Out of $50$ consecutive numbers, what is the probability that the absolute difference between the two numbers is $10$ or less?

Question

Fifty tickets numbered with consecutive integers are in a jar.

• Two are drawn at random and without replacement.
• What is the probability that the absolute difference between the two numbers is $10$ or less ?.
• Express your answer as a common fraction.

This was from mathcounts, and their answer was $\dfrac{89}{245}$, which I think is wrong.

it should be $\dfrac{30\cdot20 + 19\cdot10}{\binom{50}{2}}=\dfrac{790}{1225}$, can be simplified further.

but the basic idea should be including both negative and positive differences, not only one side. Thanks !.

Answer 1

$$\frac{10+11+\cdots+19+20+20+\cdots+20+20+19+\cdots+11+10}{50\cdot49}=$$$$\frac{10\cdot29+30\cdot20}{50\cdot49}=\frac{89}{245}$$

It is the summation $$\sum_{n=1}^{50}P(|X_1-X_2|\leq10\mid X_1=n)P(X_1=n)$$

Answer 2

When picking 1, 10 chances to pick the second.

When picking 2, 10 chances, as 1 will not be considered as that was already included in the last one.

When picking 3, 10.

.

.

.

.

When picking 40, 10.

When picking 41, 9.

When picking 42, 8.

.

.

When picking 49, 1.

Total = 10(40) + 9 + 8 + ....1 = 445

445/1225 will be the answer which is 89/245.

Now, why both positive and negative differences won't be included? Because in your 50C2, you're only bothering with the selection of the tickets and not the order in which they came in. If you pick {1,2} or {2,1}, it will be counted only once in your sample space but twice in your event probability which will give us the wrong answer.

Answer 3

Consider all $2$-element subsets of $50$ consecutive integers—there are $_{50}C_2$ of these. Without loss of generality, let's make these integers $\{0,1,\ldots,49\}$.

• In how many two-element subsets does the greater number exceed the lesser by exactly one? Well, there's $\{\{0,1\},\{1,2\},\ldots,\{48,49\}\}$, so we see there are $(48-0)+1=50-1$ such subsets.
• In how many two-element subsets does the greater number exceed the lesser by exactly two? Well, there's $\{\{0,2\},\{1,3\},\ldots,\{47,49\}\}$, so we see there are $(47-0)+1=50-2$ such subsets.
• ...
• In how many two-element subsets does the greater number exceed the lesser by exactly ten? Well, there's $\{\{0,10\},\{1,11\},\ldots,\{39,49\}\}$, so we see there are $(39-0)+1=50-10$ such subsets.

In how many two-element subsets does the greater number exceed the lesser by at most ten? From above, count is given by $(50-10)+(50-9)+\cdots+(50-1)$. By the summation formula for arithmetic series, we see that there are $\tfrac{1}{2}(10)(50-1+50-10)=\tfrac{1}{2}(10)(2\cdot 50-10-1)$.

Thus the desired probability is $$\tfrac{1}{2}(10)(2\cdot 50-10-1)/_{50}C_2\text{.}$$

Answer 4

Let $a$ be the first number that is picked. Except the numbers $1,2,3,4,5,6,7,8,9,10$ and $40,41,42,43,44,45,46,47,48,49,50$ there are $20$ numbers in the range $[a-10,a+10]$. For the exceptions, there are $10,11,12,13,14,15,16,17,18,19$ and $19,18,17,16,15,14,13,12,11,10$ numbers in the range of $[1,50]$, respectively. Therefore the answer is $$\frac{2(10+11+12+...+19)+30\cdot20}{50\cdot49} = \frac{290+600}{2450} = \frac{89}{245}$$

Answer 5

It isn't completely clear how you got the numerator of $$\dfrac{30\cdot20 + 19\cdot10}{\binom{50}{2}},$$ but if that expression is supposed to be number of successful outcomes divided by total number of outcomes, the denominator seems to contradict your claim that you are considering both positive and negative differences.

For example, the absolute difference between $20$ and $25$ is $5.$ In order for those two tickets to produce both of the differences $5$ and $-5$ (so that both are counted in the numerator), we must count two different events in the denominator for those two tickets: $25 - 20 = 5$ for one event, and $20 - 25 = -5$ for the other. Whether we get $5$ or $-5$ depends on the order in which we list the two tickets when we do the subtraction.

But when we are counting the ways to draw two tickets from $50$ tickets, $\binom{50}{2}$ is the number of ways to draw the tickets ignoring the order of the two tickets. That is, "$20$ then $25$" and "$25$ then $20$" are counted as a single event in the denominator. If we want to get a correct result, the numerator likewise must count only one way to get either $5$ or $-5$ as the difference between those two tickets.

It appears, then, that you have mistakenly double-counted some events in your numerator. It also appears you have undercounted some events, since merely doubling the denominator gets your result close to but not equal to the correct answer.

