# 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.

• Comments are not for extended discussion; this conversation has been moved to chat. – Jyrki Lahtonen May 16 '18 at 5:01
• You need to provide a little more information to actually answer this question correctly. Is the poison fast acting? If the poison is fast-acting, then being available to drink the third cup is added information that the first two cups were not poison. – jerrylagrou May 17 '18 at 16:27
• Possible duplicate of Probability of being poisoned – Xander Henderson May 17 '18 at 19:12
• @EricTowers not to mention the inference from Murphy's Law, which tells you that the probability is 1. – Michael Kay May 18 '18 at 7:13
• As someone really bad at math, gonna ask a dumb question here: Why is the probability not just 75%? – john doe May 19 '18 at 13:41

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 $$\Pr(\text{poisoned}) = \Pr(P_1) + \Pr(P_2) + \Pr(P_3) = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{3}{4}$$

• I was attempting to use Method 3 (which you've shown is the most tedious). Thank you for your very detailed answer, it helped me view the problem in different ways and understand it more clearly. Thank you. – Eli May 14 '18 at 13:37
• To be honest, Methods 1 and 3 seem like a distraction from Method 2, the only really reasonable way of calculating this. – jwg May 16 '18 at 8:51
• Both of the other methods are eminently reasonable. – Matthew Read May 17 '18 at 3:37
• Why is it that in method 3 it is necessary to include the probability that the preceding cups were not poisoned, but in method 2 the preceding cups are not necessary? Subtleties like that are what make probability both difficult and fascinating IMHO. – Jim W says reinstate Monica May 17 '18 at 20:43
• @JimW In method 2, the probability that the preceding cups were not poisoned is taken into account. The probability that the first cup is not poisoned is $3/4$. Observe that the probability that the second cup is not poisoned given that the first cup was not poisoned is $2/3$ since if the first cup was not poisoned only two of the three cups that remain are not poisoned. The probability that the third cup is not poisoned given that the first two cups were not poisoned is $1/2$ since if the first two cups were not poisoned then only one of the two cups that remain is not poisoned. – N. F. Taussig May 17 '18 at 20:52

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.

• @NicNic8 This version is clearer. – N. F. Taussig May 14 '18 at 2:30
• No idea why this isn't the accepted answer - it's the way I'd do it but it's also much clearer than the other. – arboviral May 16 '18 at 8:07
• @arboviral The other answer has better math, and can be applied generally. Consider instead that there was a fifth glass containing poison; which approach would you use? – Drunk Cynic May 16 '18 at 16:16
• Think of it this way: What is the prob that the poison cup is still full after drinking three cups? – DWin May 17 '18 at 2:23
• If there are three glasses of water and two of poison, the method analogous to this answer is you pick two glasses not to drink, and you survive if you happen to pick the two poisoned glasses, which occurs just one way out of the ten possible pairs of glasses you could pick. – David K May 20 '18 at 23:09

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.

• Agreed. Part of the SE ethos is to provided detailed additional information that could be used for other, less direct problems. (And to show off) – MarsJarsGuitars-n-Chars May 14 '18 at 22:08
• This is essentially the same as NicNic8’s solution. – wgrenard May 15 '18 at 0:56
• From a baysean perspective there is something fishy about the answer. After drinking three cups, she is either dead or alive. Talking about chance is futile! – Albert van der Horst May 20 '18 at 18:40
• Yes I agree. These top voted answers are a bit extreme it seems, even with my zero experience with any advanced level math and using only my high school school math logic. Yet sometimes you'll see some crazy "sounding" problem given by the questioner yet a few simple lines for a response that seems to solve everyone's issues, confusions. Good ol' SE – Hunter Frazier May 21 '18 at 10:38

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 $1-\frac{3!}{4!} = \frac34$.

• Just read Taussig's answer. This solution is more or less the same approach. – suneater May 14 '18 at 8:35
• This solution is more or less the same approach well I would not say so. This one has a different approach / methodology (and ultimately relies on similar concepts). – WoJ May 14 '18 at 10:34

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*}

• "Ask yourself the opposite question" is the important lesson here. +1 – joeytwiddle May 16 '18 at 5:34
• Now if one of them is the antidote... – John P May 18 '18 at 16:50

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

• This is a valid approach, under a different set of assumptions (namely that the poison is debilitating and acts faster than you can drink another cup) – Ben Voigt May 14 '18 at 23:55
• This is saying it is more a language issue than a math issue. 'If you were to drink' could mean the action is done or to be done. Interesting. – David May 15 '18 at 16:22
• There is a revolver handgun with 4 chambers, three are empty and one has a live round. If you put it to your head and squeezed the trigger three times, what is the probability of having a bullet in your head? Your answer addresses the gun question better than the poison question. – user1717828 May 15 '18 at 16:36
• hahaha! So funny! If I press three times the trigger means first and second was without bullet. Pressing third time maybe I got the bullet in my head, maybe not (I'm not lucky man! so for sure I have!). So 50% probability... – German Rodriguez May 15 '18 at 18:41
• Just make sure it is a revolver and not an automatic! – Eric Lippert May 15 '18 at 19:02

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.

