Conditional Probability - Relativity Confusion

Question

This may be too basic and am just missing a big fundamental block in probability. Suppose there are 4 people in a family. Each one gets picked randomly everyday for doing dishes. Let's say one of them is Bill.

Observer 1:

He starts predicting from day one. Let B=1 be Bill getting picked up and 0 otherwise.

Day 1:

$p(B_1=0) = \dfrac{3}{4}$

Day 2 (and day 1):

$p(B_2=0, B_1=0) = \dfrac{3}{4}\dfrac{3}{4}$

Day n (and earlier days):

$p(B_n=0, B_{n-1}=0, \cdots ,B_1=0) = \Big(\dfrac{3}{4}\Big)^n \tag{1}$

Observer 2:

He just pops up on nth day and predicts the event. According to him then,

$p(B_1=0) = \Big(\dfrac{3}{4}\Big) \tag{2}$

where, for observer 2 its the first day, which was nth day for observer 1. The events are random every day. They are neither in any way depend on observer 1 or 2 , or to one another (that is, event happening on day k is independent of day (k-1)).

So both observers are right in their own way. Then, what is the correct probability for that day, for Bill not getting picked up? Or which is better and why?

Answer 1

It seems like the observers are actually answering a slightly different question than what you stated $B$ to be. For any given day, the probability that Bill isn't picked is always $\frac{3}{4}$. Observations of previous days do not matter.

It seems like the observers are answering this question: "Wow, Bill hasn't had to do the dishes at all! What is the probability of that happening?"

It makes perfect sense that the observers conclude different things. This is because they have different knowledge about what has happened. Let me try to make an analogy.

Let's say there is a geyser that very rarely erupts. On day, I see it erupt - a rare event. A minute later, my friend joins me and sees it erupt again - back to back eruptions are even more rare. My friend says "wow, that was a rare event we witnessed" and I say, "actually, it's rarer than you know!". There is difference because what we are referring to by "this event" are different. My friend is referring only to the eruption he saw, but I am referring to both eruptions.

In your example, Observer 1 has witness more things happening, so he concludes that the situation is more rare than Observer 2. To avoided this error, be careful about what exactly $B$ refers to. Try defining $B_i$ to be the probability that Bill is picked on day $i$ (and day $i$ only).

Answer 2

From what you wrote, each observer is stating the probability of the event he/she observed. And they're both correct, if we assume the random choice is uniformly and independently distributed.

-Observer 1 saw Bill not getting picked $n$ time in a row. -Observer 2 saw Bill not getting picked only once.

Conditional probability does not really come into play until now. However, there are two curious question that could be asked in such scenario:

1: If observer 1 tells observer 2 that he saw Bill not getting picked $n$ times in a row. And observer 2 sees Bill not getting picked up again, does he claim a different probability for the event he witnessed?

The answer is no because of the (assumed) independence of events. Which causes the conditional probability (that observer 2 states after being informed by observer 1) to be the same as the unconditional probability (when no prior information is given to observer 2). In equations:

Observer 1 claims: $$ P(B_{n}=0 \wedge B_{n-1}=0 \wedge \cdots \wedge B_1=0) = \prod_{k=1}^{n}P(B_k=0)= (3/4)^n $$

Observer 2 claims: $$ P(B_{n+1}=0|B_{k<n}=0)=P(B_{n+1}=0) = 3/4 $$

2: This one is much trickier. Assume the random choice was made by a 4-sided dice, which is not necessarily fair. Observer 1 estimated the probability of the possibly unfair dice picking Bill. Observer 2 used this estimated probability to claim the probability of the event he witnesses (Bill getting picked again), what is his claim?. In this case, since all observations yield Bill not getting picked, observer 2 would claim the probability was 1, not due to the conditioning of the probability, but rather because ditching the fairness assumption leads to a different model of the choice process.

Answer 3

You can add a bit more detail in the notation to help you understand. For example Instead of using B = 1, let's say $B_n = 1$ means on the nth day, Bill got picked.

Therefore from Observer 1 on day 1

$P(B_1 = 1) = \frac{1}{4}$

on day 2, the probability of bill got picked Both on day 1 and day 2 is

$P(B_2 = 1 ,B_1=1) =P(B_1=1)P(B_2=1) = \frac{1}{4}^2$ by your independence and identical distribution assumption

this result will be consistent across all the observers that observe this event in the 2nd day.

Because whether $B_n = 1$ does not depends on the observer, the above conclusion will not depend on observer either.