# The frog puzzle

So here's the puzzle. You're poisoned in the jungle and the only way to save yourself is to lick a special kind of frog. To make matters worse, only the female of that species will do. Licking the male frog doesn't do anything. The male and female frogs look identical. The only difference is that the male frog makes a sound and the female is silent.

So you run through the jungle and spot a frog in front of you. Before you could start running towards it you hear a sound behind you. You turn around and spot two frogs there. There's only time to run to one side.

Now, the best course of action is to run towards the two frogs and lick both. The reasoning is that there are 4 possible combinations of two frogs and knowing that one of them is male eliminates only one of those possibilities. Of the remaining three, two of them have at least one female frog. This gives you a $\frac 2 3$ chance of survival as opposed to a $\frac 1 2$ with the single frog.

Now here's my problem. The reason this works is because you don't know which frog made the sound. If you did, you'd have a $50\%$ chance with the other one. But wouldn't that imply that, if you for some reason turned around earlier to see which one made the sound, you would decrease your chances of survival? What's the explanation here?

• (Note that your calculation of the $\frac23$ assumes that the two frogs' sexes are independent of each other, which may or may not be accurate....) – Greg Martin Mar 5 '16 at 1:39
• [The only difference is that the male frog makes a sound and the female is silent.] It's usually the reverse. // How are you modeling the sound making? What is the underlying process? If making a noise makes a frog male, then not making noise over time increases likelihood of it being a female. The chances for the group of 2 frogs are not 2/3 since the fact that you heard the noise from the back puts extra weight on male/male possibility as you are approximately twice as likely to hear from a pair of males than from one – A.S. Mar 5 '16 at 2:11
• @A.S. The problem statement really wasn't that vague. I take slight offense to that. Any kind of story-based statement of a problem is inherently open to nit picks. Picking the most obvious model is assumed. – Luka Horvat Mar 5 '16 at 2:29
• The statement IS vague as probability problems are quite sensitive to formulation and interpretation. I almost immediately thought of a male frog emitting a sound at some low intensity $Poisson(\lambda)$. Then probability of hearing a sound in time $t$ given MF is $e^{-\lambda t}\lambda t$ and given MM is $e^{-2\lambda t}2\lambda t$. MF:MM a priori odds are 2:1, so odds after hearing a sound are $e^{\lambda t}:1$. In the other direction, F:M odds are $e^{\lambda t}:1$ too - so you are indifferent. Same if you know exactly which frog made a sound. Did you mean my "radar gun" interpretation? – A.S. Mar 5 '16 at 2:48
• @flea This is not a math problem per se as there are no frogs in math. This is a story to be modeled as a probability question. "You know that one of two of them is male" is highly unspecified because the process that yielded this knowledge matters. A LOT. I answered this straightforward question assuming a very reasonable process. You are under false impression that questions come from a limited bank of theoretical questions you have already encountered and are able to match to a new story. You must think that $x^2+1=0$ has no solutions because, well, it's straightforward that $x^2\ge 0$. – A.S. Mar 5 '16 at 4:57

In the two-frog scenario, the event "One croak was heard" is not the same as the event "at least one frog is male".

There are eight possibilities: the "left" frog can be female or male, the "right" frog can be female or male, and exactly one croak was emitted or not. Assuming males croak with probability $p$ while you are in their presence, and everything is independent, the following table enumerates the eight possibilities and their probabilities: $$\begin{array}{c|l|c|c|c|c} \text{Outcome} & \text{Probability} &\text{One croak?}&\text{F present?}&\text{At least one M?}\\ \hline FF0 & \frac14 & & Y&\\ FF1 &0&Y&Y&\\ FM0 &\frac14(1-p) &&Y&Y\\ FM1&\frac14p&Y&Y&Y\\ MF0&\frac14(1-p)&&Y&Y\\ MF1&\frac14p&Y&Y&Y\\ MM0&\frac14(p^2+(1-p)^2)&&&Y\\ MM1&\frac14\cdot 2p(1-p)&Y&&Y\\ \end{array}$$ Using the above table, the probability that you survive given you heard a croak is $$P(\text{F present}\mid\text{one croak})=\frac{P(FF1,FM1,MF1)}{P(FF1,FM1,MF1,MM1)}=\frac{0+\frac14p+\frac14p}{0+\frac14p+\frac14p+\frac142p(1-p)}=\frac1{2-p}.$$ This makes intuitive sense, because if males croak all the time ($p=1$), then for sure the other frog is female; if males croak but very rarely, then it's a coin toss whether the other frog is female. Note that the above probability is never smaller than $\frac12$, so the two-frog lick is always a better strategy than the one-frog lick.

