Really interesting question! Thanks for posting. I don't think you have received an answer yet so I will take a bit of time to talk you through it all. Let us use a little formal notation and define some sets that will help us solve this problem. We are going to define and examine 3:
$\bf{\Omega:}$
Let us define $\Omega$ as our sample space of $\textbf{all}$ outcomes. I.e $\Omega$ is the set of all different sequences of $30$ coin flips. Clearly the size of $\Omega$ is $2^{30} $
$\bf{\mathcal{S}}: $
We define $\mathcal{S} \subset \Omega$ As the set of $\bf{all} $ successful sequences. i.e $HHHHHHHTHHHHTHHHHHHHHHHHHHHHTTT \in \mathcal{S} $ Clearly the probability we are after is $\frac{|\mathcal{S}|}{|\Omega|}$ and as we already have $|\Omega|$ all we need is $|\mathcal{S}|$ and we are done. Eyes on the prize.
$\bf{\mathcal{U}:}$
Now call $\mathcal{U}$ this set of unique ways to reach the position where #H = #T $+2$ for the first time. Elements of $\mathcal{U}$ are a sort of truncation of element from $\mathcal{S}$
For example: $HH , THHH, $and the example you gave are all some elements of $\mathcal{U} $
Notice that $\mathcal{U} \not \subset \Omega $ as $\mathcal{U} $ is $\textbf{not}$ a set of $\textbf{full}$ sequences of 30 flips, it is just the opening moves to reach this net $+2$ heads. Elements of $\mathcal{U} $ will correspond somewhat to elements of $\mathcal{F}$ eventually. For example:
$ \mathcal{U} \ni HH \not \in \Omega $ but $HHHHHHHTHHHHTHHHHHHHHHHHHHHHTTT \in \mathcal{S} \subset \Omega$ Every element of $\mathcal{S}$ has one unique corresponding element in $\mathcal{U}$
We are going to solve this problem in 5 distinct steps.
Firstly we will construct a partition of $\Omega$ into $3$ pieces
We will show that $2$ of these pieces belong to $\mathcal{S}$
We will then construct more elements of $\mathcal{S}$ that are not in our initial partitions of $\Omega$
We will then show that $\mathcal{S}$ contains only elements of these three sets, the two partitions of $\Omega$ and the final set we constructed in step 3
The sum of the size of these $3$ pieces make up the size of $\mathcal{S}$ (what we are after)
$\textbf{Step 1:}$
Call $\Omega_{16}, \Omega_{>16} $ and $\Omega_{<16}$ respectively the elements of $\Omega$ that contain 16, more than 16 and less than 16 heads. This is clearly a mutually disjoint partition.
Examples: 16 heads followed by 14 tails is in $\Omega_{16}.$ All heads and all tails a respectively in $\Omega_{>16}$ and $\Omega_{<16}$
$\textbf{Step 2:}$
Clearly elements of $\Omega_{16}$ and $ \Omega_{>16} $ are all elements of $\mathcal{S}$ as these are sequences with at least 16 heads and thus at most 14 tails and as such there exists (at least one) point in which we have 2 more heads than tails. Hence $\Omega_{16} \cup \Omega_{>16} \subseteq \mathcal{S} $
$\textbf{Step 3:}$
What I intended to do now is, for $\textbf{every single element of} $ $\bf{\Omega_{>16}}$, generate another element of $\mathcal{S} $ that is in $\Omega_{<16}$
For each element $x \in \Omega_{>16} \subset \mathcal{S} $ we shall call $u$ the corresponding element in $\mathcal{U}$ For example if $ x = HTHTHTHHHHHHHTTTHHHHHHHHHHHHHHH$ then $u = HTHTHTHH$ or if $ x = HHHHHHHHHHHHHHHHHHHHHHHHHHHHH$ then $u = HH$
Now call $l$ the "leftover" of $x$ when we take away $u$
Examples: if $ x = HTHTHTHHHHHHHTTTHHHHHHHHHHHHHHH$ then $u = HTHTHTHH$ and $l = HHHHHTTTHHHHHHHHHHHHHHH$
Now define a function "flipped", $f:$ Sequences $\to$ Sequences by flipping each tail to a head and each head to a tail.
