Implied probability of this combined sequence: A, NOT A, A, NOT A

neilhardy

New member
Joined
Feb 26, 2018
Messages
3
Hi can someone cleverer than me assits with an implied probability query?

If the implied probability if an even lets call it A is 7-1 or14% (1.16 decimal)

I want to work out the implied probablity of this combined sequence

A, NOT A, A, NOT A

Therfore the % prob of A is 14% (100/7)
Therefore the % prob of NOT at is 100 - (100/7) = 86

Therefore the implied prob of " NOT A" as a decimal is = 100/86 = 1.16

is its 7 *.86 * 7 *.86 = 36 -1 ???? Is this right

Thanks everyone

 

j-astron

Junior Member
Joined
Jan 10, 2018
Messages
181
Hi can someone cleverer than me assits with an implied probability query?

If the implied probability if an even lets call it A is 7-1 or14% (1.16 decimal)

I want to work out the implied probablity of this combined sequence

A, NOT A, A, NOT A

Therfore the % prob of A is 14% (100/7)
Therefore the % prob of NOT at is 100 - (100/7) = 86

Therefore the implied prob of " NOT A" as a decimal is = 100/86 = 1.16

is its 7 *.86 * 7 *.86 = 36 -1 ???? Is this right

Thanks everyone

Hi neilhardy, welcome to the forum :)


This post is a little bit all over the place, but I'll do my best to help. I'm not sure what you mean by "implied" probability. Also the decimal value of 1.16 is just completely wrong. Probability is a number between 0 and 1 by definition, where 0 means no chance of event A occurring, and 1 means complete certainty of event A occurring.

I'm going to stick to numbers in this range [0,1], rather than percentages, in order to avoid confusion.

So, if probability of A = 1/7 = ~0.14, then assuming either A or NOT A has to occur, their probabilities should add up to 1, and you are correct that

probability of NOT A = 1 - (probability of A) = 1 - (1/7) = 6/7 = ~0.86.

Now here's where you have to be more specific/clear in order for us to help you. Are you saying that you now have a sequence of four independent trials, each of which can result in either A or NOT A? If so, bear in mind that there are actually 16 possible outcomes. Also, you probably need to use the binomial distribution to compute the answer.
 

neilhardy

New member
Joined
Feb 26, 2018
Messages
3
Hi neilhardy, welcome to the forum :)


This post is a little bit all over the place, but I'll do my best to help. I'm not sure what you mean by "implied" probability. Also the decimal value of 1.16 is just completely wrong. Probability is a number between 0 and 1 by definition, where 0 means no chance of event A occurring, and 1 means complete certainty of event A occurring.

I'm going to stick to numbers in this range [0,1], rather than percentages, in order to avoid confusion.

So, if probability of A = 1/7 = ~0.14, then assuming either A or NOT A has to occur, their probabilities should add up to 1, and you are correct that

probability of NOT A = 1 - (probability of A) = 1 - (1/7) = 6/7 = ~0.86.

Now here's where you have to be more specific/clear in order for us to help you. Are you saying that you now have a sequence of four independent trials, each of which can result in either A or NOT A? If so, bear in mind that there are actually 16 possible outcomes. Also, you probably need to use the binomial distribution to compute the answer.



Thanks astron for your help,

yes they are four seperate events where to possible outcome is either A (7/1) or NOT A (0,86), by not connected I mean that the result of the previous test has no bearing on the next (or prev)... Thanks again for your help...
 

neilhardy

New member
Joined
Feb 26, 2018
Messages
3
Hi neilhardy, welcome to the forum :)


This post is a little bit all over the place, but I'll do my best to help. I'm not sure what you mean by "implied" probability. Also the decimal value of 1.16 is just completely wrong. Probability is a number between 0 and 1 by definition, where 0 means no chance of event A occurring, and 1 means complete certainty of event A occurring.

I'm going to stick to numbers in this range [0,1], rather than percentages, in order to avoid confusion.

So, if probability of A = 1/7 = ~0.14, then assuming either A or NOT A has to occur, their probabilities should add up to 1, and you are correct that

probability of NOT A = 1 - (probability of A) = 1 - (1/7) = 6/7 = ~0.86.

Now here's where you have to be more specific/clear in order for us to help you. Are you saying that you now have a sequence of four independent trials, each of which can result in either A or NOT A? If so, bear in mind that there are actually 16 possible outcomes. Also, you probably need to use the binomial distribution to compute the answer.
Thanks for your help,

yes tehre are 4 unconnected tests ( where two outcomes are possible A (7/1) and Not A (0.84) the result of the last has no bearing on the next and so on.

Thanks again..
 

j-astron

Junior Member
Joined
Jan 10, 2018
Messages
181
Thanks astron for your help,

yes they are four seperate events where to possible outcome is either A (7/1) or NOT A (0,86), by not connected I mean that the result of the previous test has no bearing on the next (or prev)... Thanks again for your help...
TBH I'm not sure why you are writing the probability of A as 7/1. As I said, a probability is a number less than or equal to 1. The probability of A is 1/7, expressing the fact that it tends to occur, on average, only 1 out of every 7 times.

Okay, the result of previous trials having no bearing on the current one means that the trials are independent of each other. It's as though you were flipping a very unfair coin. One that only lands on heads 1/7th of the time instead of 1/2 the time.

It turns out that probabilities of independent events multiply together. To see why this is, imagine you want to get heads (event A) twice in a row. What's the chance you'll get it the first time? It's 1/7, because the coin tends to come up heads only 1/7th the time. Assuming that that happened, what's the chance it will also come up heads a second time? Well it's only in 1/7th of the cases that it comes up heads in the first place, and it's only in 1/7th of those cases that it will come up heads again. So it's only in 1/7th of 1/7th of the cases that it will come up heads twice, which is (1/7)*(1/7) of the cases. That's why you multiply.

The fact that probabilities of independent events multiply together is all the information you need to solve this problem.
 
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