A game similar to Roulette has 360 slots. 20 of these slots are 'scoring' positions. Each slot will only accomodate 1 ball.

If a number of balls are introduced (Say 3)

What are the chances that any 1 of these 3 balls will settle into a scoring slot What are the chances that any 2 of these 3 balls will settle into scoring slots What are the chances that all 3 of these 3 balls will settle into scoring slots

I am hoping somebody might take the time to demonstrate the equation used to determine the above and perhaps also give a brief explaination and use the statistical phraseology which might describe the mathematics.

Many thanks

  • $\begingroup$ If this homework, you should probably add the self-study tag in order to get better guidance. $\endgroup$
    – chl
    Nov 9, 2020 at 11:13
  • $\begingroup$ No this is not homework - this is a real world archaeology question which I have presented in the format of a game for simplification. If it were homework, I likely would have been given some idea of how to approach this, but I have no mathematical or statistical skills beyond defining the question. $\endgroup$ Nov 9, 2020 at 12:08
  • $\begingroup$ Does "any 1 of these" mean "only one of these"? i.e. an arbitrary ball scores, others not. $\endgroup$
    – gunes
    Nov 9, 2020 at 15:24
  • $\begingroup$ Any ball in a scoring slot is a score. One ball per slot. What is the chance of achieving a single score when all balls are in play. What are the chances of achieving multiple scores when all balls are in play $\endgroup$ Nov 9, 2020 at 16:14
  • 1
    $\begingroup$ Do you want to share something closer to the actual archelogy question, just to make sure your 'game' has exactly the same probability structure? $\endgroup$
    – BruceET
    Nov 10, 2020 at 4:50

1 Answer 1


The number $X$ of balls in scoring positions follows a hypergeometric distribution. There are 20 scoring positions; 340 non-scoring positions.

In R, we can make a PDF table of the distribution (rounded to five places) as shown below. (You can ignore row numbers in brackets [ ].)

x = 0:3;  pdf = round(dhyper(x, 20,340, 3), 5)
cbind(x, pdf)
     x     pdf
[1,] 0 0.84201
[2,] 1 0.14947
[3,] 2 0.00838
[4,] 3 0.00015

Here is the computation for $P(X = 2)$ in terms of binomial coefficients:

$$P(X = 2) = \frac{{20\choose 2}{340\choose 1}}{{360\choose 3}} = 0.008377.$$

choose(20, 2)*choose(340, 1)/choose(360,3)
[1] 0.008377295
[1] 0.008377295

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