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Subject: Re: I'm wrong about 10-0 vs 60-40

Author: Amir Ban

Date: 02:27:27 02/02/01

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On February 01, 2001 at 22:58:59, Uri Blass wrote:

>On February 01, 2001 at 17:26:30, Amir Ban wrote:
>
>>On February 01, 2001 at 17:18:46, Uri Blass wrote:
>>
>>>On February 01, 2001 at 17:08:36, Amir Ban wrote:
>>>
>>>>On January 31, 2001 at 20:17:17, Bruce Moreland wrote:
>>>>
>>>>>I expressed very forcefully that a 10-0 result was more valid than a 60-40
>>>>>result.
>>>>>
>>>>>I've done some experimental tests and it appears that I'm wrong.
>>>>>
>>>>
>>>>No, you were right the first time. Check again.
>>>
>>>The question is what is the meaning of a more valid result.
>>
>>Valid in the sense of demonstrating who is stronger.
>>
>>
>>>
>>><snipped>
>>>>10-0 gets better than 99.9% confidence for the winner to be better.
>>>>
>>>>60-40 has about 95% confidence.
>>>
>>>I agree but the word confidence is misleading because you may ask the question
>>>what is the probability that the winner is the better player and the confidence
>>>does not give an answer to it.
>>
>>That's exactly what it answers.
>>
>>Amir
>
>From your previous post:
>
>"you assume the null hypothesis, which is that the
>result is NOT significant and is a random occurrence between equals."
>
>You cannot calculate the probability that the winner is the better player by
>assuming a model that does not exist.
>

This is what is taught in universities and is written in textbooks. If it
doesn't work, then statisticians have been talking nonsense for centuries.

It makes simple sense: You find the significance of an event by calculating the
probability that the event is insignificant.


>I can give a simple example:
>
>Suppose that the better program has 51% chance to win and 49% chance to lose
>when the results of games are independent and the only missing data is which
>program is better.
>

This is a completely artificial assumption. Where does it come from ?

What you show is that if the two opponents are almost equal, then both have
about the same probability to win 10-0. This is true, but not relevant, because
the question is whether the opponents are equal to start with.

Amir


>Suppose you also see 10-0 result.
>
>you need to calculate p(the winner is better /the result is 10-0)
>
>You do it by base rule
>
>You know that:
>
>1)p(the result is 10-0)=0.51^10+0.49^10
>2)p(the better player is the winner and the result 10-0)=0.51^10
>
>The probability that the winner is the better player after you see 10-0 result
>is not  the level of confidence but 0.51^10/(0.51^10+0.49^10).
>
>Uri



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