I think this can be answered like dis...
let us say that the persons have decided amongst themselves that if the the
number of people wearing white in front of dem is even he wud say white and
if odd he wud say black....
Now suppose the 100th person counts the number of hats and finds it to be
even... he wud say white...
now the 99th person will do the same...if he still finds the number to be
even and since the 100th person sed white(i.e even) he would say black...now
if the 100th person had sed black (ie odd white) and the count comes out to
be even thus 99 wud be wearing a white hat...
Now that 98th person knows dat 99 had sed the correct hat and using the same
method can say the correct hat color...thus all can be saved except the
100th prisoner...
Also note dat the 100th prisoner also has a 50% chance to survive...

Hope dis helps :)

On Fri, Jul 22, 2011 at 10:05 PM, Shubham Maheshwari
<[email protected]>wrote:

> could some1 plz post the xplainations ...
>
>
> On Fri, Jul 22, 2011 at 8:04 PM, Pankaj <[email protected]> wrote:
>
>> Chetan,
>>
>> No. How could you relate this problem with that? Do you find something
>> similar?
>>
>> ~
>> Pankaj
>>
>>
>> On Fri, Jul 22, 2011 at 8:01 PM, chetan kapoor <[email protected]
>> > wrote:
>>
>>> josehus problem???
>>>
>>>
>>> On Fri, Jul 22, 2011 at 7:57 PM, Pankaj <[email protected]>wrote:
>>>
>>>> Skipp Riddle,
>>>> Yes.
>>>> 100th prisoner will risk his life. Similar puzzle was discuss recently.
>>>> Does anyone remember the name or thread?
>>>>
>>>>
>>>> ~
>>>> Pankaj
>>>>
>>>>
>>>> On Fri, Jul 22, 2011 at 7:55 PM, SkRiPt KiDdIe 
>>>> <[email protected]>wrote:
>>>>
>>>>> Worst case 99 get released.
>>>>> Is that correct..?
>>>>>
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-- 
Aditi Garg
Undergraduate Student
Electronics & Communication Divison
NETAJI SUBHAS INSTITUTE OF TECHNOLOGY
Sector 3, Dwarka
New Delhi

9718388816

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