Thread: 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.

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 .

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 .

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 .


See the pattern?

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!

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.
Here is your first posting:

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 .

If is the number of 'rolls' on which the first seven appears then
.

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

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

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 is the number of ways that the sum of two dice is equal

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

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.

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%.

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.

Yes I figured that out.