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Scenario: Assume you are an interested party who would like to file a protest against an agency. You intend to start with an agency protest, then if you lose file a GAO protest, then if you lose file a COFC protest, then if you lose file an appeal at the CAFC. You have a 20% chance of winning at each forum.

Question: What is the probability that you will win in at least one forum?

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Guest Vern Edwards
8 hours ago, Don Mansfield said:

Carl, 

Did you use math to arrive at your answer?

Don, you did not ask for a mathematical calculation.

Question: Do you consider the COFC decision and the CAFC decision to be independent events? If so, please explain.

13 hours ago, Don Mansfield said:

Question: What is the probability that you will win in at least one forum?

Less than 50-50.

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Mathematically:

Probability of winning at agency: 20%

Probability of winning at GAO conditional on losing at Agency: 20% * (1 - 20%) = 16%

Probability of winning at COFC conditional on losing at Agency and GAO: 20% * (1 - 20% - 16%) = 12.8%

Probability of winning at CAFC conditional on losing at Agency, GAO, and COFC: 20% * (1 - 20% - 16% - 12.8%) = 10.24%

Combined probability of winning: 20% + 16% + 12.8% + 10.24% = 59.04%

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3 minutes ago, Matthew Fleharty said:

Mathematically:

Probability of winning at agency: 20%

Probability of winning at GAO conditional on losing at Agency: 20% * (1 - 20%) = 16%

Probability of winning at COFC conditional on losing at Agency and GAO: 20% * (1 - 20% - 16%) = 12.8%

Probability of winning at CAFC conditional on losing at Agency, GAO, and COFC: 20% * (1 - 20% - 16% - 12.8%) = 10.24%

Combined probability of winning: 20% + 16% + 12.8% + 10.24% = 59.04%

This is one way of calculating "at least one" probability. A simpler way is to calculate the probability of the outcome NOT occurring, then subtracting from 1.

So here, you could get the response with 1 - (.80^4). 

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6 hours ago, Vern Edwards said:

Question: Do you consider the COFC decision and the CAFC decision to be independent events? If so, please explain.

Yes. You have a 20% chance of winning your case at the COFC, and a 20% chance of winning your case at the CAFC. I'm saving conditional probabilities for a future problem.

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2 minutes ago, Don Mansfield said:

Yes. You have a 25% chance of winning your case at the COFC, and a 25% chance of winning your case at the CAFC. I'm saving conditional probabilities for a future problem.

25% or 20%?

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1 minute ago, Don Mansfield said:

You have a 25% chance of winning your case at the COFC, and a 25% chance of winning your case at the CAFC.

Don - You mean 20%?

Can you establish whether this is intended purely as a mathematical problem, as the title implies, or whether there are other factors you want considered?

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Oops. I meant 20%.

4 minutes ago, FrankJon said:

Can you establish whether this is intended purely as a mathematical problem, as the title implies, or whether there are other factors you want considered?

Purely mathematical.

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Just now, Don Mansfield said:

Purely mathematical.

So I think a few of us have confirmed the answer, but we haven't heard a direct response from OP. Do you disagree? 

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I want to see if Vern is going to change his answer, based on my response to him. I'll post the answer by the end of the day.

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Guest Vern Edwards
11 hours ago, Don Mansfield said:

I want to see if Vern is going to change his answer, based on my response to him. I'll post the answer by the end of the day.

I'm wondering if you can consider an appellate ruling on the COFC's decision as an independent event, since the appellate court does not review the protest de novo.

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Late to party here , but wouldn't it be 20% ? This is like a coin flip, the probability doesn't increase or compound as it goes through each appeal. Combining them assumes the results are connected.

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sdvr,

While it's true that each event occurs separately,  since the question is what is the chance that "at least one win occurs" you have to looks at all of the probabilities combined.

As has been mentioned there are two ways to do this if you think in simple coin flips terms....  Chance of 1 heads in 4 flips, or chance of NOT all 4 flips being tails. 

Most statisticians will say the 2nd is easier then the first... here's why. 

  • One could work out all the separate cases – one head in four tosses, two heads, or three or four.    H,H,H,H or H,H,H,T, or H,H,T,T .... etc until T.T.T.H
  • But it's much easier to find the probability of no heads, and subtract that from one: or not T,T,T,T
    • In our case Pr (at least one win in 4 cases) = 1 – Pr (no wins in 4 cases) = 1 – (0.8)4 = 59.04%.

Since you have to look at all events, the number isn't just 20% as you start losing cases and then move to level, your next 20% chance is basically taking 20% of the 80% failure to get a L,W scenario.

Hope that helps a bit.

 

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35 minutes ago, sdvr said:

Late to party here , but wouldn't it be 20% ? This is like a coin flip, the probability doesn't increase or compound as it goes through each appeal. Combining them assumes the results are connected.

This is exactly why I showed the long form explanation...see above.

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