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An
guest

Apr1603, 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 casebefore 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 rolls my big problem here is putting all the information together!


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wagner
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Apr1703, 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 infinity wagner


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golland
guest

Apr1703, 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 ?


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Larry Turnbaugh
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Apr1803, 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. 

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jet
guest

Apr2503, 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 6 so average number of roll is exactly 6/6=1 2nd roll you expect any 5 values out of 6 so average number of roll is 6/5 3rd roll you expect any 4 values out of 6 so ... 6/4 etc... until you are looking for the last number and average number of roll is 6/1 SO the total average number is 1 + 6/5 +6/4 + 6/3 + 6/2 + 6 = 14.7


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Graham C
Member since Feb503

Apr2603, 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, or P(1) = 1/6 P(2) = 11/36 P(3) = 91/216 P(4) = 671/1296 and 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. 

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RicBrad
Member since Nov1601

Apr2703, 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 = 14.7 So 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 noone does) {Sorry about the excessive length of this post.} All the best, Rich 

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Graham C
Member since Feb503

Apr2803, 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)^(n1) one of the rest is a 3 is then 1(5/6)^(n2) 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. 

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mr_homm
Member since May2205

Mar2306, 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 get sum(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 

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mr_homm
Member since May2205

Mar2406, 07:07 PM (EST) 

12. "RE: probability of faces on a die"
In response to message #11

>Thanks Stuart. You're very welcome. >I stand corrected and enlightened. I >suppose it must be infuriating to mathematicians to have >amateurs commenting when they don't know technical >terminology. Not at all! Most mathematicians are delighted when ANYONE shows an interest in their subject. I know I am always pleased to see that others are interested. Also, most of us ARE amateurs in most fields  for example, my training is primarily in algebraic topology and functional analysis, I am somewhat competent in abstract algebra and linear optimization theory, and at everything else, I am just as much an amateur as anyone else. Here's an interesting true story about mathematicans: Many years ago, before we were married, my wife as looking for a date in the personals advertisements. She answered an ad from a man who gave the usual personal information in the ad, but instead of saying what his occupation was, he said "my work is mysterious and beautiful." When they met, the first thing she asked him was "are you a mathematician?" He was astounded, because of course she was right. He asked her how she had guessed, and she told him that only a mathematician would describe his work as beautiful and at the same time be so sure others would be put off by it, that he would be afraid to mention the subject by name. He told his colleagues about this conversation, and they were delighted by her Sherlock Holmes style deduction and her incisive understanding of how mathematics is regarded by (most) nonmathematicians in our culture. Fortunately for me, things did not work out between them subsequently, and she ended up with me. Stuart Anderson


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Mark Huber
guest

Mar2706, 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 

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thil003
guest

Apr1406, 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*p we 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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