I was a street game on TikTok. I don't know the answer. I am posting it here so, I can get some thoughts from experts.

There are 24 colored small marbles. Eight marbles are in red, another eight marbles are in green, and the remaining eight are in blue. The dealer put these 24 marbles in a bag. You randomly grab 12 marbles (you can grab them in once or several times, but the total number you grabbed is 12).

The dealer wins if the number of marbles in three colors are 3, 4, and 5. For example, 3 red, 4 blue and 5 green marbles (or 5 red, 4 green, 3 blue). The order is does not matter. So you will pay 30$ to the dealer.

In any other cases, the dealer will pay you 10$. For example, if you get 8 red, 4 blue, 0 green, you win.

I want to calculate the probability of getting 3,4,5.

Some thoughts:

  1. How many unique cases (combination of numbers)? For example, (3r,4b,5g) is the same as (5r,4b,3g). I found 19 (Maybe I am wrong). How to calculate the number of unique cases using some formula or expressions?
  2. What is the probability of getting k,m,n? $p(k,m,n) = ?$
  3. If the game is true than $p(3,4,5)$ is larger than other possibility.

The question: $p(k,m,n) = ?$

  • 1
    $\begingroup$ It's not clear what you mean by "unique cases". If all marbles of the same color are indistinguishable then there are only $6$ ways to get the winning pattern. If, for computation, we are treating all the balls as unique there are $6\times \binom 83 \times \binom 84\times \binom 85$. $\endgroup$
    – lulu
    Commented Apr 1 at 22:30
  • $\begingroup$ Yes, the same colored marbles are indistinguishable. For example, the case (8, 4, 0) includes (8r, 4g, 0b), (8g,4b,0r)... So you get it right. $\endgroup$
    – Ablet Imin
    Commented Apr 2 at 2:43
  • $\begingroup$ But the number you got is for the case(3,4,5). The number of unique cases means (3,4,5), (0,4,8), (2,4,6),....etc, I am counting how many (k,m,n) exist. $\endgroup$
    – Ablet Imin
    Commented Apr 2 at 2:54
  • $\begingroup$ I get that the probability is about 48.7%, so the game's expectation is negative. $\endgroup$ Commented Apr 2 at 7:00
  • $\begingroup$ @ParclyTaxel that probability is right, but the expectation is positive since the dealer gets a much higher payout. $\endgroup$
    – lulu
    Commented Apr 2 at 7:38

1 Answer 1


To summarize the discussion in the comments:

For problems like this, it's best to regard all the marbles as distinct. You can, of course, treat all marbles as indistinct up to color, but in that case you have to deal with the fact that the various results are not equally probable. $\{3,4,5\}$ is considerably more probable than $\{6,6,0\}$ or even $\{4,4,4\}$ as the latter can only be obtained in one way, while the former can be obtained in six.

If the marbles are thought of as distinct, there are $\binom {24}{12}$ ways to choose $10$ of them.

There are $6\times \binom 85\times \binom 84\times \binom 83$ ways to choose $5$ of one color, $4$ of another, and $3$ of the third. The answer we seek is the ratio $$6\times \binom 85\times \binom 84\times \binom 83\Bigg / \binom {24}{12}\approx \boxed {.487}$$

Note that this makes the game very attractive to the dealer, as the dealer's expected payout is $$30\times .487-10\times (1-.487)\approx 9.483$$

As a side note, a similar computation shows that the probability of getting $\{4,4,4\}$ is $$ \binom 84\times \binom 84\times \binom 84\Bigg / \binom {24}{12}\approx \boxed {.1268}$$ Much smaller, as claimed. Of course, collections like $\{6,6,0\}$ or $\{8,2,2\}$ are much less probable still.

  • $\begingroup$ Nice calculation, thanks! There is a way to increase the dealer's expectations. Instead of using all 19 cases, the dealer chooses (3,4,5) and the other 9 cases with the most negligible probability, like (8,4,0), (8,2,2),(7,5,0), .... In this way, the dealer's chance of winning will be higher than 0.487. This strategy is used in the game, as I saw on TikTok. $\endgroup$
    – Ablet Imin
    Commented Apr 2 at 12:46

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