The fair meat bazaar: Who has right?

Question

In a local town the girl scouts have been selling tickets, and now it is the time to pick winners. The number of winners is yet to be decided. The names are collected in $3$ books. With respectively

1. $1000$
2. $ 600$
3. $ 400$

names in them. The question is without computer aid (eg the names are in physical books, too much hazzle to digitalize the results) how can one create a fair draw?

My solution is as follows (please try before looking)

Note that all books are divisible by 200. Hence pick $5$ names from book 1, $3$ names from book $2$ and $2$ names from book $3$. Amongst these $10$ names do another draw. Note one apparent problem with this solution is if you want more than $2$ winners. Then the smallest book is at an disadvantage. If you need $n$ winners then draw $\lceil n/2 \rceil \cdot (5,3,2)$ names$...\phantom{..............}\\$ Another solution is to have the names 1-2000 ready, and then draw $n$ names from this list. But in a practical sense, is not my solution easier than writing up $2000$ notes? (Remember no fancy technology allowed, the villages does not believe in this).

The locals does not agree this is a fair solution. Instead they mean that it should not matter what book you are in. If a winner is picked from book 1 and you are in book 3, this should not affect you. They mean it is fair to draw between the books, then pick a a number in the range 1-1000. I think they are wrong in saying that is a fair draw.

Who is right? And how can one explain the correct solution in laymans terms?

Answer 1

Certainly a fair solution would be to pick $k$ numbers from $\{1,2,\ldots,2000\}$ and then

\begin{array}{c|c|c} \text{number} & \text{book} & \text{person in the book} \\\hline 1-1000 & \mathrm{I} & n \\ 1001-1600 & \mathrm{II} & n - 1000 \\ 1601-2000 & \mathrm{III} & n - 1600 \end{array}

That is completely doable using pen and paper, if you don't have a way to pick a random number, throw a coin $11$ times: $2^{11} = 2048$ (if you happen to pick over $2000$ or a person picked previously, just try again; if you don't have a fair coin, throw the coin twice and interpret TH and HT as heads and tails respectively, while disregard any TT or HH as invalid throws).

As for whether you solution is fair: in general it is not. Assume there are to be two winners, and we would like to calculate the probability that some particular pair wins, which is $$\binom{2000}{2}^{-1} = \dfrac{2}{2000\cdot 1999}.$$

Observe that $1999$ is a prime number, so whatever calculation method we use, the denominator must has it as a factor. However, in your case the denominator can't have factors bigger than 1000 (unless you put it there artificially), so the prime number $1999$ does not happen, and so these numbers can't be equal.

I hope this helps $\ddot\smile$

Answer 2

dtldarek gave a good explenation why my logic was flawed. Here i present a slightly different view why I was wrong.

The good

The draw is fair under the assumption every person has bought exactly one ticket.

What do I mean by the last statement? Assume that the game only allows each person one ticket. Let $s$ be some integer. We pick then pick 2$s$ tickets from Book$_1$, $3s$ from book$_2$ and $5s$ from book$_3$. There will then be a total of $10s$ tickets picked. So you have a $1/10s$ chance of winning after getting picked out. The chance of getting picked out is simply $s/200$. We can see this by thinking about the desired outcomes versus the possible outcomes. Eg $$ \frac{2s}{400}=\frac{3s}{600}=\frac{5s}{1000}=\frac{s}{200} $$ For respectively book$_1$, book$_1$, and book$_3$. The chance of winning the final draw is done similarly. in total we have $2s+3s+5s = 10s$ tickets. You have one of those. So the chance of winning the final draw is $1/10s$. Combining these means that the total probability of winning is given as $$ P_{1,2,3} \ = \hspace{-1.5cm}\overbrace{\frac{1}{10s}}^{\text{Chance of winning the final draw}}\hspace{-1.5cm} \cdot \hspace{-1.8cm} \underbrace{\frac{s}{200}}_{\text{Chance of getting picked from book$_n$}} \hspace{-1.6cm} = \ \ \frac{1}{2000} $$ Which is independent of the scaling factor $s$. Picking $v$ winners instead of $1$ changes nothing. Everyone has an equal chance of getting in the final draw $s/200$. Changing the number of winners changes only the chance of winning the final draw. Say that we pick $v$ winners (note that we have $10s$ tickets in the final draw, hence $10s \geq v$. The total chance of winning will then be $$ P_{1,2,4}(s,v) = \frac{s}{200} \binom{1}{1}\binom{10s-1}{v-1}\Big/\binom{10s}{v} = \frac{s}{200} \cdot \frac{v}{10s} = \frac{v}{2000} $$ Hence the draw depends purely on the number of winners, and scales linearly.

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The Bad

The draw is unfair as long as any player buys more than one ticket, and we have more than one winner.