Using the above table, the probability $P(\text{survive}\mid\text{at least one M})$ is properly calculated as $2/3$, but this result is neither here nor there, because you didn't observe that event.

As for your other question, we can modify the outcome space to specify whether the left frog croaked or the right one. Using a similar enumeration to the above, the prob that you survive given you heard a croak from only the left frog is: $$P(\text{F present}\mid\text{only left frog croaked}) ={P(M1F0)\over P(M1F0,M1M0)}={\frac12p\cdot\frac121\over\frac12p\cdot\frac121+\frac12p\cdot\frac12(1-p)}=\frac1{2-p},$$ exactly the same probability as when you didn't know which frog croaked. The probability of survival given that only the right frog croaked is the same. Conclusion: knowledge of which frog croaked does not lower your probability of survival.

• This is what I got as well under similar model. The final probability is also equal to the probability that the frog in front is $F$ making you indifferent between 1 frog no croaks and 2 frogs one croak. – A.S. Mar 5 '16 at 6:26
• @A.S. Great observation. The conditional probability that a frog is female given that you have not heard it croak is $P(F0)/P(F0,M0)=(\frac12\cdot1)/(\frac12\cdot1+\frac12\cdot(1-p))=1/(2-p)$. – grand_chat Mar 5 '16 at 6:48
• This is always the best approach to probability questions that have seemingly non-intuitive features in them. As your last equation shows, once we know which frog did croak, probability that other frog is female is higher than that it is a male; because we're looking for not just any male frog but a non-croaked male frog, which has less probability than former one. This is part-reason why turning back earlier and see the frog that makes the sound didn't affect our chance of survival. – Alistair Mar 5 '16 at 10:59
• I think this is approaching the problem at the wrong time. A frog croaked, you turned around and now you're calculating your current chances of survival. – Luka Horvat Mar 5 '16 at 13:43
• @Luka In simplest terms: if you randomly sample families with 2 children, ask a question: "is at least one kid male?" and get an affirmative, then chances of the other being a girl are $2/3$. If you, on the other hand, randomly sample children from 2-children families and that child is a male, then chances of his subling being a girl are $1/2$. In both cases, you just KNOW that one of the kids in the family is a guy, but information comes to you from different processes. Same with Monte Hall. – A.S. Mar 5 '16 at 20:06

There are (at least) two ways of looking at this question, which can lead to different answers and has led to discussions that mix the two different viewpoints:

1. The view as in grand_chat's answer where the premise is "You are in a forest with one frog left, two frogs right" and the question "Given that you hear a croak on the right, what is the chance that you survive if you lick the frogs on the right". Here, the probability of croaking plays an essential role in the probability determination.

2. The view that takes all information in the question as a given, where the premise is "You are in a forest with one frog left, two frogs right, and you have heard a croak on the right" and the question is "What is the chance that you survive if you lick the frogs on the right?". Here, the croaking is a given and does not play a role in the probability determination; it is equivalent to the given of 'there is a male on the right'.

In a physicists language, the ensembles that are considered are different: view 1 has the ensemble of all systems with a frog left and two frogs right with the question of the correlation between a measurement of a croak on the right with finding a female on the right; view 2 has the ensemble of all systems with a frog on the left and a two frogs on the right, at least one of which is male, with the question of the probability of a female on the right.

View 1 is more closely related to the real world, but it does require some assumptions/knowledge about the croak probability distributions, and is explained well in grand_chat's answer. View 2 is more abstract and probably more in line with the intent of the 'puzzle', where no knowledge of croaking probabilities is needed.

Now to answer your question in view 2, by knowing which frog croaked, you are filtering your ensemble, which is populated by 'MM', 'MF', and 'FM' pairs, to the new ensemble where there are only 'MM' and 'MF' pairs (assuming the left of the pair is the male). So, the knowledge of knowing the position of the male filters away the also favourable situation of 'FM', therefore your chance inside this ensemble (this premise) is reduced with respect to the original.

