I roll two dice n times and sum all of them together. What is the probability that the sum is divisible by 7?
What is the limit of the probability as n approaches infinity?
2hard4sci?
If youre rolling TWO dice n number of times youre always going to get at least 2 as the sum. The further you roll and the more n times you roll these numbers will become greater and the probability increases that the sum will be divisible by 7. In other words using probability, the probability will be n*pi/(2n^2)*7 that its divisible by 7. As the limit of the probability as n approaches infinity also approaches infinity. Do your homework pleb.
>>7670095
1/6 obviously
>>7670043
>What is the limit of the probability as n approaches infinity?
1 because expected value of two dice is 7
>>7670095
> As the limit of the probability as n approaches infinity also approaches infinity
>probability also approaches infinity
probabilities are bound between 0 and 1 pleblord
>>7670111
so to n times you say that the chance is nearly 100% ?
that doesnt make sense.
1/7
>>7670043
(1 - (-1/6)^(n-1))/7
1/7
>>7670043
In a classical probability the answer to your question is 1/6, since every time you roll two dices it has nothing to do with the previous roll. But if you look at statistical probability, if you can do a large enough ammount of expiriments(i.e n approaches infinity, but not really, smth like 500 times would be enough) u`ll get number around 1/6, which means that as n approaches infinity statistical probability approaches classical.
>>7670149
1/6 cuz u have 6 events that satistfy your condition, and 36 events overall, so 6/36 = 1/6.
>>7670149
Nope, the answer is 1/7. The question is equivalent to asking for the number of closed walks of length n on the complete graph with 7 nodes from a given node and dividing this by 6^n.
>>7670166
Woah - this is correct. I need to brush up on my graph theory...how many closed walks of length n are there in a complete graph?
>>7670191
((p-1)^n+(p-1)(-1)^n)/p where v is the number of nodes
>>7670223
*where p is the number of nodes
1/12 for every roll.
So 1:12
>>7670231
Nope. Try again.
>>7670239
Nah it's incorrect.
What I don't understand is how, if I roll the dice 3 times, my potential sums are 8-36. There're three numbers divisible by 7 (14, 21 and 35) out of 28 numbers (36-8). The chances for that instance = 3:28
So if 1/7 = a unanimously applicable solution, why is 3/28 nowhere near that amount?
Mind you, I've never come across the words 'graph theory' before in my life.
>>7670251
Autism intesifies
You can view this as a Markov chain problem.
Consider vectors [math]p^{(n)} = (p_1^n, p_2^n, \ldots ,p_7^n)[/math] whose [math]i[/math]-th entry is the probability that the sum is equal to [math]i[/math] modulo [math]7[/math] after throwing the dice [math]n[/math] times. Then you can calculate the transition matrix [math]A[/math] and use it to explicitly calulate the probabilites.
[math]p^{(n+1)} = p^{(n)} A[/math]
[math]p^{(n)} = p^{(0)} A^n[/math]
[math]p^{(0)} = (0,0,0,0,0,0,1)[/math]
To find [math]A^n[/math] you just have to do the usual Eigenvalue shit.
Okay, i think i will write a simulation of this with n around 10 000 to see what is looks like.
>>7670251
Sorry, 6 out of 36. 4 potential sums equaling 7, 14, 21 and 35. 4:30's a helluva lot closer to 1:7.
>>7670260
6-36 god damn it.
>>7670251
The 7 vertices each represent a value of the sum of the dice mod 7. Each time your roll the dice, you move from one vertex to another along one of the lines (for example if you roll a 1 you move to the node that represents 1 mod 7). Since you start at 0 when n = 0, you want to take a path that leads you back to your starting vertex. That is known as a closed walk.
>>7670251
>What I don't understand is how, if I roll the dice 3 times, my potential sums are 8-36
You mean 6-36
>There're three numbers divisible by 7 (14, 21 and 35) out of 28 numbers (36-8)
No, there are four (7,14,21,35) out of 31 (36 through 6 including 6)
>The chances for that instance = 3:28
No, no, no. You are more likely to get a sum close to the average like 14 than you are to get a number closer to the extreme like 35. There are more ways to sum to 14 than there are to sum to 35 (there is only one way to sum to 36 = 6+6+6+6+6+6).
>So if 1/7 = a unanimously applicable solution, why is 3/28 nowhere near that amount?
1/7 is the limit as the number of dice rolls approaches infinity. The actual formula is (1 - (-1/6)^(n-1))/7 where n is the number of dice.
3 pairs of dice means n=6, which means the probability is 1111/7776 which is very close to 1/7
>>7670261
Still I'm confused because four rolls = the potential sums of 8-48. 14, 21, 28, 35, 42. 40 numbers and 5 possible solutions fitting the x/7 criterion.
That's a 1:8 chance and the chances after that I'm guessing stabilize to something closer to 1:7 due to the disparity between the rise of the lower end of the spectrum (2) and rise of the higher end of the potential sum's spectrum (12).
