Re: Layman's version of the Pitman discussion

In article <1135685557.759701.31410@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Seanpit <seanpitnospam@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
>
>Kim G. S. Øyhus wrote:
>
>< snip >
>
>> >> Giving your rate, it would
>> >> have a probability of existing of 1e-1169, which is impossible.
>> >
>> >Not true. It is very unlikely that a sequence chosen at random will be
>> >beneficial, but it is not impossible.
>>
>> Your probability of 1e-1200 would be correct if there was only one
>> functioning 4000 base sequence. So your probability is wrong.
>
>If there were only 1 functioning 4000bp sequence in that level of
>sequence space, the ratio would be 1 in 1e2408 (i.e., not 1 in 1e1200).
> Big difference.

A square root of difference.
And a square root is all that is needed for the birthday paradox to happen.


>> >> You do not have a 4000 bp sequence which cannot tolerate a single mutation.
>> >
>> >I never said otherwise.
>>
>> Wrong. When you supplied a probability of 1e-1200 you implied that.
>> That probability implies a single beneficial sequence of
>> 4000 bp, and that single sequence would not tolerate a single
>> mutation, because that would allow more beneficial sequences, which
>> would make that probability higher.
>
>Again, a ratio of 1 in 1e1200 would produce well over 1e1200 beneficial
>sequences in sequences space of 4000bp - not just 1 as you assume. Your
>calculations are just a bit off.

So, you claim there are clusters of 1e1200 beneficial sequences?
Well then, the number of neighbors get enourmous, thus supporting
my conclusion.


>> >However, the size of the island of tolerable variability is
>> >extremely tiny relative to the surrounding ocean of non-toleration.
>>
>> And that makes it extremely unlikely for the surrounding ocean to be
>> totally empty, which I have shown through simple calculations, which
>> you did not understand.
>
>Your calculations are wrong.

Then you better show how my calculations are wrong, instead of
focusing on some other calculation, like you did above.


>< snip >
>
>> >Oh, and by the way, you can indeed model sequence space in
>> >2-dimensions. All you have to do is draw out a grid and divide up each
>> >square with lines that go through the middle of each square. The
>> >resulting triangles will each by surrounded on all sides by whatever
>> >number of triangles where drawn in that square plus 3 triangles from
>> >the next square. Multiple dimensions are not needed to model sequence
>> >space.
>>
>> Wrong.
>> Writing lots of triangles is not the same as having more dimensions.
>> You have not understood dimensionality.
>
>I can be done just like I said. It seems to me that your calculations
>and understanding of this problem are just a bit off base.

If so, you should be able to map these 64 coordinates of the 64
dimensional hypercube into your 2 dimensional grid in such a way that
distances are preserved:

{0,0,0,0,0,0}, {0,0,0,0,0,1}, {0,0,0,0,1,0}, {0,0,0,0,1,1},
{0,0,0,1,0,0}, {0,0,0,1,0,1}, {0,0,0,1,1,0}, {0,0,0,1,1,1},
{0,0,1,0,0,0}, {0,0,1,0,0,1}, {0,0,1,0,1,0}, {0,0,1,0,1,1},
{0,0,1,1,0,0}, {0,0,1,1,0,1}, {0,0,1,1,1,0}, {0,0,1,1,1,1},
{0,1,0,0,0,0}, {0,1,0,0,0,1}, {0,1,0,0,1,0}, {0,1,0,0,1,1},
{0,1,0,1,0,0}, {0,1,0,1,0,1}, {0,1,0,1,1,0}, {0,1,0,1,1,1},
{0,1,1,0,0,0}, {0,1,1,0,0,1}, {0,1,1,0,1,0}, {0,1,1,0,1,1},
{0,1,1,1,0,0}, {0,1,1,1,0,1}, {0,1,1,1,1,0}, {0,1,1,1,1,1},
{1,0,0,0,0,0}, {1,0,0,0,0,1}, {1,0,0,0,1,0}, {1,0,0,0,1,1},
{1,0,0,1,0,0}, {1,0,0,1,0,1}, {1,0,0,1,1,0}, {1,0,0,1,1,1},
{1,0,1,0,0,0}, {1,0,1,0,0,1}, {1,0,1,0,1,0}, {1,0,1,0,1,1},
{1,0,1,1,0,0}, {1,0,1,1,0,1}, {1,0,1,1,1,0}, {1,0,1,1,1,1},
{1,1,0,0,0,0}, {1,1,0,0,0,1}, {1,1,0,0,1,0}, {1,1,0,0,1,1},
{1,1,0,1,0,0}, {1,1,0,1,0,1}, {1,1,0,1,1,0}, {1,1,0,1,1,1},
{1,1,1,0,0,0}, {1,1,1,0,0,1}, {1,1,1,0,1,0}, {1,1,1,0,1,1},
{1,1,1,1,0,0}, {1,1,1,1,0,1}, {1,1,1,1,1,0}, {1,1,1,1,1,1},

Kim0

.

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