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A protagonist is set with four cups, three containing water and one containing poison and proceeds to drink three cups. (yes, the setup is similar to another question. However, the question is unique)

It is easy to agree that the chances of drinking poison are three in four simply.

Case: The odds are far, far worse if after drinking each glass the protagonist discovers that they are NOT poisoned and THEN continues.

Will someone please find the flaw in my method below:

The chance of drinking poison on the first glass is simply one in four. After discovering that they are not poisoned, the protagonist takes another glass. This time the odds are one in three. Still surviving unpoisoned the protagonist reaches forth again to take the third glass. This time the odds are one in two that the glass will contain poison.

The summation of events leads to the horrific probability of $$\frac{1}{4} + \frac{1}{3} + \frac{1}{2} = \frac{13}{12}$$ in favour of being poisoned on the third glass. Where strictly the probability of just the third glass being poison is 50%.

It is not correct to evaluate the four glasses in the second round in this case, since a) we have removed one glass, and b) strictly one of the three remaining glasses in the second round are poisoned. For this reason, it is a reverse Monty Hall problem.

Edit: It seems I have made a fundamental error in the above. Let me introduce the second example in the hopes that I may be able to find a correction.

A person contacts a sports betting agency and makes a multi-bet over four matches. For the sake of the discussion let each team be evenly matched so that the odds of each team winning is approximately 50%, that is, for each separate match the probability of bet win on that match is 50%. Let us assume in advance that the fourth game is a nil-all draw so that it is equivalent enough to not playing, so that the setup is comparable with the first example. The probability of picking the first three matches correctly is 12.5% since there are eight possible outcomes of the three matches in total, only one of which results in my winning the bet. If any of the first three games are picked incorrectly, then my bet fails to progress.

In the case that I choose correctly in the first two matches and am down to the third match then, the chances of completing the bet are 50% at that point.

It seems like we can arrive at 12.5% with $$\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}$$ However, if we apply this method to the first example we find the probability of success would be $$\frac{1}{4}\times\frac{1}{3}\times\frac{1}{2}=\frac{1}{24}$$ which seems a little off? Is it? (actually, I think that may be the probability of drinking poison in each round if there is always one glass with poison regardless how many time we drink it.)

So, may we calculate the probability of not drinking poison at all as: $$\frac{3}{4}\times\frac{2}{3}\times\frac{1}{2}=\frac{1}{4}$$ ?

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    $\begingroup$ These are not the probabilities of events, they are conditional probabilities that condition on different things. $\endgroup$ May 14 '18 at 22:19
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    $\begingroup$ Think about what probabilities being greater than $1$ means. (It doesn't mean anything). Think about this. If there is a 50% chance of rain on saturday and 50% chance rain on sunday is there a 100% chance of rain on the weekend? If there is a 75% chance of rain on saturday and a 75% chance of rain on sunday is there an 150% of rain on the weekend? Can you work this out yourself. There are four cases Rain on Saturday, Rain on Sunday, Rain on both, or rain on neither. What are those probabilities? Can you work this out. $\endgroup$
    – fleablood
    May 14 '18 at 23:12
  • $\begingroup$ @fleablood The example you give is not a direct comparison. If it will rain either Saturday or Sunday and it does not rain Saturday then there is not 100% chance it will rain Sunday. $\endgroup$
    – Willtech
    May 15 '18 at 2:55
  • $\begingroup$ "The example you give is not a direct comparison. If it will rain either Saturday or Sunday and it does not rain Saturday then there is not 100% chance it will rain Sunday." That IS entirely the point! The probability that it will rain on the weekend is not the sum. And the probability of that the second is poison is not the probability of what is left. $\endgroup$
    – fleablood
    May 15 '18 at 3:16
  • $\begingroup$ Check out this related question: Three white and one red ball probability $\endgroup$
    – farruhota
    May 15 '18 at 4:53
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To do the calculation, you need to include the probability of surviving to the $n$th round, then add up each probabilities for the various possible $n$s. You'll have three terms: probability of dying on first drink, second drink, and third, like so:

$$\frac{1}{4} + \frac{3}{4}\cdot \frac{1}{3} + \frac{3}{4}\cdot \frac{2}{3}\cdot\frac{1}{2} $$

which works out to $\frac{3}{4}$ again.

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"This time the odds are one in three" (emphasis mine)

"This time" is not all times.

The probability of the first glass being poison is $1$ in $4$.

The probability that the second glass is poisoned is $1$ in $3$ !!!!!!!IF!!!!!!! the first glass was not poisoned. If the first glass was poisoned the probability is $0$.

So the probability that the second glass was poisoned is $\frac 34 \times \frac 13 + \frac 14 \times 0$.

The probability that the third glass is poison is $1$ in $2$ if the probability of the that the first and second glasses were not poison. And $0$ if either of them were.

