More statistics of dice rolls

[Marsal]

Found some time to look into the problem, gave up and applied brute force. You were right (partially). It seems that the answer is 29 (bit less, if you don't look at rolls as integers). Try to code it yourself, maybe I fucked something up. Tested on 10M sample groups, a couple of times, with different seeds, it gives ~50.4% probability.

Now you just have to come up with the correct formula

EDIT: "Right, so how many times do you flip a coin to get two heads in a row?" Tried it and the answer is 4. Didn't read your or Elwro's long posts, is this wrong? If it is, disregard the above post.

Why are you even helping this faggot Marsal?

Are you missing a comma, or am I the faggot in your post? I like a challenge and this seemed a pretty straightforward problem. I guess I should have remembered that I suck at this part of mathematics. Too late now.

 

[Elwro]

EDIT: "Right, so how many times do you flip a coin to get two heads in a row?" Tried it and the answer is 4. Didn't read your or Elwro's long posts, is this wrong?

No, it's correct if your goal is to have probability=1/2 and it's correct because 8 out of 16 4-element 01-sequences fulfill the success criteria.

I'll probably have time to go back to the mathematics needed to generalize my reasoning with Z_n and B_n (because the original was wrong and I should've seen the stupitidy of the conclusion - chance growing over 1 - from the start) Tuesday morning.

 

[Hobbit Lord of Mordor]

I think the confusion might be we are looking for different things *. Maybe what I am looking for is the median and you are looking at the mean. I also suck at conceiving probabilities though I like finding the solutions.

Did you look at that paper? It looks at it as a Markov Chain problem, the nature of the outcomes change depending on whether you are looking for the first success or the second. I put it into words in my last post because I don't know how to express it mathematically.

* According to graphs it seems that way. The (# trials to cover area 0.5) seems to be 4 flips, but the average is 6.

 

[Marsal]

To be honest, I don't care that much about the subject at hand to read long posts or papers. Sorry.

I'll just lurk in the shadows and point out obvious mistakes Is your goal to answer the question or to learn how to answer the question? IMO 15 min of coding is much easier (and less mistake prone) to do, than fiddle with formulas you are not comfortable with (at least for "simple" problems).

 

[Hobbit Lord of Mordor]

I am sure there is a simple formula for this problem and I want to know what it is.

Solving it by simulation would be pointless.

 

[Hobbit Lord of Mordor]

Elwro's solution is wrong.

My original reasoning was wrong; still, I think I'm on the right track with my last post.

Yes I think you are.

The discrete probabilities (of getting double heads in x flips) goes like:

0
0
25
12.5
12.5
9.375
then I don't know.

 

[Marsal]

Okay, this is the last time I read this kind of topic. I just spent an hour scribbling on a piece of paper. I think I got it.

Imagine the dice have already been thrown n times (n is the answer to the OP). Keep in mind that some terminology may be off, because I don't know exact English terms for everything.

I counted all the combinations that are invalid (don't contain any two or more sixes in a row):

1. Just numbers 1 to 5 = 5^n
2. 1 six, all other 1 to 5 = n * 5^(n-1)
3. 2 sixes, not in a row, all other 1 to 5 = (n-1)(n-2)/2 * 5^(n-2)
3. 3 sixes, not in a row, all other 1 to 5 = (n-2)(n-3)(n-4)/6 * 5^(n-3)

...and so on ...

Let's write that in a sum:

SUM (r=0 to n/2) { (n-r+1)! / [ r! * (n-2r+1)! ] * 5^(n-r) }

r is number of sixes allowed. Why does it go to n/2? It actually goes to n/2 for even n and (n+1)/2 for odd n. You must have at least one "space" (number that is not 6) between 2 sixes (ie. for n=9, r=5: 6x6x6x6x6), so you can't have more than half +1 sixes and still not have two in a row.

The above SUM is the total number of combinations we don't want. There are 6^n possible combinations. So the result should be SUM / 6^n >= 1/2.

We feed this to Wolfram Alpha (the URL tag doesn't work or I'm doing it wrong, so just c/p the links):

http://www.wolframalpha.com/input/?i=SU ... 1%29!+%2F+[+r!+*+%28n-2r%2B1%29!+]+*+5^%28n-r%29+%29%2C+r%3D0+to+n%2F2

and then plot:

http://www.wolframalpha.com/input/?i=pl ... %29%2F2%29^x%2B%285%2F2%2B%283+sqrt%285%29%29%2F2%29^x%29%2F+6^x+*2+%3D1+from+x%3D20+to+35

And there you have it, little over 28.