On 11/13/2022 7:07 AM, riverman wrote:
> On Monday, September 26, 2022 at 7:59:52 AM UTC+8,
henh...@gmail.com wrote:
>> On Friday, September 23, 2022 at 12:15:13 PM UTC-7,
henh...@gmail.com wrote:
>>> ------ pls wait 3+ days (Longer if you find it easy or trivial) before posting answers or hints.
>>>
>>>
>>> 1. in a Room of 100 people, 99% of them are Spinkos.
>>> How many ppl must leave the room
>>> to bring down the percentage of Spinkos in the room to 98% ?
>>>
>>>
>>> 2. in a Room of 1000 people, 99% of them are Spinkos.
>>> How many ppl must leave the room
>>> to bring down the percentage of Spinkos in the room to 98% ?
>> the real puzzle is... Can you tweak the problem(s) slightly so that
it becomes much more tricky and interesting ?
> Yes. What if the people leave randomly, rather than exclusively
selecting non-Spinkos?
Or exclusively selecting Spinkos, rather.
Anyway, the minimum number possible is the same as if only non-Spinkos
are selected, and the maximum number possible is whatever brings the
number of non-Spinkos down to exactly 1. It's also possible to fail to
hit the desired new percentage entirely, if the last non-Spinko is
removed before it's hit.
What's the average number, if we discard those failures? If the minimum
and maximum are equal, then obviously it's that number. Otherwise, it's
a question of which specific totals (of Spinkos and non-Spinkos removed)
are possible, how many ways there are to reach each one, and how many of
those ways also hit the percentage at some earlier point first.