# Another coin problem: a super-fair coin making patterns?

Here's the follow-up I promised to:
http://answers.yahoo.com/question/index;_ylt=AjUPi5vJAy_o_BTk83ybJdHsy6IX;_ylv=3?qid=20080825193804AAlxo3x
You'll want to refer to an old question of...
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Here's the follow-up I promised to:

http://answers.yahoo.com/question/index;...

You'll want to refer to an old question of mine:

http://answers.yahoo.com/question/index;...

http://tech.groups.yahoo.com/group/mathf...

Basically, we have a coin which, given that we've tossed it n times and gotten X heads, the probability for heads in the next toss is 1-X/n (and p=1/2 in the first toss).

It's been shown that, for n tosses, X has a distribution that follows

Pr(X=x) = E(n-1,x-1)/(n-1)!

for n>=2, x=1, 2, ..., n-1 and is zero otherwise. Here E(n, k) is an Eulerian number:

http://en.wikipedia.org/wiki/Eulerian_numbers

However, that may not be too useful here, as I care about sequences.

In sync with the previous few coin questions here, I want to know this: If we flip the coin until the combination THT or TTH shows up, which sequence is more likely to be the end of our coin flipping?

http://answers.yahoo.com/question/index;...

You'll want to refer to an old question of mine:

http://answers.yahoo.com/question/index;...

http://tech.groups.yahoo.com/group/mathf...

Basically, we have a coin which, given that we've tossed it n times and gotten X heads, the probability for heads in the next toss is 1-X/n (and p=1/2 in the first toss).

It's been shown that, for n tosses, X has a distribution that follows

Pr(X=x) = E(n-1,x-1)/(n-1)!

for n>=2, x=1, 2, ..., n-1 and is zero otherwise. Here E(n, k) is an Eulerian number:

http://en.wikipedia.org/wiki/Eulerian_numbers

However, that may not be too useful here, as I care about sequences.

In sync with the previous few coin questions here, I want to know this: If we flip the coin until the combination THT or TTH shows up, which sequence is more likely to be the end of our coin flipping?

Update:
I am equally disappointed...from what I've done though I'm not sure that one exists (or at the very least it shouldn't be easy). I have a backwards recursion, but since I need to recurse backwards from infinity, I don't really have anything.
:)
BTW, are the numbers 52500 and 47500 actually the...
show more
I am equally disappointed...from what I've done though I'm not sure that one exists (or at the very least it shouldn't be easy). I have a backwards recursion, but since I need to recurse backwards from infinity, I don't really have anything.

:)

BTW, are the numbers 52500 and 47500 actually the averages you arrived at, or did you round for convenience? It would be remarkable if they actually came out to even hundreds...I suppose I could create my own program, but I'll make you work for your 10 points.

:)

BTW, are the numbers 52500 and 47500 actually the averages you arrived at, or did you round for convenience? It would be remarkable if they actually came out to even hundreds...I suppose I could create my own program, but I'll make you work for your 10 points.

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