7:50 the next roll, but why is the solution 1/7 if before the sixth roll the chances're (far) below 1:7?
>>7670267
Ah, yeah, thank you! It's been a while since I've done math xD
>>7670278
7/x*
>>7670255
The matrix A just has 6/36 on the diagonal and 5/36 off the diagonal this means the Eigenvalues are 1/36 with the Eigenspace span((1,-1,0,0,0,0,0),...,(1,0,0,0,0,0,-1)) and 1 with the Eigenspace span((1,1,1,1,1,1,1)). From this you immediately see that p^(n) converges to the Eigenvector with the Eigenvalue 1 namely
(1/7,1/7,1/7,1/7,1/7,1/7,1/7)
>>7670166
Okay, you are clearly wrong, no way its near 1/7.
Here is a source code if you are interested http://pastebin.com/nTnM70G7
>>7670336
Value in calculator is a result of 1/6
>>7670336
This is with n = 100000000. As n increases you only get more equal digits after the point. i dunno why i`m posting it though, its clearly a bait.
>>7670336
Is this not the case where n=2 repeated 100000 times? We want there case where n=big repeated 10000 times.
>>7670343
Op said "roll dices 2 times" and then repeat it.
>>7670343
Sorry, i`m autistic. You are right.
>>7670346
No worries, I totally hear you. I can see why you might think that, but I did not mean it. Thank you for pointing that out, I imagine other people were confused as well.
For clarity: Roll 2*n dice, sum them all. What is the probability that that sum is 0 mod 7?
Yes, dude with graphs was correct. /thread
>>7670360
nice! what was n in this case?
>>7670361
100000000.
>>7670043
The probability that the sum of 2n rolls of a die (or n rolls of a pair of dice) is r is given by the coefficient [math]a_r[/math] in the generating function [eqn]\left(\tfrac{1}{6}x + \tfrac{1}{6}x^2 + \tfrac{1}{6}x^3 + \tfrac{1}{6}x^4 + \tfrac{1}{6}x^5 + \tfrac{1}{6}x^6 \right)^{2n} = \sum_{r=1}^{12n}a_rx^r[/eqn]A standard trick to extract every 7th term in sums like this is to employ the 7th roots of unity in the following manner. Set [math]x = e^{2\pi i/7}[/math]. Observe then that [math]\sum_{s=0}^6x^{rs} = 7[/math] iff r is a multiple of 7 while [math]\sum_{s=0}^6x^{rs} = 0[/math] otherwise. Hence by considering the following slight modification of the foregoing generating function [eqn]\sum_{s=0}^6 \left(\tfrac{1}{6}x^s + \tfrac{1}{6}x^{2s} + \tfrac{1}{6}x^{3s} + \tfrac{1}{6}x^{4s} + \tfrac{1}{6}x^{5s} + \tfrac{1}{6}x^{6s} \right)^{2n} = \sum_{r=1}^{12n} \sum_{s=0}^6 a_r x^{rs}[/eqn]we see that the right side of the expression is [math]7\sum_{7|r,r \le 12n}a_r[/math] while the left side simplifies to
[eqn]\sum_{s=0}^6\left( \frac{x^s}{6} \frac{x^{6s}-1}{x^s-1} \right)^{2n} = \sum_{s=0}^6\left( \frac{e^{2 \pi i s/7}}{6} \frac{\sin(6\pi s/7)}{\sin(\pi s/7)}\frac{e^{ 6 \pi i s/7}}{e^{\pi i s/7}} \right)^{2n} = \sum_{s=0}^6\left( \frac{(-1)^s}{6} \frac{\sin(6\pi s/7)}{\sin(\pi s/7)}\right)^{2n} [/eqn]Therefore the probability that the sum of 2n rolls of a die is a multiple of 7 is precisely [eqn]\frac{1}{7}\sum_{s=0}^6\left( \frac{(-1)^s}{6} \frac{\sin(6\pi s/7)}{\sin(\pi s/7)}\right)^{2n} [/eqn]
>>7670043
It is just a recursive pattern. Suppose you know the probability of getting a sum divisible by 7 for n dice (call this number p_n). Then, you know the probability of getting a sum divisible by 7 for n+1 dice. You can't get a number divisible by 7 for the sum of n+1 dice if you got a number divisible by 7 for the sum of the first n of them, so, ignore this possibility. On the other hand, getting a number that isn't divisible by 7 for n dice is 1-p_n. Regardless of what number it is, you have a 1/6 chance of getting a sum divisible by 7 after adding the n+1 die, so p_n+1 = (1-p_n)/6. To determine that a limit exists as n approaches infinity, p_n+2 - p_n = (p_n + 5)/36 - p_n = (5 - 35p_n)/36. p_n+2 - p_n is negative if p_n is bigger than 1/7, and it is positive if p_n is less than 1/7, so, along with,
abs((5 - 35p_n)/36)=abs((35/36)(1/7 - p_n))≤abs(1/7 - p_n), hence the odd terms of the sequence converge to 1/7 from below, and the even terms converge to 1/7 from above.