• It isn't any more complicated, as NicNic8 and Aganju already answered. – Dronz May 15 '18 at 3:26
• It is more complicated if you want to solve very similar problems generally. Laying out all the scenarios and counting them quickly becomes unreasonable for larger numbers of cups with more than 1 left untouched. – Matthew Read May 17 '18 at 3:47

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.

• I think you are wrong. You are not calculating the probability of the first two cups not be poisoned. – sawa May 17 '18 at 3:25
• @sawa If we know that both (a) the poison acts instantly and (b) you drink 3 cups, the poison cannot have been in either the first 2 (they would have killed you before you drank the third). That means the probability of either of the first two cups being poison is 0. Obviously, this is intentionally misinterpreting the problem for the sake of humor; it's a joke. – Matthew Read May 17 '18 at 3:42
• I am not saying that your interpretation is wrong. That is not my point. My point is that you are only calculating the conditional probability (that the third one is poison) under the assumption that the first two were not poison. I am saying that you have to calculate the conditional part to get to that situation, which Floris does correctly in my opinion. – sawa May 17 '18 at 3:48
• @sawa: The probability of surviving until the third cup does not have to be calculated, it is postulated by the problem formulation, therefore 100% (P(x | x) = 1). – Ben Voigt May 18 '18 at 3:31
• Okay, I understand. – sawa May 18 '18 at 4:31

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.

• The OP stated that the person drinking the water was proving that they can't die, so I don't think it's fair to make your conclusion that drinking three cups itself is amount to showing that the first two cups are not poisoned. – Green May 16 '18 at 23:45
• Your interpretation of the question is the same as richard1941 and German Rodriguez, but I think they gave the wrong answer under that assumption, and you are the only one to give the correct answer under the assumption. – sawa May 17 '18 at 3:29
• This is not correct; it's almost like the reverse of the Monty Hall misunderstanding. In this version of the problem, the first two cups don't matter. They might as well not exist. The problem is reduced purely to two cups, one of which is poisoned. You drink only one of them -- what are your odds? 50%, not 25%. – Matthew Read May 17 '18 at 3:59
• Or to fully dive into the scenarios you wanted to lay out here -- there are four situations: (1) cup 3 is poisoned and you drink it (2) cup 4 is poisoned and you drink it (3) cup 3 is poisoned but you drink cup 4 (4) cup 4 is poisoned but you drink cup 3. 2/4 of those are deadly; 50%. – Matthew Read May 17 '18 at 4:02
• A potentially more intuitive reason it can't be 25%: In this version of the problem, you've gained definitive knowledge of absolute safety about the first two choices made. That MUST affect the odds. It is obviously not true that drinking two safe cups leaves you with the same odds as drinking two cups that could have included the poisoned cup. – Matthew Read May 17 '18 at 4:05

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.

• NicNic8's answer is even more simple, check that top answer. – user061703 May 19 '18 at 6:02
• For your drinking 3 out of 6 with 2 poisoned, just think of six spots for glasses in a row. The first three are to be drunk. The first poisoned glass has 3 safe locations (the last three) out of 6, so we're safe with probability 0.5. The second poisoned glass has 2 safe locations to land on out of 5, so we'll stay safe with probability 0.4. Combined: 0.2 probability of a safe three drink binge... – DJohnM May 19 '18 at 8:12
• Ooh, interesting. Yes, simpler than what I had in mind, which was: number of safe combinations divided by total number of combinations = 4C3 / 6C3 = 4 / 20 = 0.2. Same answer. – Jay May 19 '18 at 23:17
• @TrầnThúcMinhTrí I though NicNic8's answer was a shade more complicated than mine as it requires reversing the probability, but whatever. Not something that I would challenge someone to a duel to the death over. – Jay May 19 '18 at 23:19
• So the problem grows as more people think about it. Now we have sampling with replacement as a possibility. The probability of surviving on a single draw is 3/4. Because the non-poisoned cup is replaced, the overall probability of survival, at the beginning of the trial, is 3/4 x 3/4 x 3/4 = 27/64. It does not matter if the poison if fast or slow because the conditional probabilities are the same as the initial probability. – richard1941 May 21 '18 at 17:27