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A fair die is rolled 100 times. Which of the following has a probability of at least 95%?

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1.) Sum of the rolls is greater than 322

2.) Sum of the rolls is less than 392

3.) Number of rolls with even outcomes is exactly 56

4.) Number of rolls with odd outcomes is less than 53

5.) Number of rolls with outcomes are fully divisible by 3 are exactly 38

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I'm not looking for a full solution, I'm looking for hints as to how to approach this question.

I have an idea for 3 and 4. Since the probability is asking "odd or "even" could I use a a binomial distribution to calculate the probability? (even is success, and odd is otherwise)

For 1 and 2, after modelling 100 tosses in wolfram it seems to be a normal type distribution. Am I correct?

For the other ones I'm still thinking of what to use to prove them. Again, hints are more useful to me than full solutions. Thanks!

RonaldB
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  • What's your intention here? Finding the true statement is easier than proving it, which is in turn much easier than proving the others false. (Assuming that the singular "has" is meant to imply that only one is true.) – joriki Jun 21 '16 at 11:06
  • Finding the one event, but I'd like to know how to go through the list and proving each one right/wrong. – RonaldB Jun 21 '16 at 11:07
  • if we assume the "fair" die gives numbers with equal probability (0.16 for each number) then the sum comes around 336...is this right? – boxed__l Jun 21 '16 at 11:07
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    @boxed: A fair die has an average of $3.5$ on a single roll, and thus an average of $350$ on $100$ rolls. – joriki Jun 21 '16 at 11:08
  • If you just wanted to answer this correctly as a multiple choice question, an efficient approach would be: For $100$ dice, it's safe to assume that the probability that anything takes any particular exact value is less than $95%$. That excludes $3$ and $5$. We expect $50$ odd outcomes, and it seems implausible that it's less than $5%$ likely to get at least $3$ more than that. That leaves only $1$ and $2$, and if only one of those is right, it must be the one with the higher probability, i.e. larger distance of the bound from the average. – joriki Jun 21 '16 at 11:11
  • Just to reiterate and makes sure I under what you're saying.

    The expected value is 350, since for a single fair roll of a dice the EV is 3.5, and the EV for getting an odd or even outcome is also 50/50, correct?

    We're left with 1 and 2, and since 392 is farther from the EV, would that mean that we're left with 1? Or do we want the furthest from the EV?

     – RonaldB Jun 21 '16 at 11:16
  • @EricTressler I would think nearer to it's expected value since its approx. a normal distribution. – RonaldB Jun 21 '16 at 11:21
  • @EricTressler: That last clause isn't necessary -- it's always more (or at least as) likely to be at least $308$ than to be at least $322$, independent of the expected value. – joriki Jun 21 '16 at 13:55

2 Answers2

1

The correct should be $2)$. Those kind of questions can be dealt by finding the mean and variance of your distribution and then say it approximates a normal distribution, and then use facts known for normal distributions.

Specifically define $X=\sum_{i=1}^{100}X_i$ where $X_i$ are i.i.d., and $X_i=i$ with probability $1/6$ for $i\in \{1,2,\cdots,6\}$.

Then $E(X)=100E(X_1) = 350$ and $V(x) = 100V(X_1)\sim 100(2,75)=275$. And $\sigma = \sqrt{275} \sim 17$. Now notice that $392 -350 = 42 \sim \frac{5}{2} \sigma$.

Therefore $\mathbb{P} (X \leq 392)$ is around equal to the probability that you observe something less than $\mu +2.5 \sigma$ which is more that $97\%$.

1) You can exclude this because $350 - 322 = 28$ which is less that $42$, since the distribution is symmetric, the probability there is less than 2). Aside from this useful observation this probability could have been again estimated by using the same analysis as in 2).

3) This sounds really rare.

4) A similar argument as in 1) by defining $X$ the sum counting the number of odd rolls. Gives $E(X)=50$ and $V(X)= 100 \frac{1}{4}=25$. Therefore the probability of seeing something less than $53$ is observing something less than $\mu +\frac{3}{5}\sigma$.

5) This again sounds very restrictive.

clark
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Since you wrote you want to prove each option right or wrong and so far only heuristics have been offered, here are some ideas for proofs, roughly in order of increasing difficulty:

For $3$) through $5$) use the binomial distribution. The probability of an even result is $\frac12$. Let $K$ be the number of even results. We have $P(K=k)=2^{-100}\binom{100}k$; that lets you immediately exclude $3$). Also $P(K\lt50)=P(K\gt50)$ and thus $P(K\lt50)=\frac12\left(1-P(K=50)\right)$. Then you just have to add in $P(K=51)$ and $P(K=52)$ to decide $4$). For $5$), the success probability is $\frac13$; proceed as for $3$).

To prove $2$), you could try using a concentration inequality, but I'm not sure whether any of them are strong enough in this case. Disproving $1$) might be the hardest. For both $1$) and $2$), if you can't do it with inequalities, you can always calculate the exact probabilities; see Probability to obtain more than X with 3 dice.

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joriki
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