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Four dice are thrown, what's the probability that:

a) None of them fall higher than three?

b) None of them fall higher than four?

c) That four is the highest number thrown?

So for a, I wanna think about the denominator first. There are 6 possible outcomes for each dice, and we have four dice. So we're technically picking r out of n objects, aka picking 4 possible values out of 6. So the denominator/sample space should be $6^4$ right?

Now this is where I'm getting tripped up. Order technically shouldn't count, because we only care about the quadruplets that don't have a value higher than 3. I might be wrong, maybe I am, but could someone explain a bit more? My professor stated that typically in the sample space/denominator, we want order to count.

So I'm trying to think of the numerator now, so we want to find the probability that no values appear higher than 3, so some events that can occur are: $(1, 1, 2, 3), (1, 2, 2, 3), (1, 1, 1, 1)$ etc. Again, I can't see why order should count here, because if we only are concerned about what appears rather than how they appear, then we can say $(1, 1, 2, 3)$ is equal to $(1, 2, 1, 3)$? So if order does not count, and we have replacement, then does this lead the case where we use $$\binom{n-1+r}{r}$$to find out the probability?

I'm pretty sure if I can do a), I could prob do b), but if someone could maybe lead me in the right direction for c) I would appreciate that too!

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  • $\begingroup$ Use the same reasoning to calculate the numerator as you did for the denominator. In a) there are $3$ possibilities for the numbers on a single die, so $3^4$ in all. $\endgroup$ – saulspatz Feb 16 at 18:33
  • $\begingroup$ Sorry, could you explain a bit more? So you're saying we're going to omit the other 3 possible values? And we will only roll a die that conceptually has 3 face values? So if that's the case, the the denominator would be 3^4 wouldn't it? But I'm really not sure what you were saying for your first statement. $\endgroup$ – Stuy Feb 16 at 18:37
  • $\begingroup$ No, I'm saying that we only count the acceptable rolls. When I said there are $3$ possibilities, I meant that there are $3$ numbers that don't exceed $3$. If you look at what you were doing, counting events like $(1,1,2,3)$ as successes, how many are there? $3^4$ right? So you divide the number of successes by the number of events and you get $3^4/6^4.$ $\endgroup$ – saulspatz Feb 16 at 18:43
  • $\begingroup$ Okay I see now. Perhaps I overthought the problem. This is part of the chapter Independence and Conditional Probability. Thanks again. $\endgroup$ – Stuy Feb 16 at 18:45
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a) The probability that a die does not fall higher than $3$ is given by $$P(X\le3)=\frac{3}{6}=\frac{1}{2}$$ So as each event is independent we can find the probability that no die falls higher than three by raising this result to the power of $4$ $$(P(X\le3))^4=\frac{1}{2^4}=\frac{1}{16}$$

b) Similarly the probability that a die does not fall higher than $4$ is given by $$P(X\le4)=\frac{4}{6}=\frac{2}{3}$$ So the final probability is $$(P(X\le4))^4=\frac{2^4}{3^4}=\frac{16}{81}$$

c) If $4$ is the highest rolled value then every die rolled has a value $\le 4$. But we also need at least one $4$ to be rolled - so we need to subtract the probability of rolling every dice $\le3$. The answer is then $$(P(X\le4))^4-(P(X\le3))^4=\frac{16}{81}-\frac{1}{16}=\frac{175}{1296}$$ which is the difference of the answers from a) and b).

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    $\begingroup$ If you compare the results for b) and c), they are equal, which they shouldn't be! $\endgroup$ – Ingix Feb 16 at 18:48
  • $\begingroup$ So is your answer correct or is @Jonas De Schouwer's correct? Both have good explanations but I can't reason which is the actual answer. $\endgroup$ – Stuy Feb 16 at 19:07
  • $\begingroup$ @Stuy Please see my comments under Jonas De Schouwer's answer. You can see that I've got the answer 175/1296 quicker than this one from the edit history. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 16 at 19:12
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Let $D_i$ be the outcome of the $i$-th die. Denote $D = (D_1,\dots,D_4)$.

