The number of bottles of beer one can buy with $10, after exchanging bottles and caps My answer to this question is 15, but my dad insists I am wrong. Who is right?

$2 can buy 1 bottle of beer.
  4 bottle caps can be exchanged for 1 bottle beer.
  2 empty bottles can be exchanged for 1 bottle of beer.

How many bottles of beer can you buy with $10?
I feel like some of the answers I've gotten so far... don't really make sense or really follow what I am trying to ask.
So I think I will clarify what I am trying to ask for.
YOU CAN (have to) TRADE / EXCHANGE bottle caps and empty bottles to get new bottles.
So how many can you buy with $10?
The answer he got is 20, which is correct. I just don't understand why.
 A: I would show your dad the following table which should be easy to follow line by line:
$$
\begin{array}{|c|c|c|}
\hline
\text{Full Beers} & \text{Empty Bottles} & \text{Caps} & \text{Action}\\
\hline
\color{red}{5} & 0 & 0 & \text{DRINK} \\
0 & 5 & 5 & \text{Buy more}\\
\color{red}{3} & 1 & 1 & \text{DRINK}\\
0 & 4 & 4 & \text{Buy more}\\
\color{red}{3} & 0 & 0 & \text{DRINK}\\
0 & 3 & 3 & \text{Buy more}\\
\color{red}{1} & 1 & 3 & \text{DRINK}\\
0 & 2 & 4 & \text{Buy more}\\
\color{red}{2} & 0 & 0 & \text{DRINK}\\
0 & 2 & 2 & \text{Buy more}\\
\color{red}{1} & 0 & 2 & \text{DRINK}\\
0 & 1 & 3 & \text{Insufficient funds}\\
\hline
\end{array}
$$
$$\color{red}{\text{Total beers} : 5+3+3+1+2+1=15}$$
Depending on how we interpret the rules there are some caveats to this solution:


*

*(Suggested by Ross Millikan) If we can exchange $2$ caps + $1$ bottle (which has the same monetary value as $4$ caps or $2$ bottles) then we can increase the total to $16$.

*(Suggested by Peter Shor) If we are able to borrow one cap then we can buy one more beer from our $4$ caps and return the borrowed cap immediately. This also gives us two empty bottles which can be used to buy a new beer which brings the total to $17$. If we are then allowed to borrow a bottle we can buy even another beer and return the bottle we borrowed giving us a total of $18$ beers.

*(Suggested by Henry) Combining the two suggestions above, i.e. allowing borrowing and using $1$ bottle + $2$ caps as currency to buy a new beer we can get up to $20$ which is the theoretical maximum. This is shown below where the first line is the last line of the table above.


$$
\begin{array}{|c|c|c|}
\hline
\text{Full Beers} & \text{Empty Bottles} & \text{Caps} & \text{Action}\\
\hline
\ldots & \ldots & \ldots & \ldots \\
0 & 1 & 3 & \text{Borrow cap}\\
0 & 1 & 4 & \text{Buy more}\\
\color{red}{1} & 1 & 0 & \text{DRINK}\\
0 & 2 & 1 & \text{Return cap}\\
0 & 2 & 0 & \text{Buy more}\\
\color{red}{1} & 0 & 0 & \text{DRINK}\\
0 & 1 & 1 & \text{Borrow bottle}\\
0 & 2 & 1 & \text{Buy more}\\
\color{red}{1} & 0 & 1 & \text{DRINK}\\
0 & 1 & 2 & \color{black}{\text{Buy more}}\\
\color{red}{1} & 0 & 0 & \text{DRINK}\\
0 & 1 & 1 & \text{Borrow cap}\\
0 & 1 & 2 & \color{black}{\text{Buy more}}\\
\color{red}{1} & 0 & 0 & \text{DRINK}\\
0 & 1 & 1 & \text{Return bottle and cap}\\
0 & 0 & 0 & \text{Insufficient funds}\\
\hline
\end{array}
$$
$$\color{red}{\text{Total beers} : 15 + 1 + 1 + 1 + 1 + 1 = 20}$$

