1, 2, 3... Probability!

In order to solve this problem, first we need to establish how many different ways there are of people attempting a single problem to produce 1%, 2%, and 3% of people getting the answer right in 3 consecutive successful attempts with the percentages rounded to the nearest whole number.

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1% (Rounding to the nearest whole number)

The lowest number of people attempting a problem to produce 1% with 1 solver is 67 $100 \times \dfrac{1}{67} = 1.492$

The highest number of people attempting a problem to produce 1% with 1 solver is 200 $100 \times \dfrac{1}{200} = 0.5$

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2% (Rounding to the nearest whole number)

The lowest number of people attempting a problem to produce 2% with 2 solvers is 81 $100 \times \dfrac{2}{81} = 2.46$

The highest number of people attempting a problem to produce 2% with 2 solvers is 133 $100 \times \dfrac{2}{133} = 1.503$

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3% (Rounding to the nearest whole number)

The lowest number of people attempting a problem to produce 3% with 3 solvers is 86 $100 \times \dfrac{3}{86} = 3.48$

The highest number of people attempting a problem to produce 3% with 3 solvers is 120 $100 \times \dfrac{3}{120} = 2.5$

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Ranges

• 1% [67, 200]
• 2% [81, 133]
• 3% [86, 120]

With each progressive range being a subset comfortably inside the previous percentages range, we can conclude that the number of ways to get 1%, 2%, 3% in 3 consecutive successful attempts is the size of the final range [86, 120] = 35 $120 - 86 + 1 = 35$

It is important to check how the ranges compare with each other on problems like this, because the ranges may not necessarily overlap (or overlap enough). In which case, the number of outcomes could be smaller than the final range.

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Probability

Now that we've established that there are 35 ways to create 1%, 2%, 3% with 3 consecutive successful attempts on a single problem, we can determine by the Rule of Product (because they are independent events) that there are 1225 combinations of outcomes for the number of people attempting the problem by Aporia and the number of people attempting the problem by Tyche (ie: 86|86, 86|87, 86|88, ... , 120|118, 120|119, 120|120). $\large \dfrac{1}{35} \times \dfrac{1}{35} = \dfrac{1}{1225}$

35 of the 1225 outcomes are outcomes where the number of people that attempted each problem is the same (ie: 86|86, 87|87, 88|88, ..., etc.), so the probability of Aporia and Tyche having the same number of people attempt their problem after 3 consecutive successful attempts is 1/35.

$\large 35 \times \dfrac{1}{1225} = 35 \times \dfrac{1}{35} \times \dfrac{1}{35} = \huge\boxed{\dfrac{1}{35}}$