So the denominator/sample space should be 6464 right?

  • Sample space $\Omega = \{1,\dots,6\}^4$
  • We use the cardinality of $\Omega$ as the denominator, and that of the desired event $E$ as the numerator for (a) and (b).
  • (a): take $E = \{1,2,3\}^4$, so $P(D \in E) = |E| / |\Omega| = 3^4/6^4 = 1/2^4 = 1/16$
  • (b): take $E = \{1,2,3,4\}^4$, so $P(D \in E) = |E| / |\Omega| = 4^4/6^4 = 2^4/3^4 = 16/81$
  • (c): take $E = \{1,2,3,4\}^4 \setminus \{1,2,3\}^4$, so $P(D \in E) = (4^4-3^4)/6^4 = 175/1296$

Edit in response to OP's question in comments:

$$\begin{aligned} & \{\max(D_1,\dots,D_4) = 4\} \\ &= \{\forall i \in \{1,\dots,4\}, D_i \in \{1,\dots,4\} \text{ and } \exists i \in \{1,\dots,4\}, D_i = 4\} \\ &= \{\forall i \in \{1,\dots,4\}, D_i \in \{1,\dots,4\} \text{ and } \neg (\forall i \in \{1,\dots,4\}, D_i \ne 4) \} \\ &= \{\forall i \in \{1,\dots,4\}, D_i \in \{1,\dots,4\} \text{ and } \neg (\forall i \in \{1,\dots,4\}, D_i \in \{1,2,3\}) \} \\ &= \{\forall i \in \{1,\dots,4\}, D_i \in \{1,\dots,4\}\} \cap \{(\forall i \in \{1,\dots,4\}, D_i \in \{1,2,3\}) \}^\complement \\ &= \{1,2,3,4\}^4 \setminus \{1,2,3\}^4 \end{aligned}$$

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  • $\begingroup$ Can you please explain c? $\endgroup$ – Stuy Feb 16 at 18:51
  • $\begingroup$ @Stuy It's done! $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 16 at 19:04
  • $\begingroup$ Why downvote? I get $175/1296$ the quickest. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 16 at 19:10
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    $\begingroup$ I appreciate your efforts, but what I was really looking for was why you did what you did instead of just throwing a bunch of math notations. I want to know why you need to take the probability of one 4 being rolled and subtract it with the probability that every other dice is less than or equal to 3. I understand the notation, but I don't understand why you did that. $\endgroup$ – Stuy Feb 16 at 19:16
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    $\begingroup$ Yes I understand that, but this is not an exam or a homework assignment, it's not a competition, so it doesn't matter who gets the answer the fastest. I'm on here because I'm trying to learn and understand the problem, not just have the answer. Sure I can read a bunch of symbols and notations, doesn't mean I'll understand why you did it in the first place. $\endgroup$ – Stuy Feb 16 at 19:23
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I'm trying to address your metholodigal doubts, GNUSupporter 8964民主女神 地下教會 already gave a (very brief) answer to your mathematical questions.

Probably it's an urban legend, but there was a dispute between two statisticians: If you throw 2 "normal", 6-sided fair dice and add the values, will you get (in the long run) the sum $11$ or the sum $12$ more often, or do they occur with the same probability?

Because addition is commutative, the order of the dice (or any other way to distinguish between them) doesn't matter. So one statistician argued that because both for $11$ and $12$ there is (up to exchange of summands) only one way to express them as the sum of 2 integers in the range of $1$ to $6$ ($11=5+6, 12=6+6$), both $11$ and $12$ will occur with the same probability.

The other argued that even though the dice may not be distinguishable, they ares still 2 different entities, so it makes a difference that $11=5+6=6+5$ can be summed in two ways, while $12=6+6$ has only one possible sum.

If you think about the dice as distinguishable (say one is red, the other green), then it is clear that the 36 outcomes that can happen upon those 2 dice being thrown are

(1 on red, 1 on green), (1 on red, 2 on green), ... (1 on red, 6 on green),(2 on red, 1 on green),......(6 on red, 6 on green).

