## Poll

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**31 members have voted**

Quote:gordonm888Quote:ssho88

I used a very basic method(and not a so straightforward way) to solve it, if we arrange the 15 dice(with sum of 53) in ascending order. you can reduce the total no of different combinations to 521 as below :-

comb1 : 1,1,1,1,1,1,1,4,6,6,6,6,6,6,6, total pemutations1 = 15!/7!/1!/7! = 51480

comb2 : 1,1,1,1,1,1,1,5,5,6,6,6,6,6,6, total pemutations2 = 15!/7!/2!/6! = 180180

comb3 : 1,1,1,1,1,1,2,3,6,6,6,6,6,6,6, total pemutations3 = 15!/6!/1!/1!/7! = 360360

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comb519 : 3,3,3,3,3,3,3,3,3,4,4,4,4,5,5, total pemutations519 = 15!/9!/4!/2! = 75075

comb520 : 3,3,3,3,3,3,3,3,4,4,4,4,4,4,5, total pemutations520 = 15!/8!/6!/1! = 45045

comb521 : 3,3,3,3,3,3,3,4,4,4,4,4,4,4,4, total pemutations521 = 15!/7!/8! = 6435

All those 521 combinations can be obtained by combinations analysis.

Grand total permutations = pemutations1 +pemutations2 + . . . .pemutations520 + pemutations521 = 27981391815

So Prob = 27981391815/6^15 = 0.059511453

Very nice work. I had thought of this approach. but I had assumed there would be many more combinations - I am surprised there are only 521.

So 15 dice that have 1-6 as possibilities and that add to 53 have 521 distinct combinations. Could we have predicted 521 based on the numbers 15, 6 and 53?

Let me mention something about this.

When rolling a normal 6-faced die, the average number of points will be 3.5. If you roll a die 15 times, the average sum of the dice will be 3.5*15 = 52.5. Thus, a sum of 53 and 52 would be the most probable sum of 15 dice rolls and will ( I think) have the most combinations of any sum.

Thus, both 52 and 53 are the most probable sums of 15 rolls of a 6-faced die and each will have 521 distinct combinations that sum to either 52 or 53.

Therefore, let us define D(n,m) as the number of combinations for the most probable sum when rolling n dice that are m-sided. (this can probably be stated more generally without reference to dice.)

And we know that D(15,6) = 521 (which happens to be prime.)

Finding an analytical formula (or approach) for D(m,n) would be quite an achievement.

I would say that with probability one the two sums are equal infinitely often. But I have no intuition about this at all as far as it being finite or infinite. We do know that one sixth of the time they agree after the first toss.Quote:WizardI get it. My hunch is the answer is infinity. The paradoxical thing is it must happen eventually.

Quote:WizardI get it. My hunch is the answer is infinity. The paradoxical thing is it must happen eventually.

37896.6

Limit exceeded 182 times!

I didn't debug or verify this code. Just hacked it out in a couple of minutes. But it does seem to confirm your hunch.

#include <stdio.h>

#include <stdlib.h>

#include <time.h>

int main() {

int d1, d2, d1tot, d2tot;

int a;

int tr, trlim;

double trtot;

d1tot = 0;

d2tot = 0;

trlim = 0;

trtot = 0;

srand(time(NULL));

for (a = 0; a < 1000000; a++) {

if (a%10000 == 0) {

printf("%d\n",a);

fflush(stdout);

}

d1 = rand()%6 + 1;

d2 = rand()%6 + 1;

d1tot = d1;

d2tot = d2;

tr = 1;

while (d1tot != d2tot) {

if (tr == 100000000) {

trlim++;

break;

}

d1 = rand()%6 + 1;

d2 = rand()%6 + 1;

d1tot += d1;

d2tot += d2;

tr += 1;

}

trtot += tr;

}

printf("%1.1f\n", trtot/1000000.0);

printf("Limit exceeded %d times!\n", trlim);

return 0;

}

This is just a waiting-time problem on a simple symmetric random walk on the difference x[n] = d1Total[n] – d2Total[n]. If x[1] != 0 (the dice don’t show the same first value) then the expected waiting time until the walk returns to 0 is infinite.

Just Google "waiting-time random walk" for sources. I used this:

https://people.bath.ac.uk/ak257/36/Markov.pdf

Quote:teliotI think I got it, the answer is that it’s infinite.