The answer is 50% regardless of if you know which frog croaked. If you don't know which frog croaked, then we must consider that the state MM has a multiplicity of 2, ie it is twice as likely as either MF or FM, because there are two ways that this result can exist. It will be more clear if we assign the croaking frog its own notation. Lets call the croaking frog Mc, this gives the states: McM, MMc, FMc, and McF, which are all equally as likely, and of which 50% contain a female.

This problem is equivalent to having a friend flip 2 coins behind your back and telling you that at least coin has come up heads, clearly the probability of the second coin being heads as well is 50%, as if it is the only coin flipped. If your friend shows you one of the coins and it is heads, this changes nothing about the probability of the other coins state.

Why is this a problem to you?

Suppose you had a lottery with 4 people. There are 3 blue pieces of paper and 1 red. Whoever draws the red will be killed. You figure your chances of surviving are 3 in 4. Person A (not you) is called forward and draws a paper. It's blue. So now you figure your chances of surviving have dropped to 2 in 3. How does knowing something lower your chances? Well, because it eliminates possibilities.

So why is this a problem with the frogs?

There are 8 possibilities for the gender of 3 frogs.

MF|F Back: GOOD, Front GOOD

FM|F Back: GOOD, Front GOOD

FF|F Back: GOOD, Front GOOD

So if you lick the front your chances of surviving are 4/8 = 1/2. If you lick the back your chances are 3/4.

If you hear a croak you eliminate 2 possibilities and are left with:

MF|F Back: GOOD, Front GOOD

FM|F Back: GOOD, Front GOOD

FF|M Back: IMPOSSIBLE

FF|F Back: IMPOSSIBLE

Licking the front gives your chances at 3/6 = 1/2. Licking the back gives your chances at 4/6 = 2/3.

If you turn and see the first frog croak you've eliminated 4 possibilities and are left with:

MF|F Back: GOOD, Front GOOD

FM|M IMPOSSIBLE

FM|F Back: IMPOSSIBLE

FF|M Back: IMPOSSIBLE

FF|F Back: IMPOSSIBLE

Now either you lick the back and your chances are 2/4 =1/2 or you lick the front and your chances are 2/4 = 1/2.

• If I had a $1/4$ chance of survival and now I have a $1/3$ then surely my chances haven't dropped but increased. – Luka Horvat Mar 5 '16 at 2:11
• Your chance of dying have increased. Your chance of survive was 3/4 Now it's only 2/3. – fleablood Mar 5 '16 at 2:12
• Right. I still don't think this is a valid argument. A person picking a ticket removes it from the pool and alters the system. The frogs don't change gender and my turning around certainly doesn't either. – Luka Horvat Mar 5 '16 at 2:16
• In other words. If I knew that the first ticket was blue, it would only serve to benefit me since I could pick it. Yes, seeing that the ticket it blue adds information that would benefit me, but removing a ticket from the pool has a much worse effect. – Luka Horvat Mar 5 '16 at 2:20
• We don't know which one was the male. We only know there is at least one male. There are three equally likely ways that could occur. Two have female frogs. If you hear 99,999 croaks there are 100,001 thousand equally likely ways this could occur. 100,000 involve females. It is a near certainty that if you hear 99,999 croaks there's a female. That isn't faulty reasoning. There is only one way the remaining frog can be male but there are 100,000 ways it can be female. – fleablood Mar 5 '16 at 6:59

The riddle restriction that you do not know which frog has crocked is unnecessary and misleading.

Even if you had seen clearly that it was the left frog that crocked. The chances of the right frog to be a female would be 67%!. How can it be? …. Because at a MM situation the chances that the left frog will crock are only ½ of the chance (100%) it will crock at a MF situation. So if the left frog has crock. It is twice as probable that its partner is a female.

To show it clearly, lest make a additional arbitrary rule: When 2 males are sitting together, the older frog will crock and the young will stay silent. Now we have 6 possible scenarios. M-old-crock / F-young M-young-crock / F-old F-young / M-old-Crock F-old / M-young-crock M-old-Crock / M-young M-young / M-old-Crock

If the left frog had crocked then it have equal probability to be: M-old-crock / F-young M-young-crock / F-old M-old-Crock / M-young

And 2/3rd of the possibilities involve the 2nd frog to be a female.