Loophole on theorem related to Conditional Probability

[sanpokhrel]

The theorem says
The probability that an event B occur after A has already occurred is given by
P(B/A) =P(A intersection B) /P(A)

But applying thus to a problem like the probability of occurrence of all 3 tails on 3 coins when tossed if 1 tail has already occurred is
P(B/A) =(1/8)/(7/8)=1/7
Since, intersection means all three tail and one tail it is 1/8, occurrence of 1 tails means excluding all heads so 7 remains, 7/8

But looking at this problem from intuitive sense after 1 tail occurs there are 4 possibilities and only 1 case gives all tail. Therefore, probability is 1/4.Where am i mistaken or the conditional probability gives the wrong probability?

 

[PeroK]

The theorem says
The probability that an event B occur after A has already occurred is given by
P(B/A) =P(A intersection B) /P(A)

But applying thus to a problem like the probability of occurrence of all 3 tails on 3 coins when tossed if 1 tail has already occurred is
P(B/A) =(1/8)/(7/8)=1/7
Since, intersection means all three tail and one tail it is 1/8, occurrence of 1 tails means excluding all heads so 7 remains, 7/8

But looking at this problem from intuitive sense after 1 tail occurs there are 4 possibilities and only 1 case gives all tail. Therefore, probability is 1/4.Where am i mistaken or the conditional probability gives the wrong probability?

In this example $P(A)$ is the probability that the first coin is a tail. That is not $7/8$.

 

[sanpokhrel]

In this example $P(A)$ is the probability that the first coin is a tail. That is not $7/8$.

There is nothing that mentions P(A) is first toss.

 

[PeroK]

There is nothing that mentions P(A) is first toss.

What do you think event A is?

 

[sanpokhrel]

That one tail should occur, which is 7 out of total sample space.

 

[Stephen Tashi]

Where am i mistaken or the conditional probability gives the wrong probability?

This is an interesting question.

Assume the 3 coins have labels so we can tell them apart. Let the labels be C1, C2, C3.

In your calculation using conditional probability, you have interpreted the event "A" to mean "At least one coin of C1,C2,C3 comes up tails".

In you intuitive calculation, you are thinking that you know that a particular coin has come up tails. For example, you are thinking of the event "A" to be: Coin "C2 comes up tails".

The difference between the two calculations is explained by the fact that they use two different interpretations of the event "A".

Try using conditional probability to compute the probability of the event "All three coins come up tails given that coin C2 comes up tails".

 

[PeroK]

That one tail should occur, which is 7 out of total sample space.

How do you know you have one tail? Without perhaps having looked at more than one coin and found a head already?

 

[sanpokhrel]

This is an interesting question.

Assume the 3 coins have labels so we can tell them apart. Let the labels be C1, C2, C3.

In your calculation using conditional probability, you have interpreted the event "A" to mean "At least one coin of C1,C2,C3 comes up tails".

In you intuitive calculation, you are thinking that you know that a particular coin has come up tails. For example, you are thinking of the event "A" to be: Coin "C2 comes up tails".

The difference between the two calculations is explained by the fact that they use two different interpretations of the event "A".

Try using conditional probability to compute the probability of the event "All three coins come up tails given that coin C2 comes up tails".

Why does it be any different, why the one specific coin matters?

 

[sanpokhrel]

How do you know you have one tail? Without perhaps having looked at more than one coin and found a head already?

Already occur tail can be in anyone of the three. But one tail is sure.

 

[PeroK]

Already occur tail can be in anyone of the three. But one tail is sure.

Okay, so here's where it goes wrong.

You start to look at the coins, you see a head and two tails. So, you say we have event A. in this case the probability of B/A is zero, because you've already seen two heads. Whereas, you would like P(B/A) to be 1/4 in this case.​

This trap caught out some of the pioneers of probability theory. Here's a variation:

What is the probability all three coins are the same?

They can't all be different, so there must be at least two the same. Focus on them. Now the third coin is the same with probability 1/2.

But, this contradicts the correct result of 1/4.

This argument has the same flaw that you cannot find the two that are the same without perhaps already having looked at the third and seen that it is different.

You can't magically find that there is at least one tail without checking through the coins one by one.

 

[sanpokhrel]

Okay, so here's where it goes wrong.

You start to look at the coins, you see a head and two tails. So, you say we have event A. in this case the probability of B/A is zero, because you've already seen two heads. Whereas, you would like P(B/A) to be 1/4 in this case.​

This trap caught out some of the pioneers of probability theory. Here's a variation:

What is the probability all three coins are the same?

They can't all be different, so there must be at least two the same. Focus on them. Now the third coin is the same with probability 1/2.

But, this contradicts the correct result of 1/4.

This argument has the same flaw that you cannot find the two that are the same without perhaps already having looked at the third and seen that it is different.

You can't magically find that there is at least one tail without checking through the coins one by one.

Thanks for your concern. But i couldn't grasp fully what you are saying.

 

[PeroK]

Thanks for your concern. But i couldn't grasp fully what you are saying.

Try looking at two coins to make it faster. Event A is "there is at least one tail" and event B is "both are tails".

There are four equally likely outcomes:

HH
HT: A
TH: A
TT: A, B

So:

P(B/A) = 1/3
P(A) = 3/4
P(A and B) = 1/4

And the formula holds.

Your intuitive version that if you find a tail, the probability of the other being a tail equals 1/2 is incorrect.

It is correct if event A is changed to "the first coin is a tail".

 

[Stephen Tashi]

Why does it be any different, why the one specific coin matters?

Intuitively, you can think in terms of how much information each statement provides. For example, which of these statement is more informative?:

1) I went to the grocery store this week.

2) I went to the grocery store on Thursday this week.

In your intuitive calculation, to get the answer 1/4, you are imagining that you have seen a specific coin come up tails.