1
$\begingroup$

I've been mulling over a problem that has something like the following form. I don't have a math or stats background so advice and answers at various levels, from terminological to strategic, would be helpful. I'll start off with the example that I've been thinking about and then move on to the specific case I'm working towards.

Suppose you have a jar full of coins, say pennies, nickels, dimes, and quarters. They are worth, respectively, $1$, $5$, $10$, and $25$ cents.

You're holding a particular coin, in this case let's say a dime.

You draw a coin from the jar, write down its value, and put it back in the jar. You do this some number of times and take the average (arithmetic mean) of the values.

What is the probability that the average of the values of the coins you've drawn is strictly greater than the coin you're holding (a dime)? What about for the case when you're holding any other coin?

For the simplest case of drawing a single coin, it's easy enough. If you're holding a penny, it's the probability that any single coin is greater in value than a penny. So, it's the probability of drawing a nickel, dime, or quarter. If you're holding a nickel, it's the probability of drawing a dime or a quarter. If you're holding a dime, it's the probability of drawing a quarter. If you're holding a quarter, there's no coin that you can draw that is greater in value than the one you're holding.

For the case of two coins, it's also easy enough. If you're holding a penny, it's the probability of drawing at least one coin that isn't a penny. If you're holding a nickel, it's the probability of drawing at least one dime or quarter. If you're holding a dime, it's the probability of drawing at least one quarter. If you're holding a quarter, then no coins will yield a higher value.

As we increase the number of coins drawn, however, the number of different combinations increases exponentially. Let's call the number of coins in the jar $C$, the number of coins you draw from the jar $n$, and the probability of drawing a penny $p_P$, nickel $p_N$, dime $p_D$, or quarter $p_Q$. Where $p_i = \frac{\text{number of i coins}}{C}$. Is there a simple, but general way to express the probability that the average of your draws will be greater than the coin you're holding?

Now, instead of monetary values, let's say the coins are worth $-b$, $0$, $a - b$, and $a$. Where $a > b > 0$. Again, is there a nice way of expressing this?

$\endgroup$
  • $\begingroup$ "the probability that the average of your draws will be greater than the coin you're holding." But "the coin you're holding" is an additional draw? I don't get why "the simplest case of drawing a single coin [...]it's just the probability of drawing a quarter" Why? And what woudl be, in this case, the "average of the drawn values"?? $\endgroup$ – leonbloy Jan 16 '14 at 0:13
  • $\begingroup$ That does not make it clear to me. $\endgroup$ – leonbloy Jan 16 '14 at 14:50
  • $\begingroup$ Thanks for asking for clarification. We can think of the coin in your hand as either being an additional draw or not. If it is, then once we have an expression for the probability that $n$ draws will yield an average value greater than a particular coin, we can just weight them by the probability of that coin being drawn. I've outlined the conditions that will satisfy the conditions for the case where $n=1$ and $n=2$. In the case of $n=1$ the arithmetic mean is well-defined and just yields the value of the coin drawn. $\endgroup$ – wdnsd Jan 16 '14 at 14:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.