For the general problem: each beer you drink can be used to buy $\frac{3}{4}$ more beers so in general we expect that if you start with $B$ beers then you will be able to drink a total of
$$B\left(1+\frac{3}{4}+\left(\frac{3}{4}\right)^2 + \ldots\right) = 4B \text{ beers}$$
A quick check on the computer gives us the formula $4B - 5 \text{ beers}$ (without bending the rules) and $4B$ when bending the rules (as in 3. above) holds for all $B\in [2,1000]$.
A: If $m$ is the number of beer bottles initially purchased and $N$ is the total number of beer bottles that can be bought or exchanged, then we have $m+\left\lfloor \frac{N-1}2\right\rfloor+\left\lfloor\frac{N-1}4\right\rfloor \geq N$.  That is, $N\leq 4m-5$ or $N=4m-3$.  The case where $N=4m-3$ happens only when there are $1$ empty bottle and $1$ bottle cap left, which can only happen when $m=1$.  (To show this, suppose that the beer is drunk immediately after a purchase or an exchange.  Then, at no point in time after the first purchase do we have the number of empty bottles or the number of bottle caps to be $0$.)  Hence, if $m>1$, $N\leq 4m-5$.  For $m=5$, we have $N\leq 15$.  (It can be proven by induction on $m>1$ that this bound is sharp.)
Now, if borrowing empty bottles and bottle caps is allowed, then we need $m+\left\lfloor{\frac{N}{2}}\right\rfloor+\left\lfloor{\frac{N}{4}}\right\rfloor \geq N$.  That is, $N\leq 4m$.  We can show by induction on $m$ that this bound is sharp.  Perhaps nonsurprisingly, you only need to borrow $1$ empty bottle and $3$ bottle caps in all cases.
However, if an empty bottle can be returned for $\$1.00$ and a bottle cap can be returned for $\$0.50$, then $\frac x2+\frac{N-1}2+\frac{N-1}4\geq N$, where the initial fund is $\$x$.  Therefore, $N\leq 2x-3$.  This bound is sharp for all integers $x \geq 2$ and for all half-integers $x\geq \frac32$.  (In this scenario, if you can also borrow empty bottles and empty caps, then $N\leq 2x$ for all nonnegative integers $x$ and for every half-integer $x\geq \frac{1}{2}$.  The bound is again sharp and you need to only borrow $1$ empty bottle and $1$ bottle cap in all cases.)
A: If you are allowed to borrow stuff, then you can get to drink $20$ bottles of beer:


*

*Buy $5$ bottles of beer for $\$10$

*Drink $5$ bottles of beer, giving you $5$ empty bottles and $5$ caps

*Borrow $15$ empty bottles and $15$ caps, giving you $5+15=20$ empty bottles and $5+15=20$ caps in total

*Swap the $20$ empty bottles for $\frac{20}{2}=10$ bottles of beer and swap the $20$ caps for $\frac{20}{4}=5$ bottles of beer, giving you $10+5=15$ bottles of beer

*Drink $15$ bottles of beer, giving you $15$ empty bottles and $15$ caps

*Repay the $15$ empty bottles and $15$ caps that you previously borrowed


In total you have drunk $5+15 = 20$ bottles of beer. So your dad may therefore be correct even if he is raising an alcoholic.
If you cannot borrow, than your final bottle of beer will leave you with an empty bottle and and a cap which combined are worth about $\$0.375$, which you cannot turn into beer, and so $20$ bottles of beer at an average cost of $\frac{\$10}{20}=\$0.50$ would not be possible.
A: Similar as in other answers, a state is described by a tuple $(z,m,b,e,c)$ where $z=$ number of bottles drunk, $m=$ available money in dollars, $b=$ full bottles in inventory, $e=$ empty bottles i inventory, $c=$ caps in inventory. The followig steps can be applied, each adding a specific tuple to the current tuple:


