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nick
nick
Joined: Jan 1, 2010
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January 18th, 2010 at 7:52:20 AM permalink
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
pacomartin
pacomartin
Joined: Jan 14, 2010
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January 18th, 2010 at 9:56:57 AM permalink
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.
DJTeddyBear
DJTeddyBear
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January 18th, 2010 at 10:23:00 AM permalink
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?
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
cclub79
cclub79
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January 18th, 2010 at 10:26:30 AM permalink
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.
DJTeddyBear
DJTeddyBear
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January 18th, 2010 at 10:31:46 AM permalink
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.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
cclub79
cclub79
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January 18th, 2010 at 10:36:19 AM permalink
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.
DJTeddyBear
DJTeddyBear
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January 18th, 2010 at 10:39:42 AM permalink
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....
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
DJTeddyBear
DJTeddyBear
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January 18th, 2010 at 10:52:56 AM permalink
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.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
stephen
stephen
Joined: Jan 5, 2010
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January 18th, 2010 at 11:13:57 AM permalink
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.
DorothyGale
DorothyGale
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January 18th, 2010 at 12:54:23 PM permalink
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
"Who would have thought a good little girl like you could destroy my beautiful wickedness!"

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