# Math Help - Dice 7 Odds Theory

1. ## Dice 7 Odds Theory

Does anyone know how to graph this phenomena on a computer? I graphed with dice how many rolls occur after a 7 before another 7 occurs. I understand that this may not work with computer generated numbers. The odds of 7 is 1 in six but this may be an additional way to predict the odds. The graph by hand showed that the probability after a seven of another 7 on the first roll is 1 in 9 and that proceeds to dip down to 1 in 3 on something like the 5th roll without a 7 and then proceeds exponentially up to approaching 100% after for example 100 rolls without a 7. At this point the dice become unpredictable.

2. ## Re: Dice 7 Odds Theory

Do you understand that "how many rolls occur after a 7 before another 7 occurs" is exactly the same as "how many rolls occur before the first 7 occurs"?
A 7 can occur on the first roll with probability 1/6 so P(x= 0)= 1/6, NOT "1 in 9".

Anything other than a 7 on the first roll has probability 5/6 and then a 7 on the second roll has probability 1/6 so $P(x= 2)= \frac{5}{6^2}$.

Any thing other than a 7 on the first roll has probability 5/6, anything other than a 7 on the second roll has probablity 5/6, and then a 7 on the third roll has probablity 1/6 so $P(x= 3)= \frac{5^2}{3^3}$.

Any thing other than a 7 on the first roll has probability 5/6, anything other than a 7 on the second roll has probablity 5/6, anything other than a 7 on the third roll has probability 5/6, and then a 7 on the fourth roll has probablity 1/6 so $P(x= 4)= \frac{5^3}{3^4}$.

See the pattern?

3. ## Re: Dice 7 Odds Theory

You are wrong in a number of ways, first- there are a variety of odds with 2 dice the odds of there being a 7 is how many possibilities with 2 dice, there is: 1/6, 1/5, 1/4, 1/3, 1/2, 1/1, 2/6, 2/5, 2/4, 2/3, 2/2, 2/1, 3/6, 3/5, 3/4, 3/3, 3/2, 3/1, 4/6, 4/5, 4/4 - 21 possibilities. 3 add up to 7, that is 3/21 and 33%. You fail to graph the number of rolls that occur before the first seven. I have a picture of the graph here, it shows how 7 has a probability of more than 33% then dips down to less than 33% then goes back up past 33% and increases exponentially as its asymptote approaches 100% probability of rolling a 7. I lost the actual graph years ago when I figured this out. Everyone drive safely to the Casino and I want a fair cut of the profit!

4. ## Re: Dice 7 Odds Theory

Originally Posted by phillipanthonybiondo
You are wrong in a number of ways, first- there are a variety of odds with 2 dice the odds of there being a 7 is how many possibilities with 2 dice,

If indeed you think that, then you are seriously uninformed.
Originally Posted by phillipanthonybiondo
Does anyone know how to graph this phenomena on a computer? I graphed with dice how many rolls occur after a 7 before another 7 occurs.
Rolling two dice. There are six ways of getting a sum of seven.

So $\mathcal{P}(S=7)=\frac{1}{6}$.

If $X$ is the number of 'rolls' on which the first seven appears then
$\mathcal{P}(X=k)=\left( {\frac{5}{6}} \right)^{k - 1} \left( {\frac{1}{6}} \right) = \frac{{5^{k - 1} }}{{6^k }}$.

Now assume that it takes $K$ rolls to get the first seven, how many more to get the next seven?

5. ## Re: Dice 7 Odds Theory

6 ways? How? 1/6, 3/4, 5/2, what else?

6. ## Re: Dice 7 Odds Theory

Originally Posted by phillipanthonybiondo
6 ways? How? 1/6, 3/4, 5/2, what else?
This question shows how poorly informed you are.
Look at the web calculation.

The coefficient of $x^k$ is the number of ways that the sum of two dice is equal $k~.$

7. ## Re: Dice 7 Odds Theory

You fail to tell me how there are more than 3 possibilities to roll a 7.

8. ## Re: Dice 7 Odds Theory

6-1, 1-6, 5-2, 2-5, 4-3, 3-4.
There are 36 possible outcomes with two dice. Six possibilities with the first to go with six possiblities for the second.

9. ## Re: Dice 7 Odds Theory

you are repeating a six and a one, it doesn't matter which dice has the 6 or 1 and I only found 25 possible combinations so 3/25 is 12%.

10. ## Re: Dice 7 Odds Theory

Try thinking it out with dice of different colours, say one red one and one blue one.
There are two ways of throwing 6-1.
6 with the red die and 1 with the blue die or,
6 with the blue die and 1 with the red die.
There is only one way of throwing, say, double one.
1 with the red die and 1 with the blue die.
6-1 is twice as likely as 1-1.

11. ## Re: Dice 7 Odds Theory

Yes I figured that out.