0
$\begingroup$

Variation to this problem

What if I tossed $100$ different coins with a different bias towards the head?

(For example, the biases are $p_1, p_2, \ldots, p_{100}$ and the coin are not necessarily tossed in order of the subscripts of $p$.)

What would be the probability that heads is flipped exactly $65$ times?

I couldn't even get the number of ways in which $65$ heads are got, any help here?

$\endgroup$
  • $\begingroup$ The question you linked to gives a formula for a general probability $p$. Is that not sufficient? Note: this is just the standard formula for any binomial process. $\endgroup$ – lulu Sep 19 '17 at 15:24
  • $\begingroup$ I meant the coins are biased with $p_1, p_2, \ldots, p_{100}$.@lulu $\endgroup$ – Random-generator Sep 19 '17 at 15:26
  • $\begingroup$ Ah, that's not clear. You should edit your question. That said, I doubt there is a useful answer. Sampling isn't a bad approach. $\endgroup$ – lulu Sep 19 '17 at 15:33
  • $\begingroup$ If I had to attack it, I'd do it by backwards induction. If the first toss is $H$ then you need $64$ out of the rest, if the first toss is $T$ then you still need $65$, and so on. But I'd almost certainly just sample. $\endgroup$ – lulu Sep 19 '17 at 15:36
  • 1
    $\begingroup$ Sure, though those numbers are so huge that this isn't going to help much. $\endgroup$ – lulu Sep 19 '17 at 16:29
0
$\begingroup$

By analogous reasoning to proving the formula for the pmf of a Binomial random variable, we get $$\sum_{\substack{S\subseteq[100]\\|S|=65}}\prod_{i\in S}p_{i}\prod_{i\in[100]\setminus S}(1-p_{i}),$$ where the sum is over all subsets $S$ of $[100]=\{1,2,\ldots,100\}$ with size $65.$ It's not terribly clear how useful this formula will be, but if you had some bounds on the $p_{i},$ you might be able to use this to prove that the probability lies in some interval.

For instance, if all $p_{i}\in[1/2,3/4],$ then $\prod_{i\in S}p_{i}\in[(1/2)^{65},(3/4)^{65}],$ and $\prod_{i\in[100]\setminus S}(1-p_{i})\in[(1/4)^{35},(1/2)^{35}]$, giving a product in $[(1/2)^{65}(1/4)^{35},(3/4)^{65}(1/2)^{35}]$ for all $S\subseteq[100].$ Then the probability lies in the interval $[\binom{100}{65}(1/2)^{135},\binom{100}{65}(3/4)^{65}(1/2)^{35}]\approx[2.514\times 10^{-14},1]$, which admittedly is not super helpful, but with tighter bounds, this could be improved.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.