Let x = E[n=3].

We have an equation of the form x = 2 + x/3.

Solve for x.

Paul Smith

[email protected]


On Mon, May 9, 2011 at 3:10 PM, Eagle <[email protected]> wrote:
> @Pedro,
>
>      In your last two steps,
>
>> E[n=3] = 2 + 1/3 * E[n=3]
>>
>> E[n=3] = 3
>
> there is something wrong. How are you eliminating E[n=3] on the right
> hand side of the equation?
>
> Eagle
>
>
>
> On May 9, 12:28 am, Pedro Osório <[email protected]> wrote:
>> Hi Ricardo,
>>
>> For your example regarding 3, you have the following possibilities:
>>
>> 1 2 3 - perfect
>> 1 3 2 - one correct
>> 2 1 3 - one correct
>> 2 3 1 - wrong
>> 3 1 2 - wrong
>> 3 2 1 - one correct
>>
>> As you can see, if goro hits without holding anything, 1/6 of the time
>> it will be sorted, 1/2 of the time there will be 2 out of place and
>> 1/3 of the time there will be 3 out of place.
>>
>> This means that:
>>
>> E[n=3] = 1 + (1/6 * 0  + 1/2 * E[n=2] + 1/3 * E[n=3])
>>
>> As you have shown, E[n=2] = 2, so:
>>
>> E[n=3] = 2 + 1/3 * E[n=3]
>>
>> E[n=3] = 3
>>
>> On May 8, 7:57 pm, werneckpaiva <[email protected]> wrote:
>>
>> > I read too and it doesn't make any sense for me.
>>
>> > I understand that this is a geometric distribution where if P(X)=p so
>> > E(X)=1/p. So, if you have 2 numbers unsorted and Goro hits the table,
>> > there is 0.5% chance to stay in the same position and 0.5% chance to
>> > swap positions. But, if you have 3 unsorted elements, there are 6
>> > different permutations, so, P(X)=1/6 and the E(X)=6. My solution is
>> > hold 1 number, swap the other 2; hold the sorted element and swap the
>> > other 2 remaining, so 2 + 2 = 4 hits
>>
>> > 3 1 2
>> > 1 3 2
>> > 1 2 3
>>
>> > But this doesn't seem to be the correct answer. The developers
>> > solutions say that for 3 unsorted numbers needs only 3 hits. Anyone
>> > knows how to explain that?
>>
>> > Regards,
>>
>> > Ricardo
>>
>> > On May 7, 8:24 pm, SwiftCoder <[email protected]> wrote:
>>
>> > > I looked at some of the solutions, as per that umber of hits are same
>> > > as the count of numbers which are not at their correct sorted
>> > > position. Is that so?
>
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