# Draw nickels and dollar coins until $p$ dollars, expectation

Here's a question from my probability textbook:

A bag contains $$m$$ dollar coins and $$n$$ nickels. A man is allowed to draw coins one by one until he has drawn at least $$p$$ dollars. Show that the value of his expectation is $${{np}\over{20(m+1)}} + p$$ dollars.

Here's what I did. We space our $$m$$ dollar coins equally so that they partition our $$n$$ nickels into $$m+1$$ parts of $${n\over{m+1}}$$ nickels each. And each of those parts will have value of $${n\over{20(m+1)}}$$ dollars. So then our desired expectation $${{np}\over{20(m+1)}} + p$$ dollars is equal to drawing coins up until the $$p$$th (evenly spaced) dollar coin.

But these are just my observations, I'm not sure if I have even showed what we wanted to show. Have I just merely asserted what we want to show without explanation? How can I conclude the result? What am I missing? Is the problem statement even correct?

Update: To clarify, I realize the problem statement means at least $$p$$ dollar coins, hence the solution method I gave and also Yuri Negometyanov's solution below. But I am more interested in the answer when it's at least $$p$$ dollars in money (and not at least $$p$$ dollar coins).

• What is a nickle? Oct 27, 2021 at 3:44
• @GrahamKemp A nickel is a 5 cent coin. There are 100 cents in every dolllar. Oct 27, 2021 at 3:45
• Is the question asking for the expected value until he has drawn p dollar coins, or just p dollars in value? Oct 27, 2021 at 4:15
• The question seems wrong, or I need some clarification. If $n = 0$ then $p$ has to be an integer for the formula to make sense. His expectation is $p$, everything is fine; but this suppose that $p$ is an integer, otherwise it should be $p+1$ and that is not the case. if now $m = 0$ then his expectation is again... $p$ because $p$ is an integer; and clearly it can be divided by nickels. But this is not what the formula gives you... Oct 29, 2021 at 11:17
• I understand this as an expectation in value (i.e. what the man ends up with at the end of the game) and not in the number of times he picks a coin inside the bag. But perhaps I am mistaken ? Oct 29, 2021 at 11:19

Let $$\;P(p,q)\;$$ is the possibility to get $$\;p\;$$ dollars and $$\;q\;$$ nickels of $$\;(p+q)\;$$ coins, when the last coin is a dollar.

Also, let $$\;Q(p,q)\;$$ is the possibility to get $$\;p\;$$ dollars and $$\;q\;$$ nickels of $$\;(p+q)\;$$ coins, when the last coin is a nickel.

Then $$P(p,q)=\dfrac{\dbinom{m}{p-1}\dbinom{n}{q}}{\dbinom{m+n}{p+q-1}}\,\dfrac{m-p+1}{m+n-q-p+1}$$ $$=\dfrac{m-p+1}{m+n-q-p+1}\, \dfrac{m!n!(p+q-1)!(m+n-p-q+1)!}{(p-1)!(m-p+1)!q!(n-q)!(m+n)!}$$ $$=\dfrac{m!n!(p+q-1)!(m+n-p-q)!}{(p-1)!(m-p)!q!(n-q)!(m+n)!} =\dfrac{\dbinom{p-1+q}{p-1}\dbinom{m+n-p-q}{m-p}}{\dbinom{m+n}{n}},\,$$ $$Q(p,q)=\dfrac{\dbinom{m}{p}\dbinom{n}{q-1}}{\dbinom{m+n}{p+q-1}} \,\dfrac{n-q+1}{m+n-p-q+1}$$ $$=\dfrac{n-q+1}{m+n-p-q+1}\, \dfrac{m!n!(p+q-1)!(m+n-p-q+1)!}{p!(m-p)!(q-1)!(n-q+1)!(m+n)!}$$ $$=\dfrac{m!n!(p+q-1)!(m+n-p-q)!}{p!(m-p)!(q-1)!(n-q)!(m+n)!} =\dfrac{\dbinom{p+q-1}{q-1}\dbinom{m+n-p-k}{m-p}}{\dbinom{m+n}{n}},\,$$ $$\sum\limits_{q=0}^n P(p,q)=1$$

and similarly $$\sum\limits_{p=0}^m Q(p,q)=1.$$

If $$\;n<20,\;$$ then the expectation of the money sum is $$E_{\tilde p}\left(p+\dfrac{q}{20}\right)= \tilde p +\dfrac1{20}\sum\limits_{q=0}^n qP(\tilde p,q)=\tilde p+\dfrac{n\tilde p}{20(m+1)}.$$

In the common case, $$\;Q(p,q)\;$$ does not influence to the fractional part of the sum, and $$E_{\tilde p}\left(p+\dfrac{q}{20}\right)= \tilde p +\left\{\sum\limits_{q=0}^{\min(n,10\tilde p)}\genfrac\{\}{}0q{20}P(p,q)\bigg|_{p=\tilde p- {\small\genfrac\lfloor\rfloor{}0q{20}}}\right\}.$$

• Thanks @Yuri. See my update where I give some clarification as to what I'm looking for. Nov 3, 2021 at 5:47
• @EmperorConcerto Thanks! Tried to fix. Nov 3, 2021 at 12:44