A coin is thrown until it comes up tails. What is the probability of making at most $3$ throws?

Question

A coin is thrown until it comes up tails. What is the probability of making at most $3$ throws?

The answer is :$\frac{7}{8}$.

I don't understand why. What are the possible cases and what are the favorable ones?

Let 1=Tail 0=Head then an event is for example: 001 or 00001 or 01 or 1 but this events are not equiprobable we have more chances to appear 1 than to appear 00000000000000000001.

In these conditions we can say :$\color{blue}1)$the probability to appear 1 is 1/2 (because we 0 and 1 possible cases) $\color{blue}2)$the probability to appear 01 is 1/4 (because we have 4 possible cases 11, 00, 10, 01) and $\color{blue}3)$ the probability to appear 001 is 1/8 (from the same reason)

Then the answer is 1/2+1/4+1/8=7/8. what is unclear for me is why actually in $\color{blue}1),\color{blue}2),\color{blue}3)$ we have prob. 1/2, 1/4, resp. 1/8 I think the explanation I gave in brackets is not rigorous.

$\color{red}{------------------------------}$

I rephrased the question.

Let $\mathcal{E}$ be an experiment. Let $Ω$ be the sample space on $\mathcal E$, and let $Σ$ be the event space of $\mathcal E$.

We know that: $\Omega=\{\underbrace1_{\omega_1},\underbrace {01}_{\omega_2},\underbrace{001}_{\omega_3},...\}$

Probability: $P:Σ=\mathscr{P}(\Omega)\rightarrow[0,\infty)$ respect Kolmogorov Axioms. Now how to prove $P(\omega_1)=1/2,P(\omega_2)=1/4,P(\omega_3)=1/8$ using just axioms.

Answer 1

$\def\Om{\Omega}$ $\def\om{\omega}$ $\def\si{\sigma}$ $\def\cF{\mathcal{F}}$ $\def\bN{\mathbb{N}}$

The axioms of probability are analogous to the rules of inference in formal logic. The rules of inference cannot prove anything by themselves. They must be applied to a set of premises. Without premises, there can be no proof. Similarly, the axioms of probability must be applied to premises, or assumptions.

The assumptions in your probability model are that

1. The coin flips are fair.
2. The coin flips are independent.

Unfortunately, these assumptions cannot be formulated in the probability space you wrote down in your reformulation of the question. For example, in your space, there is no event that corresponds to "The second coin flip is tails." (You have an event corresponding to "The second coin flip is tails and the first is heads.", but not "The second coin flip is tails and the first is tails." Without the latter event, there is no way to construct the event which denotes "The second coin flip is tails.")

To properly formulate the assumptions of this model, we need a more robust probability space, such as $$ \Om = \{\om = (\om_1,\om_2,\ldots):\om_j\in\{0,1\}\}, $$ with $\cF$ being the smallest $\si$-algebra containing sets of the form $$ A_{a_1,\ldots,a_n} = \{(a_1, \ldots, a_n, \om_{n+1}, \om_{n+2}, \ldots): \om_j \in \{0,1\} \text{ for $j>n$}\}, $$ where $(a_1,\ldots,a_n)\in\{0,1\}^n$. With the measurable space $(\Om,\cF)$ in place, we next need to identify the events we will need. We need events related to our assumptions as well as the conclusions we are aiming for. Informally, we need

• $T_j = \text{"The $j$-th coin flip is tails."}$
• $F_j = \text{"The first tail occurs on the $j$-th flip."}$

Formally, these translate to

• $T_j = \{\om: \om_j = 1\}$
• $F_j = T_j \cap \bigg(\bigcap_{k=1}^{j-1} T_j^c\bigg)$

The assumptions can now be formulated mathematically as follows:

1. $P(T_j) = 1/2$ for all $j$.
2. The events $\{T_j\}_{j=1}^\infty$ are independent, which means that for all finite sets $J\subset\bN$ and all functions $a:J\to\{0,1\}$, we have $$ P\bigg(\bigcap_{j\in J} T_j^{a(j)}\bigg) = \prod_{j\in J} P(T_j^{a(j)}), $$ where $T_j^0=T_j$ and $T_j^1=T_j^c$.

