NextDoubleExclusive() can return exactly 1.0, so Normal/LogNormal Sample returns +Infinity #60

What's wrong

IRandomProvider.NextDoubleExclusive() is defined in Essentials/IRandomProvider.cs, around line 173:

public double NextDoubleExclusive() => ((NextUInt64() >> 11) + 0.5) * RandomHelpers.UnitScale; // UnitScale = 2^-53

When the 53-bit draw is at its maximum, 2^53 - 1, the sum (2^53 - 1) + 0.5 cannot be represented as a double. Doubles near 2^53 are spaced 1 apart, so this is an exact tie between 2^53 - 1 and 2^53. Round-half-to-even picks 2^53, and the result is 2^53 * 2^-53 = 1.0.

The method's own remarks say the half-step offset exists so that inverse-transform sampling never sees 0 or 1, because the quantile of an unbounded distribution is infinite there. The upper end is not excluded.

Consequence

The default IDistribution.Sample computes Quantile(NextDoubleExclusive()). When the draw hits this value:

• Normal(0, 1).Sample(random) returns +Infinity.
• LogNormal behaves the same way.
• Every other distribution that uses the default Sample receives u = 1, which the method's contract excludes.

Reproduction

A stub IRandomProvider whose NextBytes fills the buffer with 0xFF was run against the net10.0 build:

NextDoubleExclusive = 1 (== 1.0? True)
Normal(0,1).Sample: Infinity

The chance per draw is 2^-53. However, a seeded Xoshiro or Pcg state that produces this word will produce it every time it is replayed. The existing test at RandomProviderTests.cs:152 only checks random samples, so it can never hit this boundary.

Suggested fix / acceptance criteria

• Use 52 bits so that the sum is always exact:

  ((NextUInt64() >> 12) + 0.5) * (1.0 / 4503599627370496.0) // 2^-52

  The largest result is then 1 - 2^-53 and the smallest is 2^-53, both strictly inside (0, 1).
• Add a boundary test with a stub provider that returns ulong.MaxValue and one that returns 0. Assert 0 < u < 1, and assert that Normal(0,1).Sample is finite for both.