There are 4 cups of liquid. Three are water and one is poison. If you were to drink 3 of the 4 cups, what is the probability of being poisoned? In Season 5 Episode 16 of Agents of Shield, one of the characters decides to prove she can't die by pouring three glasses of water and one of poison; she then randomly drinks three of the four cups. I was wondering how to compute the probability of her drinking the one with poison.
I thought to label the four cups $\alpha, \beta, \gamma, \delta$ with events 


*

*$A = \{\alpha \text{ is water}\}, \ a = \{\alpha \text{ is poison}\}$

*$B = \{\beta \text{ is water}\},\ b = \{\beta \text{ is poison}\}$

*$C = \{\gamma \text{ is water}\},\ c = \{\gamma \text{ is poison}\}$

*$D = \{\delta \text{ is water}\},\ d = \{\delta \text{ is poison}\}$


If she were to drink in order, then I would calculate $P(a) = {1}/{4}$.
Next $$P(b|A) = \frac{P(A|b)P(b)}{P(A)}$$ Next $P(c|A \cap B)$, which I'm not completely sure how to calculate.
My doubt is that I shouldn't order the cups because that assumes $\delta$ is the poisoned cup. I am also unsure how I would calculate the conditional probabilities (I know about Bayes theorem, I mean more what numbers to put in the particular case). Thank you for you help.
 A: You have 50%. If you drink the third cup means first and second are not poison. So when drink third you have just two cups, and one is poison.... 50%. I think this problem is about logic more than math... Thanks a Lot
A: Not sure why everybody uses such a complicated approach:
after drinking three cups, one remains. The chance that she is alive is equal to the chance that the remaining cup is the poison, which is one in four = 25%.
The sequence of drinking water and or poison is completely irrelevant.
A: The probability of not being poisoned is exactly the same as the following problem:
You choose one cup and drink from the other three.  What is the probability of choosing the poisoned cup (and not being poisoned)?  That probability is 1/4.
Therefore, the probability of being poisoned is 3/4.
A: Label the cups A, B, C, D. Now we can assume WLOG that the cups she drank are A, B and C.
There are only 4 scenarios according to which cup is the poisoned one. In 3 of the scenarios (poisoned cup = cup A, B, C respectively), she is poisoned. In 1 of the scenarios (poisoned cup = cup D), she is not poisoned. Therefore the probability is 3/4 = 75%.
Not sure why it is any more complicated than that.
A: The solution to this problem depends on how fast the poison works.   If the poison is slow acting, the previous solutions are OK.   But if the poison works instantly, there may be no opportunity to drink three cups.
The problem states that three cups are taken, so the first two cups can NOT be poison.   The third cup is chosen from a set of two with one poison and the other safe.   So the probability of survival = probability of poison = 1/2.
A: There are $4!$ permutations of $W_1W_2W_3P$. 
The only way to live is if $P$ is last, and there are $3!$ ways for this to occur.
So there are $4!-3!$ ways to die with probability $1-\frac{3!}{4!} = \frac34$.
A: NicNic8 has provided a nice intuitive answer to the question.  
Here are three alternative methods.  In the first, we solve the problem directly by considering which cups are selected if she is poisoned.  In the second, we solve the problem indirectly by considering the order in which the cups are selected if she is not poisoned.  In the third, we add the probabilities that she was poisoned with the first cup, second cup, or third cup.  
Method 1:  We use the hypergeometric distribution.  
There are $\binom{4}{3}$ ways to select three of the four cups.  Of these, the person selecting the cups is poisoned if she selects the poisoned cup and two of the three cups of water, which can be done in $\binom{1}{1}\binom{3}{2}$ ways.  Hence, the probability that she is poisoned is 
$$\Pr(\text{poisoned}) = \frac{\binom{1}{1}\binom{3}{2}}{\binom{4}{3}} = \frac{1 \cdot 3}{4} = \frac{3}{4}$$ 
Method 2:  We subtract the probability that she is not poisoned from $1$.  
The probability that the first cup she selects is not poisoned is $3/4$ since three of the four cups do not contain poison.  If the first cup she selects is not poisoned, the probability that the second cup she selects is not poisoned is $2/3$ since two of the three remaining cups do not contain poison.  If both of the first two cups she selects are not poisoned, the probability that the third cup she selects is also not poisoned is $1/2$ since one of the two remaining cups is not poisoned.  Hence, the probability that she is not poisoned if she drinks three of the four cups is 
$$\Pr(\text{not poisoned}) = \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{4}$$
