# Probability of increasing order of a $3$ rolls of a die. [closed]

A six-sided die is rolled three times. What is the probability that the value of every next roll is greater than the value of the previous one?

## closed as off-topic by Holo, Jendrik Stelzner, Namaste, Leucippus, DidSep 15 '18 at 3:10

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• What are your thoughts about that? – Taroccoesbrocco Sep 14 '18 at 22:32
• If I had any, I would outline them. – V. Spitsyn Sep 14 '18 at 22:35
• You can do something. Suppose the first roll is a $6$, how many good sequences are there? How about if it is $5$? What about $4$? And so on. – lulu Sep 14 '18 at 22:37
• V.Spitsyn This site is not intended to be a "solve this for me", nor a "do my work for me" service. No one here is paid a dime, not even the moderators of this site. It is considered rather rude to dump your work, expecting others to do it for you. Next time you come here, come prepared. Where did this question arise? What course? From what text? What have you learned up till now? Where are you stuck? (Not resorting to the helpless: "I can't do anything by myself". You need to put time in thinking, as lulu tried to guide you to do, before posting a question so you can participate – Namaste Sep 14 '18 at 22:55
• Should be $\binom 63=20$. The idea is that you can choose any three distinct values and then arrange them uniquely in ascending order. – lulu Sep 15 '18 at 10:36

So the total possible rolls of the dice are $6^3$. Among these the way to see the cases in which the numbers are "strictly" increasing are choose any three numbers then there is one case corresponding to it.

Therefore the probability is $$\frac{^6C_3}{6^3}=\frac{5}{54}$$

• I have assumed that by the statement more than the previous one, you mean strictly greater than. – Vishnu N Sep 14 '18 at 22:45
• You might better understand the answer from this user as $$\frac{\binom 63}{6^3}$$, though, I'll admit, Vishnu never explains anythin reasoning except that there are $6^3$ possible outcomes from rolling the die three times in a row. – Namaste Sep 14 '18 at 22:59
• Well, the idea is that there is a one to one correspondence between an increasing sequence of three numbers and choosing three numbers. Hence there are $^6C_3$ ways to do it. – Vishnu N Sep 15 '18 at 8:58
• @VishnuN, $C_6 ^3$ is just the number of ways to choose any three distinct number from 1 to 6, isn't it? – V. Spitsyn Sep 15 '18 at 9:14
• yes I am sorry i did not know to type the other notation in tex – Vishnu N Sep 15 '18 at 12:44