Different probabilities depending on meaning?

[disregardthat]

If you have already observed those four sixes in a row, and are starting calculations from that point, you treat $P(C) = 1$

Why would you use the same event name? C is an event defined by me which does not assume any prior events, hence your new event could more properly be described as C|C, that is, "C given C", who's probability obviously is 1. Naming this C only adds to the confusion.

In the second case $A$ can be interpreted as "What is the probability that, continuing on, we finish the next roll and have five sixes in a row?" The only way that can happen is for the next roll to be a six, since we've seen four. In this sense, all calculation refers to one roll only, but we still obtain $1/6$ as the answer; in this case, however, $P(A) = 1/6 = P(A \mid C)$ because the setting of the problem is different from our first case.

Again, define your own events if you want to give them "different meaning", and don't misuse mine. A is a single unambiguously defined event which does not assume any prior events, the event you are talking about could more properly be described as A|C. If you want to change the setting where we want five 6es when we already have four, one will only generate confusion by calling the event of getting the last 6 A.

 

[SW VandeCarr]

Why would you use the same event name? C is an event defined by me which does not assume any prior events, hence your new event could more properly be described as C|C, that is, "C given C", who's probability obviously is 1. Naming this C only adds to the confusion.
Again, define your own events if you want to give them "different meaning", and don't misuse mine. A is a single unambiguously defined event which does not assume any prior events, the event you are talking about could more properly be described as A|C. If you want to change the setting where we want five 6es when we already have four, one will only generate confusion by calling the event of getting the last 6 A.

Please reread statdad's last post. He gave two cases. P(A|C)=1/6 and P(A|C)=P(A)=1/6. We can't use your definition of P(A) in the conditional probability because P(A) is independent. It's always 1/6. I asked you to go back and read the wiki article on conditional probability. It clearly states that for independent probabilities such as P(A): P(A|C)=P(A). Your definitions only apply to two strings of length four and length five before any rolls occur in each case.

 

[disregardthat]

Please reread statdad's last post. He gave two cases. P(A|C)=1/6 and P(A|C)=P(A)=1/6. We can't use your definition of P(A) in the conditional probability because P(A) is independent. It's always 1/6. I asked you to go back and read the wiki article on conditional probability. It clearly states that for independent probabilities such as P(A): P(A|C)=P(A).

No, please reread my definitions of the events. You are apparently too worked up by your presupposition that A and C are independent events, clearly they are not, which leads you to false conclusions about this trivial problem. It is as simple as this, and can be summed by this: rolling five 6s in five rolls has a probability of 1/6^5 (P(A)). Given that you have rolled four 6s(C), the probability becomes 1/6(P(A|C)). Period.

 

[SW VandeCarr]

No, please reread my definitions of the events. You are apparently too worked up by your presupposition that A and C are independent events, clearly they are not, which leads you to false conclusions about this trivial problem. It is as simple as this, and can be summed by this: rolling five 6s in five rolls has a probability of 1/6^5 (P(A)). Given that you have rolled four 6s(C), the probability becomes 1/6(P(A|C)). Period.

Yes, given that you rolled four sixes, the probability of P(A|C)=1/6 conditioned on the given result that you have rolled four 6's in a row. As a given, P(C) now has a probability of 1 because it has already occurred. You don't assign probabilities (other than 1) to events that have already occurred and the result is known. You only assign probabilities to outcomes that are not known. You can say that the probability of 10 consecutive heads in 1/1024; but once such an event has happened, the probability of a string of 11 heads is now 1/2. But the probability of any single toss being a head is 1/2. That has nothing to do with any particular set of coin tosses. That is, the probability of a head with any single toss of a fair coin is an independent probability.

 

[disregardthat]

As a given, P(C) now has a probability of 1 because it has already occurred. You don't assign probabilities (other than 1) to events that have already occurred and the result is known. You only assign probabilities to outcomes that are not known.

"C, when C is given", or "C given C" reads C|C - not C - as C is defined in the prior sample space. It's problematic not to differentiate notation between events when you are changing your sample space. It has certainly caused the confusion in this thread where no confusion ought occur.

 

[SW VandeCarr]

"C, when C is given", or "C given C" reads C|C - not C - as C is defined in the prior sample space. It's problematic not to differentiate notation between events when you are changing your sample space. It has certainly caused the confusion in this thread where no confusion ought occur.

It's problematic because you are using simple probabilities but calling them conditional probabilities. The proper sample space is the one I'm using for the conditional probabilities of contingent independent events.

 

[Aleksey]

"The fifth roll of the die is not independent" - exactly. It depends of my opinion regarding randomness."

Not an opinion - you would need to put forth a description of how one result could possibly be influenced by previous rolls.

It is simple P(AB) != P(A)P(B).

 

[schip666!]

There is a 1/7776 probability of rolling ANY specific sequence in 6 tries. 6-6's are one of the 7776 possibilities, it just seems special to some species of animals with a propensity for classifying things.

 

[statdad]

"It is simple P(AB) != P(A)P(B)."

You have simply given a statement of what it means for events to be dependent - you have neither given your explanation for how you see this happening nor how this might apply to the issue under discussion.

You say in #33 "Right: the fifth roll of the die is not independent".

Give an example of two events, $A, B$, where $A$ refers to some specific result of the first four rolls and $B$ refers to a specific result of the fifth roll, but where

$$P(A \cap B) \ne P(A)\,P(B)$$

 

[chiro]

Correct me if I wrong, but wouldn't true randomness imply that any conditional property of an event given any history of event (ie P(A | X) where X is any composition of events that occur prior to A) is essentially the same for all X?

Basically it seems that if X does affect the outcome of A, then it seems to imply that there is some order to how A occurs.

If the experiment is done a significant number of times and all conditional probabilities show the above, then through law of large numbers, the real sample probabilities will reflect the population probabilities with greater and greater accuracy.

If conditional probabilities did not end up being the same, it would indicate that there is more of a pattern than one that is truly random.

 

[Aleksey]

"It is simple P(AB) != P(A)P(B)."

You have simply given a statement of what it means for events to be dependent - you have neither given your explanation for how you see this happening nor how this might apply to the issue under discussion.

You say in #33 "Right: the fifth roll of the die is not independent".

Give an example of two events, $A, B$, where $A$ refers to some specific result of the first four rolls and $B$ refers to a specific result of the fifth roll, but where

$$P(A \cap B) \ne P(A)\,P(B)$$

For example, changing the center of gravity of the cube during testing.

 

[statdad]

"For example, changing the center of gravity of the cube during testing."

Nope. Changing a characteristic of the die would result in a new experiment: you would not be justified doing probability calculations from the "old" die to predict results from the "new" one.

 

[Aleksey]

"For example, changing the center of gravity of the cube during testing."

Nope. Changing a characteristic of the die would result in a new experiment: you would not be justified doing probability calculations from the "old" die to predict results from the "new" one.

Sorry, did not catch. Not old not new one, just the die (one die) whose properties you don't know exactly.