Wikipedia:Reference desk/Archives/Mathematics/2013 March 14

= March 14 =

Probability
Hello all. I am having a brain fart on this seemingly very easy probability problem, probably (pun intended) due to lack of sleep. It goes: if I roll a fair six-sided die n times, what is the probability that the highest result will be 3? (it is not necessary that three occur only once, merely that 4, 5, 6 do not occur). My reasoning is: the first n-1 times, the die can come up to anything less than or equal to three: $$(\frac{3}{6})^{n-1}$$. However, I need to guarantee that 3 occurs at least once! Suppose it occurs the nth time: $$\frac{1}{6}$$ But I don't like that supposition--it can really occur any kth time, k<=n; in other words, how many ways can I permute these n numbers: n! Multiplying it all together, I've got $$(\frac{3}{6})^{n-1}(\frac{1}{6})\cdot n!$$ which for some n is greater than 1! (exclamation point, not factorial!) Not quite sure where my math is sketchy, any help will be appreciated. 72.128.82.131 (talk) 01:45, 14 March 2013 (UTC)


 * One way to do it is splitting in two steps: What is the probability that the highest result in n tries is at most three? And assuming it is at most three, what is the probability that it is at most two? Then combine to get the final result. PrimeHunter (talk) 02:40, 14 March 2013 (UTC)


 * Agreed, but you have to be careful here, as the second probability is dependent on the first. The probability of each roll being 3 or under is (3/6)^n, or (1/2)^n, and the probability of each roll being 2 or under is (2/6)^n, or (1/3)^n, if those two are independent.  But, once we've established that all rolls are 3 or under, the probability of them also being 2 or under is now (2/3)^n.  Therefore, the probability of that not all rolls are 2 or under is (1-(2/3)^n).


 * So, I believe your final result should be (1/2)^n × (1-(2/3)^n). Let's run the first couple cases of n:

n  (1/2)^n   (2/3)^n   (1-(2/3)^n)   (1/2)^n × (2/3)^n =  =======   =======   ===========   ================= 1     1/2       2/3        1/3               1/6 2     1/4       4/9        5/9               5/36


 * Make sense ? StuRat (talk) 03:26, 14 March 2013 (UTC)


 * Or, since (at most 2) is a subset of (at most 3), you can ignore the conditional probability and just subtract. (1/2)^n - (1/3)^n.--80.109.106.49 (talk) 08:27, 14 March 2013 (UTC)


 * Agreed. That would yield the following:

n  (1/2)^n   (1/3)^n   (1/2)^n - (1/3)^n =  =======   =======   ================= 1     1/2       1/3           1/6 2     1/4       1/9           5/36


 * StuRat (talk) 15:02, 14 March 2013 (UTC)


 * Highest result is 3 = all die roll is 3 or less - all die roll is 2 or less.


 * all die roll is 3 or less is (3/6)^n


 * all die roll is 2 or less is (2/6)^n


 * So the answer is (3/6)^n - (2/6)^n 202.177.218.59 (talk) 22:03, 14 March 2013 (UTC)


 * Which reduces to (1/2)^n - (1/3)^n, the same answer given above. StuRat (talk) 04:57, 15 March 2013 (UTC)