# In a village, $90\%$ of people drink Tea, $80\%$ Coffee, $70\%$ Whiskey, $60\%$ Gin. Nobody drinks all four. What percentage of people drinks alcohol?

In a small village $$90\%$$ of the people drink Tea, $$80\%$$ Coffee, $$70\%$$ Whiskey and $$60\%$$ Gin. Nobody drinks all four beverages. What percentage of people of this village drinks alcohol?

I got this riddle from a relative and first thought it can be solved with the inclusion-, exclusion principle. That the percentage of people who drink alcohol has to be in the range from $$70\%$$ to $$100\%$$ is obvious to me

When $$T$$, $$C$$, $$W$$, and $$G$$ are sets, and I assume a village with $$100$$ people, then what I am looking for is $$\lvert W\cup G\rvert = \lvert W\rvert+\lvert G\rvert-\lvert W\cap G\rvert$$ I know that $$\lvert T \cap C \cap W \cap G \rvert = 0$$ and also the absolute values of the singletons.

But I do not see how this brings me any closer, since I still need to figure out what $$\lvert W\cap G\rvert$$ is and that looks similar hard at this point

On the way there I also noticed that $$\lvert T\cap C\rvert \ge 70$$ and similar $$\lvert W\cap G\rvert \ge 30$$

By now I think there is too little information to solve it precisely.

• There is too little information, as it is phrased--namely, without knowing the population of the village, there's no way to answer "How many people...?" At best, we may answer questions of the form "What percent of the villagers...?" Aug 7, 2018 at 11:54
• Well, I suppose we can answer "How many people of the village drink all four beverages?" Aug 7, 2018 at 11:55
• Clearly the village is in the UK, most likely Ireland (since otherwise people would drink whisky). So the answer is 100%. And everybody drinks ales, in addition to whatever drinks from the list they normally have. Aug 7, 2018 at 22:54
• @AsafKaragila Not sure the Irish would take kindly to being told they were in the UK lol
– Jam
Aug 9, 2018 at 14:38
• @Jam: I am pretty sure that Belfast is both in Ireland and in the UK. But that's just me. ;) Aug 9, 2018 at 14:39

If you add up the percentages, they come out to $300\%$. This means that the average number of beverages per person is $3$. No one drinks more than that, so no one can drink less than that, either. Since everyone drinks exactly three beverages, everyone has exactly one beverage that they don't drink. So no one doesn't drink both whiskey and gin, i.e. everyone drinks alcohol.

• (+1) Out of the current 4 answers, this is the simplest/clearest explanation. (A direct corollary from this explanation is that everyone drinks either tea or coffee, too.)
– Matt
Aug 7, 2018 at 21:59
• I'll note that this technique of summing over a certain domain of interest is also how one can prove the measure-theoretic pigeonhole principle, namely that if $S_{1..n}$ are measurable subsets of $[0,1]$ and $\sum_{i=1}^n |S_i| > k$ then some real in $[0,1]$ is in more than $k$ of the sets $S_{1..n}$. (Hint: Summing corresponds to integrating.) Aug 9, 2018 at 3:19
• An answer on Math.SE I can understand, never thought I'd live to see that day. Aug 9, 2018 at 7:21

Yes, there is enough information. Clearly the overlap between whiskey drinkers and gin drinkers is at least $30\%$, being exactly $30\%$ iff $40\%$ drink whiskey but not gin, and $30\%$ drink gin but not whiskey. Similarly at least $70\%$ drink both tea and coffee. Since nobody drinks all four, the number who drink both whiskey and gin is exactly $30\%$, and the number who drink alcohol is $30\% + 30\% + 40\%$, i.e. everybody.

