Solution to The Impossible Bet

Solution to The Impossible Bet


Ok, so you remember the setup: 100 people
have had marked dollar bills put randomly into 100 boxes, and every person, one at a
time, gets to look in fifty of the boxes, and you all collectively win only if every
single person finds their own bill. $100 dollar payout to each of you if you win, nothing
if you lose. In the last video I asked whether or not you should take the bet. Did you make the right choice? I mean, clearly, if all 100 of you gamblers
just choose 50 boxes at random when you go into the room, the likelihood of each of you
finding your bill is one half, so the probability of ALL of you finding your bills and winning
the whole game is one half times one half times one half… a hundred times. And one
half to the power of 100 is about 0.0000000000000000000000000000008. Basically zero. You’re more likely to roll
9 one-roll yahtzees in a row – and then pick the king of hearts at random out of a deck
of cards. You could do a little better by trying to
correlate which boxes you pick – like, if there were only two people and two boxes,
they should each pick different boxes, because if they pick the same box only one of them
can possibly find their own bill, and they’ll automatically lose. But this kind of strategy
only improves your odds a tiny little bit, and it has less benefit the more people there
are. However, despite all this, it turns out that
there’s a strategy that results in you gamblers winning over 30% of the time! This strategy
is SO GOOD that you’re more likely to win using it than if just TWO people out of a
hundred picked randomly, since one half times one half is twenty-five percent. What we’ve forgotten is that you can obtain
information FROM the boxes themselves, because the room is exactly the same each time. The
first box will always have the same-numbered dollar in it, and so will the second box,
the third, the fifty-seventh, etc. All you have to do is use the dollar in one
box to tell you which box to go to next. This might not seem like a great advantage, but
it’s HUGE. The trick is to start with the box that corresponds
to your number. Your dollar probably isn’t IN that box, but the bills inside will take
you on a random path through the boxes, a scavenger hunt where each box tells you where
to go next. You can think of this as linking up boxes
in chains, connecting them based on the bills inside. Any random arrangement of the boxes results
in a different set of chains, some long, some small. Short chains are ones like “box 30
says go to box 82, box 82 says go to box 5, 5 says go to 30.” That’s a chain of link 3.
Long chains link up more boxes before circling back – and the LONGEST possible chain links
up all 100 boxes. But each chain ALWAYS circles back, since there are finite boxes. And of course, if you end up back where you
started within 50 boxes, you’ve won, because that means the previous box had a dollar bill
that told you to go your starting point, and you started on your number, so the previous
box’s dollar bill was YOURS! And that’s why this strategy is so good – because
when the boxes are randomly arranged, about 30% of the time there will be NO chains connecting
more than 50 boxes, so not only will you win, but so will everyone else – as long as they’re
following the same strategy! The magic of this solution isn’t that it’s
better odds for any single individual to find their dollar (it isn’t) – it’s that all of
the successes and failures are forced to happen together – everyone wins together, everyone
loses together – your fates are linked by using the numbers inside the boxes to guide
you along invisible chains… towards victory.

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About the Author: Sam Caldwell

100 Comments

  1. You know, maybe an important information would have been like… "You can try how many times you want". I feel that's a pretty important piece of information in a riddle like this.

  2. Ehm, if you can talk to the others, there is a 50% chance of winning, no matter how many people play.

    First opens 50 boxes, every time he opens a box, he says the number of the bill.
    Everybody knows now if their bill was called and is in the first 50 boxes or wasn't called and is in the second 50 boxes.

    The second guy even could open the second half but open his own box instead of one and everybody now knows exactly which box his dollar is in.

    Or did I get something wrong here?

  3. Doesnt this all go to shit as soon as someone takes the first doller and breaks a chain?

  4. the same matrix is used to decode passwords of whatever kind. just replace the boxes with letters and numbers. also it is used in logistics. where to go first and next and so on and even with conditions to comply.

  5. Except that the boxes are not numbered, and therefore must be able to agree on a numbering during the draw. But indeed, no contrary mention, so implicitly, it's possible. And maybe it is what is interesting, you have to interpolate to do math.

