(Assuming 1/2 probability for heads or tails)

Using coin 1 I flip it say $n$ times and I have 60 heads, so probability of head:

$\frac{60}{n}, n>0$

Using three coins I flip the first $m$ times and get $6$ heads, the second I flip $p$ times and get also $6$ heads, the third coin flipped $q$ times gets $7$ heads:

Coin 2: $\frac{6}{m}$

Coin 3: $\frac{6}{p}$

Coin 4: $\frac{7}{q}$

and since they are independent probability of a head with the 3 coins is

$\frac{6}{m}+\frac{6}{p}+\frac{7}{q}=\frac{6pq+6mq+7mp}{mpq}, q,m, p>0 $

and assume : $n>q>m \geq p>0$

Now I want to ask if I am to flip the coins again once, which "group" would give me the highest probability of getting a head.

Flipping only Coin 1 once or flipping all of Coins 2,3 and 4 once. Is it possible to reason without knowing $n,m,p,q$ If not what minimal information might we need to further assume.

  • 1
    $\begingroup$ If $m=12$, $p=12$, and $q=14$, then coins 2 and 3 and 4 each show heads $1/2$ of the time. Why would you want to add these together to get $1/2 + 1/2 + 1/2 = 3/2$? $\endgroup$ – angryavian Dec 7 '18 at 2:45
  • $\begingroup$ @angryavian I want a head from either of the 3. $\endgroup$ – glockm15 Dec 7 '18 at 3:18
  • $\begingroup$ Sorry, but this text doesn't make much sense. If in the very beginning you already said that we're Assuming 1/2 probability for heads or tails, then how can these probabilities be anything else ($60/n$ or whatever)? After performing an experiment, such as actually flipping coins, the observed outcomes give you the frequencies, not the probabilities -- and a posteriori frequencies don't have to agree with a priori probabilities. $\endgroup$ – zipirovich Dec 7 '18 at 3:28

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