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Update: I see now that this answer was already summarized at the bottom of another answer, but I'll leave it up since it's already written and it fleshes out some details.

Answer 6

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{\sum_{s = 1}^{10}\sum_{n_{1}\ =\ nMin}^{nMin + 49} {1 \over 50}\sum_{n_{2}\ =\ nMin}^{nMin + 49}{1 \over 49} \bracks{n_{2} \not= n1}\bracks{\verts{n_{2} - n_{1}} = s}}} \\[5mm] = &\ {1 \over 2450}\sum_{s = 1}^{10}\sum_{n_{1}\ =\ 0}^{49} \sum_{n_{2}\ =\ 0}^{49}\bracks{\verts{n_{2} - n_{1}} = s} = {1 \over 2450}\sum_{s = 1}^{10}\bracks{z^{s}}\sum_{n_{1}\ =\ 0}^{49} \sum_{n_{2}\ =\ 0}^{49}z^{\verts{n_{2} - n_{1}}} \\[5mm] = &\ {1 \over 2450}\sum_{s = 1}^{10}\bracks{z^{s}}\pars{% \sum_{n_{2} = 0}^{49}z^{n_{2}} + \sum_{n_{2} = 0}^{49}z^{49 - n_{2}} + \sum_{n_{1}\ =\ 1}^{49}\sum_{n_{2}\ =\ 0}^{49}z^{\verts{n_{2} - n_{1}}}} \\[5mm] = &\ {1 \over 2450}\sum_{s = 1}^{10}\bracks{z^{s}}\bracks{% 2\sum_{n_{2} = 0}^{49}z^{n_{2}} + \sum_{n_{1}\ =\ 1}^{48}\pars{% \sum_{n_{2}\ =\ 0}^{n_{1} - 1}z^{n_{1} - n_{2}} + \sum_{n_{2}\ =\ n_{1} + 1}^{49}z^{n_{2} - n_{1}}}} \\[5mm] = &\ {1 \over 2450}\sum_{s = 1}^{10}\bracks{z^{s}}\bracks{% 2\,{z^{50} - 1 \over z - 1} + \sum_{n_{1}\ =\ 1}^{48}\pars{% z^{n_{1}}{z^{-n_{1}} - 1 \over 1/z - 1} + z\,{z^{49 - n_{1}} - 1 \over z - 1}}} \\[5mm] = &\ {1 \over 2450}\sum_{s = 1}^{10}\bracks{z^{s}}{1 \over 1 - z}\braces{% 2 - 2z^{50} + z\sum_{n_{1}\ =\ 1}^{48}\bracks{% \pars{1 - z^{n_{1}}} + \pars{1 - z^{49 - n_{1}}}}} \\[5mm] = &\ {1 \over 2450}\sum_{s = 1}^{10}\bracks{z^{s}}{1 \over 1 - z}\bracks{% 2 - 2z^{50} + 2z\sum_{n_{1}\ =\ 1}^{48}\pars{1 - z^{n_{1}}}} \\[5mm] = &\ {1 \over 1225}\sum_{s = 1}^{10}\bracks{z^{s}}{1 \over 1 - z}\bracks{% 1 - z^{50} + z\pars{48 - z\,{z^{48} - 1 \over z - 1}}} \\[5mm] = &\ {1 \over 1225}\sum_{s = 1}^{10}\bracks{z^{s}}\bracks{% {1 + 48z - z^{50} \over 1 - z} + {z^{50} - z^{2} \over \pars{1 - z}^{2}}} = {1 \over 1225}\sum_{s = 1}^{10}\bracks{z^{s}}\bracks{% {1 + 48z \over 1 - z} - {z^{2} \over \pars{1 - z}^{2}}} \\[5mm] = &\ {1 \over 1225}\sum_{s = 1}^{10}\bracks{z^{s}} \sum_{n = 0}^{\infty}\bracks{\pars{1 + 48z}z^{n} - z^{2}{-2 \choose n}\pars{-z}^{n}} \\[5mm] & = {1 \over 1225}\sum_{s = 1}^{10}\bracks{z^{s}} \sum_{n = 0}^{\infty}\bracks{z^{n} + 48z^{n + 1} - \pars{n + 1}z^{n + 2}} \\[5mm] = &\ {1 \over 1225}\sum_{s = 1}^{10}\bracks{z^{s}} \bracks{\sum_{n = 0}^{\infty}z^{n} + 48\sum_{n = 1}^{\infty}z^{n} - \sum_{n = 2}^{\infty}\pars{n - 1}z^{n}} \\[5mm] = &\ {1 \over 1225}\bracks{\sum_{s = 1}^{10}1 + \sum_{s = 1}^{10}48 - \sum_{s = 2}^{10}\pars{s - 1}} = \bbx{89 \over 245} \approx 0.3633 \end{align}