# Probability of flipping 1 coin vs 3 to land a head given previous successes

(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.

• 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 $1/2 + 1/2 + 1/2 = 3/2$? – angryavian Dec 7 '18 at 2:45
• @angryavian I want a head from either of the 3. – glockm15 Dec 7 '18 at 3:18
• 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. – zipirovich Dec 7 '18 at 3:28