In order of miniziming the error of this method assume that we want $s$ to be as small as possible. This is only logical. We do not want to do any more draws to get $v$ winners than neccecarily. Hence we want $10s$ to be the smalest integer larger than $v$. In other words $$ s \geq \left\lceil \frac{v}{10} \right\rceil $$ Where $\lceil x \rceil$ denotes the greatest integer function (round up to nearest integer). From this point forward I define $$ s = \left\lceil \frac{v}{10} \right\rceil $$ Note that for $v \in[1,10]$ then $s_v=1$. I doubt we will have more than $10$ winners with just $2000$ tickets. This means the only variable to consider is $v$, for simplicity the notation $s_v$ is employed.To further simplify the notation we define $$ b_n = \frac12 x^2-\frac 12x+2 $$ Such that $b_1=2$, $b_2=3$ and $b_3=5$. Which means $200 \cdot b(n)$ gives the number of tickets in book$[n]$. Assume now that you are allowed to buy $\ell$ tickets. The chance of getting $k$ of your $\ell$ tickets picked from book $n$ kan be written as $$ P_n(v,\ell,k) = \binom{\ell}{k}\binom{200b_n - \ell}{s_v b_n - k}\Big/ \binom{200b_n}{s_v b_n} $$ These probabilites can be explictly written as $$ \begin{align*} P_1(v,\ell,k) & = \binom{\ell}{k}\binom{400 - \ell}{s_v b_n - k}\Big/ \binom{400}{2s_v} \\ P_2(v,\ell,k) & = \binom{\ell}{k}\binom{600 - \ell}{s_v b_n - k}\Big/ \binom{600}{3s_v} \\ P_3(v,\ell,k) & = \binom{\ell}{k}\binom{1000 - \ell}{s_v b_n - k}\Big/ \binom{1000}{5s_v} \end{align*} $$ Assume that you got $k$ out of $\ell$ tickets picked out. I now assume that you want to win on atleast one of these tickets. Assume that you got $k$ tickets into the final draw. The chance of winning on atleast one of these is given as $$ W(v,k) = 1 - \binom{k}{0}\binom{10s_v-k}{v}\Big/\binom{10s_v}{v} = 1 - \binom{10s_v-k}{v}\Big/\binom{10s_v}{v} $$ The chance to win on $k$ out of $\ell$ tickets is therefore given as $$ t_n(v,\ell,k) = W(v,k) \cdot P_n(v,\ell,k) $$ We are almost there.. This means that the chance to win on atleast one ticket is $$ \begin{align*} T_n(v,\ell) & = \sum_{k=1}^{\large \ell} W(v,k) \cdot P_n(v,\ell,k) \cdot W(v,k) \\ & = \sum_{k=1}^{\large \ell} \left[ 1 - \binom{10s_v-k}{v}\Big/\binom{10s_v}{v} \right] \binom{\ell}{k}\binom{200b_n - \ell}{s_v b_n - k}\Big/ \binom{200b_n}{s_v b_n} \end{align*} $$ This can of course also be written out explicitly for $n=1,2,3$. Without making it any prettier. $$ \begin{align*} T_1(v,\ell) & = \sum_{k=1}^{\large \ell} \left[ 1 - \binom{10s_v-k}{v}\Big/\binom{10s_v}{v} \right] \binom{\ell}{k}\binom{400 - \ell}{2s_v - k}\Big/ \binom{400}{2s_v} \\ T_2(v,\ell) & = \sum_{k=1}^{\large \ell} \left[ 1 - \binom{10s_v-k}{v}\Big/\binom{10s_v}{v} \right] \binom{\ell}{k}\binom{600 - \ell}{3s_v - k}\Big/ \binom{600}{3s_v} \\ T_3(v,\ell) & = \sum_{k=1}^{\large \ell} \left[ 1 - \binom{10s_v-k}{v}\Big/\binom{10s_v}{v} \right] \binom{\ell}{k}\binom{1000 - \ell}{5s_v - k}\Big/ \binom{1000}{5s_v} \\ \end{align*} $$ There are of course some restrictions on the variables. Obviously they must all be integers.

The probability

The draw is fair under the assumption every person has bought exactly one ticket. Otherwise the error scales with both the number of tickets bought $\ell$ and the number of prices.

As an thought experiment say you go crazy over the cute girl scouts and buy $400$ tickets. Which girl should you buy the tickets from, and does it matter how many prices there are? Say you buy the tickets from Book$[1]$, this means you are guaranteed to participate in the draw, but you only have $b_1 s_v = 3\lceil v/10\rceil$ tickets and there are $v= 10 \lceil v/10 \rceil$. Assume instead you buy from the biggest books, becuase bigger is better right? Now you only have a $400/1000$ chance to get the first ticket, but a better chance in the final draw.

The question is what weights more, being sure to get into the second round or being much better prepared if you get to the second round? Numerical analysis gives the following tables

Note that you can follow this link if the image is too small.

Conclusion

The draw is not fair, but in a normal lottery where the prices range from $1-10$, it does not matter how many tickets a person buys. Even if someone did go crazy and bought $400$ tickets the three books differ only by a couple percent. However the draw is only fair if each person is given one and only one ticket.

So if you want to have a infinitesmal better chance at winning go for the smallest book. Or if you want to be a smarty pants, suggest to pick numbers from 1-2000 instead.