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long odds

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January 18th, 2010 at 7:52:20 AM permalink
nick Member since: Jan 1, 2010 Threads: 1 Posts: 4	My question is based on dice odds. I know that there are 6 ways to get 7 and 1 way to get 12 but what are the chances of getting 6 sevens before 1 twelve. Are they even and if not how many 12s should be added to the equation to make it an even proposition? Thank you NIck

January 18th, 2010 at 9:56:57 AM permalink
pacomartin Member since: Jan 14, 2010 Threads: 215 Posts: 1501	Unfortunately I don't have time to figure that out right now. However, you must set up a Markov transition matrix to cover the situation. You can't do this problem with an algebraic formula. It's much more complex than that. Wine loved I deeply, dice dearly -Edgar, betrayed son of Gloucester in King Lear

January 18th, 2010 at 10:23:00 AM permalink
DJTeddyBear Member since: Nov 2, 2009 Threads: 39 Posts: 1681	Intuition says the odds of hitting a 6 in 36 event six times before hitting a 1 in 36 event just once should be 50%. I GOTTA be missing something. Or am I? Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood?

January 18th, 2010 at 10:26:30 AM permalink
cclub79 Member since: Dec 16, 2009 Threads: 21 Posts: 642	I think it would be the chances of a 7 aren't 6/36, but 6/7, since you are ignoring everything but 7s and 12s.

January 18th, 2010 at 10:31:46 AM permalink
DJTeddyBear Member since: Nov 2, 2009 Threads: 39 Posts: 1681	cclub is commenting on a rather long post that I had up for about 2 minutes before I realized that I was completely out of my mind. Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood?

January 18th, 2010 at 10:36:19 AM permalink
cclub79 Member since: Dec 16, 2009 Threads: 21 Posts: 642	You weren't out of your mind; I think you had it exactly right in your equation except replace (6/36) ^ 6 with (6/7) ^ 6.

January 18th, 2010 at 10:39:42 AM permalink

DJTeddyBear

Member since: Nov 2, 2009
Threads: 39
Posts: 1681

Yeah, ignore everything but 7 and 12, and the odds of a 7 are 6/7 = 85.7% while the odds of a 12 are 1/6 = 14.3%. That adds up to 100%. Cool.

---

The odds of any 7 six times in a row seems to be 85.7% ^ 6 = 39.7%
The odds of hitting a 12 once in 6 tries seems to be 14.3% * 6 = 85.7%

The obvious problem is that it adds up to more than 100%

Back to the drawing board....

Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood?

January 18th, 2010 at 10:52:56 AM permalink

DJTeddyBear

Member since: Nov 2, 2009
Threads: 39
Posts: 1681

Quote: DJTeddyBear
The odds of hitting a 12 once in 6 tries seems to be 14.3% * 6 = 85.7%

That's where my error is.

Since the problem is resolved prior to six throws once a 12 shows up in the first 5 throws, you have to either create a very complex formula that takes that into account, or calculate for the event that requires all six throws and subtract from 100%

I.E. The odds of hitting six 7s in a row, an event that requires all six throws to resolve, is ( 6 / 7 ) ^ 6 = 39.7%
Therefore the odds of NOT doing that, i.e. hitting AT LEAST one 12 in those six throws, is 100% - 39.7% = 60.3%

THEREFORE: Bet on 12.

Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood?

January 18th, 2010 at 11:13:57 AM permalink
stephen Member since: Jan 5, 2010 Threads: 1 Posts: 28	DJTeddyBear is correct. I verified it with a simple script that ran through a few million iterations and my experimental results matched his derived ones perfectly.

January 18th, 2010 at 12:54:23 PM permalink

DorothyGale

Member since: Nov 23, 2009
Threads: 18
Posts: 252

Quote: nick
My question is based on dice odds. I know that there are 6 ways to get 7 and 1 way to get 12 but what are the chances of getting 6 sevens before 1 twelve. Are they even and if not how many 12s should be added to the equation to make it an even proposition?

Thank you
NIck

Interesting question. I didn't feel like tacking it direcctly, because a simple simulator was about a two minute programming project. Here are the results of a simulation of one billion (1,000,000,000) rolls of the dice for four separate cases.

Here are the simulation results for four 7's vs. one 12 --
Number of times four 7's came up before a 12: 32581424
Number of times a 12 came up before four 7's: 27768376

Here are the simulation results for nine 7's vs. two 12's --
Number of times nine 7's came up before two 12's: 13442654
Number of times two 12's came up before nine 7's: 10107307

Here are the simulation results for five 7's vs. one 12 --
Number of times five 7's came up before a 12: 23914613
Number of times a 12 came up before five 7's: 27781426

Here are the simulation results for six 7's vs. one 12 --
Number of times six 7's came up before a 12: 18256704
Number of times a 12 came up before six 7's: 27781741

Here is the sloppy C code for the latter of these four simulations. I did not audit or double check this code:


#include 
#include 
#include 

main() {
 int d1, d2, s; // dice and dice sum
 int noS = 0, noT = 0; // counters for sevens and twelves
 int ctr, x = 0, y = 0; // other counters

 srand(time(NULL));

 x = 0;
 y = 0;

 for (ctr = 0; ctr < 1000000000; ctr++) {
 if (ctr%1000000 == 0)
 fprintf(stderr, "%d\r", ctr);

 d1 = rand()%6+1;
 d2 = rand()%6+1;
 s = d1+d2;
 if (s == 7)
 x++;
 if (s == 12)
 y++;

 if (x == 6) {
 noS++;
 x = 0;
 y = 0;
 }
 if (y == 1) {
 noT++;
 x = 0;
 y = 0;
 }
 }

 printf("\n");
 printf("Number of times six 7's came up before a 12: %d\n", noS);
 printf("Number of times a 12 came up before six 7's: %d\n", noT);
}

--Dorothy

Resident OZ-like entity ...

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