Two people randomly select two different 1-digit numbers (including 0 of course) and calculate their absolute difference. Then They share what they get, What is the probability of having the same absolute difference for these two?

For example, Player 1 take "0", then take "3" (he cannot choose 0 again), the absolute difference is 3. Players 2 takes "9" then takes "6", the absolute difference becomes 3. and they share their result of the difference. since both has the same result they yell "yay"... :)


closed as off-topic by Anthony Pham, Engineer Toast, Spencerkatty, Zandar, AJL Mar 5 '16 at 12:12

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    $\begingroup$ At some point this site will need guidelines to distinguish math puzzles from math problems. $\endgroup$ – user1717828 Feb 27 '16 at 13:24
  • $\begingroup$ @user1717828 puzzle means "a game, toy, or problem designed to test ingenuity or knowledge." this kind of probability question is actually a puzzle! $\endgroup$ – Oray Feb 27 '16 at 14:00
  • $\begingroup$ So suppose the original numbers were 1 and 2. The absolute difference is 1. What does "They share what they get" mean? $\endgroup$ – Lawrence Feb 27 '16 at 14:15
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    $\begingroup$ @Oray, Consider this math problem (or is it a puzzle?): A red person and a blue person each have an urn filled with 80 red balls and 20 blue balls and alternate drawing one ball at a time. If a person draws a ball of their own color, they can remove 4 random balls from their urn. What is the probability the red person will empty their urn first? IMO, this is just an undergraduate probability question; not a puzzle. $\endgroup$ – user1717828 Feb 27 '16 at 14:26
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    $\begingroup$ @user1717828 since this is not just a straight forward probability question and it requires not only checking probabilities but also some understanding, counting and logic thinking (especially if this problem was prepared with 2 or 3 digits), I believe it requires more than just being an undergrad. I intentionally ask this question as simple as possible (1 digit) to get people's attention and their interest... still there is no right answer has come yet. $\endgroup$ – Oray Feb 27 '16 at 14:30

Each person picks 2 numbers and finds their absolute difference.

There are 9 ways to get an absolute difference of 1 (1-0, 2-1, ..., 9-8).

There are 8 ways to get an absolute difference of 2 (2-0, 3-1, ..., 9-7).


There are $w(d) = 10-d$ ways to get an absolute difference of $d$, with $1 \leq d \leq 9$.

Total number of digit combinations is ${10 \choose 2} = 45$.

This is also the total number of 'ways': $\sum_{d=1}^{9} (10-d) = 90 - \frac{9 \cdot 10}{2} = 45$.

We require both to pick the same absolute difference. The probability of this is the sum of the probabilities of both picking the same absolute difference, which works out to be $\frac{19}{135}$.

$$\begin{align} \sum_{d=1}^{9} \left( \frac{w(d)}{45} \cdot \frac{w(d)}{45} \right) &= \sum_{d=1}^{9} \frac{(10 - d)^2}{45^2} \\ &= \frac{900 - 20(45) + \frac{(9)(9+1)(2 \cdot 9 + 1)}{6}}{45^2} \\ &= \frac{19}{135} \end{align}$$


The probability either player has a absolute difference of $k$ is $\dfrac{10-k}{\binom{10}{2}}$. For a match, both need to be the same, and so we consider:

$$\sum_\limits{k=1}^9 \Big[\dfrac{10-k}{\binom{10}{2}}\Big]^2$$

The sum of the first $n$ squares is $\dfrac{n(n+1)(2n+1)}{6}$, and so we have:



The absolute difference $|a_1-a_2|$ for a player has the following distribution of probability:



$$P(|a_1-a_2|=k)=2\frac{10-k}{100}$$, for $k=1, \cdots ,9$.


$P(|a_1-a_2|=|b_1-b_2|)=\frac 1 {100}+\sum_{k=1}^9 \left( 2\frac{10-k}{100} \right)^2=12.4\%$

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    $\begingroup$ sorry but this is not the right answer :( $\endgroup$ – Oray Feb 27 '16 at 14:00
  • $\begingroup$ @Oray, so each person selects numbers whose difference isn't 0? $\endgroup$ – humn Feb 27 '16 at 14:12
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    $\begingroup$ @hmmn they share results of abs difference with each other only, not the numbers. the absolute difference cannot be 0 since the numbers they have to choose are distinct numbers from each other. $\endgroup$ – Oray Feb 27 '16 at 14:19

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