L*****e
发帖数: 169 | 1 flip a coin, p for heads, 1-p for tails, what is the probability
that the number of heads equal the number of tails, assuming infinite
number of flipping? | s***t
发帖数: 49 | 2 0
【在 L*****e 的大作中提到】
: flip a coin, p for heads, 1-p for tails, what is the probability
: that the number of heads equal the number of tails, assuming infinite
: number of flipping?
| b********u
发帖数: 63 | | L*****e
发帖数: 169 | 4 how did you get it?
【在 b********u 的大作中提到】
: 1-sqrt(1-4pq)
| s***t
发帖数: 49 | 5 貌似是2次方程的解
【在 L*****e 的大作中提到】
: how did you get it?
| b********u
发帖数: 63 | 6 I'm not sure whether it's correct, but you can convert it to a random walk,
and compute the prob. returning to the origin, similar to the method below,
http://www.mitbbs.com/article/Quant/31228851_0.html
【在 L*****e 的大作中提到】
: how did you get it?
| c*******d
发帖数: 255 | 7 assume 2*N flips, and we'll let N->infty
the probability of having N heads and N tails is
Pr = C(2N, N)*p^N*(1-p)^N
= (2N)!/(N!)^2 *[p(1-p)]^N
using sterling formula, we have
Pr = [sqrt(2 pi 2N)*(2N/e)^(2N)*(1+O(1/N))] /
[sqrt(2 pi N) * (N/e)^N*(1+O(1/N))]^2 * [p(1-p)]^N
~= 1/sqrt(pi N) * [4p(1-p)]^N
since 4p(1-p) <= 1, we have [4p(1-p)]^N <=1
thus Pr -> 0 as N-> infty
【在 L*****e 的大作中提到】
: flip a coin, p for heads, 1-p for tails, what is the probability
: that the number of heads equal the number of tails, assuming infinite
: number of flipping?
| r*****d
发帖数: 44 | 8 if p!=1/2, it should be 0 since non-typical sequences have vanishing
probabilty of occurance asymptotically.
【在 L*****e 的大作中提到】
: flip a coin, p for heads, 1-p for tails, what is the probability
: that the number of heads equal the number of tails, assuming infinite
: number of flipping?
| z****i
发帖数: 406 | 9 MS的电面吧？ 是2*min(p,1-p).
http://www.mathproblems.info/working.php#s11
第11题。
我当时就挂在这个题上了，唉。。。。 |
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