Only now can we begin the process of applying the axioms of probability to these assumptions in order to derive $P(F_j) = 2^{-j}$.

The first thing we need to do is prove that there exists a probability measure $P$ on $(\Om,\cF)$ that satisfies our assumptions. This can be done using the Kolmogorov extension theorem. Once we know that such a $P$ exists, the proof is very easy and follows immediately from our assumptions and the fact that $P(A^c)=1-P(A)$ (which is easily derived from the axioms of probability): \begin{align} P(F_j) &= P\bigg(T_j \cap \bigg(\bigcap_{k=1}^{j-1} T_j^c\bigg)\bigg)\\ &= P(T_1^c \cap \cdots \cap T_{j-1}^c \cap T_j)\\ &= P(T_1^c)\cdots P(T_{j-1}^c)P(T_j)\\ &= \left({\frac12}\right)\cdots\left({\frac12}\right)\quad\text{($j$ factors)}\\ &= \left({\frac12}\right)^j = 2^{-j}. \end{align}

EDIT:

I do not know what the level of your probability knowledge is. Based on the original formulation of your question, I would have guessed that your experience is at the level of a basic introductory course in non-rigorous probability, like the kind we teach to undergraduates here in the US. But your reformulation of the question mentioned Kolmogorov axioms and asked for proof. This is why I posted this answer.

It may be surprising that such a seemingly simple situation (a sequence of coin flips) requires such elaborate mathematics, involving concepts usually taught only in math graduate school. But I assure you, this is what is required, almost from the very beginning, if you want to study the rigorous proofs of probability theory.

I am sorry if this post is too complicated to be helpful, but perhaps someday you will want to return to it.

Answer 2

There are three possibilities: T, HT, HHT with probabilities $1/2$, $1/2\cdot1/2$, and $1/2\cdot1/2\cdot1/2$, resp. They sum up to $7/8$.

Answer 3

You will need more than three throws only when head comes out thrice.

Hence

$$p=1-\frac1{2^3}.$$

More generally,

$$p=1-\frac1{2^n}$$ for the probability of making at most $n$ throws.

Answer 4

Let $T$ be the event that a tail is obtained within 3 flips. We want to find $pr(T)$. It is a bit difficult to compute this because we have to consider obtaining tails within 1 flip, 2 flips, AND 3 flips, to which there is overlap. It is a bit easier to compute $pr(T')$ or the probability $T$ does not happen.

For event $T'$ to occur, we would need to not obtain tails within 3 flips, i.e. get heads three times in a row. Since the probability of throwing heads is $\frac{1}{2}$, getting three heads in a row is $\frac{1}{2^3}=\frac{1}{8}$. Hence, $pr(T')=\frac{1}{8}$. Since any given event either happens or doesn't, we have in general for event $A$, $$pr(A')=1-pr(A).$$ Therefore, $$\frac{1}{8}=pr(T')=1-pr(T)\implies pr(T)=1-\frac{1}{8}=\frac{7}{8}$$.

Answer 5

A really simple way to see this is just to list out all the possible outcomes of flipping a coin three times. There are only $8$ of them: $$ HHH, HHT, HTH, HTT, THH, THT, TTH, TTT$$

Because the coin flips are independent of each other, each of these combinations are equally likely. So the probability of each is $\frac 18$. As you can see, $7$ of them include at least one tail, so the probability of getting a tails in three coin tosses is $\frac 78$.

Answer 6

the answer is 7/8 because you are asking the probability that it will take at most three throws before a coin turns up tails so for this the probability it will take one throw is (1/2)^1 probability it will take two throws is (1/2)^2 and so on. if you solve this infinite series you get 1. next you find the sum of the first three numbers which is 1/2+1/4+1/8=7/8, and then you subtract this number from the sum of the entire series which is 1 so you have 1-7/8=1/8 and then you take 1 and do 1-1/8=7/8