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February 24th, 2017, 06:49 AM   #1
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Simple or complex?

Hi, I'm new here, but have been a lifelong math enthusiast. I came across this problem the other day and believe I have the correct solution but have never been entirely sure of myself when it comes to probability.

A box contains five identical balls, one of which is marked as "the winner." Five people draw one ball each, and of course only one can win. The question is, does the order of drawing matter? My knee-jerk response was, "Of course it doesn't." The probability is one in five, whether you draw first or last.

But wait, my friend pointed out. The first person to draw has a one in five chance. But then there are only four balls left, then three, then two, then one, and either the "winner" is still in there at each drawing, or it isn't. How do you account for that? If there were five balls to draw from every time, then the probability would always be one in five, but the balls don't get replaced—the drawing isn't finished until everyone has a ball. It's more complex than you think.

This confused me, I admit. I started thinking about the Monty Hall problem and whether this might fall into that category, whether I was blind to some counterintuitive reality, etc. I went down the rabbit hole a bit. So I decided to write out the problem. Here's what I came up with, and I hope someone can either confirm my thinking or show me where I'm in error:

The first to draw has, of course, a 1/5 chance. The second has a 1/4 chance, but also a 4/5 chance that the "winner" ball is still in the box: (1/4 * 4/5) = 4/20, so, again: 1/5. The third has a 1/3 chance, but only a 3/5 chance that the "winner" is still in the box: (1/3 * 3/5) = 3/15, and so, again: 1/5. And so on. Therefore the probability is, in fact, 1/5, no matter whether you draw first or last.

Does this work as a solution? Or is it right but the method wrong? Or am I missing something else?

Thanks in advance for giving this your attention.
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February 24th, 2017, 07:34 AM   #2
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One way to consider this is to line the balls up in the order in which they are picked. (You may consider a "machine" to be putting a ball under the hand of the person if you like).

Where is the winner in this sequence of 5 balls? Well, one way to arrange the balls randomly is to put the "winner" into one of the 5 positions and then throw the rest of the balls into the remaining places. Clearly the "winner" can be in any one of the five places with equal probability.

Your analysis is correct, but not necessarily the easiest way to go about it.

In general, but especially in probability, it is a good idea to avoid reasoning along the lines of "of course..." because intuition can be really, really bad at mathematics. If you have a question, always reason it out.
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February 24th, 2017, 07:43 AM   #3
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Thanks for the quick reply, v8archie!
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