What is the probability?

Question

Say there are one hundred hamburgers sitting in front of you. One of the burgers contains a poison that will kill you within twenty four hours of eating the burger, but your face will develop spots instantly so you know you've been poisoned. Another one of the burgers contains the antidote, which will cure you of all poisonous effects. However, it is a quick working antidote and it must be taken only after you have eaten the poisonous burger for it to work. Your friend, who is sitting next to you, eats ninety-nine of the burgers for you. Your back is turned, so you cannot see if he is poisoned while he is eating the burgers. When he is done eating the burgers, you can see he is has no spots and is not currently poisoned (if he ever was). There is one burger left. What is the probability it contains poison?

Answer 1

There are four cases possible (before information about your friend is known)...
I. Friend drinks antidote only, leaves poison for you.
II. Friend drinks poison only, leaves anitidote for you.
III. Friend drinks antidote, then poison.
IV. Friend drinks poison, then antidote.

The probability of case I is 1/100
The probability of case II is 1/100
The probabilities of cases III and IV must be the remaining 98/100 (since the cases listed are mutually exclusive and exhautive, if you need to get technical), and since both cases are equally likely, the probability of case III is 49/100, and the probability of case IV is 49/100.

Once you know your friend is not poisoned, this excludes cases II and III, and therefore excludes 50 instances from the probability calculation. In the remaining 50 instances, 49 can occur by your friend drinking the poison then the antidote (case IV), and 1 can occur by your friend leaving the poison for you (case I). The probability of your burger containing poison is therefore 1 in 50, or 0.02.

Technically, using conditional probability, you need to know the probability of the last burger being poisoned (event A), given that your friend is not poisoned (event B).

The probability of A is 1/100 (just case I)
The probability of B is 50/100 (cases II and III)

By rule, P(A|B) = P(A)/P(B)
= (1/100)/(50/100) = 1/50 = 0.02

One thing is for certain, the answer is NOT 1 in 100. The whole point of these problems is that you have to update a probabilty once you receive additional information.

Answer 2

Let's define the following events:

L = last burger is poisoned
H = friend is healthy after eating 99 burgers

We are looking for p(L|H), the probability that the last burger is poisened given that our friend is healthy after eating 99 burgers. We know a few things:

p(L) = 1/100.
p(H|L) = 1. (friend won't eat the last burger for sure)
p(H|~L) = 98/99 * 1/99 = 98/99² (poisoned burger has to be among the first 98 burgers, and the antidote, which is one of the 99 remaining burgers, must comme immediately after. ***

p(H) = p(H|L)*p(L) + p(H|~L)*p(~L) = 1 * 1/100 + 98/99² * 99/100 = 197/9900.
p(H&L) = p(H|L) * p(L) = 1/100.

So p(L|H) = p(H&L) / p(H) = 1/100 ÷ 197/9900 = 99/197 = about 50%.

*** This assumes the antidote must be eaten immediately after the poisonous burger. If the antidote can work even if it isn't in the next burger (it only has to be eaten sometime after the poison) then p(H|~L) becomes 98/99 * 1/2 = 49/99 because 1) the antidote must not be the last burger (p=98/99 given that the poison isn't last either) and 2) given that both special burgers are among the first 99, one has equal probability of being ahead of the other. Then we have p(H) = 1 * 1/100 + 49/99 * 99/100 = 1/2, so p(L|H) = 1/100 ÷ 1/2 = 1/50 - same answer as Ethan below me. So this makes a huge difference because ther's a much greater chance that the friend will have taken the poison and be saved by the antidote.

Answer 3

1/50.
It's an exercise in calculating a posterior probability from the conditional probabilities and the prior probabilities. I could write out the formula, but this can also be explained in plain language. The key point is that, a priori, the probability that the last one is normal is 98/100, but half of those cases (49) get excluded because we know that *if* he has eaten both the poison and the antidote, then he did so in the right order. This is what increases the probability that the last one is poisonous: there are 49 cases left of the last one being normal and 1 case of the last one being poisonous (the last one can't be the antidote). So, 1 in 50.

Answer 4

There are 100*99 = 9900 ways that the hamburgers could have one with the poison and one with the antidote. Since the friend cannot have failed to eat the one with the antidote, that leaves out 99 of the possible outcomes, or 9900 - 99 = 9801. Therefore, since there are 99 outcomes with the last burger poisoned, the answer is 99 / 9801, or 1 in 99, not 1 in 100. Tricky problem. I wonder if anyone else is going to get this right?

Addendum: If the antidote has to be taken after the poison is eaten, not before, then there are just 4950 outcomes, not 9801, and the answer is 99/4950, or 1 in 50.

Answer 5

More data is needed. For antidote to work properly when should it be taken? Immediately after eating the poisoned burger or after eating how many burgers?

Answer 6

More clarity is needed. As poised the question cannot be solved:
It's not clear whether the antidote shows any side effects, or how quickly should the two be eaten one after the other. Nor is it clear how fast does the poison develop symptoms.

Answer 7

1 in 100.
It was from the beginning until the end.
If, in his 99 burgers, he got the poisoned one and the antidote one, you still wouldn't know if the 100th one was poisoned or not, since he'd show no signs of poisoning.
If, however, the 100th one is poisoned, you're screwed since your friend has already eaten the antidote...

Answer 8

probability = number of poised burgers / total number of burgers
= 1/100