Hence, the probability that she is poisoned is 
$$\Pr(\text{poisoned}) = 1 - \Pr(\text{not poisoned}) = 1 - \frac{1}{4} = \frac{3}{4}$$
Addendum:  We can relate this method to the first method by using the hypergeometric distribution.
She is not poisoned if she selects all three cups which do not contain poison when selecting three of the four cups.  Hence, the probability that she is not poisoned is 
$$\Pr(\text{not poisoned}) = \frac{\dbinom{3}{3}}{\dbinom{4}{3}} = \frac{1}{4}$$
so the probability she is poisoned is 
$$\Pr(\text{poisoned}) = 1 - \frac{\dbinom{3}{3}}{\dbinom{4}{3}} = 1 - \frac{1}{4} = \frac{3}{4}$$
Method 3:  We calculate the probability that the person is poisoned by adding the probabilities that she is poisoned with the first cup, the second cup, and the third cup.
Let $P_k$ denote the event that she is poisoned with the $k$th cup.
Since there are four cups, of which just one contains poison, the probability that she is poisoned with her first cup is 
$$\Pr(P_1) = \frac{1}{4}$$
To be poisoned with the second cup, she must not have been poisoned with the first cup and then be poisoned with the second cup.  The probability that she is not poisoned with the first cup is $\Pr(P_1^C) = 1 - 1/4 = 3/4$.  If she is not poisoned with the first cup, there are three cups remaining of which one is poisoned, so the probability that she is poisoned with the second cup if she is not poisoned with the first is $\Pr(P_2 \mid P_1^C) = 1/3$.  Hence, the probability that she is poisoned with the second cup is 
$$\Pr(P_2) = \Pr(P_2 \mid P_1^C)\Pr(P_1) = \frac{3}{4} \cdot \frac{1}{3} = \frac{1}{4}$$
To be poisoned with the third cup, she must not have been poisoned with the first two cups and then be poisoned with the third cup.  The probability that she is not poisoned with the first cup is $\Pr(P_1^C) = 3/4$.  The probability that she is not poisoned with the second cup given that she was not poisoned with the first is $\Pr(P_2^C \mid P_1^C) = 1 - \Pr(P_2 \mid P_1^C) = 1 - 1/3 = 2/3$.  If neither of the first two cups she drank was poisoned, two cups are left, one of which is poisoned, so the probability that the third cup she drinks is poisoned given that the first two were not is $\Pr(P_3 \mid P_1^C \cap P_2^C) = 1/2$.  Hence, the probability that she is poisoned with the third cup is 
$$\Pr(P_3) = \Pr(P_3 \mid P_1^C \cap P_2^C)\Pr(P_2^C \mid P_1^C)\Pr(P_1^C) = \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{4}$$
Hence, the probability that she is poisoned is 
$$\Pr(\text{poisoned}) = \Pr(P_1) + \Pr(P_2) + \Pr(P_3) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$$ 
A: A good way to think about such problems is to ask yourself the opposite question: what is the probability that I will not get poisoned?
\begin{align*}
 \Pr(\text{not poisoned}) &= \Pr(\text{not poisoned on first glass}) \cdot \Pr(\text{not poisoned on second glass}) \cdot \Pr(\text{not poisoned on third glass}) \\
 &= \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} \\
 &= \frac{1}{4} \\
\end{align*}
It follows that
\begin{align*}
 \Pr(\text{poisoned}) &= 1 - \Pr(\text{not poisoned}) \\
 &= 1 - \frac{1}{4} \\
 &= \frac{3}{4} \\
\end{align*}
A: I thought it would be instructive to try a different approach. 
If you drink three cups, that tells me for sure that two of the cups you drank (the first two) were not poisoned - otherwise you would not have gotten to the third cup.
So the poison must be in either cup 3 or cup 4, and since it is equally likely to be in any of the cups, there is a 50% chance of that happening.
Having survived the first two cups, you now have a 50-50 chance that cup 3 is poisoned (because it's either cup 3 or cup 4). To survive drinking cup 3, you need to beat those odds as well. 
Multiplying these probabilities, you once again get 0.25 for the chance of surviving.
Of course it's the same answer - but I thought this would give additional insight.
A: Wow, people are making this complicated.
There are 4 cups. One is poisoned. She picks 3. There are 3 chances that she will pick the poisoned cup out of 4. Therefore, the probability is 3/4.
This assumes that she does not pick a cup, drink it, then put it back and someone refills it before she picks another.
I'm also assuming that she either picks 3 cups before drinking any of them, or that if she dies before picking 3 cups, that we treat that as if she picked enough to fill out the 3 at random. Otherwise the probability is impossible to calculate, because if the first or second cup is poisoned, she doesn't "pick 3 cups".
You can do all the permutations and bayesian sequences, but as others have shown, they all come to the same answer.
If she picked, say, 3 out of 6 cups and 2 are poisoned, I don't see how to do it other than with combinatorics. But maybe I'm missing an easy way.
