Independent events or conditional probability?

Question

In my daily morning walk there is a 20% chance I drop my pouch. Every day I follow the exact same straight path, to the district square and then back. Assuming that my walk is on a straight segment from A to B and then back from B to A, what is the probability I dropped it from A to B?

I know this seems to be a very easy problem, but I am a little confused: Initially, there is equal probability for me to drop my pouch in any of the two directions. So for AB, this probability is 50% of 20%?

But for the second part of my walk back home, the probability is conditional. If I have already dropped it in AB, the probability is zero. If I haven’t, the probability is 100% of the total 20%, that is, 20%.

Any clue?

Edit : There are two answer which have been updated, but they sort of conflict each other. Any help clearing it out would be greatly appreciated.

Answer 1

Assuming there's nothing special about going one way or the other, then you can say that the probability you drop your wallet going from A to B is $p$, and the probability that you drop it going from B to A, conditional on not dropping it from B to A, is also $p$.

Then the probability that you drop the wallet going from B to A is given by $P(BA) = P(BA \land \lnot AB) = P(BA | \lnot AB)P(\lnot AB) = p(1-p)$, i.e. it's the product of the probabilities of (a) not dropping the wallet going from A to B, and (b) dropping the wallet going from B to A, having not dropped it going from A to B.

So the overall probability you drop your wallet somewhere on the walk is the sum of the two events - $P(AB \lor BA) = P(AB) + P(BA \land \lnot AB) = p + p(1-p) = p(2-p)$.

Then since we know this overall probability, we can say $p(2-p) = 0.2$ and solve for $p$, getting $p = 1 - \frac{2\sqrt{5}}{5} \approx 0.10557$.

Answer 2

Let $E_{1}$ be the event of non-dropping on segment $AB$ and let $E_{2}$ be the event of non-dropping on segment $BA$.

Then: $$P\left(E_{2}\mid E_{1}\right)=P\left(E_{1}\right)\tag1$$(any objections against $(1)$? Please let me know)

or equivalently: $$P\left(E_{1}\cap E_{2}\right)=P\left(E_{1}\right)^{2}$$

This with: $$P\left(E_{1}\cap E_{2}\right)=1-0.2=0.8$$ so that: $$P\left(E_{1}^{\complement}\right)=1-P\left(E_{1}\right)=1-\sqrt{0.8}\approx0.105573$$

Not an essentially different answer but a bit more concise. It all rests on the concept of a distribution that has no memory.

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Edit (inspired by the answer of @user2661923)

Let us look at the following experiment in order to compare.

There are $10$ marbles in a bag and exactly one of them is red. We pick $2$ marbles one by one without replacement (which is essential here). Analogously let $E_1$ be the event that the first picked marble is not red and let $E_2$ be the event that the second picked marble is not red. Then easily we find that:$$P(E_1)=P(E_2)=0.9\text{ and }P(E_1^{\complement}\cup E_1^{\complement})=P(E_1^{\complement})+P(E_2^{\complement})=0.2$$ Note that in this situation:$$P(E_2\mid E_1)=\frac89\neq0.1=P(E_1)\tag2$$ This because under condition $E_1$ the second pick is from a bag containing $9$ (not $10$) marbles of which exactly one is red. In short: "things have changed". This however is definitely not the case for the walk on segment $BA$ under the condition that there was no dropping during the walk on segment $AB$. Under that condition things are exactly the same as they were by the start of the walk on segment $AB$. That is an essential difference.

I am convinced that in the situation sketched in your question statement $(1)$ is correct and statement $(2)$ is incorrect.

Answer 3

To be added to other already good answers.

It seems that your model corresponds to a failure model with constante rate - that is, the probability of "fail" (dropping your pouch) in the next (small) walking step, given that you have not yet failed, is a constant, only depending on (proportional to) the length (in time or space) of the step. Let's use space walked, so that the walk goes from $x=0$ to $x=2$, with $x=1$ being the point of return.

Then, the conditional probability of fail in the interval $[x, x+\delta x)$ is $\lambda\, \delta x$ , for small $\delta x$. In the limit this tends to a exponential distribution.

Then, the probability of fail over some interval $L$ is $P(L)=1- \exp(-\lambda L)$

We know that $P(2)=0.2$, then $\lambda = -\log(0.8)/2 = 0.11157$

Hence, the probability of failing over the first half of the trajectory is $P(1)=1-\exp(-\lambda)= 0.1055729 \approx 10.56\%$

Also, without solving for $\lambda$:

$$P(1)= 1- \exp(-\lambda) = 1 - \sqrt{\exp(- 2 \lambda )}=1-\sqrt{1+P(2)}$$

which gives the same result and agrees with @ConMan.

