# Drawing red and blue balls from a box by $2$ people,

A box contains five blue and eight red balls. Jim and Jack start drawing balls from the box, respectively, one at a time, at random, and without replacement until a blue ball is drawn. What is the probability that Jack draws the blue ball?

The books says the answer is:

$$8\cdot5\cdot11!+8\cdot7\cdot6\cdot5\cdot9!+8\cdot7\cdot6\cdot5\cdot4\cdot5\cdot7!+ 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 5 \cdot 5! = 2, 399, 846, 400$$

Therefore, the answer is $$\frac{2, 399, 846, 400}{13!} = 0.385$$.

I do not understand the solution the book provided!

This is how I approached the problem:

There are $$13!$$ factorial ways of picking out all the balls. But there are $$\frac{13!}{5!\cdot8!}$$ distinguishable ways of picking out all the balls. So I don't see how we got different denominators. Second, I was hoping to calculate the probability that the first $$8$$ balls are red, the complement of which would give me an answer that at least one blue ball would be picked in the first $$8$$ picks, and divide that by two for the probability of Jack. There is $$1$$ way of picking $$8$$ red balls in a row out of $$\frac{13!}{5!\cdot8!}$$.

$$P[$$Jack picking blue$$] =\frac{1-\frac{1}{13\cdot 11\cdot 9}}{2}\approx 0.499$$

How did I go wrong? And what was the thinking of the author?

• First of all, order matters here since the game ends when either one draws a blue ball, thus there are $13!$ distinguishable ways of picking out all the balls. Feb 19, 2020 at 6:52
• Secondly, since Jim picks first, why does Jim & Jack has the same probability of picks the first blue ball? (I assume that's where your $1/2$ comes from?) Feb 19, 2020 at 6:57
• So if I instead of picking red and blue balls, I were picking letters of the word Mississippi, and suppose the game stops when Jack picks an $S$. the total number of ways to pick out Mississippi is $11!$ not $\frac{11!}{4!\cdot 4! \cdot 2!}$ ? for more context see my problem here math.stackexchange.com/questions/3551255/… @MoonKnight
– Kam
Feb 19, 2020 at 6:58
• can you explain your logic using 3 red balls and 2 blue balls? That may be easier for us to find our discrepancy? Feb 19, 2020 at 7:13
• If you understand the book's solution I would rather you explain that :) I may have over complicated the issue with my previous comment @MoonKnight
– Kam
Feb 19, 2020 at 7:16

It is absolutely irrelevant to take $$13!$$ or $$\frac{13!}{5!8!}$$ events. You can either consider all permutations of balls of the same color as different or not. This does not effect on the ratio since both coefficients $$5!8!$$ in denominator and numerator are cancelled.

And what is relevant is how you calculates number of events where Jack picks blue ball. How the probability that first $$8$$ balls are red relates to this? Yes, if the first $$8$$ balls are red, then Jack cannot win. But this is not the unique case when he lost. If the first drawn ball is blue, he also lost. If the first two balls are red and the third is blue, Jack also lost. And so on.

First Jim picks ball. He must pick red ball ($$8$$ variants). Next Jack can pick blue or red. If he picked blue ($$5$$ variants) then the game is over and the other permutations of $$5+8-1-1=11$$ balls are irrelevant. If we will not distinguish the balls of the same color, the number of elementary events which correspond to the case when Jack wins by his first step is $$\frac{\color{red}8\cdot \color{blue}5\cdot 11!}{5!8!}.$$

If Jack picked red ($$7$$ variants), then Jim should also pick red ($$6$$ variants) and then Jack can pick blue or red. If he picked blue ($$5$$ variants), then the game is over and the potential $$5+8-4=9$$ balls can permute in any order, so if we will not distinguish the balls of the same color, the number of elementary events which correspond to the case when Jack wins by his second step is $$\frac{\color{red}8\cdot \color{red}7\cdot \color{red}6\cdot \color{blue}5\cdot 9!}{5!8!}.$$

Next, there are $$\frac{\color{red}8\cdot \color{red}7\cdot \color{red}6\cdot \color{red}5 \cdot \color{red}4 \cdot \color{blue}5\cdot 7!}{5!8!}$$ variants where Jack wins by its third step. And there are the $$\frac{\color{red}8\cdot \color{red}7\cdot \color{red}6\cdot \color{red}5 \cdot \color{red}4 \cdot \color{red} 3 \cdot \color{red}2 \cdot \color{blue}5\cdot 5!}{5!8!}$$ variants to win by fourth step of Jack.

All the other variants are impossible since there is not enough amount of red balls.

• I am awed by the time you generously gave to answer me! Thank you so much :) colors and all! Now I'll read it :) mega upvote!
– Kam
Feb 19, 2020 at 7:26
• BEAUTIFUL! I am eternally grateful, thanks again :)
– Kam
Feb 19, 2020 at 7:32
• You are welcome!
– NCh
Feb 19, 2020 at 7:33