*

*Buy a bottle: $(0,-2,+1,0,0)$

*Drink a bottle: $(+1,0,-1,+1,+1)$

*Trade two empty bottles: $(0,0,+1,-2,0)$

*Trade four caps: $(0,0,+1,0,-4)$


We start with the tuple $(0,10,0,0,0)$ and are allowed to apply the above steps in any order, provided that no component ever becomes negative, and our goal is to maximze the $z$ component. 
Among all step sequences that lead to the maximal possible $z$, consider one that minimizes the sum of the index positions in the sequence where we execute a "drink" step.
Then before each drink step we have $b=1$: Certainly $b\ge 1$ as otherwise we cannot drink. On the other hand, if for some dringk steps $b\ge 2$, consider the first drink step applied to a state with $b\ge2$. Then the preceding step (which must exist because initially $b=0$) must be a buy or trade step, and it is possible to swap the order of these two steps without changing the final result, contradicting the minimality of drink-index sum.
On the other hand, we certainly have $b=0$ after the last step as otherwise we could have yet another drink, contradicting optimality of the sequence.
Since the only possible way to reach $b\le 1$ from $b\ge2$ would be drinking and drinking occurs only at $b=1$, we conclude that $b\ge 2$ never holds. Therefore, the four possible steps occur only in the following combinations:


*

*Buy and drink: $(1,-2,0,1,1)$

*Trade bottles and drink: $(1,0,0,-1,1)$ 

*Trade caps and drink: $(1,0,0,1,-3)$


By repeating the argument with a sequence that minimizes the sum of indices where we "buy and drink" among all double step sequences leading to optimal $z$, we see that "buy and drink" can always be swapped before any preceeding "trade  bottles/caps and drink" (simply because the only thing it decreases, $m$, is not changed by the other double steps). 
Thus we may assume that we start with a few  (by optimality: five) "buy and drink" double steps and after that only "trade and drink". So essentially, we start with $(5,0,0,5,5)$ and can apply only $(1,0,0,-1,1)$ (provided $e\ge2$) or $(1,0,0,1,-3)$ (provided $c\ge 4$) from then on.
By induction, we conclude that $e\equiv c\pmod 2$ throughout, because neither of the two trading double steps changes this fact.
Likewise by induction, $z+2e+c=20$ and $e\ge1$ and $c\ge 1$.
Using $e\ge1, c\ge1$ the next thing we show by induction is $e+c\ge 3$: It holds initlyy when $c=e=5$ and it holds after each trade doublke step because one of te numbers is increased from being $\ge1$ before the double step and the other is $\ge1$ anyway.
After the final step we have $e=1$ by optimality because $e\ge2$ would allow us to get yet another drink. Similary, $1\le c\le 3$ after the final step. 
As $e\equiv c\pmod 2$ and $e+c\ge 2$, we conclude $c=3$.
Therefore in the end $$z=18-e-c=15.$$
A: $10 can get you 5 bottles. From 5 bottles you get 5 caps and 5 empty bottles. Using 4 caps and 4 empty bottles you can get 3 new bottles and you are left with 1 cap and 1 empty bottle(8 bottles bought so far).3 new bottles give you 3 caps and 3 empty bottles and with leftovers you have 4 caps and 4 empty bottles in total. 4 caps and 4 empty bottles give you 3 new bottles (11 bottles so far).3 new bottles provides you 3 caps and 3 empty bottles, so you can get 1 new bottles and be left will 3 caps and 1 empty bottle(12 bottles so far). Now that one bottle provides 1 cap and 1 empty bottle and with leftovers you have 4 caps and 2 empty bottles. These provide you with 2 new bottles (14 bottles so far). 2 new bottles give you 2 caps and 2 empty bottles. 2 empty bottles give you 1 new bottle (15 in total so far). So now you are left with 3 caps and 1 empty bottle which cannot buy you a new bottle or bottles.
