# 100 chocolate bars! [duplicate]

A boy was taken to a magical land of wonder and chocolate, but sadly, a disease broke out in this land! He has to face off the evil Dr. Vanscosh, who caused the disease to break out, in a probability test.

Dr. Vanscosh gives the young boy 100 chocolate bars, 50 injected with deadly poison. He allows the boy to sort them in 2 bags, and the boy knows which has been poisoned. The boy can place as many of the 100 in each bag, as long as neither bag is empty!

Then the Dr. plays a dirty trick: he blindfolds the boy and then he swaps the bags around (either 1 or 2 times, meaning there's a 50/50 chance of them being in the same spot or swapped)

The boy, still blindfolded, then needs to take a bar and eat it, he cannot feel the bags, he has to stick his hand in the bag, and take a chocolate bar. If he dies, all chocolate in the world does. If he lives, chocolate will be saved forever and the Dr will be forced to eat his own poison.

What would have been the best way for the boy to organize them? What will his chances be afterwards..?

• I can't resist linking to my favorite puzzle answer ever (CTRL+F "Marble Jars".) Warning; page contains some coarse language. – Lopsy Dec 21 '14 at 16:31
• One bit could still be made clearer in this puzzle: the method of bag selection. The solution "safe bar alone in one of the bags, then find the bag with one bar" immediately comes to mind but should be ruled out by the puzzle in order to keep it within the realm of probability. Perhaps change the containers to boxes, and make it clearer that he has to choose, e.g., either the (closed and shuffled) left or the right box first, then stick with that selection when drawing a bar. – jscs Dec 21 '14 at 19:55
• @Josh He's blindfolded when he picks... – warspyking Dec 21 '14 at 20:16
• Blindfolded, he could feel through the bags, or pick one up and see how heavy they are. The blindfold alone doesn't explain how the bag selection is made. – jscs Dec 21 '14 at 20:36
• Well @Josh Caswell Edited :D – warspyking Dec 21 '14 at 22:24

I assume neither of the bags may remain empty and all chocolate bars must be sorted in one of the bags.

The boy should put one of the good chocolate bars in one bag and all the other chocolate bars in the other bag. This will give him (and all chocolate in the world) maximum chances to survive, namely $\frac{74}{99}$.

• If one bag contains 1 and the other contains 99, why can't he just choose the 1 and be safe? – Rand al'Thor Dec 21 '14 at 16:25
• @rand al'thor he's blindfolded – warspyking Dec 21 '14 at 16:28
• @randal'thor good point... reading the question, my impression is that the boy, being blindfold, cannot choose a bag. But if he could feel it with his hands... :-) – GOTO 0 Dec 21 '14 at 16:28
• Feel it with his hands - exactly! @warspyking Maybe edit the question to clarify? – Rand al'Thor Dec 21 '14 at 16:31
• @rand Done, it should be clear now! :D – warspyking Dec 21 '14 at 16:37

Well, he could of course put half of the good chocolate in each bag, but at the bottom and cover it with the poison-free chocolate. No matter how the bags are swapped around, if he picks one from the top it will always be safe to eat! This is probably considered cheating, however.

Another possibility: put all bars except a poison-free one in one bag, then the boy can determine by touch which bag has only one bar. That bar is safe to eat. Apparently, this was already mentioned and deemed illegal in the comments.

• It wasn't tagged lateral-thinking but +1 for creativity/cleverness – warspyking Dec 21 '14 at 18:40
• Your first part seems the most obvious solution to me – erdekhayser Dec 22 '14 at 22:29
• @erdekhayser To me too, hence the answer. – 11684 Dec 22 '14 at 22:30

It is not given explicitly that the total of chocolates within the jar should be 100. Why not put only the good chocolate bars in both the jars(25 each or whatever combination) and keep all the other bars in his hand?

This way, he can discard the poisoned chocolate bars when he is given a chance to choose one from any bar right? His chance of survival will be 100% in that case.

Assuming the doctor plays 'dirty tricks' - I'd like to elaborate on the GOTO 0's answer.

While is he correct that with a RANDOM choice of 1 or 2 swaps you end up at odds of 74/99 (1/2 x 1 + 1/2 x 49/99), you can also assume that the doctor might have figured out that scenario as being the optimal for you...

Being into 'dirty tricks', his choice of swaps might not be entirely random (meaning: despite what the OP said, the 50/50 odds might not actually be 50/50). In that case, with the (admittedly fairly obvious) scenario with the 1 'clean' bar in a seperate bag, you may get the tables turned on you, and only end up with 25/99 odds.

Also, the riddle does not state the doctor doesn't see what the boy does, so in any case with a non 50-50 rate, any evil guy with half a brain will be able to figure out which is the least favorable option and give the boy that bag. Given the 'Dr.' title, I'll assume he's clever enough.

Therefor the best achievable rate is 50%, which is formed by any combination with equal amounts of good and poisoned candy in both bags.