Re: Expected number of trials to first success

From: PalmBug (palmbuddy_at_springs.com)
Date: 08/01/04 

Date: Sun, 01 Aug 2004 01:27:30 GMT

"Stephen J. Herschkorn" <herschko@rutcor.rutgers.edu> wrote in
news:410ACAF1.2060902@rutcor.rutgers.edu:

> We discuss the experiment of tossing a fair die until a 6 appears.
>
> Guaneri wrote:
>
>>"Stephen J. Herschkorn" <herschko@rutcor.rutgers.edu> wrote in
>>news:410A9062.90804@rutcor.rutgers.edu:
>>
>>
>>
>>>[Guaneri:]
>>>
>>>
>>>
>>>>Surely something's amiss. But I keep coming to this situation where
>>>>I see the sequence of no six appearing as impossible, but the
>>>>expected value of the game is infinite.
>>>>
>>>>
>>>>
>>>There is nothing wrong with what you noted. In summary:
>>>1. The number of tosses until the first 6 can be *any* positive
>>>integer. 2. With probability 1, a 6 will appear at *some* toss.
>>>I.e., the event of a six never appearing (though in the probability
>>>space) has probability 0 of happening.
>>>
>>>
>>
>>Ok, let me expose some potential stupidity. Is it wrong to say that
>>with probability 1 the game is finite?
>>
>
> Yes.

hmmm. Here is where I'd like to be enlightened. So what is the
probability of the game being finite?

>
>>When a 6 appears the procedure ends,
>>making the sequence of tosses finite. Yeah, I know we don't know WHEN
>>that will happen, but we do know that it is certain to happen.
>>
>>
>>
>>>3. The payoff scheme you suggest *does* have inifinite expectation.
>>>There is nothing pathological about a random variable having
>>>inifinite expectation.
>>>
>>>
>>
>>>From my, probably faulty, reasoning above, it is if the sequence from
>>which it is calculated is finite. The fact that the variable is
>>random, seems to me, guarantees that an infinite sequence won't occur,
>>
>
> Though the sequence will almost surely end with a finite number of
> throws, to calculate the expected value, you must sum over the
> infinitely many possibilities.
>

This seems equivical. How can the sequence end with an infinite number
of throws?

I'm not trying to be obstinate. I'm trying to understand this!

Thanks

Richard A.

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