According to the math (A fair coin is tossed until a head comes up for the first time. The probability of this happening on an odd number toss is?), it seems perfectly clear that the probability that the first toss of a coin being heads occurring on an odd flip is $2/3$ and thus on an even flip $1/3$.

However, it seems intuitive that half of the sample set is composed of odd flips and half of the sample set are even flips -- why is it the case that odd flips have a higher probability (2/3)? Is it because the first flip is odd?

In a finite sample space, say $N=10$, there are $5$ even flips ${2, 4, 6, 8, 10}$ and $5$ odd flips ${1, 3, 5, 7, 9}$. In the case where $N=11$, of course, there are more odd flips. Is there a sense in which the infinite case is more like $N=11$ than $N=10$? Is it merely because in the infinite case the odds have gotten "a head start"?

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    $\begingroup$ It is unclear how the finite sample space mentioned in the last paragraph is related to the first heads experiment. Do you have in mind an experiment somehow related to the 'first heads' question where you would get, say, the '$N=10$ sample space' and all flips would be equally likely? $\endgroup$ – Juho Kokkala Oct 1 '15 at 17:40

Up until your last paragraph I'm totally with you. You are totally correct in saying the reason that the probability the first head is odd is $2/3$ is because the first toss is odd. There is a 50% chance that not a single even toss will occur! I think your intuition might be aided by studying another question.

Q: If the first toss is tails, what is the probability that the first heads toss will be an even toss?

A: $2/3$

Intuitively this should make a lot of sense. Since the first one missed, now even has the advantage, we're in the same situation we were in a second ago, but now even even has switched places with odd. How about the math? Let's let $O$ be the event that the first head is odds, $E$ for even, and $T$ be the event that the first toss is tails. What we know is $$P(T) = 1/2, P(O)=2/3, P(E)=1/3$$

We can pretty easily show that $P(O, T) = P(O)-P(T')=1/6$, and that $P(E, T) = P(E) = 1/3$. Dividing both by $P(T)=1/2$ we end up with $$P(E|T) = 2/3, P(O|T)=1/3 .$$

One last bit of intuition:

Suppose we're playing a game where we're taking turns trying to make a certain basketball shot. Whoever makes the shot first wins. If the probability of successfully making the shot is very high, the player who goes first has a huge advantage. The harder the shot is to make the less of an advantage the first player has. The example we just talked about is the example where there is a 50% chance of making the shot.

  • $\begingroup$ I like it. Some of the notation is unfamiliar, such as $P(T')$ as well as P(some event, some other event), but the intuition is clear. $\endgroup$ – compguy24 Oct 1 '15 at 19:34
  • $\begingroup$ @compguy24 The apostrophe means "not". So $P(T') =$ The probability that event $T$ does not happen. The comma in this case can be interpreted as "and". $\endgroup$ – jlimahaverford Oct 1 '15 at 20:22

Look at the outcomes of the repeated trials in pairs (1st,2nd), (3rd,4th), (5th,6th), and so on.

Of the four possible outcomes $(H,H), (H,T), (T,H)$ and $(T,T)$ on the first pair of trials, your desired event (that $H$ occurred on an odd-numbered trial) is settled in three of the four cases, with a Yes answer in 2 cases and a No answer in 1 case. In the fourth case, the determination of whether $H$ occurred on an odd-numbered trial is deferred to the next pair of trials. So, in two cases out of three, $H$ occurred on an odd-numbered trial, and in one case out of three $H$ occurred on an even-numbered trial. If we kicked the can down the road, then we look at trials #3 and #4 and repeat the analysis above, keeping in mind that what is past is prologue and "History is bunk" and that we can forget the past because we are doomed to repeat the whole "toss the coin twice" routine again.

So, even though there are "equal numbers" of Heads and Tails in an infinite sequence of coin tosses, whenever we actually get to determine whether the first occurrence of Heads was on an odd-numbered toss or an even-numbered toss, the only possible (equally likely) outcomes are $(H,H), (H,T), (T,H)$: no more kicking the can down the road; the answer is known when we are done. So looking on this pair of trials on which only three outcomes are possible, Heads first occurs on an odd-numbered toss with probability $\frac 23$. Is it possible that we never get to determine the answer? Well, yes, it is possible in the sense that we do not deny the logical possibility that every toss results in Tails, but we remedy this mental defect ofours by insisting that the probability of $T$ being the outcome forever and ever and a day is $\lim_{n\to\infty} 2^{-n} = 0$. (This is actually an equivalent form of the third axiom of probability: for a telescoping sequence of events, the probability of the limiting event is the limit of the probability.)

Now, if you want to have some more fun, consider that we could exchange Heads and Tails (and $H$ and $T$) everywhere in what is written above and arrive at the result that the probability that the first Tails occurs on an odd-numbered trial is also $\frac 23$. Now, the events $A$ = "First Heads on an odd-numbered trial" and "$B$ = First Tails on an odd-numbered trial" can possibly occur simultaneously, and this compound event $A\cap B$ occurs exactly when the outcomes are \begin{align} &(H,H,T)\\ &(H,H,H,H,T)\\ &(H,H,H,H,H,H,T)\\ & \cdots\\ or~~~~ \\ &(T,T,H)\\ &(T,T,T,T,T,H)\\ &(T,T,T,T,T,T,T,H)\\ & \cdots \end{align} Thus, $P(A\cap B) = 2\left(\frac 18 + \frac{1}{32}+\frac{1}{128}+\cdots\right) = \frac 14\times\frac{1}{1 - \frac 14} = \frac 13$ of occurring. This makes perfect sense: the union $A\cup B$ is the event that either a Head _or a Tail occurs for the first time on an odd-numbered trial, and this has probability $1$. Indeed, since $$P(A\cup B) = P(A) + P(B) - P(A\cap B),$$ we could have deduced that $P(A\cap B) = \frac 13$ from the knowledge that $P(A) = P(B) = \frac 23$, and $P(A\cup B) = 1$ without needing to sum a geometric series.


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