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I played Rummikub with my family last night and we couldn't figure out how to calculate the probability of drawing 4 consecutive numbers of same colors in the initial draw. would appreciate help..

Rummikub is a Rummy-style game played with tiles. There are 2 sets of 4 colored tiles numbered 1 through 13, plus 2 jokers for a total of 106 tiles.

The game starts with each player drawing 14 tiles.

what is the probability that a player will have at least 1 set of 4 consecutive numbers of the same color ?

I assume the hypergeometric distribution should be used, but I'm not sure how to apply it.

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Computing the exact probability will be complicated. It's easier to compute the expected number of such sets.

Consider a particular set, let's say red 1234. The probability that you get $n_1$ red 1's, $n_2$ red 2's, $n_3$ red 3's and $n_4$ red 4's is $$ \dfrac{{2 \choose n_1} {2 \choose n_2} {2 \choose n_3} {2 \choose n_4}{98 \choose 14 - n_1 - n_2 - n_3 - n_4}}{106 \choose 14}$$ If $n_1 = n_2 = \ldots = n_4 = 2$, you have two sets of red 1234, otherwise if all $n_i \ge 1$ you have one. Thus the expected number of red 1234's is $$\eqalign{ \sum_{n_1=1}^2 \sum_{n_2=1}^2 \sum_{n_3=1}^2 \sum_{n_4 = 1}^2 &\dfrac{{2 \choose n_1} {2 \choose n_2} {2 \choose n_3} {2 \choose n_4}{98 \choose 14 - n_1 - n_2 - n_3 - n_4}}{106 \choose 14} + \dfrac{{2 \choose 2}{2\choose 2}{2\choose 2}{2 \choose 2}{98 \choose 6}}{106 \choose 14}\cr &={\frac{1850962}{702982725}} }$$ There are $4 \times 10 = 40$ possible sets, so the expected number of sets is $$40 \times \frac{1850962}{702982725} = \frac{14807696}{140596545} \approx 0.10532$$

The probability of having at least one set is less than this, but probably not much less because it's rather rare to have multiple sets.

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  • $\begingroup$ thank you for the quick response. $\endgroup$ – KMARSAL Aug 22 '17 at 2:26

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