Consider the following tables:

enter image description here

Given a number of independent trials $n=2,3,4\ldots$ each of these tables represents the list of all the possible $\binom{k+n-1}{n}=\binom{n+2}{n}=6,10,15\ldots$ ways (rows) in which a certain random process may occur (downward arrows) in correspondence of $k=3$ distinguishable and equiprobable states (here defined as three colors: red, green and blue).

In particular, I underline the following events:

  • $L$: "To get at least one red state and at least one blue state". This event occurs in $\binom{n}{2}=1,3,6\ldots$ different cases (half-red-half-blue dots on the left of each table).

  • $E$: "To get always the green state", which occurs only once for each $n$ (green dot on the right of each table).

The problem is to

Find a physical system such that the ratios $$p_L=\frac{\binom{n}{2}}{\binom{n+2}{n}} \text{ and } p_E=\frac{1}{\binom{n+2}{n}} $$ represent the probabilities of two events, with the above characteristics, related to that system.

The key should be that the order of the "trials" does not affect the occurrence of these events (I have the suspect that there should be some game which is described by such probabilities, but I don't know which one).

The context and the motivation of this problem can be found in this post, together with many useful and interesting comments.

Thanks for your suggestions!


Does this game work?

Shuffle a deck of $n+2$ cards numbered $1$ to $n+2$ and deal out two cards.
Let $L$ be the event that neither $1$ nor $n+2$ is drawn.
Let $E$ be the event that both $1$ and $n+2$ are drawn.

Edit In order to see the process more clearly, say that you have a bag with $n+2$ marbles, $n$ of which are white, while two of them are black. You shuffle the bag and dump the marbles into a linear trough, so they are lined up in a random order. The two black marbles function as dividers; the white marbles to the left of the left divider is considered red, the white marbles between the red dividers are considered green, and the white marbles to the right of the right divider are blue.

Note that this is the same as the card process described above; we are just shuffling a deck of marbles instead of cards.

Perhaps what you want is more of a random process where the colors are chosen one at a time. To do this, you need buckets and some "magnetic" marbles. Specifically, there is a red, blue and green bucket. One by one, you randomly throw a marble in one of the buckets. However, the probability of each bucket is uneven and increases as it gets more marbles. Specifically, if the red, green and blue buckets each have $r,g$ and $b$ marbles, then the probability the next marble falls in the red, green or blue bucket is respectively $$ \frac{r+1}{r+b+g+3},\qquad \frac{g+1}{r+b+g+3},\qquad \frac{b+1}{r+b+g+3}.$$

This also produces the same distribution.

  • $\begingroup$ Thanks Mike! I have some difficulty to see the three equiprobable states, which are represented with the three colors in the tables. $\endgroup$ – user559615 Aug 7 '18 at 23:18
  • $\begingroup$ Good question. If you draw cards numbered $p$ and $q$, where $p<q$, then this corresponds to the situation where there are $p-1$ Red marbles, $q-p-1$ Green marbles, and $n+2 - q$ Blue marbles. Adding up these number, you do get $n$ marbles total, and every distribution of marbles is equally likely. $\endgroup$ – Mike Earnest Aug 7 '18 at 23:23
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    $\begingroup$ $p$ and $q$ are random, so of course the numbers of marbles of each color are sometimes different. Look at the second row in the $n=3$ case of the table you provided. There, there are 2 red, 1 blue and 0 green marbles, which are all different numbers. $\endgroup$ – Mike Earnest Aug 7 '18 at 23:33
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    $\begingroup$ @AndreaPrunotto (1) Without replacement. (2) Sort of. There is a process with marbles, but It doesn’t quite make physical sense. I’ll edit $\endgroup$ – Mike Earnest Aug 11 '18 at 17:07
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    $\begingroup$ @AndreaPrunotto See edits $\endgroup$ – Mike Earnest Aug 13 '18 at 17:03

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