The probability of being poisoned in three glasses is

$\frac 14 + (\frac 34*\frac 13 + \frac 14*0) + (\frac 34*\frac 23*\frac 12 + \frac 34*\frac 13*0 + \frac 14*0) = ...????...$ well, do the math and see.

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  • $\begingroup$ I understand what you are saying in the case that three glasses are taken without knowledge of the result thereby not removing any glasses from the set. In the case used, when the second round is begun we have survived the first round with a probability of 3/4, however, the probability of surviving the second round since the result of the first round is known is only 2/3, leaving the inverse case for the probabilities of being poisoned. Simplify, I perform three seperate tasks with know probabilities of failure each. $\endgroup$
    – Willtech
    May 15 '18 at 3:03
  • $\begingroup$ Since we are not poisoned in the first round there is 0% probability that the first glass contains poison when we begin the second round. $\endgroup$
    – Willtech
    May 15 '18 at 3:16
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    $\begingroup$ But you can't throw away the $1$ in $4$ cases where the first mug was poisoned!. This is akin to saying that if you are alive by the time you are $95$ years old, the probability of you being alive when you are $96$ is 75%. Therefore you conclude that a baby's probability of living to $96$ is $75\%$. If the baby lives to be $95$ then the probability of dying in the first $95$ years is $0$!. That's true, but to get there, you have to get there first! And the probability of that is less than $10\%$. $\endgroup$
    – fleablood
    May 15 '18 at 3:22
  • $\begingroup$ So yes, if you get past the first two round the probability of being poison is $1/2$. But you don't also add the $1/4$ and $1/3$ because you already got past it. And if you haven't gotten past it, you don't count the probabilities as $1/3$ or $1/2$ because those are only the probabilities AFTER you get past it. $\endgroup$
    – fleablood
    May 15 '18 at 3:24
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    $\begingroup$ Before you begin drinking. You probability of being poisoned in three glasses is $\frac 34$. But IF you survive the first glass the probability of being poisoned by the third glass is $\frac 23$. But IF you survive the first two glasses the probability of being poisoned in the third glass is $\frac 12$. That is not a contradiction as we are taking the probabilties with different known information. What we can't do is assume we know something before we do. (The probability that glass 2 is poison is not 1/3 unless we know glass one is not)... tbc... $\endgroup$
    – fleablood
    May 15 '18 at 3:31
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Initial Probability:

Probability of being poisoned by the third glass = probability of being poisoned in glass 1 + probability of being poisoned in glass 2+probability of being poisoned in glass 3=$\frac 14 + \frac 14 + \frac 14=\frac 34$.

Takes glass one and survives.

Adjusted probability:

Now that we know glass one is safe our adjusted probability is:

Probability of being poisoned by the third glass = probability of being poisoned in glass 1 + probability of being poisoned in glass 2+probability of being poisoned in glass 3=$0 + \frac 13 + \frac 13=\frac 23$.

Things got better. But that's makes sense because we have ruled out $\frac 14$ of the outcomes. (All bad but we haven't ruled out any of the good outcomes).

Takes glass two and survives.

Adjusted probability:

Now that we know glass one and two are safe our adjusted probability is:

Probability of being poisoned by the third glass = probability of being poisoned in glass 1 + probability of being poisoned in glass 2+probability of being poisoned in glass 3=$0 + 0 + \frac 12=\frac 12$.

Things got better. But that's makes sense because we have ruled out $\frac 12$ of the outcomes. (All bad but we haven't ruled out any of the good outcomes).

....

Okay, so suddenly an ignorant time traveler bursts onto the scene and says:

"I took the probability of the first glass being poisoned from the time before he drank it. It was $\frac 14$. I took the probability of the second glass being poisoned from the time after he drank the first glass but before he drank the second. It was $\frac 13$. And I took the probability of the third glass being poisoned from the time after he drank the second glass but before he drank the third. It was $\frac 12$. If you add them up you get $1 \frac 1{12}$ he gets poisoned."

So, we say, you can't add up probabilities from different times and conditions and have it mean anything.

"You can if you are an ignorant time traveler" says the ignorant time traveler.

You can, but you would be wrong, we say.

