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# Dice Challenge!

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8 fair 6-sided dice are rolled 8 times. Their sum totals were 28, 29, 30, 31, 32, 33, 34 and 35, but not necessarily in that order. What is the overall probability of this occurring? This is a very tough one to crack! Any help would be great. Thank you.

Oct 24, 2018

### 1+0 Answers

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It is not as tough as it looks! It is just a lot of work! Where in the world did get this from? At any rate, you can calculate the individual permutations of each sum using this complicated-looking formula and then sum them all up:
S=28 -35; N=8; sum(k, 0, ((S - N)/6), ((-1)^k * (N nCr k) * (S - 6*k - 1) nCr (N - 1))

However, fortunately, there is an elegant shortcut to doing all the above work, thanks to Wolfram/Alpha!!.
Simply Expand this polynom: [n + n^2 + n^3 + n^4 + n^5 + n^6]^8 and then just add up all the coefficients of n^28 to n^35 inclusive, because the formula above calculates exactly the same coefficients individually. Here is the expansion of the above polynom:

n^48 + 8 n^47 + 36 n^46 + 120 n^45 + 330 n^44 + 792 n^43 + 1708 n^42 + 3368 n^41 + 6147 n^40 + 10480 n^39 + 16808 n^38 + 25488 n^37 + 36688 n^36 + 50288 n^35 + 65808 n^34 + 82384 n^33 + 98813 n^32 + 113688 n^31 + 125588 n^30 + 133288 n^29 + 135954 n^28 + 133288 n^27 + 125588 n^26 + 113688 n^25 + 98813 n^24 + 82384 n^23 + 65808 n^22 + 50288 n^21 + 36688 n^20 + 25488 n^19 + 16808 n^18 + 10480 n^17 + 6147 n^16 + 3368 n^15 + 1708 n^14 + 792 n^13 + 330 n^12 + 120 n^11 + 36 n^10 + 8 n^9 + n^8.
Now Just add up the coefficients: =135,954+133,288+125,588+113,688+98,813+82,384+65,808+50,288 =805,811 - total permutations of all the 8 tosses of the dice!!
Therefore, the overall probability is: 805,811 / 6^8 =47.9759%. And that is the END!.

Oct 25, 2018