  6. I don't understand how this "strategy" is useful. Each individual still pulls a chain of 50. If they hit a cycle, it doesn't help them. What am I missing?

  7. Is there any other strategy with a higher chance to win? Or equivalently, is 31% the highest possible probability for any n>1?

  8. I've seen this riddle tons of times before, but this is the first video that really explained how it worked well enough.

  9. if you are allowed to speak to each other before each of you picks a box you have a chance of 50%. the first guy does that chain thingy and then he tells the second one if he saw his number. if yes then number two knows where 2 is and he then looks up all left over boxes nobody looked in. then one of number one and number two tells number 3 where the 3 is and so on. only if the first guy didnt get his number(wich has a chance of 50%) they lose, otherwise they win.

  10. But if you find out a Empty box and if you cant contact or see with other. Or simple if number in cash but not in box.

  11. It all comes down to how you arbitrarily choose to number the boxes in the beginning. If the players are not allowed to communicate with each other once the game begins (so you can't just open a box with one of your 50 guesses, announce "this box has 4 in it" and then the person who needs 4 comes forward and picks that box), then how could you all agree on a numbering system, how could you all agree on which box is box #4 and which is box #7? This strategy won't work because individually each and every one of them will be individually just picking 50 at random if the boxes are unlabeled and it is up to the players to arbitrarily decide on a numbering of them. Though the odds are not .5^100, it's actually 50^100*50!/100! or 50/100 times 50/99 times 50/98 times 50/97…. times 50/51 if on finding the right box, a player can take it out of circulation for the others.

  12. This is really useful information for the next time time I'm stuck with a 100 people searching boxes for money as a bet. #relatable

  13. I dont understand the bet so the solution makes no sense to me. And if you can look inside the boxes why not just look for your own marked bill and take it? What am I missing here?

  14. I love this channel because I love being the only person I know to know how to solve and win at theoretical gambling scenarios, riddles, and puzzles, that I will literally never be asked to be a part of. 😀

  15. i have a better strat 50 % win guaranted -_-".
    Since the box are the same for everyone the first seek the number 1 and 2 and open the 50 first box.
    If he don't find his number = lose.
    If he find his number he then tell the second participant if he found his number is the first 50 box. If not the second go to the 50 last box.
    The second go grab his number and seek the number 3 …

    Who need 31% when you can have 50% ?

  16. But what if there is one chain of boxes with 49 boxes so they all win but also one chain with 51 boxes. You said "you either win all together or lose all together" but that's simply not true?

  17. It would be 0.3118% you forgot that the first person has to choose randomely from 100 boxes. If they choose correctly, then it is a 31.18% chance.

  18. But if you open a box and its not your bill then everyone losses. How are you going to know what is inside the box so you can follow the number and go to the other boxes ?

  19. Thas alotta bills man, 100 uniquely marked notes each time a person goes so i can keep track? They sure are generous… What do you mean i have to put them back in the exact order they came in?

  20. if too random, add rules

    its a bonkers concept at first sight, but it can be thought of as adding your own neat and tidy scaffolding to a problem, if the problem is too messy by itself to navigate around. it will be orderly, based on your rules, even when you're generating something from random junk numbers

  21. You get all 100 people to take turns looking at 1 unique box, then one at a time call out the number on the bill and give the box to the person that owned it. Works 100% of the time and they don't have to look at the rest of the 49 boxes.
    You see, the problem with your description is you didn't mention that they had to find the bill on their turn through the method of looking in the box themselves on one of their 50 tries.

  22. Isn't there a problem though? As soon as a bill is taken out of a box, the chain is broken, and how can the next person (since it's just one at a time) know how to follow the chain?

  23. Wouldnt it be 30% of 30% of 30% …. 100 times ?? Which is equal to( 3/10)^100 < (1/2) ^100??

    This method makes the chance for one person as 30%, for everyone to win it should be (3/10) ^100 ? I may be wrong xd

  24. So you take the note away because you need to, then theres nothing in the box. How is another person gonna contimue the chain if theres no note there

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