Example: $f(HTHTTT) = THTHHH $
Now finally for each $x \in \Omega_{>16} $ define $g(x): \mathcal{S} \to \mathcal{S} $ by $g(x) := u +f(l) $
Example: If $x = TTHHHHHHHHHHHTHTHHHHHHHHHHHHHH $ then $g(x) = TTHHHH + f(HHHHHHHTHTHHHHHHHHHHHHHH) = TTHHHH + TTTTTTTHTHTTTTTTTTTTTTTT = TTHHHHTTTTTTTHTHTTTTTTTTTTTTTT$
Properties of g:
- If $x \in \mathcal{S} \implies g(x) \in \mathcal{S} $ as g does not change the inital elements $u$ that result in a net $+2$ heads
- Notice that $g$ is its own inverse as g never changes the beginning terms, it simply flips the last and the "flipping" function is clearly its own inverse. Hence $g(g(x)) = x $
- g is a bijection.
- $g : \Omega_{>16} \to \Omega_{< 16}$ or $g : \Omega_{<16} \to \Omega_{> 16}$
As if $x$ has $h$ heads and $t$ tails with $h_u$ heads in $u$ and $h_u - 2$ tails in $u$ with $h$ > 16 (i.e it is in $\Omega_{>16} $ ) then $g(x) $ has $h_u$ heads in $u$ and $t-(h^u-2) = t + 2 - h^u$ heads in $l$ Hence has $t + 2$ heads in total. And as $h = 30 - t > 16 $ we have $t +2 < 16 $ Hence $g(x) \in \Omega_{<16} \iff x \in \Omega_{>16} $ and visa versa
Define $ \Omega_{<16} \supset \Omega_{G} := g(\Omega_{>16}) \subset \mathcal{S} $ as everything in $ \Omega_{>16} $ is in $\mathcal{S} $ and property 1) of g shows that it will keep everything in $\mathcal{S} $
$\textbf{Step 4:}$
We now have $3$ disjoint sets: $\Omega_{16} , \Omega_{>16} , \Omega_{G} $
Define $\mathcal{B} := \Omega_{16} \cup \Omega_{>16} \cup \Omega_{G}$
We will prove $\mathcal{B} = \mathcal{S}$
We have already show that everything in $\Omega_{16} , \Omega_{>16} , \Omega_{G} $ is in $\mathcal{S}$ Hence $\mathcal{B} \subseteq \mathcal{S}$ Thus all we wish to show for equality is : $\mathcal{S} \subseteq \mathcal{B}$
Finally assume $ x \in \mathcal{S}$ we have $3$ options as to which of $\Omega_{16} , \Omega_{>16} , \Omega_{<16} $ $x$ belongs to. If $x$ is in one of the first two then great! As this would imply $x \in \mathcal{B} $
If $ x \in \Omega_{<16} \cap \mathcal{S} \implies g(x) \in \Omega_{>16} \implies g(g(x)) = x \in g(\Omega_{>16}) = \Omega_{G} \implies x \in \mathcal{B} $
Hence $\mathcal{S} \subseteq \mathcal{B}$ and as $\mathcal{B} \subseteq \mathcal{S}$ we have $\mathcal{B} = \mathcal{S}$
$\textbf{Step 5:}$
$|\mathcal{S}| = |\mathcal{B}| = |\Omega_{16} \cup \Omega_{>16} \cup \Omega_{G}| = |\Omega_{16}|+ |\Omega_{>16}|+|\Omega_{G}| $ (as they are disjoint sets.) $ = {30 \choose 16} + \sum\limits_{i=17}^{30}{30 \choose i} + |\Omega_G|$
However $\Omega_G$ is the image of an bijective map from a finite set and as such its size is equal to the size of its domain and as thus$ |\Omega_{>16}|=|\Omega_{G}| = \sum\limits_{i=17}^{30}{30 \choose i} $
And thus : $|\mathcal{S}| = {30 \choose 16} + 2\sum\limits_{i=17}^{30}{30 \choose i} $
And hence probability to have won this game, $\frac{|\mathcal{S}|}{|\Omega|} $ = $\frac{{30 \choose 16} + 2\sum\limits_{i=17}^{30}{30 \choose i}}{2^{30}} \approx 0.720100131817 $
I hope this helped.
Oskar
Best Answer
No that is incorrect. The probability of the second toss does not depend on the outcome of the first toss, therefore it is not a conditional probability problem. So whether you were right or wrong on the first toss the probability of the second toss is ($\frac{1}{2}$). If you were to guess the next two tosses then the probability of getting them both right would be $(\frac{1}{2})^2$ no matter what your guess was for each toss, same goes for the nth case.