long odds
January 18th, 2010 at 7:52:20 AM
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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
Thank you
NIck
January 18th, 2010 at 9:56:57 AM
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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.
January 18th, 2010 at 10:23:00 AM
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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 GOTTA be missing something. Or am I?
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/ —————————————————————————————————————
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
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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
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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/ —————————————————————————————————————
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
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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
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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.
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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....
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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/ —————————————————————————————————————
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
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That's where my error is.Quote: DJTeddyBearThe odds of hitting a 12 once in 6 tries seems to be 14.3% * 6 = 85.7%
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/ —————————————————————————————————————
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
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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
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Quote: nickMy 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!"