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  • $\begingroup$ Okay, so, I go to some sports betting agency and take a four-game multi. Let us assume all teams are even so the probability of each match being picked correctly is almost 50%. Let us assume in advance that the fourth game is a nill-all draw so that it is equivalent enough to not playing it so that the setup is comparable albeit with a different probability. Is the probability of my picking the first three matches correctly not 12.5%? (in which case I accept that I have probably made a fundamental error in the OP.) If any of the first three games is picked incorrectly my bet fails to progress. $\endgroup$
    – Willtech
    May 15 '18 at 21:46
  • $\begingroup$ "If any of the first three games is picked incorrectly my bet fails to progress. " I do not know what you mean. You have a $\frac 18$ probability of doing this successfully. If you first game is incorrect we through it out. There is a $\frac 12$ chance of doing that. And there is $\frac 12$ chance of not doing that. If we do that $\frac 12$ of not happening then we have a $\frac 12$ chance of that game being wrong. That is a $\frac 12*\frac 12 = \frac 14$ and so we have a $\frac 12 + \frac 14=\frac 34$ chance of failing in the first two games. .... tbc..... $\endgroup$
    – fleablood
    May 15 '18 at 22:26
  • $\begingroup$ If that didn't happen then we have a $\frac 12$ of the third game failing. But that can only happen if the first two games were successful and there is only a $1-\frac 34 = \frac 14$ chance of that. So the chance of the third game failing but the first two passing is $\frac 14*\frac 12 = \frac 18$ So the chance of me failing =chance failing first game + chancepassing first game failing second + chance passing first two failing third = \frac 12 + \frac 14 + \frac 18 =\frac 78$. We simply don't calculate whether I pass or fail the second game UNLESS I pass the first game. $\endgroup$
    – fleablood
    May 15 '18 at 22:31
  • $\begingroup$ And my chances of passing to first game are only $\frac 12$. So we have a fifty percent chance of not calculating it at all. So we only have a 1 in 4 chance of both calculating it and failing. $\endgroup$
    – fleablood
    May 15 '18 at 22:32
  • $\begingroup$ That's pretty much what I was thinking, I think. $$12.5\% probability = \frac{1}{8}$$ $\endgroup$
    – Willtech
    May 16 '18 at 4:00
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To elaborate on the comment by @spaceisdarkgreen, indeed after surviving the first drink the odds of being poisoned in the next round are 1 in 3, but you also have to compute the odds of arriving at this situation, which are 3 in 4.

(Odds of dying after second drink)=(odds of dying after surviving first drink) X (odds of surviving first drink)=$(1/3)\times (3/4)=1/4$

(Odds of surviving two drinks)=(odds of surviving second drink after surviving first drink)X(Odds of surviving first drink )=$(2/3)\times(3/4)=1/2$

(Odds of surviving three drinks)=(Odds of surviving third drink after surviving two drinks)X(Odds of surviving two drinks)=$(1/2)\times(1/2)=1/4$.

Why do you think that knowing you have survived would change the odds? The argument given does not depend on how quickly you die

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  • $\begingroup$ Nitpick. Odds $\ne$ probability. $\endgroup$
    – fleablood
    May 14 '18 at 23:13
  • $\begingroup$ Yeah, but it kind of went with the argument $\endgroup$
    – Philip Roe
    May 14 '18 at 23:16
  • $\begingroup$ I fail to see how the probability of surviving two drinks can be greater than one. $\endgroup$
    – Willtech
    May 15 '18 at 22:08
  • $\begingroup$ "I fail to see how the probability of surviving two drinks can be greater than one. " They aren't. Where on earth do you think anyone claimed they were???? $\endgroup$
    – fleablood
    May 15 '18 at 22:40
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As you have obtained a probability greater than one, you should recognise that you have done something wrong, somewhere and somehow.   Let us look closer.

Label the events that a cup is poisoned $A,B,C$ for the first three and, if you need it $D$ for the fourth.   We are assured that only one will be, and that there is no bias as to which.

Note that $\tfrac 13$ is the conditional probability for the second cup being poisoned given that the first is not; ie. $\mathsf P(B\mid A^\complement)=\tfrac 13$. Et cetera.

Now, to employ the additive rule, we need to apply it to a union of disjoint events.   Then we may use the definition for conditional probability to apply the product rule for non-independent events.

$$\begin{split}\mathsf P(A\cup B\cup C) &= \mathsf P(A\cup (A^\complement \cap B)\cup (A^\complement \cap B^\complement\cap C)) \\ &= \mathsf P(A)+\mathsf P(A^\complement \cap B)+\mathsf P(A^\complement \cap B^\complement\cap C)\\ &=\mathsf P(A)+\mathsf P(A^\complement)\cdotp\mathsf P(B\mid A^\complement)+\mathsf P(A^\complement)\cdotp\mathsf P(B^\complement\mid A^\complement)\cdotp\mathsf P(C\mid A^\complement\cap B^\complement)\\ &= \tfrac 14+\tfrac 34\cdotp\tfrac 13+\tfrac 34\cdotp\tfrac 23\cdotp\tfrac 12\\&=\tfrac 34\\[2ex] \mathsf P(D^\complement) & = 1-\mathsf P(D)\\&= 1-\tfrac 14\\ &=\tfrac 34\end{split}$$

Why, now, of course we should anticipate that these probabilities would have the same value; they do measure the same event.

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  • $\begingroup$ If the probability of being poisoned is greater than 1.0, then the probability of anything else happening is negative! $\endgroup$
    – Philip Roe
    May 16 '18 at 13:24

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