A 6 sided die is thrown P times and you are told the result is a monotonic sequence?

Question

What is the probability that all numbers appear at least once in the sequence?

Answer 1

M3's approach gave me some ideas, and I gave him thumbs up.



Let p(n, k) be the number of ways to write n as the sum of exactly k terms irrespective of order ( http://mathworld.wolfram.com/PartitionFunctionP.html , see after (54)). Usually it is uppercase, but the letter P is already used for the number of times the die is thrown.



The ways to have a sequence satisfying the desired condition (monotonic, all numbers represented) is 2 * p(P, 6). Take that as numerator. Then as denominator take 2 * ( sum_(i=2)^6 ( C(6, i) * p(P, i) ) ) + C(6, 1) * p(P, 1) = 2 * ( sum_(i=2)^6 ( C(6, i) * p(P, i) ) ) + 6, where C(n, k) is binomial coefficient. This quotient is the desired probability.



I often make mistakes with questions like these, so if anyone wants to verify my answer or point out a flaw, by all means do so.



Edit: Oh wait, we need order to matter... In this case we want compositions rather than partitions, and then I think it would work as desired. Would have to dig around to find a compositions function taking (n, k). (Further edit: Ah my memory is like Swiss cheese. Finding the number of compositions of n into exactly k pieces is easily handled by "stars and bars" technique, it's just C(P - 1, k - 1). I'd like to see how this compares with gianlino's solution, but I'm getting a little tired of looking at this problem and will set it aside for now.)



@M3, for your example, "p(5,3) = 2, since partitions of 5 of length 3 are {3 1 1} & {2 2 1}" -- this would correspond to die sequences involving three distinct face values, and this count would be multiplied by C(6, 3) specifying the face values, and further by 2 to distinguish increasing and decreasing. So, let's pretend for now I didn't mess up the order part -- then if we choose face-values {1, 2, 3}, the partition 5 = 3+1+1 would represent sequences 11123 and 33321. Of course for 11123 we also need 12223 and 12333, which is the order part that I missed. When we break into 6 parts (i.e., k = 6 instead of k = 3), we must have all 6 face values represented. For P = 5, this will result in 0.



Can't say I understand your solution though. You seem to have r1, ... , r6 appearing in your final solution, but it should be only in terms of P because we are not given r1, ... , r6. Wouldn't we have to sum over all possible values of r1, ... , r6? Maybe I just read it wrong. Thanks for the feedback.



Further edit: Okay, I checked my answer against gianlino's and they are almost the same. The only thing I don't understand is why gianlino changed the 6 to a 3. It seems it should stay a 6.



So this expression:



[((P+5) C 5) - 6 ((P+4) C 4) + 15 ((P+3) C 3) - 20 ((P+2) C 2) + 15 ((P+1) C 1) - 6] / [((P+5) C 5) - 3]



is equal to this expression:



2 * C(P-1, 5) / (2 * sum(i=2, 6, C(6, i) * C(P-1, i-1)) + 6)



Or, factoring out the two,



C(P-1, 5) / (sum(i=2, 6, C(6, i) * C(P-1, i-1)) + 3)



By the way, the modified expression of gianlino can be written this way as well:



sum(i=0, 5, (-1)^i * C(6, i) * C(P+5-i, 5-i)) / (C(P+5, 5) - 3)



Furthermore! The numerators and denominators are identical. That is,



C(P-1, 5) = sum(i=0, 5, (-1)^i * C(6, i) * C(P+5-i, 5-i))



and



sum(i=2, 6, C(6, i) * C(P-1, i-1)) + 3 = C(P+5, 5) - 3



So I don't think we'll get any simpler than this:



C(P-1, 5) / (C(P+5, 5) - 3)



:D

Answer 2

Since it's a conditional probability problem, we can assume that sequences are inreasing. The decreasing case yielding the same result.



The number of increasing monotonic sequences valued in [1,6] and of length P is ((P+5) C 5)



The number of hose who omit one given value is ((P + 4) C 4).



And so forth and so on. So the number of increasing sequences which take all values is, by inclusion exclusion



((P+5 C 5) - 6 ((P+4) C 4) + 15 ((P+3) C 3)

- 20 ((P+2) C 2) +15 ((P+1) C 1) - 6



Here the constant sequences are counted as increasing, but they are also decreasing. so in the probability count, they should be listed as 3.



Now you divide by ((P+5)C 5) and you get your probability.



1 - 6*5/(P+5) +15*5*4/(P+5)(P+4) - 20*5*4*3/(P+5)(P+4)(P+3)

+ 15*5*4*3*2/(P+5)(P+4)(P+3)(P+2) - 3/ ((P+5 C 5)



edit There is a mistake connected to those constant sequences which are counted twice. So the denominator should be ((P+5) C 5 ) - 3 instead of ((P+5) C 5).So the probabilty I found must be multiplied by



1 + 3 / [((P+5) C 5 ) - 3 ]



@ Edit Siamese my mistake, just forgot what we were trying to prove =)

Answer 3

Strictly speaking, each throw is equal to or greater than the previous if this is an increasing monotonic sequence. It could be a decreasing sequence as well.



If it is an increasing sequence, the only time you would get all the elements is if the sequence starts at 1 and each throw in greater than (rather than equal to). So the probability given an increasing sequence is as follows:



1/6 that the first is a 1.

1/2 that each of the next is an increase rather than equal.



This means the probability given an increasing monotonic series is:



1/6 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/(64*3) = 1/192



The decreasing monotonic possibility has the same probability, 1/192.



1/192 + 1/192 = 1/96



So, if you are given that the sequence is monotonic, there is a 1/96 chance that it contains every number on the die. A strictly increasing, or strictly decreasing monotonic series would indicate a 1.0 probability.



Your answer is 1/96.

Answer 4

a monotonic sequence would either be something like

11123344445666 or

66544433332111



let the sequence contain r1 1's, r2 2's,......r6 6's , r1+r2+...r6 = p , r_n > 0

then required probability = 2*r1! *r2! *.....*r6! / 6^p



edit:

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no, it is more complex. the above gives P[all #s appear monotonically]

what has been asked for is P[all #s appear monotonically | monotonic]

i 'll see if i can come back to it



edit2:

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for the denominator, we have to use the expression for the numerator with 6c1 ways with only 1 #, 6c2 ways with 2 #s,..... 6C6 ways with 6#s,

(the last term being the same as the numerator)



as for siamese attempt to simplify the solution using p(n,k),

to start with, i don't see how p(P,6) guarantees that all 6 #s appear.

as stated in the reference cited, eg,

p(5,3) = 2, since partitions of 5 of length 3 are {3 1 1} & {2 2 1} , and partitions adding up to 5 with maximum element 3 are {3,2} & {3 1 1}.

neither interpretation guarantees that 1,2 & 3 are all there,

or that the LENGTH of partitions is 5



i give siamese TU for the attempt. may be there is some other integer sequence that can encapsulate the appropriate formula more neatly than my attempt.

Answer 5

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