Can you solve the giant iron riddle? – Alex Gendler

Can you solve the giant iron riddle? – Alex Gendler

The family of giants you work for is throwing a fancy dinner party, and they all want to look their best. But there’s a problem – the elder giant’s favorite
shirt is wrinkled! To fix it, you’ll need to power up… the Giant Iron. The iron needs two giant
batteries to work. You just had 4 working ones and 4 dead ones in separate piles, but it looks like the baby giant
mixed them all up. You need to get the iron working
and press the giant shirt, fast – or you’ll end up being the
main course tonight! How can you test the batteries so that you’re guaranteed to get
a working pair in 7 tries or less? Pause the video now if you want to
figure it out for yourself Answer in 3 Answer in 2 Answer in 1 You could, of course, take
all eight batteries and begin testing the 28
possible combinations. You might get lucky within
the first few tries. But if you don’t, moving the giant
batteries that many times will take way too long. You can’t rely on luck – you need to assume the worst
possibility and plan accordingly. However, you don’t actually need
to test every possible combination. Remember – there are four
good batteries in total, meaning that any pile of six you choose will have at least two
good batteries in it. That doesn’t help you right away, since testing all six batteries could
still take as many as 15 tries. But it does give you
a clue to the solution – dividing the batteries into smaller
subsets narrows down the possible results. So instead of six batteries, let’s take any three. This group has a total of three
possible combinations. Since both batteries have to be
working for the iron to power up, a single failure can’t tell you whether
both batteries are dead, or just one. But if all three combinations fail, then you’ll know this group has either
one good battery, or none at all. Now you can set those three aside and repeat the process for
another three batteries. You might get a match, but if every
combination fails again, you’ll know this set can have
no more than one good battery. That would leave only two
batteries untried. Since there are four good
batteries in total and you’ve only accounted for two so far, both of these remaining ones must be good. Dividing the batteries into
sets of 3, 3, and 2 is guaranteed to get a working result
in 7 tries or less, no matter what order
you test the piles in. With no time to spare,
the iron comes to life, and you manage to get the shirt
flawlessly ironed. The pleased elder and his family show up
to the party dressed to the nines … well, almost.


  1. Get the solution to the bonus riddle here:! Also, the first 833 of you who sign up for a PREMIUM subscription will get 20% off the annual fee. Riddle on, riddlers!

  2. I solved it differently:
    I divide the group into four pairs and try out each pair: AB, CD, EF and finally GH. If all of these fail, it means that all the pairs have exactly one good battery. They can't have two dead batteries, because then another pair would have had two good batteries and I would have already found it and solved the puzzle. So if all pairs fail, I take any two pairs, say AB and CD and try all combinations of them to find the two good batteries. Since I've already tested the pairs themselves I only need to try AC, AD and BC. One of those remaining combinations has to have two good batteries. So there you go, done!

  3. If you set 4 group of 2 batteries
    You can get it by within 5 tries …

    Let group…
    "AB CD EF GH " each letter is battery …

    So trying
    AB … If works you are saved, If doesn’t work then try
    BC … If works you are saved, If also doesn’t work try
    AC… If works you are saved, If also doesn’t work try
    BD … If works you are saved, If also doesn’t work then this 4 are bad batteries… Try any of this E F G H … and you saved ….

  4. Hmmm correct me if im wrong but is my solution okay:
    how about grouping them into four sets and trying them all out. Worst case scenario would be none of them would work. This means that each set will have one good and one dead battery. Just take two sets and try plugging them in until the iron works. 🙂

  5. The specific name for this technique of guaranteeing good batteries in the sets is called the Pigeon Hole principle. This is just one example of a myriad of problems you can solve using pigeon hole.

  6. My solution: I have 8 batteries total, 4 good and 4 bad. I make 4 piles at random. Worse case scenario, each pile has 1 dead and 1 good. If that's not true, at least one of the pairs will work (e.g. 01, 01, 00, 11[the working pair]. But the worse case scenario explained in numbers is 01, 01, 01, 01. If I get to the last pair and the iron still doesn't work, I know that each pile has a good battery and a dead one. Now I work with only two piles. I swap out a battery for one from the other pile. Worse case scenario I get 00, which doesn't work, so I swap out the "newer" battery: 01. The newest battery has to work, which means the newest and the one I didn't use is my solution.

