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I've been going in circles with this question which belongs to certainty about something. The original source of this problem is unknown. I found it in a textbook who doesn't have an author but rather a collection of riddles about combinatorics and differents problems.

The problem is as follows:

A porcelain jar is filled with a set of sphere candies. The baker made them in a peculiar condition, so that one half of the ball shaped candy has a color different from its other half. $24$ of those candies has one side colored blue (due a blueberry flavor) and the other half is colored green (due a pear flavor), $30$ has one half green and the other half red (due strawberry flavor), $28$ has one half red and the other white (due mint flavor), $40$ has one half white and other half orange (due orange flavor) and $35$ candies has one half orange and the other half blue. How many of these candies must be taken out from the jar at random and at least to affirm that $7$ candies share the same color?. (Assume that you are not allowed to have a peek inside the jar).

The alternatives given on the source are:

$\begin{array}{ll} 1.&\textrm{13}\\ 2.&\textrm{14}\\ 3.&\textrm{15}\\ 4.&\textrm{16}\\ \end{array}$

For this particular problem I'm very confused as I don't understand the phrase from the passage "to affirm that 7 candies share the same color". Does it mean should I count other colors whose one half also coincides with the first group?.

Before to explain what I tried to do to solve this problem I must say that I'm aware that in such kinds of situation when there is some uncertainty about something the procedure is to assume the most difficult to happen scenario or event and from then on rulling out possibilities until we can assure that our next pick will guarantee what we need.

But in different examples I have seen and are well documented they do show this for objects which share entirely one color or shape. But what to do if they have split colors?

Should I understand that they ask to match the identical ones or match the objects (the candies in this case) which have the same set of colors i.e red and white, with ones which one side red and other which have the other have white, but may have its opposing side with a different color?. Can somebody clarify this for me?

In my attempt to solve this problem what I tried to do was this:

Since they ask for $7$ candies, then the hardest possibly to happen scenario might be:

$\textrm{6 white and orange} +\\ \textrm{6 green and red} + \\ \textrm{1 of any of the remaining group}\\ \textrm{(could be blue and green or red and white or orange and blue)}$

But this is kind confusing shall I count the other half as color to meet the asked condition for 7 candies sharing the same color?.

I hope somebody can help me with this riddle. Supposedly the answer is $16$ but I don't know how to get there. But more importantly how to justify it?.

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  • $\begingroup$ Could you also provide where this puzzle comes from? $\endgroup$ – Conifers Oct 23 '19 at 15:05
  • $\begingroup$ @Conifers I wish to know who had created this riddle, in the original source does not indicate who did it. It is from a practice sheet which I had from a collection of riddles. I wish this doesn't influence much on the method of solving this problem. $\endgroup$ – Chris Steinbeck Bell Oct 25 '19 at 10:23
  • $\begingroup$ the source to provide is the rule for the PSE if the puzzle is not created by your own. Take easy :) $\endgroup$ – Conifers Oct 25 '19 at 10:38
  • $\begingroup$ @JMP Okay! I will revert it back to the original question, but the question is still on hold and I'm unsure how can I reword it to fit in the criteria of this stack site. $\endgroup$ – Chris Steinbeck Bell Oct 25 '19 at 10:52
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The question seems to ask that you need to ensure that in the N candies, you always find some 7 candies have same shared colors, and find the minimum N.

The answer should be $16$ due to you can take 3 candies in each type(total 5 types) by considering the worst case. This cause you have all colors with 6 candies shared for each. And plus any candy you will get a 7th shared color for any. Thus the total minimum candies is $3 \times 5 + 1 = 16$

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  • $\begingroup$ I tried to follow your steps but I still don't understand why should it be 3 candies from each of those 5 types and not as I described it?. Can you please offer a detailed explanation or tell what part in my reasoning was not correct?. Wouldn't the worst case scenario be almost reaching 6 of the same colors in both sides and the next pick not getting that color but instead of other?, but with that other having a totally different color combination from the preceding group?. $\endgroup$ – Chris Steinbeck Bell Oct 25 '19 at 10:25
  • $\begingroup$ @ChrisSteinbeckBell Check the candy distribution, there are just 5 colors will be involved 2 candy types for each(type1:B+G, type2:G+R, type3:R+W, type4:W+O, type5:O+B), so the worst case you just picked 3 candies for each type. This leads to you get total 15 candies and all 5 colors will appear in 6 candies for each. Thus plus one more candy you can assure that you will find a color involved in 7 candies out of total 16 candies. $\endgroup$ – Conifers Oct 25 '19 at 10:36
  • $\begingroup$ I did added in my question the source of my confusion. I don't know if I made myself better clear with that statement, I hope you can help me with that as I'm confused of what's exactly been asked here. Again, why should it be the worst case five of each type and not just six?. What makes $3$ the rationale to choose? Can you add this in a clarification please?. Because this is the part where I'm stuck. $\endgroup$ – Chris Steinbeck Bell Oct 25 '19 at 11:02
  • $\begingroup$ @ChrisSteinbeckBell The question ask to affirm there must have 7 candies sharing the same color. So the unluckiest case is you draw 3 candies for each type, and there are 5 types, each type has 2 different color, so that every color(B,G,R,W,O) appears only in 6 candies separately. And plus any one candy no matter what the type is, you can get 2 colors sharing in 7 candies. If the answer is 13, then I can make a counterexample that I pick 3(B+G), 3(G+R), 3(R+W), 3(W+O), and 1(O+B), however I still can't get a same color sharing in 7 candies, only 6. $\endgroup$ – Conifers Oct 25 '19 at 14:56
  • $\begingroup$ I'm so sorry to ask again, but can you add some picture or drawing to understand this?. The more I look at it, I can't find exactly a justification for using only 3 and not 6. The way how I see it is okay, he worst case is grabs $6$ but because the objects are split in color then take half of that value hence is $6$, so if the objects would have had three colors then I would consider to use $2$ instead?. But what if it had been $5$ colors?. $\endgroup$ – Chris Steinbeck Bell Nov 2 '19 at 0:07
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I think that where you're getting confused is in the framing of the question. The question is asking for the fewest number of candies we need to take in order to guarantee that 7 candies share a colour.

The scenario you gave results in 7 candies sharing a colour after picking 13 candies out. However, if I can find a scenario where after picking 13 or more candies out of the bag, we do not have 7 candies sharing a colour, then 13 cannot be the correct answer.

If I pick out 15 candies:

  • 3 blue/green
  • 3 green/red
  • 3 red/white
  • 3 white/orange
  • 3 orange/blue

And then we count the number of candies that have each colour, we get:

  • Blue candies: 6 (3 blue/green + 3 orange/blue)
  • Green candies: 6 (3 blue/green + 3 green/red)
  • Red candies: 6 (3 red/white + 3 green/red)
  • White candies: 6 (3 white/orange + 3 red/white)
  • Orange candies: 6 (3 orange/blue + 3 white/orange)

So we now know that any answer less than or equal to 15 can't be correct, as we have given a scenario involving 15 candies which does not result in 7 candies sharing a colour.

Given the multiple choice question, we know that the answer must be 16. If we pick out any further candy, no matter the colour, there will be 7 candies of both of the colour groups involved. For example if we now pick out an orange/blue candy, there will be 7 candies sharing orange and 7 candies sharing blue.

Hope this is helpful.

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