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?

  • $\begingroup$ 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

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.

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