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An guest
Apr-16-03, 08:36 PM (EST)

"probability of faces on a die"

 Q: What is the expected # of rolls it will take for a single, 6 sided die to show each face atleast once?i'm pretty sure this is a geometric distribution so to count the # of failures before a success( in this case-before rolling all faces) you need to know how many trials there are, however, this is not given in the question and therefore equations aren't very helpful. this could be quite simple or it could be as complicated or long as doing individual cases for finding the probability of each face showing, and using that to find the expected # of rolls. information:-there are 6 faces-each face has a 1/6 probability of being rolled-# of rolls expected is obviously eqaul to or greater than 6 but for the answer it'should show as greater than 6-each face must be rolled atleast once-trying to find the expected # of rollsmy big problem here is putting all the information together!

Subject     Author     Message Date     ID probability of faces on a die An Apr-16-03 TOP RE: probability of faces on a die wagner Apr-17-03 1 RE: probability of faces on a die golland Apr-17-03 2 RE: probability of faces on a die Larry Turnbaugh Apr-18-03 3 RE: probability of faces on a die jet Apr-25-03 4 RE: probability of faces on a die Graham C Apr-26-03 5 RE: probability of faces on a die RicBrad Apr-27-03 6 RE: probability of faces on a die Graham C Apr-28-03 7 RE: probability of faces on a die junglemummy Mar-22-06 9 RE: probability of faces on a die mr_homm Mar-23-06 10 RE: probability of faces on a die junglemummy Mar-24-06 11 RE: probability of faces on a die mr_homm Mar-24-06 12 RE: probability of faces on a die Mark Huber Mar-27-06 13 RE: probability of faces on a die thil003 Apr-14-06 14

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wagner guest
Apr-17-03, 12:42 PM (EST)

1. "RE: probability of faces on a die"
In response to message #0

 I think there's something missing here. The number of rolls must be related to a probability, I mean, you can roll the die forever and don't get a 1 for example. So you are looking for the number of rolls "n" where you get the probability "P" of having each face once.This probability must increase with the number of rolls. To get it more complicated one can define the function P=f(n). Of course P=0 for n<6 and P tends to 1 as n goes to infinitywagner

golland guest
Apr-17-03, 09:01 PM (EST)

2. "RE: probability of faces on a die"
In response to message #0

 Hi, it is rather simple: we are looking for an AVERAGE number of trials to get the first face (any of 1,2,3,4,5,6 ) 1 trial is enough we need 6/5 trials on average to get second face (any of 1,2,3,4,5,6 except the result of the first trial ) and so on ..... and the total is 6 * (1/6 + 1/5 + 1/4 + 1/3 + 1/2 + 1 )= 6*143/60=14.3 The big question now is what to do with .3 ? Larry Turnbaugh guest
Apr-18-03, 06:04 PM (EST)

3. "RE: probability of faces on a die"
In response to message #2

 In this case, the way you are using probobility is wrong, the way you have it'set up, is if each of the faces of the die were removed after being rolled. As if one were removing marbles from a paper bag. This however is not the case. One would think that the chances of rolling a certain number would increase over time, but in reality, it does not, there is a 1 in 6 chance to roll any number on the die no matter how many times youve rolled it in the past. jet guest
Apr-25-03, 03:40 PM (EST)

4. "RE: probability of faces on a die"
In response to message #3

 There is 1/6 probablility to obtain a SPECIFIC number say 5. But this is now what your are looking for.First roll: you expect any of the 6 values out of 6so average number of roll is exactly 6/6=12nd roll you expect any 5 values out of 6so average number of roll is 6/53rd roll you expect any 4 values out of 6so ... 6/4etc...until you are looking for the last numberand average number of roll is 6/1SO the total average number is 1 + 6/5 +6/4 + 6/3 + 6/2 + 6 = 14.7

Graham C
Member since Feb-5-03
Apr-26-03, 02:56 PM (EST)    5. "RE: probability of faces on a die"
In response to message #0

 This one is tougher than it looks, so this is just an intermediate post.You're looking for the expected number of rolls before each face has come up once. I'm not totally sure what you mean by the 'expected number' here: I'll assume that if P(n) is defined as the probability that in n rolls each face comes up at least once, then the 'expected number' is n where P(n)>=0.5. I.e. the number of rolls that gives you an even or better chance of getting that distribution.Taking the simpler but related question: 'What is the expected number of rolls to get a six?' then we would have P(n)=1-(5/6)^n, orP(1) = 1/6P(2) = 11/36P(3) = 91/216P(4) = 671/1296and the answer would be 4.So far though I can't figure an expression for P(n) for the problem you're posing, though I'll think about it. Member since Nov-16-01
Apr-27-03, 04:16 PM (EST)    6. "RE: probability of faces on a die"
In response to message #5

