I was trying the following puzzle from this link :

You randomly withdraw 3 balls out of a lot containing identical red, identical green and identical blue balls. What is the probability that you get 1 red, 1 green and 1 blue ball given the number of

a) Red balls = 3, Green balls = 3, Blue balls = 3
b) Red balls = p, Green balls = q, Blue balls = r such that p,q,r>3
c) Red balls = infinity, Green balls = infinity, Blue balls = infinity

My solution for the puzzle is as follows:
Number of ways of selecting ‘r’ things from ‘n’ identical things is ‘1’.
Therefore, first answer would be 1.1.1/(number of solutions of x+y+z=3, where x,y,z>=0) i.e. 1/10
For second part : 1/10, since p,q,r > 3
For third part : 1/10. Same logic.

But if I apply the same logic to solve : "What is the probability to select a red ball out of a lot of 1 red ball and infinite blue balls?", my answer is 0.5 which intuitively is wrong(It should be 0).

Please tell me where am I wrong.

  • $\begingroup$ Unless you are doing quantum mechanics, "identical" things don't exist. $\endgroup$ – celtschk Sep 22 '12 at 20:26
  • $\begingroup$ You case c) is undefined, since it does not tell you the probability that selecting one ball it will be red respectively green or blue. Contrary to the other cases you cannot assume a uniform probability: such a distribution does not exist; so some balls are necessarily more likely to be picked than others! $\endgroup$ – Marc van Leeuwen Sep 25 '12 at 8:00

Other answers have explained how to work the problem correctly, but not what’s wrong with your solution. You correctly calculated that there are $10$ distinguishable combinations of colors, all of which are possible in every version of the problem, and observed that only one of them contains one ball of each color. If the $10$ different combinations were equally likely to be drawn, your answer would be correct. However, they are not equally likely to be drawn.

This is probably easiest to see in the first version, in which we have specific numbers of balls of each color. Imagine that the nine balls have invisible labels: the red balls are labelled $1,2$, and $3$, the green balls are labelled $4,5$, and $6$, and the blue balls are labelled $7,8$, and $9$. You reach into the box and without looking grab three balls. If the numbers were suddenly made visible, you’d find that you’d chosen a $3$-element subset of $\{1,2,3,4,5,6,7,8,9\}$. Each of these subsets is equally likely to be chosen. Thus, the probability of getting one of your $10$ color combinations can be found by dividing the number of subsets representing that combination by the total number of subsets, which is $\binom93=84$.

The all-red combination, for instance, is obtained only when you draw the set $\{1,2,3\}$, so the probablity of getting this color combination is only $\frac1{84}$. If all $10$ of the possible combinations were equally likely, the probabilities of getting them would add up to only $10\cdot\frac1{84}=\frac5{42}$, which is absurd. Thus, they cannot be equally likely, and this fact immediately invalidates your reasoning.

You can of course go on to use this approach to calculate the correct probability. In order to get one ball of each color, you must get a set containing exactly one member of $\{1,2,3\}$, one member of $\{4,5,6\}$, and one member of $\{7,8,9\}$. There are $3^3=27$ such sets, so the probability of getting balls of all three colors is $\frac{27}{84}=\frac9{28}$.

Similarly, you can calculate that the probability of getting two red balls and one green ball is $\frac9{84}=\frac3{42}$, since there are $\binom32\binom31=9$ $3$-elements subsets of $\{1,2,3,4,5,6,7,8,9\}$ that result in that color combination. Each of the other two-color combinations of course has the same probability.

This is enough to show that you can’t safely assume that the different combinations of colors are equally likely; we don’t have to demonstrate it anew for each of the other versions of the problem.


In this case, for infinite balls it seems you should consider the chance of any given color to be 1/3, and not to change as balls are drawn. The chance of one of each would be $1$(you can draw any one first)$\cdot \frac23$ (can't match the first)$\cdot \frac 13=\frac 29$

  • $\begingroup$ I'd rate this answer below 52.8k. $\endgroup$ – Christian Blatter Sep 22 '12 at 21:13

Ross Millikan's elegantly simple answer works whenever the number of each ball is the same. The same approach to version A of your questions yields: $1\cdot \tfrac{6}{8}\tfrac{3}{7}$, or $\tfrac{9}{28}$.

A couple of less elegant approaches work in more cases.


DETERMINE THE PROBABILITY OF A REPRESENTATIVE FAVORABLE OUTCOME. This works when every favorable outcome is equally likely. Let's determine the probability that you will draw first red, then green, then blue. The probability that you will draw red on your first draw is $$ \frac{p}{p+q+r}. $$ If you draw red first, the probability that you will next draw blue is $$ \frac{q}{p+q+r-1}. $$ If you draw red and blue on your first two draws, the probability that you will next draw green is $$ \frac{r}{p+q+r-2}. $$ The probability that you will draw first red, then green, then blue is the product of these probabilities: $$ \frac{pqr}{(p+q+r)(p+q+r-1)(p+q+r-2)}. $$ You can also express this as $$ \frac{pqr}{(p+q+r)!/(p+q+r-3)!}. $$ MULTIPLY THIS PRODUCT BY THE NUMBER OF EQUALLY LIKELY FAVORABLE OUTCOMES. In this case there are $3!$ or $6$ equally likely favorable outcomes; bgr, brg, gbr, grb, rbg, rgb. So the probability is $$ \frac{6pqr}{(p+q+r)(p+q+r-1)(p+q+r-2)}. $$ In this case, that's $\tfrac{6\cdot 3\cdot 3\cdot 3}{9\cdot 8\cdot 7}$ or $\tfrac{9}{28}$.


This method requires basic knowledge of combinatorics, which I'm going to assume. Count the number of ways to get one of each color: $p$ ways to get red, $q$ ways to get blue, $r$ ways to get green gives you $pqr$ ways to get one of each color. Count the number of ways to choose $x$ different balls out of $p+q+r$ balls irrespective of order. That's $$ \frac{(p+q+r)!}{(p+q+r-x)!x!}. $$

Divide the number of favorable outcomes by the total number of outcomes, $$ \frac{pqr}{ (p+q+r)!/[(p+q+r-x)!x!]}. $$ In this case that means $\tfrac{3\cdot 3\cdot3}{9! / (6!3!)}=\tfrac{9}{28}$.

  • $\begingroup$ Is there a difference between picking all 3 balls simultaneously and picking them one by one ? $\endgroup$ – damned Sep 22 '12 at 21:02
  • $\begingroup$ No difference in the result, damned. Sometimes it's easier to think about it as a series of decisions, though. $\endgroup$ – Michael Schwartz Sep 22 '12 at 21:08
  • $\begingroup$ You are trying to grab $3$ balls. Does it make any difference whether your hand touches the three balls simultaneously, or with a delay of a few microseconds? $\endgroup$ – André Nicolas Sep 22 '12 at 21:10

It can not be 1/10, because if then the answer will be 1/10 even if there are 3R 10B and 20W balls. If you think, it's not possible that the probability remains same, for increasing numbers of balls of certain colours only. Hence 3C1*3C1*3C1/9C3=9/28 is correct. AND probability of getting same combination in three turns without replacement will be different=3/9*3/8*3/7=3/56 with replacement in three turns it shall be=3/9*3/9*3/9=1/27


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