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We need to find the probability that all the red balls are exhausted before the white or black balls are exhausted. So, we can still pick white or black balls, but we cannot pick ALL of the white/black balls.

I tried going case by case but quickly realized there are way too many cases for that. I'm not very good at permutations and combinations so I can't figure out how to get in all the cases.

Few example cases can be: R R R W B W B R, B B B W B R R R R, etc.

TheMultiRounderGamer
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    Who cares that you have $3$ white and $5$ black balls? You have $4$ red and $8$ other coloured balls, am I right? – Dominique May 07 '24 at 08:35
  • So we can exhaust white balls as long as black balls are exhausted after red ones? Or do you want the red balls to be the first colour of the three to be exhausted? – Julio Puerta May 07 '24 at 08:39
  • @Dominique it matters because you can't pick all the 8 coloured balls first, but you CAN pick 4 black and 2 white balls – TheMultiRounderGamer May 07 '24 at 08:41
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    @JulioPuerta yes, the red balls have to be the first color exhausted – TheMultiRounderGamer May 07 '24 at 08:41
  • So we cannot pick all the black balls, then all the red and all the white? – Julio Puerta May 07 '24 at 08:42
  • So your question becomes: "Given $4$ red, $3$ white and $5$ black balls. You pick balls one after another without replacement until one colour is exhausted. What is the probability that that exhausted colour is red?" – Dominique May 07 '24 at 08:44
  • No we cannot, since that would mean the black balls are exhausted BEFORE the red ones, but the question requires that the red balls are to be exhausted first – TheMultiRounderGamer May 07 '24 at 08:45
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    @Dominique yeah i think thats much better worded than my question – TheMultiRounderGamer May 07 '24 at 08:45
  • @lulu Does this mean the second last pick and the last pick are reserved for black/white or white/black combo? Because only then we can exhaust red before the two colours – Gwen May 07 '24 at 09:18
  • @Gwen Yes, you are right. I misread the problem as requiring only that the last one wasn't red. I'll delete my unhelpful comment. – lulu May 07 '24 at 09:19
  • @lulu then shouldn't possible combinations in which red balls can be put is $2×\binom{10}{4}$ right? Isn't this getting too simple :") I'm really scared of combinatorics – Gwen May 07 '24 at 09:27
  • @Gwen I have posted a solution along the lines a sketched. It's harder than my (now deleted) comment suggested because of the three color issue, but the approach still works. – lulu May 07 '24 at 09:31

3 Answers3

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Note that each ball has an equal chance of appearing last.

We distinguish two cases, according to the color of the last ball drawn.

Case I: Last ball white (probability $\frac 3{12}=\frac 14$).

Now, consider the cards other than the last (white) one. We want the conditional probability that the reds were exhausted before the blacks, at this point the whites are irrelevant. There are $4$ reds and $5$ blacks in that collection, so the probability that the last of those to be drawn was black is $\frac 59$. Thus the probability of success with a white final ball is $$\frac 14\times \frac 59=\frac 5{36}$$

Case II: Last ball black (probability $\frac 5{12}$).

As before we get that the probability of success with a black final ball is $$\frac 5{12}\times \frac 37=\frac {5}{28}$$

The final result is the sum $$\boxed {\frac 5{36}+\frac {5}{28}=\frac {20}{63}}$$

Sanity Check: if there were $4$ of each color, then symmetry would apply and the answer would be $\frac 13$. As the distribution is close to uniform, we'd expect the answer to be close to $\frac 13$, as it is. Note: I find it hard to have intuition as to whether the result should be greater or less than $\frac 13$. Red is more likely to finish first than Black is, but White is more likely still.

lulu
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    "I find it hard to have intuition as to whether the result should be greater or less than 1/3" Just to take an intuitive stab: in the series 3-4-5, 4 is a 33% increase over its predecessor, but 5 is only a 25% increase over its predecessor => imbalance. Humans intuitively see additive equidistance as equality, but since probabilities are generally a multiplicative affair, it's the multiplicative equidistance that matters more. Or, to phrase it as a question: what number is in the middle of 1 and 9? – Flater May 08 '24 at 01:01
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  • To get the desired result for the red ball, the three colors must survive longer either in the order $WBR$ or $BWR$.

  • Consider the sequence $WBR$
    P($W$ survives max) = P($W$ drawn last)
    = P($W$ put first)$=\frac{3}{3+4+5}$

  • Having computed the survival probability of $W$, to compute the survival probability of the next color , $B$ we just imagine that white balls have vanished from the row, so again we have only two types to deal with and compute in a similar way !

  • Thus P($W$ survives longest followed by $B$) $=\frac{3}{3+4+5}\cdot \frac{5}{4+5} = \frac5{36}$
    and ($B$ survives longest followed by $W$)
    $= \frac{5}{3+4+5}\cdot\frac{3}{3+4} = \frac5{28}$

  • Adding the two cases,
    P(red is first to "die") $=\boxed{\frac5{36}+\frac5{28} =\frac{20}{63}}$

true blue anil
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Recurse on the color of the first ball. Let $p(a,b,c)$ be the probability of exhausting reds first when there are amounts $a,b,c$ of red, white and black balls respectively. We have for $a,b,c>0$ that

$$ p(0,b,c)=1 \\ p(a,0,c)=p(a,b,0)=0 $$

and

$$ p(a,b,c) = \frac{ap(a-1,b,c) + bp(a,b-1,c)+cp(a,b,c-1)}{a+b+c}. $$

These are tedious to tabulate by hand, but computer calculation gives

$$ p(4,3,5)=\frac{20}{63}. $$

ploosu2
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