I'm trying to calculate the probability of guessing 3 numbers on a set of 5. The range is $[1, 6]$, the order is not relevant and numbers can repeat.
I did a small script to calculate it by brute force, and I got something unexpected. Choosing 3 identical numbers (e.g. 1, 1, 1) has a lower probability of drawing than 2 identical numbers, and this is also lower than 3 different numbers. I can't figure out why.
On
Guessing $1,1,1$ will match the sets that have $3$ or more $1$. There is $1$ set of $5$ ones, there are $5*5=25$ sets that have $4$ ones ($5$ positions for the non-one to be in and $5$ options for the value of the non-one), there are ${5\choose 2}*5 = 50$ ways to have $3$ ones and two of another value. And ${5\choose 2}5*4 = 200$ ways to have $3$ ones and two other values. This is $255$ ways.
Guessing $2,2,1$ will match sets that have $4$ twos and $1$ one: $5$ sets. $3$ twos and $2$ ones: ${5 \choose 2} = 10$ sets. $3$ twos, $1$ one and something else: ${5\choose 3}{2\choose 1}*4 = 80$ sets. $2$ twos, and $3$ ones: ${5\choose 2} = 10$ sets. $2$ twos, $2$ ones, and something else: ${5\choose 2}{3\choose 2}*4 = 120$ sets. $2$ twos, $1$ one and $2$ of something else: ${5\choose 2}*{3\choose 1}*4 = 120$ sets. $2$ twos, $1$ one and two other things: ${5\choose 2}{3\choose 1}4*3 = 360$. Or $360+120+120+10+80+10+5 = 705$ sets. Many more options!
Guessing $1,2,3$ can mean.... well, three of one of the numbers and one of the other two: $3*{5\choose 3}*2! = 60$( three choices for the triple, choices for where to put the triples, ways to place the remaining two). Or two of two of them and one of then third: $3{5\choose 2}{3\choose 2}= 90$ (Three choices for the single, where to place the first of te doubles, where to place the other doubles.) Two of one of them, one of the other two, and forth number. $3{5\choose 2}{3\choose 1}{2\choose 1}*3 = 540$. (3 chooses for the double, where to put them, where to put the singles, and what the fourth number is. Or exaclty one of each. ${5\choose 3}*3!*(3)^2 = 540$ (places to put the three, order to place the three within their posistions, 9 posible value of the remaining two numbers.) For $60 + 90 +540+540 = 1230$ possible ways.
OK, let's calculate. If you choose $3$ identical numbers (say $1,1,1$), then you are right when the 5 numbers are:
A. those three $1$'s plus two non-$1$'s: ${5 \choose 3}=10$ 'spots' for your three $1$'s and $5 \cdot 5 = 25$ possibilities for the other two, for a total of $250$
B. you get four $1$'s and one non-$1$: ${5 \choose 4} \cdot 5=25$ possibilities
C. All five are a $1$: $1$ possibility
Now compare this with $1,1,2$. You get this when:
A. you get $1,1,2$, and two numbers that are neither $1$ nor $2$: $5 \cdot {4 \choose 2} = 30$ 'spots for your three numbers times $4 \cdot 4=16$ other numbers, make $480$ possbilities
... OK, we already have more possibilities for this to happen!
OK, and we also see the basic reason why: with $1,1,2$ you get this extra factor of $3$, as there are three ways for your $3$ 'hits' to occur: you first get a $1$, then another $1$, and then a $2$, or first a $1$, then a $2$, and then a $1$, or first a $2$, and then a $1$, and then the last $1$. With three $1$'s, you have no such possible permutations.
And this also explains why picking three different numbers is even better: Now you get $6$ permutations, i.e. $6$ different orders in which your 'hits' can occur.