Stacked Probability Problem

Probability theory and statistics

Stacked Probability Problem

Postby Guest » Fri Mar 24, 2017 3:50 pm

Lets say we have three cans each having 600 golf balls. 599 are white and one is black. Blindfolded I know the probability of picking out the black ball in one try is 600:1. If I go to the second can the probability is the same for that can. But what is the probability of picking both black balls? Is there some kind of multiplier effect? And I then go to the third can and try to pick up the black ball. - - - The real question is what is the probability that I will end up with three black balls, one from each can???





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Re: Stacked Probability Problem

Postby Guest » Sat Mar 25, 2017 5:25 am

I think you are confusing your notation. The probability of choosing a black ball is 1/600 (number of successful outcomes / total number of outcomes). To use gambling notation, the odds are 1:599 (number of successful outcomes : number of unsuccessful outcomes), usually in gambling you are given the odds against in which case it would be 599:1.

To answer you question, yes there is a multiplicative effect.

If instead each can had 3 balls (one of which is black), the probability would be 1/3. To see why let us label each ball from 1 to 3, with the black ball being 1. There are three 3 outcomes:
We choose ball 1 (Success)
We choose ball 2 (Unsuccessful)
We choose ball 3 (Unsuccessful)
Each outcome is equally likely, and we have 1 successful outcome out of 3, so the probability is 1/3.

If we have two cans now, each with three balls, let us label the balls 1 to 3 in each can, (with the black balls being labelled 1), now let us consider all the outcomes:
We choose ball 1 from can 1, and ball 1 from can 2 (Success)
We choose ball 1 from can 1, and ball 2 from can 2 (Unsuccessful)
We choose ball 1 from can 1, and ball 3 from can 2 (Unsuccessful)
We choose ball 2 from can 1, and ball 1 from can 2 (Unsuccessful)
We choose ball 2 from can 1, and ball 2 from can 2 (Unsuccessful)
We choose ball 2 from can 1, and ball 3 from can 2 (Unsuccessful)
We choose ball 3 from can 1, and ball 1 from can 2 (Unsuccessful)
We choose ball 3 from can 1, and ball 2 from can 2 (Unsuccessful)
We choose ball 3 from can 1, and ball 3 from can 2 (Unsuccessful)
There are 9 outcomes now, only 1 of which was successful, giving a probability of 1/9.

In general listing all the outcomes is impractical so instead we just count them, in this case it is quite easy, there are 3 choices for the first ball and for each of those choices there was 3 choices for the second, so the total number is 3x3 = 9. Counting the successful choices is also easy, if I want a successful outcome there is only 1 choice that works for the first can, and for each of the choices that work for the first can, there is only one choice that works for the second can, giving 1x1 = 1.

In general if the events are *independent* (i.e. the result of event 1 won't affect how event 2 plays out) then we can just multiply the probabilities because, the total number of outcomes is just the product of total outcomes in event 1 and event 2, and the total number of successful outcomes is the product of successful outcomes in event 1 and event 2. So
Probability of success in event1 and success in event 2

[tex]= \frac{\text{"number of outcomes that are successful"}}{\text{"total number of outcomes"}}[/tex]

[tex]= \frac{\text{"no. of successful outcomes in event 1"} \times\text{"no. of successful outcomes in event 2"}}{\text{"total outcomes in event 1"}\times\text{"total outcomes in event 2"}}[/tex]

[tex]= \frac{\text{"no. of successful outcomes in event 1" }}{\text{"total outcomes in event 1"}}\times\frac{\text{"no. of successful outcomes in event 2"}}{\text{"total outcomes in event 2"}}[/tex]

[tex]= \text{"Probability of success in event 1"}\times\text{"Probability of success in event 2"}[/tex].

In the case of the 600 balls in cans, the probability of choosing 2 black balls is (1/600) x (1/600) = 1/360000 (because the events are independent, the chance of you picking a black ball in the second can is unaffected by what happened with the first can). Similarly the probability of picking a black ball in the third can is independent of whether you picked 2 black balls in the first two cans, so the overall probability is (1/360000) x (1/600) = 1/216000000.

Hope this helped,

R. Baber.

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Re: Stacked Probability Problem

Postby Math Tutor » Mon Mar 27, 2017 7:30 am

Mr. Rabber is a genius.

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