randint(1, 6) + randint(1, 6) == 2) else 0 ... >>> list(islice(rolls(), 100)) [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

In this code, rolls() is an infinite sequence of 0s and 1s obtained by rolling two dice and looking for a snake eyes. Here 1 means a success an 0 a failure.

Next you could use a function that transforms such a sequence by summing consecutive 1s, for example

Output:
0 0 1 1 1 0 0 1 0 1 1 0 0 0 1 1 1 1 0 0 1   becomes
0 0 1 2 3 0 0 1 0 1 2 0 0 0 1 2 3 4 0 0 1

Here is how you can write such an infinite sequence

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>>> def sum_consec(seq):                                                                     ...     s = 0                                                                                ...     for x in seq:                                                                        ...         if x: s += 1                                                                      ...         else: s = 0                                                                      ...         yield s                                                                          ...   >>> list(sum_consec([0, 0, 1, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1])) [0, 0, 1, 2, 3, 0, 0, 1, 0, 1, 2, 0, 0, 0, 1, 2, 3, 4, 0, 0, 1] >>> >>> target = 4 >>> for i, v in enumerate(sum_consec(rolls())): ...     if v == target: ...         print(i) ...         break ... 4698374 >>>

We rolled 4,698,375 times the dice before getting 4 snake eyes in a row!

[Yoriz]

You could break it down into functions.
I've given an outline below for you to fill in the gaps.

create a constant variable for the rolled dice pattern you would like to match

1	SNAKE_EYES = (1, 1)

Make a function to roll dice

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def roll_dice(num_of_dice):     ....     return # returns the rolled dice, possible result (1, 1) if called as roll_dice(2)

Make a function that counts how many rolls to find a match that can be called like this roll_till_match(SNAKE_EYES)

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def roll_till_match(match):    num_of_dice = len(match)     ....    # uses roll_dice(num_of_dice) in a loop checking if the result is the same as match     return # count of rolls till the match

Another function that can call roll_till_match x amount of times that can be called like this roll_till_match_x_times(SNAKE_EYES, 5)

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def roll_till_match_x_times(match, x_times):    ...    # uses roll_till_match(match) in a loop of x_times    return # total sum of rolls

The same functions can be used against other matches.

[deanhystad]

I like to start out by first solving the problem the same way I would solve it if I were doing so without a computer.

1. Set count and snake eye count = 0
2. Roll the dice
3. If roll is snake eyes, increment snake eye count by 1, else set count to zero
4. Increment count by 1
5. Repeat 2 through 4 until snake eye count = 5

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import random   def snake_eyes(count):     streak_cnt = 0  # Step 1     overall_cnt = 0 # Step 1     while streak_cnt < count: # Step 5         roll = (random.randint(1, 6), random.randint(1, 6)) # Step 2         if roll == (1, 1): # Step 3             streak_cnt += 1  # Step 3         else: # Step 3             streak_cnt = 0 # Step 3         overall_cnt += 1 # Step 4     return overall_cnt # Step 5   print(snake_eyes(5))

Output:
6984725

When you roll 2 dice there are 36 possible outcomes and only 1 of those is snake eyes. Your probability of rolling snake eyes is 1/36 and your probability of rolling snake eyes 5 times in a row is (1/36)^5. On average you will need to roll the dice 60, 466, 176 times to roll 5 snake eyes in a row. My second time running the program it printed 12297433 and the third time 57268606.

These numbers are all suspiciously low, so I modified the code to calculate how long it takes to roll snake eyes and see if that agrees with my estimate of 1/36.

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import random import statistics   def snake_eyes(count):     streak_cnt = 0     overall_cnt = 0     while streak_cnt < count:         roll = (random.randint(1, 6), random.randint(1, 6))         if roll == (1, 1):             streak_cnt += 1         else:             streak_cnt = 0         overall_cnt += 1     return overall_cnt   print(statistics.mean([snake_eyes(1) for _ in range(100000)]))

Running this code I consistently get numbers very close to 36. I repeated with fewer iterations for rolling snake-eyes twice in a row (1296 times expected, 1292 times measured) for three times in a row (46,656 times expected, 47,955 times measured). Finally I ran the test for 5 consecutive snake-eyes. The average of 100 tests was 56, 0711,113 times with a high of 102,260,693 and a low of 4,217,949.

[Gribouillis]

Here is a shorter version of the infinite sequences method

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>>> from random import randint >>> import itertools as itt >>> rolls = (randint(1, 6) + randint(1, 6) == 2 for _ in itt.repeat(None)) >>> sum_consec = (s := x and s + 1 for x in rolls) >>> target = 4 >>> next(itt.dropwhile((lambda t: t[1] < target), enumerate(sum_consec, 1)))[0] 2378988