I've got a deck of 36 cards... 3 are labeled with #1, 3 with #2, and so on up to 3 with #12. I'm drawing 3 cards from the deck & wondering what the probability is that I will end up with NONE of the cards being within 2 of one another (including wrap-arounds--meaning a #12 is allowed to wrap-around to a #1). For example, {1,3,5} or {4,10,12} would qualify, but {10,12,1} would not.
I know that it's 38.9% WITH replacement as I actually drew the whole thing out on a spreadsheet with every possible outcome (and got 56/144), but I know that the number will increase slightly WITHOUT replacement, and I'm just not sure how to come up with the formula to figure it out.
I came up with this formula WITH replacement (using the inverting method)... 1 - {9/36 + (27/36 * 9/36) + (25/36 * 9/36)}, but I don't know how to adjust it for WITHOUT replacement. And I may have the wrong formula altogether. I'm new at this.
Is there anybody out there who wouldn't mind helping me out on this one? I've been at it for weeks with no luck!
With replacement
With replacement, you can choose any card you like as the first card. After that there are three cases:
The three cases are mutually exclusive, so altogether the chance of a bad set is found by adding the three cases together, $$ \frac{9}{36} + \left(\frac{6}{36}\right)\left(\frac{15}{36}\right) + \left(\frac{21}{36}\right)\left(\frac{18}{36}\right) = \frac{11}{18}, $$ and therefore the probability of a good set is $1 - \frac{11}{18} = \frac{7}{18} \approx 0.38889.$
Without replacement
Without replacement, there are only $8$ ways out of $35$ for the first case to occur; the second case occurs for $6$ of $35$ second cards but only $13$ of $34$ third cards; and the third case occurs for $21$ of $35$ second cards but only $16$ of $34$ third cards. Now the sum of the three cases is $$ \frac{8}{35} + \left(\frac{6}{35}\right)\left(\frac{13}{34}\right) + \left(\frac{21}{35}\right)\left(\frac{16}{34}\right) = \frac{49}{85}, $$ and the probability of a good set is $1 - \frac{49}{85} = \frac{36}{85} \approx 0.423529.$
Without replacement - alternative method
A second way to compute the probability without replacement uses the fact that a good set cannot contain two matching numbers. That is, you must choose three different numbers from the set $\{1, 2, 3, \ldots, 12\}$ without choosing two numbers that would be adjacent on a clock face.
For the case where $12$ is one of the numbers chosen, you can choose any two distinct numbers $x < y$ from the set $\{1, 2, 3, \ldots, 8\}$; then use $x + 1$ as your first number (so $2\leq x\leq 9$ and $x$ is not too close to $12$), and $y+2$ as your second number (so $x + 1 < y \leq 10$ and $y$ is not too close to either $x$ or $12$). There are $\binom82$ choices of $x$ and $y$ in this case.
To cover all other cases you can choose any three distinct numbers $x < y < z$ from the set $\{1, 2, 3, \ldots, 9\}$; then use $x$ as your first number, $y + 1$ as the second number, and $z + 2$ as the third number; then $x + 1 < y < z - 1$, so the numbers are not too close in sequence, and the $12,1$ case also is eliminated. There are $\binom93$ choices of $x$, $y$, and $z$ in this case.
Note that for any set of three distinct numbers that does not include "adjacent" numbers (that is, for every good set), you can reverse the procedures in the previous two paragraphs in order to express the set either as $\{x,y+1,12\}$ or as $\{x,y+1,z+2\}.$ That is, the combinations in the preceding two paragraphs generate every good set of numbers once and only once.
So there are a total of $\binom82 + \binom93 = 28 + 84 = 112 = 2^4 \times 7$ good sets of three numbers.
For each good set of three numbers, there are three choices of card for each number, $3^3 = 27$ choices altogether.
That means there are $112 \times 27 = 3024 = 2^4 \times 3^3 \times 7$ good sets of three cards.
But there are $\binom{36}{3} = \frac{36\times 35\times 34}{3\times 2\times 1} = 7140$ ways to choose three cards without replacement from your $36$ cards.
As a result, the chance of drawing a good set of cards is
$$ \frac{3024}{7140} = 2^4\times 3^3\times 7\times \frac{3\times 2\times 1}{36\times 35\times 34} = \frac{36}{85}. $$
Without replacement - second alternative
Another way to count the number of "good" sets of three numbers chosen from $\{1, 2, 3, \ldots, 12\}$ is to choose any of those $12$ numbers first. This leaves a set of $9$ numbers that are consecutive around a clock face; choose $x < y$ from the set $\{1, 2, 3, \ldots, 8\}$, then choose the $x$th and $y+1$st of the $9$ consecutive numbers as the other two members of your set of three numbers; there are $\binom82 = 28$ choices for that step. This is guaranteed not to produce any two numbers that are adjacent around a clock face, but it produces each set three times (because any of the three numbers in the set can be the first one chosen). So we have a total of $$ \frac{12 \times 28}{3} = 112 $$ possible "good" sets of three numbers. For the rest of the calculation, follow the part of "without replacement - alternative method" after the step where we find that there are $112$ good sets of numbers.
With replacement - alternative method
With replacement, you are really just doing three repetitions of choosing a number uniformly from the set $\{1, 2, 3, \ldots, 12\}.$ Each number has a $\frac1{12}$ chance of being chosen each time, independent of the other two choices. So we can say the second card "ruins" the set in just $3$ ways out of the $12$ possible choices of the second card. Another way to put this is, when working out the probabilities with replacement, instead of writing $\frac{9}{36}$ we can write $\frac{3}{12}$, instead of writing $\frac{15}{36}$ we can write $\frac{5}{12}$, and so forth.