Probably of choosing 2 numbers between 0-10000 correcctly

Question

A player can choose numbers between 0000-9999, which means there are 10,000 numbers to choose from.

The probability of choosing correctly one number is 1/10000.

So the player picks three different numbers and the game picks three different numbers, what is the probability that two of them match without regard to order?

The numbers are not repeated.

Answer 1

So we have these possibilities of picking two correct numbers out of three: CCW, CWC and WCC where C denotes the event of choosing a number correctly and W denotes the event of choosing a number incorrectly.

The probability of this happening then is $$3\cdot\frac{1}{10000}\frac{1}{10000}\frac{9999}{10000}.$$

Answer 2

I suppose that your scenario is:

• A player picks 3 numbers without replacement.
• The game picks 3 numbers without replacement (obviously from the whole set).
• The player win if he/she has picked at least 2 of 3 numbers right.

So the winning scenarios are $CCW$, $CWC$, $WCC$, $CCC$ where $C$ stands for correct and $W$ stands for wrong.

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$CCW$, $CWC$, $WCC$

In the first scenario you have to take the first one correct ($\frac{3}{10000}$), the second one correct ($\frac{2}{9999}$) and the third one wrong ($\frac{9997}{9998}$).

In the second scenario you have to take the first one correct ($\frac{3}{10000}$), the second one wrong ($\frac{9997}{9999}$) and the third one correct ($\frac{2}{9998}$).

In the third scenario you have to take the first one wrong ($\frac{9997}{10000}$), the second one correct ($\frac{3}{9999}$) and the third one correct ($\frac{2}{9998}$).

As you can see, when you multiply, they have all the same value, that is $\frac{2 \cdot 3 \cdot 9997}{9998 \cdot 9999 \cdot 10000}$. You could semplify and sum up the three cases and obtain $\frac{9997}{1111 \cdot 5000 \cdot 9998}$.

$CCC$

In the fourth scenario you have to take the first one correct ($\frac{3}{10000}$), the second one correct ($\frac{2}{9999}$) and the third one correct ($\frac{1}{9998}$) so its probability is $\frac{3 \cdot 2 \cdot 1}{9998 \cdot 9999 \cdot 10000}$

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If my scenario is correct you should add the two probability, if your scenario is simply $CCW$, $CWC$, $WCC$ then don't add the $CCC$ part.