Can you solve the 4 foods puzzle?

Thanks to James W. for the suggestion! At an amusement park, 65% of visitors ate a donut, 80% ate a soft pretzel, 80% ate pizza, and 90% ate ice cream. What is the minimum percentage of visitors that ate all 4 foods?

100 soldiers riddle Math StackExchange

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[KroganCharr]

For the practical application, you actually do NOT want to target the people who bought all 4. Bundles are typically cheaper than the sum of the individual items - that's the point of the bundle. If we only sell the bundle to people who bought all 4 anyway, we just lose money.

The people we ACTUALLY want to target are the ones who bought less, but might be convinced to buy all 4. Most likely, the people who usually buy 3 out of 4 could be tempted by a bundle offer. The donut merchant would be the most likely to profit from this, as most people who only skipped out on 1 probably skipped the donut.

[thewordsmith]

I solved it by inverting the percents (100-p)%, then adding it together, to figure out what percentage at most didn’t eat at least one food

[ZekeRaiden]

Minimum percentage can always be determined by picking the lowest percentage (in this case, 🍩 at 65%), and then subtracting away all the "did not eat X" percentages. In this case, those are 20%, 20%, and 10%, for a total of 50%. 65% - 50% = 15%. The actual amount that ate all four items is probably higher, up to maximum of the lowest percentage, 65%.

[AgusSkywalker]

The easiest way is calculating the inverse. The max amount of people that didn't eat all 4. You just add the percentages of people who didn't eat each food. 10 for ice-cream, 20 for pretzels and pizza, and 35 for donut. 10+20+20+35=85, which means at most 85% didn't eat all 4 foods, so at least 15% had to eat all 4.

[Mr_Gamer_]

Those looking for a simpler method: I did this problem instantly in my head. 100-65 = 35, 100-80 = 20, 100-80 = 20 & 100-90 = 10. Add all these numbers up (35+20+20+10) = 85. 100 - 85 = 15%! Explanation: 35% didn’t eat a doughnut, 20% didn’t eat a pretzel etc. So 85% didn’t eat all 4 foods. So 15% did!

[b_ru]

Here's what I did:
65% (Donut) + 80% (Pretzel) = 145%. 145% - 100% = *45% of people who ate a Donut/Pretzel ate both of them.*
45% (Donut+Pretzel) + 80% (Pizza) = 125%. 125% - 100% = *25% of people who ate a Donut, Pretzel, or Pizza ate all 3 of them.*
25% (Donut + Pretzel + Pizza) + 90% (Ice Cream) = 115%. 115% - 100% = *15% of people who ate any of the 4 foods ate all 4 of them.*

Therefore, at least 15% of customers must have eaten all four foods.

[g.c.]

I can never solve the puzzles in the channel because when I open the video the solution is right in front of me in the top comment.

[mihailghinea]

Minimum 15%. Maximum 65%. Expected percent of people to eat all 4 foods 37.44%.

[doodlePimp]

42 seconds in. "Amusement park: What is the minimum percentage of visitors that must have eaten all 4 foods?" 0% Odds are some did not buy any food. - Solved!

[OOKIEDOKIE]

I did it by thinking of a group of 100 people, and taking it at groups of 2 at a time. If you take the first two groups, 80 people had the pretzel, which means minus the 35 that didnt have the donut you still need 45 that did. Then for pizza you take the 55 away that didnt have both donut and pretezel but you still need 25 that had both, so then you need 25 that had all 3. Then for icecream you take 75 away that didnt have all previous 3, but that still leaves you with 15 that did, so atleast 15 had to have all 4.

[howareyou]

My elementary school way is the simplest (and also proves it at the same time)
* 65% eat A, 80% eat B, so "eat A and B" is at least 65% + 80% - 100% = 45%
* now same logic, 45% eat AB, 80% eat C, therefore "eat ABC" is at least 25%
* eat ABCD is at least 15%
Each step itself also naively proved that it's a reachable lower bound (thus minimum)
The calculation is even simpler as:
1. you remove the 100th so it's smaller number
2. you can early exit if the sum is less than 100 and claim that the min is already 0 without adding all other numbers.

[chinareds]

I approached this by simplifying it to 2 foods. This can be visualized by two sliding bars (line segments) of various lengths which must be positioned in a specific space (100%). How would we minimize the overlap? Easy, we put one bar all the way to one side and the other bar all the way to the other. So for example with Ice Cream (90%) and Pizza (80%), the mininum overlap would be 70%. But now we know a minimum of 70% had both Ice Cream and Pizza, so I can repeat this with the Pretzel (80%). So mininum 50% had Ice Cream/Pizza/Pretzel. Apply this once more with Donut (65%) and the minimum overlap is 15%.

[gblargg]

I loved seeing it worked in many different ways. This is the real skill, being able to approach things in multiple ways and gauge which one will be simplest.

[Yusso]

I did the second method but then I realized it's easier:
Assuming that there are 100 people and no person ate the same item twice.
If all of them ate all 4 items there would be 400 sales.
But if we add up the numbers there has been 315 sales.
So at most 85 people did not eat all 4 items.
That means at minimum 15 people ate all 4 items.

[mark]

After my headache subsides, I'll watch this video again and see if I can follow the logic this time.

[sunnicivang]

I was mentally thinking of Method 2, but was physically trying to write it out using Method 1, and using 100 people.

[cedricdrahelia]

35% didn't eat a donut, 20% didn't eat a pretzel, 20% didn't eat a pizza, 10% didn't eat an ice cream. Worst case scenario : they are not the same people, therefore 35 + 20 + 20 +10 = 85% of people didn't eat one item, so 15% ate all 4.

[MustangKepler]

I just calculated extra after mini 65% overlapping in other three food found easy 15℅ to be minimum.

[Imperial_Squid]

I got the first 65% 80% trick to arrive at 45% due to seeing this kind of puzzle but didn't think to extend it out further to all four, cool trick!

It's also interesting to note that, so long as each set contains less than 100%, you can keep adding more and more orthogonal sets and eventually arrive at 0% minimum overlap

Eg, for sets that contain 50% of a population, you need 2 such sets to have a minimum overlap of 0%, if they cover 75% then you need 4 sets, 90% coverage requires 10 such sets, etc etc but however many ways you have of grouping people, so long as you have enough of them you'll be able to show that the minimum possible overlap is 0%

[guyhoghton]

The "No Doughnut" set contains 35% of the visitors. "No Pretzels" has 20% of them, "No Pizza" has 20% and "No Ice Cream" has 10%. Anyone not in any of those sets must have bought all four foods. To minimise these people you need to maximise the numbers in the above-mentioned four sets, and to do that you need to assume no sharing of people between them. In that case you simply sum the sets' percentages to get the total %age of visitors that did not buy all four foods, because in doing so no one gets counted twice: 35+20+20+10 = 85. Therefore 15% bought all four foods in this extreme case.