Is the following true (It seems obvious to me that it's not... but... a PhD in physics, Derek Abbott, seems to think others explanation at end of post):

Someone flips 3 coins on the table, they are then covered with paper so I can't see. Two of the coins are on the left of the table, one coin is on the right of the table. I go to the left of the table and start to slide the paper back slowly, eventually revealing the first of the two coins, if it's tails I just have the person re-flip, but if it's heads then I predict the other coin is tails and continue pulling the paper back to reveal tails 66% of the time.

Ted-ed (a TED talks division) published a video recently via Physics PHD Derek Abbott called "The Frog Riddle" - you can watch it here: https://www.youtube.com/watch?v=cpwSGsb-rTs and now I'm very confused, please tell me their video is just wrong.

I don't understand mathematically how they are arriving at their conclusion. If this were any other youtube video in my feed I would just wave it off as erroneous, but TED is a very large and very famous organization with lots of editors and is also notoriously intellectual. Also I was having a hard time finding many who disagreed in the youtube comments.

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    $\begingroup$ You should stop drinking TED-flavoured Kool-aid and look up a thread on reddit (or here) that explains why the video's reasoning is wrong. In a nutshell, $MM$ is more likely (x2) to produce a crock as a opposed to $MF$ and that makes chances the same as 1 frog and zero crocks. $\endgroup$ – A.S. Mar 5 '16 at 9:10
  • $\begingroup$ @A.S. +1, Sorry it's just that he is a PhD in physics so I assume I am wrong (as I did not even go to college (pursued business) haha). Just wanted to get multiple opinions with the scenario I posed. $\endgroup$ – Albert Renshaw Mar 5 '16 at 9:12
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    $\begingroup$ As with the Monty-Hall paradox, there is a simple way to cast out any doubt: build the tree of possible outcomes, and count the number of branches. $\endgroup$ – Graffitics Mar 5 '16 at 9:12
  • $\begingroup$ See reddit.com/r/askscience/comments/48br02/… and my answer: math.stackexchange.com/questions/1683658/the-frog-puzzle/… (see other comments as well). If you assume $\lambda t$ small, you'll get 1:1 odds of a female. $\endgroup$ – A.S. Mar 5 '16 at 9:12
  • $\begingroup$ @Graffitics but the monty hall problem relies on a guaranteed pre-set of 2 goats and 1 car, and also the pre-knowledge (by the hose) of which doors contain which objects. $\endgroup$ – Albert Renshaw Mar 5 '16 at 9:13

The coin scenario you describe differs from the frog scenario in the video in that you peek at a particular one of the coins and try to predict the other coin on that basis. You're right that this doesn't work and the probability for tails is still $\frac12$.

By contrast, in the video you hear a croak from the general direction of the two frogs, so you only know that one of them is male, but not which one. The video does make a bit of a mistake in assuming that one male and two males would have been equally likely to give a single croak while you were listening. However, if you did somehow get the information that at least one of the frogs is male in a manner that doesn't distinguish between the possibilities of one male or two males, then indeed the probability of there being one male would be $\frac23$.

  • $\begingroup$ So if I was in the forest and I heard but did not see the croak it's $2/3$, but had I turned my head a few seconds earlier and also seen which croaked all of a sudden it's $1/2$? This doesn't make much sense to me, it seems as if real world objects can behave like quantum mechanics, just by observing you change another's odds? $\endgroup$ – Albert Renshaw Mar 5 '16 at 13:03
  • $\begingroup$ @AlbertRenshaw, turning your head earlier doesn't change the probability. Also, surviving probability in the first case is $\frac{2}{3}$ only if male frog's croak probability when he sees you is $\frac{1}{2}$. Look at grand_chat's answer. $\endgroup$ – Alistair Mar 5 '16 at 14:48
  • $\begingroup$ @AlbertRenshaw: No, you're talking about a single identifiable croak, and as I wrote that's not the assumption that leads to $\frac23$. It's $\frac23$ if you get precisely the information that at least one of the frogs is male. It's conceivable that you might get that information from croaking, e.g. if you listen long enough that you'd be sure to hear croaks if there's a male, but you don't know enough about the frequency of croaking to tell whether it's one or two males croaking. In that case turning your head and looking would allow you to identify both frogs correctly. $\endgroup$ – joriki Mar 5 '16 at 17:34
  • $\begingroup$ I can't think of any reasonable non-quantum process (with assumption of independence btw frogs) generating croaks that won't make MM more likely to produce a croak than MF: math.stackexchange.com/questions/1683658/the-frog-puzzle/… for example. Hearing $1$ croak from $2$ frogs is like hearing $0$ croaks from $1$ frog. $\endgroup$ – A.S. Mar 5 '16 at 18:55
  • $\begingroup$ @A.S.: What's wrong with the process I proposed above? (By the way, the more fundamental incosistency to me seems that whatever process one imagines for the two frogs should also apply to the one frog, and would necessarily make it more likely that this frog is female since it wasn't heard croaking.) $\endgroup$ – joriki Mar 5 '16 at 18:56

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