• M0

    Sir I have a question about probability.

    Question: What is the probability that you roll a dice 100 times and the sum is a multiple of 2 after every 5 turns?

    Sir could you please explain me how to solve this problem !

    Thank you

  • MOD M0 M1 M2 M3 M4 M5


    This is a quite tricky problem! A multiple of 2 means an even number, because all even numbers are divisible by 2. In this problem, we want the sum of all numbers from each turn to be an even number after every 5 turns. For example, if the first 10 rolls turn out to be 3, 5, 2, 3, 1, 2, 6, 4, 1, 3, then the sum for each turn would be: 3, (3+5=8), (3+5+2=10), (3+5+2+3=13), (3+5+2+3+1=14) and so on.

    Now, when do we get an even number after adding two numbers?
    Let's take a simple example:
    1+1=2 (odd+odd=even)
    1+2=3 (odd+even=odd)
    2+2=4 (even+even=even)
    Here, we can see that the sum of 2 odd numbers and the sum of 2 even numbers result in even numbers. This means that the dice roll after every 5 turns should have the same parity (even/odd) as the turn just before it, in order to get an even number.

    To calculate the probability, let's use this equation:
    Probability = (Number of possible cases where the sum after every 5 turns is even) / (Total # of possible outcomes)

    The Total # of possible outcomes = \( 6^{100} \), because a dice has 6 possible outcomes per roll and we're rolling the dice 100 times.

    To calculate N, the number of possible cases in which the sum after every 5 turns:
    First, how many 5th turns are there in 100 rolls? There are 100/5=20.
    Hence, since there are 3 evens and 3 odds in a dice:
    \( N = (6\times6\times6\times6\times3)^{20}\)
    \( N = 6^{80}\times3^{20} \)
    So the Probability = \( \frac{6^{80}\times3^{20}}{6^{100}} = \frac{3^{20}}{6^{20}} = \frac{1}{2^{20}}\).

    In short, all other rolls besides every 5th roll are not important, so we only have to count the important turns. Since every 5th roll should either be even or odd, the probability=1/2. For 100 rolls, there would be 20 5th turns, so the total probability=\(\frac{1}{2^{20}}\).

    I hope this helps. Please let me know if you have any other questions!

    Happy Learning,
    The Daily Challenge Team

  • M0

    @minji Why did you multiply 6 four times followed by 3 - sorry I am quite dum and confused at the probability topic

  • ADMIN M0★ M1 M5

    @fabulousgrizzly I think minji meant that in a set of five rolls, it doesn't matter what the first four rolls are. It only matters what the parity (oddness or evenness) of the sum of the first four rolls is. See, if the sum of the first four rolls is even (e.g. \(1 + 2 + 3 +4 = 10),\) then we will want the fifth roll to be also even (e.g. \(10 + \text{even} = \text{even}\)), and if the sum of the first four rolls is odd (e.g. \(1 +2 + 3 + 3 = 9),\) then we will want the fifth roll to be odd as well (since \(9 + \text{odd} = \text{even}\)). So since we don't care what the first four numbers are, there are \(6\) choices for each of those, and since the last roll has to match the parity (oddness or evenness) of the running total, there are \(3\) choices for that number.

    Actually, I have another way of solving this question! 📒 🙂

    Each dice has three odd numbers and three even numbers. The chance of getting an even is \(\frac{1}{2},\) and the chance of getting an odd is \(\frac{1}{2}.\) How about not worrying about what the numbers are, and only keeping track of whether the number on the dice is even or odd? Then, looking at a set of five consecutive rolls, we can describe this like this:

    $$ EOEEO$$

    Now, we want the sum of these five numbers to be even, so there are some cases:

    • There are zero odds (\(O\))

    • There are two odds (\(OO\))

    • There are four odds (\(OOOO\))

    How many ways are there for each case?

    • If there are zero odds, there is \(1\) way (\(EEEEE\))

    • If there are two odds, this is like choosing \(2\) spots out of \(5\) to declare to be odds, which has \(\binom{5}{2} = \frac{5 \times 4 }{2} = 10\) ways

    • If there are four odds, this is like choosing \(4\) spots out of \(5\) to declare to be odds, or, more simply, like choosing \(1\) spot out of \(5\) to be even, so this is equal to \(\binom{5}{1} = 5\) ways

    In all, there are \( 1 + 10 + 5 = 16\) ways to get an even sum after five consecutive rolls.

    Now, the total number of ways is the ways to create a line of five letters like

    $$ EOOEO$$

    with no restriction. Each spot has \(2\) choices (\(E\) or \(O\),) so there are \(2 \times 2 \times 2 \times 2 \times 2 = 32\) ways total.

    Thus the probability of getting an even sum after five consecutive rolls is equal to \(\frac{16}{32} = \frac{1}{2},\) which might also make sense if you just think about the fact that the parity of the sums should be balanced over even and odd!

    Now we just need to stretch this to the case where we have \(100\) rolls. Well, if every five rolls has an even sum, then the total sum over \(100\) rolls should also be even! By the multiplicative rule, since we can do a set of five rolls \(20\) times, the probability overall is \(\boxed{\frac{1}{2}^{20}.}\)

    I hope that made sense! This was a fun problem. 🙂 I'm glad that you asked it!