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You are standing in a forest and have just eaten a very deadly mushroom which will kill you in a short while. Fortunately magic frogs inhabit the forest of which a lick on the female frog cure all ills, while the male frogs do nothing but croak (the females do not croak). The females and males look identical and the probability of a frog being born a male is $50\%$.

You spot a frog at a distance ahead of you. Before you start running towards it, you hear a croak behind you. As you turn around, you see two frogs. You only have time to run and lick either the one in front of you or the two behind you. Which direction should you choose?


The question was inspired by watching a TED-Ed Youtube-video on it, and the solution is obtained by conditional probability, where we get a $1/2$ chance of survival if we run to the sole frog, while the survival probability is $2/3$ if we run to the two frogs behind us (the two frogs may be Male + Female, M+M and F+M, of which two of the cases guarantee survival).


Is everything fine? Is suggested solution and arguments absolutely clear? Or do you have some kind of dual thoughts? And if you do, but everything seems to be right, why can't you fully admit it?

Let us now imagine that at the same time our lucky guy that have just eaten poisoned mushroom hear male croak another lucky guy (probably his friend) eat another poisoned mushroom and he is looking at the two frogs and he see which one is croaking - so he knows which one is male. Obviously now our sharp-sighted friend doesn't care where he gonna run, because he has a 1/2 chance of survival in both directions. And as he doesn't care he can choose to run with his friend and what happens then... one of them is going to die with probability 2/3 and the other with 1/2? And if we repeat this experiment pure second guy gonna die much more just because of knowledge about which one is male??? Jeez... Mad, mad world! =) What is wrong with my thinking here?

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  • $\begingroup$ A few comments: in English, a coin has the sides 'heads' and 'tails'. However, I fail to see what that paragraph contributes to your question. If you wanted to formulate it in a normal probabilistic way, you could consider iid $X_i \sim \operatorname{Bern}(1/2)$ for $i = 1,2,3$ the three "frogs". Going left is then a binomial random variable you have some information about. $\endgroup$ – Therkel Jan 2 '17 at 9:51
  • $\begingroup$ @G.Artem I did quite some substantial editing: added the riddle to the question, removed some of the paragraphs and a bit more. I (hopefully) preserved your actual question and thoughts. If not, you should reject or roll-back the edit. :) $\endgroup$ – Therkel Jan 2 '17 at 10:25
  • $\begingroup$ The frog example is a contrived situation unlikely to be of actual use, but I think the probability argument is fine. I prefer to get my (safe and boring) mushrooms at a super market, where there are no magic frogs. A very roughly related probability puzzle is the 'Monte Hall' problem which you might enjoy reading about (if so, google). $\endgroup$ – BruceET Jan 3 '17 at 8:02
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The answer you provide for the initial problem based on conditional probabilities is too simple. Since you hear one croak behind you, the M+M scenario is not equally likely as the M+F or the F+M scenario. Suppose one frog croaks with probability $p$ in the time interval that you are running, and that the frogs croak independently, then the probability on hearing one frog croak in the M+M scenario is $2p(1-p)$, while in the M+F or F+M scenario it is $p$. So, if you turn around, the probability that one of the two frogs is female can be computed with Bayes' rule:

$$\begin{align*} & P(\textrm{FM or MF | one croak}) \\ = & \frac{P(\textrm{one croak | FM or MF})P(\textrm{FM or MF)}}{P(\textrm{one croak|FM})P(\textrm{FM}) + P(\textrm{one croak|MF})P(\textrm{MF}) + P(\textrm{one croak|MM)P(MM)}} \\ = & \frac{p\cdot0.5}{(0.25p+0.25p+0.5p(1-p)} \\ = &\frac{1}{2-p} > 0.5 \end{align*} $$ In other words, turning around is still a good idea, but the probability of meeting a female frog could by anywhere in the interval $(0.5,1]$.

For this calculation I assumed independence of the croaking. If it is more likely that a frog croaks if he hears another frog croak, the last term in the denominator gets smaller, and the chance of meeting a female frog goes up. If it is less likely that a frog croaks if he hears another frog croak, the last term in the denominator gets larger, and the chance of meeting a female frog goes down. In the extreme case that a frog never croacks if he hears another frog croack, the probability on meeting a female is still larger than $0.5$.

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