  7. Did it with other solution, tell me if it's wrong. Try AB, CD, EA, EB, if none were succesful this means you have at least 3 dead batteries so far, then you can try FGH in every order and find 2 good batteries

  8. There is another solution to this riddle, if you group them in pairs and try each pair and none of them work it means each pair must contain a working battery and a broken one, then take 2 pairs and swap a battery from each and test the 2 new pairs, if they still don't work it means that the batteries you swapped are both broken and you can then use the last 2 batteries

  9. Or there is another solution
    We can divide it to four groups we will assume that no one of it works which mean each group has a good battery and we use four chances then we will take two groups of it and switch two of them and try, if it doesn't work that mean the other one won't work too.
    The last move is to switch another two of them and this time one group will work.

  10. Can't you just pick two random batteries and then separate the other 3 in groups of 2.
    You would need 3 tries.
    If the original batteries both work, you are set no matter if the others work or not.
    If both the original batteries dont work, you try the first 2 of the three groups. If both of those dont work, the other two groups have 2 working batteries each. If one of the batteries work an the other one doesn't you try the second group. You know there are between 3 and 4 working batteries in those 2 groups, which means one of those has 2 working batteries.
    If one is working and the other not in the original 2 then you just need 2 tries.

    Idk if that makes sense

  11. I actually solved it a different (and probably more complicated) way.
    Break the batteries into 4 groups of two, and test the first 3 groups. If all tests fail, then you know that
    (a) the last group has at least 1 good battery
    (b) if one of the groups you tested has no good batteries, then the last group has 2 good batteries

    Now, take one of the batteries from the group you didn't test, and try it against both batteries in one of your failing groups. If both tests fail, then you know that either the battery you tested was bad or it was good, but both of the batteries you tested it against were bad.
    If the battery you tested was bad, then by (a) you know the other battery from the last group must be good. But if both of the batteries you tested it against were bad, then by (b) you also know that the other battery from the last group must be good.

    Now, there are only 4 bad batteries to go around., and you just found 2 of them. Thus, the other two failing groups both must have at least one good battery. So, if you test your one battery you know to be good against both batteries from one of your other failing groups, you must get a success.

  12. After thinking about the 2nd riddle, it's just the opposite of the first, but instead of testing 2 batteries and leaving 6, we test 6 batteries and leave 2. just the same first method but in opposite 😀

  13. For the bonus riddle: do what you did for the normal riddle but instead of finding a pair of 2 working battery’s, find a pair of 2 dead ones and leave them out of the vacuum (because my brain thinks the iron is a vacuum)

  14. How can I test the batteries… and by the 7th time I test them on the iron, it has to work. Giant Ma'am, do you have copper cables, preferably wrapped in insulating material, and a light bulb? I'll test each battery on the light bulb until I find two that work. As you can see, I'm not testing them on the iron, and therefore I'm not using any of my chances for the iron.

  15. The giant batteries have a high voltage, and if it is good, it will shock you if you touch one of the terminals. So you can guarantee a good one by just touching six batteries and the ones that shocked you are the solution. The answer to the bonus question is touching seven batteries and getting shocked four times.

  16. Batteries that are working should weight more than a dead battery.
    Drop them in 3,3,2 pattern and pick up the any batteries that didn’t bounce a lot.

  17. I found an alternate solution:
    1. Try A and B
    2. Try C and D
    3. Try E and F
    4. Try G and A
    5. Try G and B
    6. Try H and C
    7. Try H and D

  18. Haven't seen the result yet.. I think dropping them, the new batteries don't bounce..? And the old ones do..?

    Lmao I was so wrong 🤣🤣🤣

  19. There's more than one explanation though, mine is different.
    In this, assume the worse.
    Step 1:
    Pair (A, B) (C, D) (E, F) (G, H)
    Assuming they are opposites.
    So I took 4 tries already, 3 left.
    Step 2:
    Pair (A, C) (B, D)
    If A and C are both failing batteries, then B and D are both working batteries and the iron works, so 2 tries taken.
    But assuming the worse, If A and C, and B and D: are opposite batteries. Then either pair up A and D, or B and C — It will guarantee success, taking 3 tries.

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