# Fallacy with indepent events and probability

Consider tossing a fair coin. If we get a head, we further randomly draw a number out of {1, 2, 3}. If we get a tail, we randomly draw a number out of {4, 5, 6, 7}. It is evident that probability of getting a head is $o.5$. However, I'm running into also sorts of troubles proving this mathematically. Let me explain.

Let me define few relevant samples spaces as:
$Overall = \{H1, H2, H3, T4, T5, T6, T7\}$
$Toss = \{H, T\}$ $Draw1 = \{1, 2, 3\}$ $Draw2 = \{4, 5, 6, 7\}$

Let me also define few events of our interest in the '$Overall$' sample space.
$\mathcal{H}$= the event of getting a head = $\{H1, H2, H3\}$
$\mathcal{D}3$= the event of getting a three = $\{H3\}$
I need to prove $ℙ$ (getting a head) = $ℙ (\mathcal{H}) = 0.5$

Argument 1:
Since the coin is fair, $ℙ (\mathcal{H}) = 0.5$. But $\mathcal{H}$ is not just $\{H\}$, it also includes the draws. To me declaring $ℙ_{toss} (\{H\}) = 0.5$ is fine as it is based on axioms of probability. But declaring $ℙ_{overall} (\mathcal{H}) = 0.5$ is a bit abrupt.

Argument 2:
Probability of an event is not affected by events of the future experiments and therefore $ℙ_{overall}$ $(\mathcal{H}) = ℙ_{toss}$ $(\{H\}) = 0.5$. This means probability of getting a head is independent of getting any value in the subsequent random draw; and this has implications as described below:
$\mathcal{H}$ is independent of $\mathcal{D}3$
$\Rightarrow$ $\mathcal{D}3$ is independent of $\mathcal{H}$
$\Rightarrow$ $\mathcal{D}3$ is independent of $\mathcal{H}^c$ - This is not true. $ℙ ($a tail and $3$ $) \ne ℙ ($a tail$) \times ℙ ($ $3$ $)$

Can anyone help me prove $ℙ (\mathcal{H}) = 0.5$
Please use symbols rather than words, if possible, for events.

• $P(H)=0.5$ is an assumption..specifically, the statement "Consider tossing a fair coin"...you cannot prove it. – user408433 Jun 30 '17 at 5:22
• @ Bey : By 'fair' we assign each of the events $\{H\}$ and $\{T\}$ a probability measure of $0.5$. But how do we derive a probability measure of $0.5$ for the event $\{H1, H2, H3\}$? – KGhatak Jun 30 '17 at 7:16

You think too compicated, maybe you should better think of random variables. You can model the experiment as follows: Let $X$ be the outcome of the coin tossing, and Y the outcome of the number drawing. Then you have: $$P(X= \mbox{head})=P(X=\mbox{tail})=0.5$$, since the coin is fair. Moreover, $$P(Y= i|X=\mbox{head})=1/3$$, for $i =1,2,3$ and $0$ else and $$P(Y= i|X=\mbox{tail})=1/4$$, for $i=4,5,6,7$. If you wish you can calculate everything else from this, e.g. the common distribution (but there is no point in that, even the modelling for the number drawing is not relevant for your question).

• @ crankk : Referring to the assignment $P(X= \mbox{head})=P(X=\mbox{tail})=\frac12$, you could assign this if $\mbox{X}$ is defined within the scope of just a toss. My question is how do you assign the same probability in the context of a toss followed by a drawing of a number. – KGhatak Jun 30 '17 at 7:35
• Write h=head,t=tail. I would say $P(H1,H2,H3)=P((X,Y) \in \{(h,1),(h,2),(h3)\})=P( (X,Y) = (h,1) ) + P( (X,Y) = (h,2) ) + P( (X,Y) = (h,3) )$. Then, $P( (X,Y) = (h,1) ) = P( X=h,Y=1) = P(Y=1|X=h)P(X=h)=1/3*1/2$ and so on. I think it depends on how you model your experiment. A fair coin toss is modeled as I wrote it by definition. – crankk Jun 30 '17 at 7:46
• @ crankk : I like the explanation using two r.v.s – KGhatak Jul 1 '17 at 18:23

What may be leading you astray is thinking the outcomes in the "overall" set are equally probable.   This is not true.   You cannot say $\mathsf P_\mathrm{overall}(\mathcal H)=3/7$, because you need to account for the differing probability weighting for each outcome.

Now, the draw is certainly not independent of the coin toss, since the possible outcomes from the draw clearly depend on the result of the toss.

The conditional probability for obtaining $3$ from the draw given a toss of $\mathcal H$ is $1/3$, because we are drawing without bias from $\{1,2,3\}$, while the conditional probability for doing so given a toss of $\mathcal T$ is surely $0$ since we are drawing from $\{4,5,6,7\}$.

As the coin is fair, then by definition of fairness we obtain that:

\begin{align}\mathsf P(\mathcal H3) ~&= \mathsf P(\mathcal H)\mathsf P(\mathcal D3\mid\mathcal H) && \neq \mathsf P(\mathcal H)\mathsf P(\mathcal D3)\\ &= \tfrac 12\cdot \tfrac 13 \\ &= \tfrac 16\end{align}

Note that by Law of Total Probability: $\mathcal P(\mathcal D3)=\mathsf P(\mathcal H)\mathsf P(\mathcal D3\mid\mathcal H) +\mathsf P(\mathcal T)\mathsf P(\mathcal D3\mid\mathcal T) = \tfrac 16$ .

And similarly for the other possible results.

Thus obtaining, albeit circularly, that $\mathsf P(\mathcal H) ~{= \mathsf P(\mathcal H1\cup\mathcal H2\cup\mathcal H3) \\= \tfrac 16+\tfrac 16+\tfrac 16 \\= \tfrac 12.}$

• @ Graham Kemp : Are $\mathcal H_1$, $\mathcal H_2$,.. same as $\mathcal D_1$, $\mathcal D_2$,.. ? In you're first derivation, you've used $\mathsf P(\mathcal H) = \tfrac 12$ , is it intentional? – KGhatak Jun 30 '17 at 6:59
• @KGhatak $\mathcal D3$ is, as used in your post, the event of drawing a three. $\mathcal H3$ is the event of obtaining a head and drawing a three. These are coincidentally the same event (since we cannot draw a three without obtaining a head). $~$ $\mathsf P(\mathcal H)=\tfrac 12$ because by virtue of being a fair coin the probability for obtaining a head is that. – Graham Kemp Jul 1 '17 at 11:20