;1" results we'll get per drawing is 0.2 +
0.2333 = 0.4, and similarly for "2" the expected number 0.7666, and for
"3" 0.7. As you can see, the proportions are off: Item 2 is NOT twice
common than item 1 as we originally desired (we asked for probabilities
0.2, 0
question is how does this result fit the desired probabilities?
>>
>> If we get [1,2] at probability 0.2 and [1,3] at probability 0.2333,
>> then the expect number of "1" results we'll get per drawing is 0.2 +
>> 0.2333 = 0.4, and similarly for "
On Wed, Jan 18, 2017 at 1:58 AM, aleba...@gmail.com
wrote:
>
>
> 2017-01-17 22:13 GMT+01:00 Nadav Har'El :
>
>>
>> On Tue, Jan 17, 2017 at 7:18 PM, aleba...@gmail.com
>> wrote:
>>
>>> Hi Nadav,
>>>
>>> I may be wrong, but I thi
On Wed, Jan 18, 2017 at 11:00 AM, aleba...@gmail.com
wrote:
> Let's look at what the user asked this function, and what it returns:
>
>>
>> User asks: please give me random pairs of the three items, where item 1
>> has probability 0.2, item 2 has 0.4, and 3 has 0.4.
>>
>> Function returns: random
On Wed, Jan 18, 2017 at 4:30 PM, wrote:
>
>
> Having more sampling schemes would be useful, but it's not possible to
>> implement sampling schemes with impossible properties.
>>
>>
>
> BTW: sampling 3 out of 3 without replacement is even worse
>
> No matter what sampling scheme and what selection
On Mon, Jan 23, 2017 at 4:52 PM, aleba...@gmail.com
wrote:
>
>
> 2017-01-23 15:33 GMT+01:00 Robert Kern :
>
>>
>> I don't object to some Notes, but I would probably phrase it more like we
>> are providing the standard definition of the jargon term "sampling without
>> replacement" in the case of
On Mon, Jan 23, 2017 at 5:47 PM, Robert Kern wrote:
>
> > As for the standardness of the definition: I don't know, have you a
> reference where it is defined? More natural to me would be to have a list
> of items with integer multiplicities (as in: "cat" 3 times, "dog" 1 time).
> I'm hesitant to