**
Aporia
**
published a difficult problem. Her problem was so difficult that it remained unsolved after many people had attempted it.

When the first solver eventually occurred, the percentage of people getting her problem right showed as follows:

1% of people got this right. (1 solver)

Then, the next person to attempt her problem solved it.

2% of people got this right. (2 solvers)

Surprisingly, the next person to attempt her problem also solved it.

3% of people got this right. (3 solvers)

Despite the difficulty of Aporia's problem, 3 consecutive attempts by 3 different people were all successful.

**
Tyche
**
had also published her own difficult problem which remained unsolved after many attempts, and (as if it were fate) eventually saw the same progression for the percentage of people getting her problem right in 3 consecutive successful attempts.

1% of people got this right. (1 solver)

2% of people got this right. (2 solvers)

3% of people got this right. (3 solvers)

What an amazing turn of fortune!

**
Question
**

Considering that all percentages shown are rounded to the nearest whole number (eg:
$0.5\% \rightarrow 1\%$
); What is the probability that
**
Aporia
**
and
**
Tyche
**
each had the same number of people
**
attempt
**
their problem at the point that they each had 3 solvers?

$1$
$\frac13$
$\frac1{35}$
$\frac1{53}$
$\frac1{134}$
$\frac1{1225}$
$0$

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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.

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$100 \times \dfrac{1}{200} = 0.5$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$100 \times \dfrac{2}{133} = 1.503$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$100 \times \dfrac{3}{120} = 2.5$RangesWith 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.

ProbabilityNow 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.

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