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### Picking the Same Ball

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Date: 07/07/97 at 01:19:58
From: Guy
Subject: Probability

I have a jar with 377 individually numbered balls in it. One ball is
selected at random each day. What are the odds I will randomly select
the same ball two days in a row?

I know the odds for selecting a given numbered ball on one day are 1
in 377. I'm stumped on how to compute the probability of selecting the
same numbered ball on two consecutive days.

So far, I have selected one ball per day for 7 days. Wouldn't the
praobablity of selecting the same ball on two consecutive days
increase if I continued to do this for 30 days?

Thank you for your help!
Guy
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```
Date: 07/09/97 at 14:14:54
From: Doctor Daniel
Subject: Re: Probability

Hi Guy,

I'm not 100 percent sure I understand your problem, so I'll answer a
couple of different problems:

1) Suppose I pick one ball today and one ball tomorrow. What's the
probability they're the same? Answer: 1/377. This is actually fairly
easy, though a little counter-intuitive.

Perhaps you could think of it slightly differently; suppose instead of
377 balls, you had a set of dice with 377 faces. Then, really, the
question is what's the probability that if I roll 2 of them, they'll
both come up on the same face?

Well, suppose that the first one comes up 1. Then there's a 1/377
chance that the second one will too. Similarly, if you get a 344 on
the first one, there's a 1/377 chance the second one will. So it
should be clear that no matter what the first die comes up, the second
one has a 1/377 chance of also coming up that number. So it's clear
that the chance that the two dice agree is 1/377. This is also the
chance that the ball I pick today is the same as the one I pick
tomorrow.

2) Suppose I pick one ball today, one ball tomorrow, one ball the next
day, and so on, for X days. What's the probability that I have two
identical balls come up on consecutive days?

This is a good time to introduce a really important rule in
probability theory, which is called the law of the complement when
people want to be formal. Basically, here it is:

The probability that something happens is exactly 1 minus the chance
that it doesn't happen. (Yes, I know.  That's something obvious, AND
you already knew it. Give me a second here!)

So I could just as easily ask this question: What's 1 minus the
probability that all of the pairs of consecutive balls are different?

Now, we know that there are X-1 pairs of consecutive balls (for
example, if X = 3, there's the first and second ball, and the second
and third ball).

We also know that no pair of consecutive balls has any influence on
any other pair, when it comes to them being the same. That is, the
chance that the second and third balls are the same is 1/377, no
matter whether the first and second balls are the same or the 457th
and 458th ball are the same.  They're _independent_.

So the chance that, say, the second and third balls are different AND
the fourth and fifth balls are different is 376/377 * 376/377, because
the probability that two independent events happen is the product of
their probabilities.

The probability that the first and second balls are different, the
second and third are different, and so on, to the X-1 and Xth balls
being different is:

376/377 * 376/377 * ... * 376/377 = (376/377)^(X-1)

So the probability that there's a duplicate ball gotten consecutively
in the first X days is: 1-(376/377)^(X-1).

So, you're right; if you keep doing this for 30 days, the chance that
you've gotten the same ball consecutively is:

1-(376/377)^29 = .0741

This is not a particularly good chance; you've only got a 7.4 percent
chance of having it happen. It turns out it's not until you've gone
for 261 days that it will be 50 percent likely to have happened.

I hope this helps answer your question!  Good luck.

-Doctor Daniel,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
High School Probability

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