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Players are pulled up to pick a ball out of a hat containing $14$ red and $1$ blue.

If the odds of drawing the blue ball are $1/15$ what are the odds of every person not drawing the blue ball and leaving it for the last person to draw?

Initially I thought you would multiply the probability of not drawing the blue ball for each person and multiplying them together so i did

$$\dfrac{14}{15} \times \dfrac{13}{14} \times \dfrac{12}{13} \times \cdots$$

But when I roughly calculate that, I get the same as $1/15$.

Is this correct?

Henry
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Kenny Hammerlund
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    Yes this is correct. – Laurent Hayez Jan 24 '18 at 09:11
  • Yep. That's fine. Assuming the balls are actually removed once they are picked and not thrown back in the bag. – Oria Gruber Jan 24 '18 at 09:11
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    It is correct but there is a more direct way to arrive at this result. Personalize with the blue ball and wonder what are your odds to be drawn as last. That makes things more easy. The difficult part is that human beings are not so eager to personalize with a ball. However, in probability it can be very fruitful. – drhab Jan 24 '18 at 09:29

4 Answers4

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Yes, the answer is $\frac{1}{15}$, and there is a much easier way of reaching it by re-thinking the question. Essentially, the $14$ people drawing balls are uniquely determining which ball the final person will choose. Though the final person really makes no choice at all, the final person recieves a randomly selected ball. And from the final person's perspective, all balls are equally likely (which is important).

There are $15$ balls that can be left in the hat, one of them is blue, therefore the probability is $\frac{1}{15}$.

5xum
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Rephrase it like this:

There is a blue ball together with $14$ red balls.

The balls will be numbered with $1,2,\dots,15$ randomly.

So for a fixed ball the numbers have equal probability to become the number of that ball.

Now what is the probability that the blue ball will be numbered with $15$?

Yes, of course: $$\frac1{15}$$

drhab
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The probability that you have calculated ($\frac {14}{15}*\frac {13}{14}*\frac {12}{13}*...*\frac {3}{4}*\frac {2}{3}*\frac {1}{2}*$) can be written rather neatly as $\frac {14!}{15!}=\frac 1{15}$ which is your answer.

For the love of maths
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The $14$ in $\frac{14}{15}$ is cancelled against the $14$ in $\frac{13}{14}$, and the $13$ in $\frac{13}{14}$ is cancelled against ... and so on. In the end, you're left with only the $15$ in $\frac{14}{15}$ and the $1$ in $\frac12$, so the final answer is indeed $\frac1{15}$.

Arthur
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