# What is the probability that the first head will appear on the even numbered tosses

## Question

$\text{Consider a coin with probability R to be heads. What is the probability}$ $\text{that the first head will appear on the even numbered tosses?}$

## My Approach

let the required Probability$=P$.

Hence we can write our eauation as-:

$$P=(1-R) \times R+(1-R) \times (1-R) \times (1-R) \times P$$

$$P(1-(1-R) \times (1-R) \times (1-R))=(1-R) \times R$$

$$P=\frac{(1-R) \times R }{(1-(1-R) \times (1-R) \times (1-R)}$$

Am i correct?

$$P=\frac{(1 - R)}{(2 - R)}$$

Alternative approach - to see the first head on an even numbered toss we need to have:

1. Toss 1 is a tail (probability 1-R)

2. Renumbering Toss 2 as Toss 1, Toss 3 as Toss 2 etc. we are now want to see first head on an odd numbered toss

So

P(first head on even toss) = P(first head on odd toss) x (1-R)

but if we denote P(first head on even toss) by p then P(first head on odd toss) = 1-p, so

$p = (1-p)(1-R)$

$\Rightarrow p + p(1-R) = 1-R$

$\Rightarrow p = \frac{1-R}{2-R}$

Sanity checks: (i) p=0 when R=1 (ii) p approaches 1/2 as R approaches 0.

• I'm having trouble seeing intuitively why p should approach 1/2 as R approaches 0. – Chris Nov 17 '17 at 17:30
• @Chris - because the advantage that odds have is that an odd comes first. Otherwise it is equally likely to occur as an even or an ood toss. But the more flips that occur before a head occurs, the less impact the odds' advantage gives. The smaller $R$, the greater the expected number of flips needed to get a head will be. – Paul Sinclair Nov 17 '17 at 18:36
• Ah, yes. That makes sense. I had a vague feeling it was something like that but I couldn't quite think why. Thanks for the explanation. – Chris Nov 17 '17 at 22:45

The probability that the first head will appear on the second toss is $(1 - R)R$.

The probability that the first head will appear on the fourth toss is $(1 - R)^3R$.

The probability that the first head will appear on the sixth toss is $(1 - R)^5R$.

In general, the probability that the first head will appear on the $2k$th toss is $(1 - R)^{2k - 1}R$.

Hence, the desired probability is $$p = \sum_{k = 1}^{\infty} (1 - R)^{2k - 1}R = (1 - R)R\sum_{k = 1}^{\infty} (1 - R)^{2k}$$ which is a geometric series with common ratio $(1 - R)^2$. If $R < 1$, we obtain \begin{align*} p & = (1 - R)R \cdot \frac{1}{1 - (1 - R)^2}\\ & = \frac{(1 - R)R}{1 - (1 - 2R + R^2)}\\ & = \frac{(1 - R)R}{2R - R^2}\\ & = \frac{(1 - R)R}{R(2 - R)}\\ & = \frac{1 - R}{2 - R} && \text{provided that $R \neq 0$} \end{align*} If $R = 1$, then heads will be obtained on the first toss, so $p = 0$. If $R = 0$, then heads will never be obtained, so again $p = 0$.

Your equation for $P$ isn't quite right. Let's write $T_k$ for the outcome of the $k$-th toss: $T_k \in \{H,T\}$, and $P(T_1 = H) = R$ and $P(T_1 = T) = 1-R$. We then need to carefully split cases: $$P(\text{first heads on even}) = P(\text{first heads on even} \mid T_1 = H)P(T_1 = H) \\ + P(\text{first heads on even} \mid T_1 = T, T_2 = T)P(T_1 = T, T_2 = T) \\ + P(\text{first heads on even} \mid T_1 = T, T_2 = H)P(T_1 = T, T_2 = H) \\ = P(\text{first heads on even} \mid T_1 = T, T_2 = T) \cdot (1-R)^2 \\ + P(\text{first heads on even} \mid T_1 = T, T_2 = H) \cdot R(1-R).$$ Now observe that $$P(\text{first heads on even} \mid T_1 = T, T_2 = T) = P(\text{first heads on even})$$ (by the Markov property, if you like), and $$P(\text{first heads on even} \mid T_1 = T, T_2 = H) = 1$$ since this is a 'success'. Hence, writing $p = P(\text{first heads on even})$, we have $$p = p(1-R)^2 + R(1-R).$$

This is the same as you had except that you had a factor $(1-R)^3$. I've put in all the details so you can see exactly why it's a power of 2, not a power of 3.

Intuitively, I think you've got the idea, except that you were thinking that if we don't get $TH$, then we need to go $TTT$ to not fail -- which is correct -- but this first $T$ it taken into account with the probability $p$, so we could write down straight away $$p = R(1-R) + p(1-R)^2;$$ the danger of this, as you have found out, however, is that if there's a slight mistake, then it's not at all clear where it came from.

Hopefully this helps! :)