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I have two four sided dice. Each die has 1, 2, 3, 4 as choices.

What the probability of the the second roll being greater than the first?

Examples:

1, 2

1, 4

3, 4

I used python to figure the probability of ~0.3750245 with 10 million trials, so close to 0.375 which is also 6/16 or 3/8

How can I get to 3/8 without programming it?

Michael Hardy
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Vader
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    There are three results: "first roll greater" (a), "second roll greater" (b), "rolls are equal" (c). For symmetry reasons, probabilities of (a) and (b) are equal, probability of (c) is $1/4$ ($1/n$ for $n$-sided die). Sum of probabilities is 1. Or you could just write down all 16 possible pairs and count those 6 corresponding to (b). – Abstraction Dec 13 '15 at 20:39

4 Answers4

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One simple way to think about problems like these is simply to count how many possible die rolls you have with two four sided die, and then count how many possible die rolls you can have where the second roll is greater than the first.

Then, $P(\text{second roll greater}) = \dfrac{\#\text{Rolls where the 2nd die is greater}}{\#\text{ Possible rolls}}$.

I claim that the possibilities for rolls where the second die is greater are: (1,2),(1,4),(1,3),(2,3),(2,4),(3,4).

Jay
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  • counting is fine in this case, but what if I have dice with more sides or add more dice to the equation? – Vader Dec 13 '15 at 20:46
  • From this user's answer, which I like, http://math.stackexchange.com/a/100684/50960, $P(\text{second roll greater})=\dfrac{{n \choose k}}{n^k}$, where $n$ is the number of sides and $k$ is the number of rolls. – Jay Dec 13 '15 at 21:02
  • What is that n with a k above it? On the top – Vader Dec 13 '15 at 21:30
  • ${n \choose k}$ is the number of ways to pick $k$ elements out of a set of $n$ elements, where order doesn't matter. See: https://en.wikipedia.org/wiki/Combination – Jay Dec 13 '15 at 22:48
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Since there are $4$ possible outcomes for each roll, there are a total of $4^2 = 16$ possible outcomes. Of these, there are four outcomes in which the second roll is equal to the first. As for the remaining $12$ outcomes, by symmetry, it is equally likely that the outcome of the first roll exceeds that of the second roll as it is that the outcome of the second roll exceeds that of the first. Hence, the probability that the second outcome exceeds the first is $$\frac{\frac{1}{2}(16 - 4)}{16} = \frac{6}{16} = \frac{3}{8}$$

N. F. Taussig
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If the first roll is $1$, what is the probability of the second roll being lower? If the first roll is $2$, what is the probability of the second roll being lower? If the first roll is $3$, what is the probability of the second roll being lower? If the first roll is $4$, what is the probability of the second roll being lower?

Since these cases are disjoint, the sum of the probabilities, times the probability that the first roll happens ($1/4$ for each, assuming a fair die) is the probability of the event.

Kevin Long
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Listing all the successful trials (i.e $2^{nd}>1^{st}$) we see there are 6 combinations. Additionally, listing all the unsuccessful trials, there are the 4 cases where $2^{nd}=1^{st}$ and a further six cases where $2^{nd}<1^{st}$. This means the success rate is 6(successful outcomes) over 16 (total outcomes, $6+4+6$)

Thus we get the probability: $$\frac{6}{16}=\frac{3}{8}$$

Rhys Hughes
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