It should also be clear that those 36 outcomes all have the same probability: $1 \over 36$, as each outcome requires the red die to have a given value (proability $1 \over 6$), the green die to have a given value (also proability $1 \over 6$) and those events are independent, so the probablity that both happens is the product of the respective probabilities.

Now you can also see that the event "one die is a 5, the other a 6" corresponds to 2 of those outcomes: (red is 5, green is 6) and (red is 6, green is 5). That means the probability of that event (which is just another description for "sum is $11$") is ${2 \over 36}= {1\over 18}$.

OTOH, the event "both dice are a 6" is just the outcome (red is 6, green is 6), so the probability is ${1 \over 36}$.

The moral of that story is that the "order does not matter" argument only affects one part of the reasoning: To calculate the sum, you do not need to know the order. What it gets wrong is that it makes you wrongly belief that the probabilities of the underlying events are the same.

To come back to your problem, this will have the same incorrect calculations when you count things as "not considering order". The moment you (corrctly) said that the denominator of all of your probabilities is $6^4$, this means you are (correctly) considering ordered (or as in my example, colored) dice. Because you only get $6^4$ outcomes if you considered ordered/colored dice.

In the case of 2 dice, there are 36 outcomes for the ordered/colored dice. 6 of them have the same value for the red and green dice, the other 30 have different values. If you consider unordered/indistinguishable dice, you get 21 outcomes: The 6 same-value pairs are the same as the ordered/colored outcomes, the 30 different-value pairs get halved to 15 (e.g. (red is 1, green is 2) and (red is 2, green is 1) becomes (one die is 1, the other 2) for the unordered/indistinguishable dice).

But as we have seen above, those 21 outcomes dont't have equal probabilities, so basing a simple counting argument on them is impossible.

So the result of all this is that in case of dice (and usually always if you have outcomes where things can repeat) it is usually imperative to count ordered/distinguishable outcomes, because they are usually the outcomes that all have equal probability. If you count unordered/indistinguishalble outcomes, you will usually make an error because those outcomes are not equally likely.

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$a)$ The probability that one dice rolls a number equal to $3$ or lower is $\frac{3}{6}=\frac{1}{2}$. Hence, the probability that all of them roll $3$ or lower is $(\frac{1}{2})^4=\frac{1}{16}$.

Can you do the same for $b)$?

$c)$ Firstly, There is a probability of $(\frac{4}{6})^4=(\frac{256}{1296})$ that all numbers rolled are $4$ or lower.

However, at least one of the numbers must equal $4$, so we subtract the probability that they are all 3 or lower. We already calculaties this in $a)$.

Hence, the probability that the highest number rolled is $4$, equals $$\frac{256}{1296}-\frac{81}{1296}=\frac{175}{1296}$$

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  • $\begingroup$ Okay, so I'm a bit confused now. Why can't I just say there's a $\frac{1}{6}$ chance of a 4 being rolled? I understood what you did with finding the probability that no 4's would be rolled, and then subtracted. But could you explain c a bit more? Thanks! $\endgroup$ – Stuy Feb 16 at 18:58
  • $\begingroup$ When we have only 1 dice, there is a 1/6 chance that a 4 is rolled. However, when there are 4 dice, the probability is somewhat higher. Unfortunately, we can’t just say the probability is 4x1/6 (according to that logic, there would be a 7/6 chance of a 4 being rolled when we have 7 dice). So instead of calculating the probability directly, it’s easier to calculate the probability that none of the dice throws a 4. $\endgroup$ – Jonas De Schouwer Feb 16 at 19:04
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    $\begingroup$ If you take an event $E$ as a subset of the sample space $\Omega$, then $P(E) = |E|/|\Omega| = ? / 6^4$, so the denominator should always be a factor of $6^4 = 1296$. However, $4374 = 2 \times 2187$, so $2 || 4374$. Therefore, your answer for (c) is wrong. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 16 at 19:09
  • $\begingroup$ @JonasDeSchouwer, Could you explain why we need to subtract by the answer in a? $\endgroup$ – Stuy Feb 16 at 19:30

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