# Welfare offer 12bet.com login_login bonus bet365 live casino review_free login baseball betting software

```On Tue, Jan 17, 2017 at 4:13 PM, Nadav Har'El <n...@scylladb.com> wrote:
>
>
> On Tue, Jan 17, 2017 at 7:18 PM, aleba...@gmail.com <aleba...@gmail.com>
wrote:
>>
>>
>> I may be wrong, but I think that the result of the current
implementation is actually the expected one.
>> Using you example: probabilities for item 1, 2 and 3 are: 0.2, 0.4 and
0.4
>>
>> P([1,2]) = P( | 1st=) P() + P( | 1st=) P()
>
>
> Yes, this formula does fit well with the actual algorithm in the code.
But, my question is *why* we want this formula to be correct:
>
>>
>> Now, P() = 0.2 and P() = 0.4. However:
>> P( | 1st=) = 0.5     (2 and 3 have the same sampling probability)
>> P( | 1st=) = 1/3     (1 and 3 have probability 0.2 and 0.4 that,
once normalised, translate into 1/3 and 2/3 respectively)
>> Therefore P([1,2]) = 0.7/3 = 0.23333
>> Similarly, P([1,3]) = 0.23333 and P([2,3]) = 1.6/3 = 0.533333
>
>
> Right, these are the numbers that the algorithm in the current code, and
the formula above, produce:
>
> P([1,2]) = P([1,3]) = 0.23333
> P([2,3]) = 0.53333
>
> What I'm puzzled about is that these probabilities do not really fullfill
the given probability vector 0.2, 0.4, 0.4...
> Let me try to explain explain:
>
> Why did the user choose the probabilities 0.2, 0.4, 0.4 for the three
items in the first place?
>
> One reasonable interpretation is that the user wants in his random picks
to see item 1 half the time of item 2 or 3.
> For example, maybe item 1 costs twice as much as item 2 or 3, so picking
it half as often will result in an equal expenditure on each item.
>
> If the user randomly picks the items individually (a single item at a
time), he indeed gets exactly this distribution: 0.2 of the time item 1,
0.4 of the time item 2, 0.4 of the time item 3.
>
> Now, what happens if he picks not individual items, but pairs of
different items using numpy.random.choice with two items, replace=false?
> Suddenly, the distribution of the individual items in the results get
skewed: If we look at the expected number of times we'll see each item in
one draw of a random pair, we will get:
>
> E(1) = P([1,2]) + P([1,3]) = 0.46666
> E(2) = P([1,2]) + P([2,3]) = 0.76666
> E(3) = P([1,3]) + P([2,3]) = 0.76666
>
> Or renormalizing by dividing by 2:
>
> P(1) = 0.233333
> P(2) = 0.383333
> P(3) = 0.383333
>
> As you can see this is not quite the probabilities we wanted (which were
0.2, 0.4, 0.4)! In the random pairs we picked, item 1 was used a bit more
often than we wanted, and item 2 and 3 were used a bit less often!
>
> So that brought my question of why we consider these numbers right.
>
> In this example, it's actually possible to get the right item
distribution, if we pick the pair outcomes with the following probabilties:
>
>    P([1,2]) = 0.2        (not 0.233333 as above)
>    P([1,3]) = 0.2
>    P([2,3]) = 0.6        (not 0.533333 as above)
>
> Then, we get exactly the right P(1), P(2), P(3): 0.2, 0.4, 0.4
>
> Interestingly, fixing things like I suggest is not always possible.
Consider a different probability-vector example for three items - 0.99,
0.005, 0.005. Now, no matter which algorithm we use for randomly picking
pairs from these three items, *each* returned pair will inevitably contain
one of the two very-low-probability items, so each of those items will
appear in roughly half the pairs, instead of in a vanishingly small
percentage as we hoped.
>
> But in other choices of probabilities (like the one in my original
example), there is a solution. For 2-out-of-3 sampling we can actually show
a system of three linear equations in three variables, so there is always
one solution but if this solution has components not valid as probabilities
(not in [0,1]) we end up with no solution - as happens in the 0.99, 0.005,
0.005 example.```
```

I think the underlying problem is that in the sampling space the events (1,
2) (1, 3) (2, 3) are correlated and because of the discreteness an
arbitrary marginal distribution on the individual events 1, 2, 3 is not
possible.

related aside:
I'm not able (or willing to spend the time) on the math, but I just went
through something similar for survey sampling in finite population (e.g.
survey two out of 3 individuals, where 3 is the population), leading to the
Horvitz�CThompson estimator. The books have chapters on different sampling
schemes and derivation of the marginal and joint probability to be surveyed.

(I gave up on sampling without replacement, and assume we have a large
population where it doesn't make a difference.)

In some of the sampling schemes they pick sequentially and adjust the
probabilities for the remaining individuals. That seems to provide more
flexibility to create a desired or optimal sampling scheme.

Josef

>
>
>
>>
>> What am I missing?
>>
>> Alessandro
>>
>>
>> 2017-01-17 13:00 GMT+01:00 <numpy-discussion-requ...@scipy.org>:
>>>
>>> Hi, I'm looking for a way to find a random sample of C different items
out
>>> of N items, with a some desired probabilty Pi for each item i.
>>>
>>> I saw that numpy has a function that supposedly does this,
>>> numpy.random.choice (with replace=False and a probabilities array), but
>>> looking at the algorithm actually implemented, I am wondering in what
sense
>>> are the probabilities Pi actually obeyed...
>>>
>>> To me, the code doesn't seem to be doing the right thing... Let me
explain:
>>>
>>> Consider a simple numerical example: We have 3 items, and need to pick 2
>>> different ones randomly. Let's assume the desired probabilities for
item 1,
>>> 2 and 3 are: 0.2, 0.4 and 0.4.
>>>
>>> Working out the equations there is exactly one solution here: The random
>>> outcome of numpy.random.choice in this case should be [1,2] at
probability
>>> 0.2, [1,3] at probabilty 0.2, and [2,3] at probability 0.6. That is
indeed
>>> a solution for the desired probabilities because it yields item 1 in
>>> [1,2]+[1,3] = 0.2 + 0.2 = 2*P1 of the trials, item 2 in [1,2]+[2,3] =
>>> 0.2+0.6 = 0.8 = 2*P2, etc.
>>>
>>> However, the algorithm in numpy.random.choice's replace=False
generates, if
>>> I understand correctly, different probabilities for the outcomes: I
believe
>>> in this case it generates [1,2] at probability 0.23333, [1,3] also
0.2333,
>>> and [2,3] at probability 0.53333.
>>>
>>> My question is how does this result fit the desired probabilities?
>>>
>>> If we get [1,2] at probability 0.23333 and [1,3] at probability 0.2333,
>>> then the expect number of "1" results we'll get per drawing is 0.23333 +
>>> 0.2333 = 0.46666, and similarly for "2" the expected number 0.7666, and
for
>>> "3" 0.76666. 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, 0.4, 0.4 for the individual items!).
>>>
>>>
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>>
>>
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