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?

Observer $1$ needs to review basic probability! The probability that Bill is picked on day $2$ is not $(3/4)(3/4)$. That would be the probability of Bill being picked on both days 1 and 2. — Mike Earnest, Aug 20 '18 at 18:53
Consider a point in time when Observer 1 has already been watching for $n-1$ days, and has seen someone else (not Bill) be picked on day $n - 1,$ and it has not yet been decided who will do the dishes on day $n$; suppose at that instant in time I ask Observer 1 to estimate the probability that Bill will be picked on day $n.$ What answer will Observer 1 give me? — David K, Aug 20 '18 at 22:17

score 2 · Accepted Answer · answered Aug 20 '18 at 19:11

2

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).

answered Aug 20 '18 at 19:11

rikhavshah

1,355

1

I have updated the notations. Can you please check and update now? yes both O1 and O2 are different, but in a sense, they both give a prediction what is probability of bill not getting picked that day? O1 fees less probability, and O2 feels higher with no prior info, so which one is more useful? For eg, if this is stock price, which one would you take for better prediction of stock that day? If so why? For eg, if O1, wouldnt it be a gambler's fallacy? – Parthiban Rajendran Aug 20 '18 at 19:48
Perfect! Your revisions look good. Now, on day $n$, the probability that O1 wants is not $P(B_1,B_2,\cdots, B_n)$. The reason for this, is that O1 already knows the outcome of $B_1,\cdots,B_{n-1}$! What O1 really wants is $P(B_1,B_2,\cdots, B_n ,|, B_1=1,\cdots,B_{n-1}=1)$. By Bayes rule, this is simply $P(B_n=1)=3/4$. – rikhavshah Aug 21 '18 at 00:11
To further clarify, it would be Gambler's fallacy: specifically, the fallacy is the belief that observing past events in any way effects the outcome of a current independent event. If you assume that the stock market randomly increases or decreases price every day, independently of previous days and external forces, then observations about days in the past tell you nothing about the price today. – rikhavshah Aug 21 '18 at 00:15
1

so, 1) you can never apply past joint probability when about to predict current event (assuming independence)? 2) when would past joint probability would be valid w.r.t time frames? only when they are dependent? – Parthiban Rajendran Aug 21 '18 at 06:37
I took the example from khan, here, where hypothesis is introduced. There Bill does not gets picked in a row, so khan calculates the probability what it would be and applies joint probability. According to us, then it is wrong?. At any day, his chance of not getting picked up is 3/4?! – Parthiban Rajendran Aug 21 '18 at 06:47
Correct; 2) Also correct. As for the Khan academy video, they are computing the probability of Bill getting or not getting picked for some days in a row without observing any of the days. Once you observe the first day, the probability that Bill won't get picked for the first three days changes. Khan is answering the question "What is the probability that this thing will happen on the first three days?" which is different from "What is the probability that this thing will happen on the first three days given that I saw what happened on the first two days already?"

rikhavshah

Aug 22 '18 at 17:17

thank you. Can you also please answer my other question here – Parthiban Rajendran Aug 22 '18 at 18:35

1

@PaariVendhan Why have you not given the check mark to this answer? It makes it look like you are still waiting for this question to be answered for you. – David K Sep 01 '18 at 17:37

sorry about that, missed it amid multiple questions. – Parthiban Rajendran Sep 03 '18 at 07:02

Mefitico · Answer 2 · 2018-08-20T19:40:56.227

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.

thank you very much, I quickly updated question so many careless mistakes in my simple question. — Parthiban Rajendran, Aug 20 '18 at 19:34
But in question 1, for observer 1, wouldn't it be $p(B_{n+1}=0)p(B_n=0)\cdots p(B_1=0)?$ — Parthiban Rajendran, Aug 20 '18 at 19:38
There was a small confusion in my question. I actually wanted to mean "predict", got the word wrongly as "observes". So on $nth$ day, both observers arrive to predict the probability that Bill would not get picked that day. If they differ, what and whose answer could be more useful? You mean, both would arrive at $frac{3}{4}$ even if one has prior info and other does not by your case 1? — Parthiban Rajendran, Aug 20 '18 at 19:41
I think I see the point in your last comment. See the update to my answer using $\wedge$ as the "and" operation. — Mefitico, Aug 20 '18 at 19:41
can you please also check and answer my other question here if possible? — Parthiban Rajendran, Aug 21 '18 at 17:18

Allen Kuo · Answer 3 · 2018-08-20T19:18:14.000

0

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.

edited Aug 20 '18 at 19:18

answered Aug 20 '18 at 19:07

Allen Kuo

268

great, will update the notations accordingly. quickly updated this question, so many careless mistakes. apologies to all. – Parthiban Rajendran Aug 20 '18 at 19:17
I understand the difference between observer 1, talking about $p(B_2=1,B_1=1)$, and observer 2 talking only about, say $p(B_2=1)$, and in both cases $p(B_2=1)$ is same. But if I have to come up with best probability that B is not picked up today, which one should I pick? – Parthiban Rajendran Aug 20 '18 at 19:31