• Is it true that 70% drink both tea and coffee? Consider the case that there are 10 inhabitants named [a b c d e f g h i j]. Let's say that [a b c d e f g] (70%) drink tea, and that [c d e f g h i j] (80%) drink coffee. The subset that drink both is (in this instance) [c d e f g] - only 50%. I think we don't have enough information: there are a lot of different subsets which satisfy all the conditions, but would give different results for "drinks whiskey or gin", or any other union/intersection question you might have. Aug 8, 2018 at 12:08
• @MaxWilliams, I think that you will find 10 inhabitants to be insufficient to meet the rest of the % requirements. When you have a population that WILL support it, then Michael Behrend's answer should hold up
– JoeG
Aug 8, 2018 at 12:50
• @MaxWilliams You read it wrong --- the percentages for tea and coffee are 90% and 80%, not 80% and 70%. Aug 8, 2018 at 13:50
• This problem looks similar to my question, math.stackexchange.com/q/2607531/521955 @Michael Behrend could you answer that please? Aug 9, 2018 at 15:47

Hint: consider what "Nobody drinks all four beverages" means in terms of the four groups

• People who don't drink tea
• People who don't drink coffee
• People who don't drink whiskey
• People who don't drink gin

and then see how big each of those groups are.

• :-) ${}{}{}{}{}$ Aug 7, 2018 at 11:47

From the problem we can say that the village is $90\%+80\%+70\%+60\% = 300\%$ saturated with liquids. From this we can say that every person should drink strictly $3$ beverages, no more no less.

Having drinking three beverages, while there are only two non-alcoholic beverages supplied to the village, you are guaranteed to get drunk. Thus, $100\%$ of the village drink alcohol.

• "... will get drunk.." they must be a bunch of lightweights, since no mention of the volume of each type of drink was mentioned! Aug 8, 2018 at 18:06
• ISTM that Acccumulation’s answer explained this. Aug 8, 2018 at 22:25

You can use inclusion/exclusion but you might not have enough information. Or then again you might.

The number of people who are in $A$ or $B$ is $A + B - (A\cap B)$ and so if $A+B > 100$ percent we can conclude $A+B - 100\le A\cap B \le \min (A,B)$

So $WHISKEY + GIN - 100 = 70+60 -100 = 30 \le(WHISKEY \cap GIN) \le \min WHISKEY, GIN = GIN = 60$.

Likewise $TEA + COFFEE - 100 = 90 + 80 100 = 70 \le(COFFEE \cap TEA)\le \min COFFEE, TEA = COFFEE = 80$.

Let $A = (COFFEE \cap TEA)$ and $B = (WHISKEY \cap GIN)$ and $70 + 30 = 100 \le A + B$ so $A+B -100 \le A\cap B$. But we know that $A \cap B = COFFEE \cap TEA \cap WHISKEY \cap GIN)=0$.

That can only happen if $A = 70$ and $B = 30$.

So the people drink alcohol $= WHISKEY + GIN - (WHISKEY \cap GIN) = 70 + 60 -30 = 100$ percent.

(We have everybody drinking alcohol and everbody drinking caffeine, and everybody drinks two of one and one of the other:

$30\%$ drink gin, tea, and coffee.

$40\%$ drink whiskey, tea, and coffee.

$10\%$ drink whiskey, gin, and coffee.

$20\%$ drink whiskey, gin, and tea.

• +1 for the first line. Aug 8, 2018 at 10:02
• Is the first paragraph meant to be a joke? If it is, it’s not very clear (to me).  It looks (to me) like a typo. Aug 8, 2018 at 22:21

A friend of mine proposed this solution which you could call a solution without words.

He showed me this picture:

And then he said: "Isn't it just $100\%$?". I said: "How?!" And then after a couple of seconds I said: "Wooow!".

Explanation

The white boxes correspond to the proportion of people that drink a specific type of drink. You see that there are no four white boxes underneath each other which makes sure that everyone drinks less than 4 types. We also see that everyone drinks exactly 3 types. So everyone is drinking alcohol.

I think this picture is beautiful, since it also could lead to the start of a nice mathematical proof.

• The key idea in the picture is that no black area's can overlap because that would mean a person drinks all four types. Aug 12, 2018 at 18:43
• @fejfo yes, my formulation was wrong, edited! Aug 12, 2018 at 19:13

Rephrase the question equivalently:

10% don't drink tea, 20% don't drink coffee, 30% don't drink whiskey, and 40% don't drink gin. This must sum to at least 100% (since nobody doesn't not drink all 4).

Clearly they sum to at most exactly 100%, so the "not-X-drinkers" are disjoint sets, and so everybody drinks at least one of whisky or gin.