 >You're looking for the expected number of rolls before each >face has come up once. I'm not totally sure what you mean by >the 'expected number' here: The definition I have met of "expected" in this context is what might be vaguely called the mean or average, that is:Sum_{n} n p(n)where p(n) is the probability of the event (all faces have been rolled) happening on the nth roll. {so Sum_{n} p(n) = 1}golland above is correct (apart from his/her arithmetic!), but I hope you won't mind if I expand on what (s)he said:When we have a series of repeated, identical, trials and count the number of tries until success, this is the Geometric distribution. An event of probability p will take, on average, 1/p tries to be achieved. (i.e. the expected number of tries is 1/p)The first roll will always be a number we haven't seen before - add 1 to our total.Next we wait until we roll a different number - this event has probability 5/6, hence we expect to wait 6/5 of a roll.Do this until we have seen 6 unique numbers and we are done:E = 6/6 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1 = 14.7So we can expect to wait 14.7 rolls. (It is not a problem that this number is a fraction, in the same way it might be average to have 2.4 children, but clearly no-one does){Sorry about the excessive length of this post.}All the best,Rich Graham C
Member since Feb-5-03
Apr-28-03, 09:46 AM (EST)    7. "RE: probability of faces on a die"
In response to message #6

 >>You're looking for the expected number of rolls before each >>face has come up once. I'm not totally sure what you mean by >>the 'expected number' here: >>The definition I have met of "expected" in this context is >what might be vaguely called the mean or average, that is: >>Sum_{n} n p(n) >>where p(n) is the probability of the event (all faces have >been rolled) happening on the nth roll. {so Sum_{n} p(n) = >1} >I take your argument and its conclusion.My definition of 'expected' was different, but gets a very similar result.Rephrase the question slightly to be 'how many dice would I expect to have to roll for the probability of there being at least one of each face to be >= 0.5'The probability that one is a 1 is 1-(5/6)^n one of the rest is a 2 is then 1-(5/6)^(n-1) one of the rest is a 3 is then 1-(5/6)^(n-2)and so on.Multiply them together to get the probability that one of each occurs.The probability of it happening with 14 dice is then 0.434178....With 15 dice it is 0.503598....Which accords with your answer.However, for the probability to be exactly 0.5 I get approximately 14.94653 dice.

junglemummy
Member since Nov-7-05
Mar-22-06, 00:58 AM (EST)    9. "RE: probability of faces on a die"
In response to message #0

 Maybe I'm missing something here, but this whole topic seems vaguely reminiscent of the recently revived Probability thread in the Early Maths section. They are both terrible misuses of the whole concept of probability. The problem in both cases seems to be the way the initial question is phrased, or set up. In the Early Maths example, we have to use the results of a previous experiement to determine the results of another one. In the current example, the use of the word "expected" is very loose. In both cases, we should not lose sight of the bigger picture, and should remember that we are dealing with what used to be called "independent events" when I was as high school (don't know if this terminology is still in use). You could roll the die till the cows come home and still not get each number. Your "expectation" depends on whether you are an optimist or a pessimist. mr_homm
Member since May-22-05
Mar-23-06, 11:39 AM (EST)    10. "RE: probability of faces on a die"
In response to message #9

 I'm afraid I must disagree with you. The term "expected" is a specialized technical term in probability. It means specifically, the average value a variable would have if you could do infinitely many trials. It is defined computationally as sum(value*probability).For example, in one of the above posts it is mentioned that if the probability of success is 5/6 then the expected time to wait is 6/5. You can arrive at this from the technical definition as follows:Let the variable be the number of trials, n. Then the probabilities are p(1)=5/6, p(2)=(1/6)*(5/6) (because you have to fail on the first try and succeed on the second, in order for your FIRST success to be at n=2), p(3)=(1/6)*(1/6)*(5/6), and in general p(n)=5/(6^n). Computing the expected time to first success by the formula, you getsum(n*5/(6^n)), which sums up to 6/5. Intuitively, it is clear that the lower the probability, the longer you must be prepared to wait, so the reciprocal relationship is quite reasonable.Hope this clears things up.--Stuart Anderson junglemummy
Member since Nov-7-05
Mar-24-06, 07:17 AM (EST)    11. "RE: probability of faces on a die"
In response to message #10

 Thanks Stuart. I stand corrected and enlightened. I suppose it must be infuriating to mathematicians to have amateurs commenting when they don't know technical terminology. In my defence, I only took maths to first year university level, quitting it to major in zoology and genetics. I wanted to pursue maths further, but there were too many timetable constraints, so I have to be content with remaining an interested amateur. mr_homm
Member since May-22-05
Mar-24-06, 07:07 PM (EST)    12. "RE: probability of faces on a die"
In response to message #11 Mark Huber guest
Mar-27-06, 07:52 PM (EST)

13. "RE: probability of faces on a die"
In response to message #12

 Since this thread has been recently resurrected, I'll just take this opportunity to add that this is a special case of a very famous problem in probability known as the coupon collector problem. In the general formulation, a collector is trying to collect a complete set of n different coupons, and at each trial, each coupon is equally likely to appear. The same logic as used above in the n = 6 problem shows that the expected time to get all n coupons is about n ln n, a result that has intersting ramifications for some probabilistic algorithms. -mark

thil003 guest
Apr-14-06, 09:21 AM (EST)

14. "RE: probability of faces on a die"
In response to message #0

 This is geometric distribution, but we have 6 cases which are not independant on each other,actually we can't find the number of trials it can vary from 6 to infinite,so we can only take expectation,consider one case E(x)=Sumation(0 to infinity) x*q^r*pwe have 6 cases,so 6*E(x)-intersections they make( interestingly sumation of 6Cn intersections are there)...so we need some approximations...to get the one answer

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