Problem 199 Answer

Let p_n be the probability you should pick n.

p₁ = (x-1+(1-x)^0.5)/x.

x=1-y².

y=z-(2/3). y is also the solution to the equation y^3+2y^2-2=0.

z=A+B.

A=(-b/2+((b^2/4)+(a^3/27))^0.5)^(1/3)

B=(-b/2-((b^2/4)+(a^3/27))^0.5)^(1/3)

a=-4/3.

b=-2

After a lot of messy math, p₁ = 0.456310987.

p_n = p_n-1 × (1-p₁).

Here are the first 34 values of p:

p₁=0.456310987308

p₂=0.24809127017

p₃=0.134884497736

p₄=0.073335219402

p₅=0.039871553032

p₆=0.021677725303

p₇=0.011785941067

p₈=0.006407886662

p₉=0.003483897573

p₁₀=0.001894156832

p₁₁=0.001029832258

p₁₂=0.000559908483

p₁₃=0.000304416091

p₁₄=0.000165507684

p₁₅=0.000089984709

p₁₆=0.000048923698

p₁₇=0.000026599277

p₁₈=0.000014461735

p₁₉=0.000007862686

p₂₀=0.000004274856

p₂₁=0.000002324192

p₂₂=0.000001263638

p₂₃=0.000000687026

p₂₄=0.000000373528

p₂₅=0.000000203083

p₂₆=0.000000110414

p₂₇=0.000000060031

p₂₈=0.000000032638

p₂₉=0.000000017745

p₃₀=0.000000009648

p₃₁=0.000000005245

p₃₂=0.000000002852

p₃₃=0.000000001551

p₃₄=0.000000000843

The probability of winning, assuming any two logicians choose this strategy, is 0.29559774. The probability of a tie is 0.113206772.