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Math Help - Am I more likely to roll at least one six if I have more dice?

  1. #1
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    Am I more likely to roll at least one six if I have more dice?

    Hi,

    I've had a debate with some friends, about whether I'm more likely to roll at least a single 6 when I have 1000 dice, than I am to roll one with only 1 dice.

    I say, the more dice I have, the more likely it is to roll at least a single 6. The others try to convince me, that the possibility to roll at least one 6 is always 1/6.

    Now my thoughts for this are as follows:
    At first, it doesn't matter, whether I roll them one after another, or one at a time.
    The chances to roll a six on the first dice that I roll a 6, is 1/6.

    Now, I don't care about the others if I did roll a 6, but I do care if I didn't (which is in 5/6 of all cases)

    So for the second dice roll a six, the probability is still 1/6
    However, I'm wondering about the probability that the first dice does NOT show a six, but the second does, so the probability for that to happen is 5/6*1/6.

    So the probability to roll at least one 6 with two dice is:
    1/6 + 1/6*5/6

    If I was to expand this for 1000 dice, I would end up with:

    $$\sum_{k=0}^{999}\left (\frac{5}{6} \right )^k \times \frac{1}{6}$$
    The probability will get ever more closely to 100%, the more dice I have, correct?
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  2. #2
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    Why get 1000 dice? Same thing if you have 1 dice: just keep rolling it.
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  3. #3
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    Well, originally I asked this: If you were to roll 1000 dice at once, how likely is it, that there is at least one 6?

    But it doesn't really matter
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  4. #4
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    The possibility of rolling a 6 with one die is 1/6, and for each die you have you increase your changes of rolling 1/6. By your friend's reasoning, you're no more liking to have "at least one head (H) in two coin flips" than in just one coin flip. Let us use this much more basic example. The probability of flipping a head of an unbiased coin is 1/2 because we split the event space {T, H} evenly into two. When we flip two coins we have a new event space, namely {TT, TH, HT, HH}. Here the events are equally likely of 1/4, but notice "at least one H" can turn out in three different events. Thus, the probability of "at least one H" is 3/4.

    The same works for dice. With one die we have an event space {1, 2, 3, 4, 5, 6}. With two dice the size of the event space is 36:

    Spoiler:
    11 12 13 14 15 16
    21 22 23 24 25 26
    31 32 33 34 35 36
    41 42 43 44 45 46
    51 52 53 53 55 56
    61 62 63 64 65 66


    Now, in how many ways can we have "at least one six?" In this small, but sufficient, example, we can just look at see. The bottom row and the right-most column all include a 6. This is 11/36, which is almost 0.31, compared to 1/6 =0.167. So yes, increasing the die increases your chances of rolling a six.
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  5. #5
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    Thanks, that would confirm my point of view.

    They: But how is that possible? The probability never changes? So it should always be 1/6?

    Also, we'd value second and third opinions Thank you very much in advance!
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  6. #6
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    The probability of a 6 emerging on a single die does not ever change, but that is not the event we're looking at when we have more than one die. When we are using, say, two dice we're looking at a different event space with each event having 1/36 chance of occurring. Of those thirty-six, eleven of them will have a 6 appear. This is undeniable. This has no impact on the probability of a six appearing on a single die. Your friends seem to be ignoring the fact that the possible ways of a 6 occurring change when you have two dice. With one die there is one, and only one, way of a 6 occurring. Thus, the probability is 1/6. With two dice we have 11 different ways, each with 1/36 = (1/6)(1/6) chance of happening. The probability of the individual dice does support this result, as we can see, but you cannot ignore the fact our event space is different when you include another die.

    I can go even further and define it this way by the inclusion-exclusion rule of probability theory. Let X be the outcome of the first die and Y the outcome of the second. What we want then is:

    P(X = 6\ or\ Y=6)=P(X=6)+P(Y=6)-P(X=6\ \&\ Y= 6)=\frac{1}{6}+\frac{1}{6}-\frac{1}{36}=\frac{11}{36}

    Do you see why these values work? The probability of the single die being six does not change, but we have to sum the two probabilities together. We then have to subtract the event that is common to both of them, otherwise we're double counting that event. This happens only once.
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  7. #7
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    Quote Originally Posted by theyThinkImWrong View Post
    Hi,

    I've had a debate with some friends, about whether I'm more likely to roll at least a single 6 when I have 1000 dice, than I am to roll one with only 1 dice.

    I say, the more dice I have, the more likely it is to roll at least a single 6. The others try to convince me, that the possibility to roll at least one 6 is always 1/6.

    Now my thoughts for this are as follows:
    At first, it doesn't matter, whether I roll them one after another, or one at a time.
    The chances to roll a six on the first dice that I roll a 6, is 1/6.

    Now, I don't care about the others if I did roll a 6, but I do care if I didn't (which is in 5/6 of all cases)

    So for the second dice roll a six, the probability is still 1/6
    However, I'm wondering about the probability that the first dice does NOT show a six, but the second does, so the probability for that to happen is 5/6*1/6.

    So the probability to roll at least one 6 with two dice is:
    1/6 + 1/6*5/6

    If I was to expand this for 1000 dice, I would end up with:

    $$\sum_{k=0}^{999}\left (\frac{5}{6} \right )^k \times \frac{1}{6}$$
    The probability will get ever more closely to 100%, the more dice I have, correct?
    Let X be the random variable 'Number of 6's rolled'.

    Then X ~ Binomial(n, p = 1/6).

    Calculate Pr(X > 0) = 1 - Pr(X = 0). Then it is obvious that as n increases so does the probability.

    I suggest you review the binomial distribution if you don't know it ( I do not plan to give a tutorial).
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