Notice that this $P(1)$ is the probability of dropping the pouch in the first part ($A\to B$), and also the probability of dropping it in the second part ($B \to A$) given that it was not dropped before.

The relationship above shows that $P(1) \ne \frac{1}{2} P(2)$ (but this is asympotically right if the probability is small).

Answer 4

+1 : also - nicely questioned.

Depends on what simplifying assumptions that you make.

As you have surmised, the following events are mutually exclusive:

• dropping the pouch when going from A to B
• dropping the pouch when going from B to A

Therefore, the two events can not be regarded as independent.

Further, absent any other information, it is reasonable to assume that there is a $10\%$ chance of dropping the pouch when going from A to B and a $10% chance of dropping the pouch when going from B to A.

What is confusing you is that the events are not being trialed simultaneously. To clarify the situation, suppose that you throw a single 6-sided fair die, and you are focusing on the following mutually exclusive events:

• the die comes up a 1 : probability $= (1/6)$.
• the die comes up a 2 : probability $= (1/6)$.
• the die comes up some other number : probability $= (2/3)$

The 3 events above are mutually exclusive, and not independent events. Never the less, the $(1/6), (1/6), (2/3)$ probabilities stand.

With respect to dropping the pouch, the analogy is apt:

• the event of dropping the pouch when going from A to B : probability $= 0.1$.
• the event of dropping the pouch when going from B to A : probability $= 0.1$.
• the event of not dropping the pouch : probability $= 0.8$.

Answer 5

I support @ConMan’s answer. Simply put, assume probability of dropping pouch is $p$ and hence not dropping it is $q = (1-p) $. So, total probability of losing it is $$ P(E) = p*1 + (1-p)*p = 0.2 $$

On solving the quadratic, $ p = 1 - 2/ \sqrt 5 = 0.1056 (approx) $. Also, the probability of losing the pouch is $higher$ in the first half AB, as $p > p(1-p) $, since $ 0<p<1 $ and this is actually intuitive.

This is because when you drop it in the first half, the second path outcome is irrelevant, so it’s basically that your do not drop your pouch with probability of 1.

On the other hand to drop it in the second half, you should not drop it in the first half, hence a condition comes into play and reduces your probability.

( From $p$ to $p(1-p)$ ). So answer = $0.1056$.
Hope it helps.

Answer 6

To add to the conflicting answers: The question mentions that OP perform their "daily morning walk" "every day". Assuming that once the pouch is dropped it is not recovered for subsequent walks (the same assumption that OP used in their question).

From other answers, given that OP starts a daily walk with the pouch,

• they might drop the pouch in one round trip with probability $20\%$;
• they might drop the pouch in one trip from A to B with probability $p \approx 10.557\%$;
• they might drop the pouch in one trip from B to A with probability $1-p \approx 9.443\%$;

but these probabilities are only correct for the first day, and do not consider the "daily" aspect.

The actual "probability I dropped it from A to B" is

$$p + (1-20\%) p + (1-20\%)^2 p + \cdots = \frac{p}{20\%} \approx 52.786\%$$

The corresponding probability that OP dropped it from B to A is

$$(1-p) + (1-20\%) (1-p) + (1-20\%)^2 (1-p) + \cdots = \frac{1-p}{20\%} \approx 47.214\%$$

Answer 7

To add to the conflict answers: Why assume that OP does not pick up a dropped pouch? After all, if OP does not pick up the pouch, how would they repeat the same morning walk "every day"?

What if OP knows how to pick up the pouch immediately when dropped, and might drop the pouch in each part of the walk independently with probability $p$?

$$\begin{align*} 1-(1-p)^2 &= 20\%\\ p&= 1-\frac{2}{\sqrt5} \approx 10.557\% \end{align*}$$

It's the same answer that OP might drop the pouch from A to B with probability $10.557\%$.

What's different here is that OP also might drop the pouch during the return part with the same probability $10.557\%$.

Answer 8

All of these comments that mention "intuitively" seem to be from a non-intuitive (calculating) perspective. The intuitive view for many people, IMO, is that the risk is the same on both parts of the journey, and therefore the probability is divided equally. Granted, naive intuition is often wrong, e.g. Monty Hall responses. But, because we have a mathematical tool, it doesn't mean we should always use it. The issue here is that "one event is dependent", so it seems that we MUST use the conditional probability formula from Bayes. The problem with that is: it ends up distorting the reality that every step, in either direction, is memoryless and contains the same amount of risk (of dropping the pouch). I don't believe those who are saying the probability is higher on the first part of the journey. If you follow that logic, every single step becomes less risky (with no reason to think that). The "independent event" is just part of a continuous, uniform probability distribution, and exactly half of that.