I agree with this.

First of all, I am assuming that the two dice are different - for example, one is red, and the other is blue - and the rolling stops when the sum of the red die rolls equals the sum of the blue die rolls.

Let E(n) be the expected number of rolls needed until the two dice have equal sums, given that their current difference is n (it does not matter which particular die has the higher sum).

The initial value = 1 + 1/6 x 0 + 5/18 x E(1) + 2/9 x E(2) + 1/6 x E(3) + 1/9 x E(4) + 1/18 x E(5)

E(1) is a sum of E(1) through E(6) multiplied by rational numbers, but the E(1) terms can be combined, so the value is in terms of E(2) through E(6).

E(2) is a sum of E(1) through E(7) multiplied by rational numbers, but the E(1) terms can be replaced by E(2) through E(6) and the E(2) terms can be combined, so the value is in terms of E(3) through E(7).

Similarly, E(n) can be expressed in terms of E(n+1) through E(n+5), but as we cannot get to a point where E(n+5) has a value (it has an E(n+10) term, which in turn has an E(n+15) term, which in turn has an E(n+20) term, and so on), the result is infinite.

A fair coin is flipped until the count of heads equals the count of tails. What is the expected number of flips?

It seems obvious this would happen sooner than in teliot's question. So if it would take infinite coin flips, it would certainly take that with the dice.

Run 6311 has length 1,897,896,779 (619,729)

Run 9454 has length 23,231,798,283 (5,128,464)

Run 13,852 has length 1,261,141,555 (2,007,408)

Run 18,352 has length 7,965,870,273 (2,324,037)

Run 18,660 has length 1,395,131,771 (2,008,340)

Run 62,289 has length 1,935,125,795 (715,661)

Run 62,729 has length 3,275,843,264 (784,303)

Run 74,935 has length 1,375,446,093 (658,213)

The numbers in parentheses are the mean number of rolls per run up to that point.

Quote:ThatDonGuyI have been running a simultaion of the problem. Through 75,000 "runs," there were eight runs that required more than 1 billion rolls:

Run 6311 has length 1,897,896,779 (619,729)

Run 9454 has length 23,231,798,283 (5,128,464)

Run 13,852 has length 1,261,141,555 (2,007,408)

Run 18,352 has length 7,965,870,273 (2,324,037)

Run 18,660 has length 1,395,131,771 (2,008,340)

Run 62,289 has length 1,935,125,795 (715,661)

Run 62,729 has length 3,275,843,264 (784,303)

Run 74,935 has length 1,375,446,093 (658,213)

The numbers in parentheses are the mean number of rolls per run up to that point.

Don, what do your simulations show as the median number of rolls for them to be equal?

Quote:unJonDon, what do your simulations show as the median number of rolls for them to be equal?

Through 500,000 runs, here are the medians after each set of 10,000 runs - yes, the numbers do jump around a bit, don't they?

10,000 runs has median 3

20,000 runs has median 2

30,000 runs has median 8

40,000 runs has median 48

50,000 runs has median 1

60,000 runs has median 2

70,000 runs has median 2

80,000 runs has median 126

90,000 runs has median 106

100,000 runs has median 9

110,000 runs has median 25

120,000 runs has median 25

130,000 runs has median 1

140,000 runs has median 62

150,000 runs has median 19

160,000 runs has median 27

170,000 runs has median 7

180,000 runs has median 5

190,000 runs has median 107

200,000 runs has median 1

210,000 runs has median 16

220,000 runs has median 8

230,000 runs has median 7

240,000 runs has median 25

250,000 runs has median 329

260,000 runs has median 5

270,000 runs has median 7,301

280,000 runs has median 28

290,000 runs has median 3

300,000 runs has median 763

310,000 runs has median 146

320,000 runs has median 6

330,000 runs has median 64

340,000 runs has median 65,260

350,000 runs has median 1

360,000 runs has median 34

370,000 runs has median 4

380,000 runs has median 20

390,000 runs has median 1

400,000 runs has median 1

410,000 runs has median 6

420,000 runs has median 49

430,000 runs has median 9

440,000 runs has median 12

450,000 runs has median 2

460,000 runs has median 2

470,000 runs has median 35

480,000 runs has median 1

490,000 runs has median 3

500,000 runs has median 46