The answer is 7.

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ns i didn't think of that

Bob Yang
- 7 years, 8 months ago

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good job :) i love your reasoning

Bob Yang
- 7 years, 8 months ago

The first sentence should read "Let us consider when five die are rolled".

Shunping Xie
- 7 years, 8 months ago

This is exactly my logic! Nice one. :)

Ivan Sekovanić
- 7 years, 8 months ago

Suppose we are rolling $n$ dice.

$\textbf{Claim:}$ The probability of rolling a sum that is a multiple of six is $\frac16$ for all integers $n\ge1$ .

$\textbf{Proof:}$

Let the result of the roll of the $n^{th}$ die be $r_n$ .

**
Case 1:
**
$n=1$
.

The probability of rolling a multiple of six is creary $\frac16$ .

**
Case 2:
**
$n>1$
.

Let $0\le a\le 5$ be the integer such that $\sum_{k=1}^{n-1}r_k\equiv a\pmod6$ Note that for $\displaystyle\sum_{k=1}^{n}r_k$ to be a multiple of six, we must have $r_n\equiv 6-a\pmod6\quad\implies\quad r_n=6-a,$ since $1\le r_n\le6$ . Hence for every $a$ there exists exactly one $r_n$ satisfying the conditions, so the probability is $\frac16$ .

Therefore, the final answer is $\boxed{\frac16}$ .

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creary

Ryan Soedjak
- 7 years, 8 months ago

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Sorry, I meant the first five dice, not two.

Frodo Baggins
- 7 years, 8 months ago

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there six multiples of 6, (6,12,18,24,30,36), than the probability is sum of number of ways of multiplies divided by $6^6$ ,.

for calculating the number of ways of multiples like $d_1+d_2+d_3+d_4+d_5+d_6=24$ i used inclusion–exclusion principle.

(this is algorithm, i know isn't a programming problem it's just from simplifying calculation)

where $n = 24, k = 6, R = 6, 1 \leq d_i \leq R$

```
def numOfWays(n,k,R):
n -= k
p = 2
i = 0
s = 0
while n - i * R >= 0:
s += ((-1)**p) * C(k,k - i)* C(n + k - 1 - i * R, k - 1)
p += 1
i += 1
return s
```

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The only dice that really matters is the very last dice rolled because the sum of the first five must be either 0 mod 6, 1 mod 6, 2 mod 6, 3 mod 6, 4 mod 6, or 5 mod 6. (any integer must be one of these 6 due to the nature of mods) Thus, for each case to get the sum to evenly divide by 6, there is only one number to get the final sum to 0 mod 6. So, our answer is 1 in 6 or 1/6 which gives us a value for a+b of 7