Ratio of infinite heads and one tail

Question

Lots of people throw coin until get a tail. It means each person should immediately stop, when get exactly one tail. The odds of someone having a head or a tail on any given throw is equal.

Each person will have a sequence of zero or more heads followed by a single tail.

That is, if "H" indicates a head and "T" indicates a tail, the sequence of results will look like one of:

• T
• HT
• HHT
• HHHT
• HHHHT
• And so on

Question

What will the ratio of heads and tails be on very large N?

My thought: we could say that sequences have the following probabilities:

P(T)     = 1/2;  Tails_Count = 1
P(HT)    = 1/4;  Tails_Count = 2
P(HHT)   = 1/8;  Tails_Count = 3
P(HHHT)  = 1/16; Tails_Count = 4
P(HHHHT) = 1/32; Tails_Count = 5
And so on

We know that each person will have exactly 1 tail. I'm confused about how to calculate heads average count.

If only two person throw coins, than one person can get sequence like HT and the second person can get sequence like HHHHHHHHHT.

It means that we have 10H and 2T.

On the other hand probability of getting HHHHHHHHHHT is $${1 \ \over 2^{10}}$$ which is almost impossible.

Staring this momment I have no idea how to caculate ratio of heads and tails.

Answer 1

Any play of the game produces one $T$, so the ratio of the number of $H$ to the number of $T$ in that play is just the number $n$ of heads.

Extrapolating from your table, that expected value of the number $n$ of heads is: $$\Bbb E[n] = P(n = 0) \cdot 0 + P(n = 1) \cdot 1 + P(n = 2) \cdot 2 + \cdots = \sum_{i = 0}^{\infty} P(n = i) \cdot i = \sum_{i = 0}^{\infty} \frac{i}{2^{i + 1}}.$$

There are many methods for evaluating that infinite sum.

Answer 2

Let $\mu$ denote the expected number of heads.

Then $$\mu=0.5\times0+0.5(1+\mu)=0.5+0.5\mu$$so that: $$\mu=1$$

This on base of:$$\mathbb EH=P(T_1)\times\mathbb E(H\mid T_1)+P(H_1)\times\mathbb E(H\mid H_1)$$where $T_1$ denotes the event that the first toss is a tail, $H_1$ denotes the event that the first toss is a head, and $H$ denotes the number of heads. Essential is that here: $$\mathbb E(H\mid T_1)=0\text{ and }\mathbb E(H\mid H_1)=1+\mathbb EH$$

So the ratio of heads and tails by large $N$ will be $1:1$.