Comments by "\/" (@joebazooks) on "TED-Ed"
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Eddie no, man. you can disregard one of the frogs in the clearing altogether because you already know that 1 out of 2 are male. the probability of the 2nd frog being female or male is 50/50 and this is independent from from 1 of the frogs being male. just because we don't know which of the 2 frogs is male, doesn't skew the probability of the 2nd frog towards female.
the only time the following possibilities and probabilities (MM25%FF25%FM25%MF25%) come in to play is if you are calculating the probabilities of 2 unknown frogs in a row.
for instance, it's like having two die. in order to survive you need two 6s, but we know already that one of the die is 6, so you're chance of rolling 6 and surviving is 1/6, not 1/36. NOW, if you had to roll two consecutive 6s, then the chance of that happening is 1/36, but we already know one of the die is a 6, so we are only rolling once and the probability and results of that roll are independent from the predetermined roll...
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Eddie Fair enough that you think it's a problem of conditional probability, and if the problem was intended to be an example of conditional probability, it's a terrible example at that, because it's not in fact conditional probability in the way that you imagine.
Further, conditional probability is no different than statistical probability, insofar as with conditional probability you're simply taking into account foreknowledge of one event (A) that effects the probability of another event (B).
Yes, the Monty Hall problem is a problem of conditional probability because knowing that the host has to reveal a door with a goat behind it (event B), after the contestant has already chosen one of three a doors (event A), influences the probability of the contestants second choice (event C). In this case, the knowledge concerning event B is a condition that effects the probability of event C.
Let's look at the conditions and events of the frog problem...
(condition A) knowing of the species 50% are male, 50% are female
(condition B) only female frogs possess the chemical antidote
(event A) spotting a frog on a stump
(event B) hearing a male croak from a group of two frogs
(event C) choosing which to go
None of these events or conditions effect the probability in the way that you imagine, nor in the way that the video proposes--so don't feel bad about yourself for screwing up because the creators screwed up too.
Let me illustrate this for you with a simplified analogous experiment for the sake of clarity and your understanding.
a. Get 4 pennies
(the 4 of which will represent the entire species of frog, which is analogous to condition A earlier in my comment because in order for the fact that 50% of the species is male and 50% of the species is female to be true, the species must contain an even number of frogs or pennies)
b. Place 2 Heads-up and the other 2 Tails-up
(this is analogous to condition B earlier in my comment because we know that 50% of the frogs are male and 50% are female)
c. Remove 1 Heads-up penny from the group (which will be your Male in the group of two)
d. Put on a blindfold
e. Mix up or have somebody else mix up the remaining 3 (2T and 1H) pennies that are leftover after step c.
d. Remove 1 penny from the leftover group of 3 while blindfolded
(the penny that you remove blindly is 66% likely to be a Tails-up penny because in the leftover group of three pennies is 2 Tails-up and 1 Heads-up, which leaves you with a group of 1T and 1H.)
e. Distribute 1 of the 2 pennies that remain in the game to the left
(50% likely to be T and 50% likely to be H)
f. Distribute the last penny to the right
(also 50% likely to be T and 50% likely to be H)
g. Remove your blindfold
h. record result
i. Repeat steps A to H indefinitely 100,000 times
The frog problem is conditional probability, but not in the way you believe it is.
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Alkaski yes, i have heard of conditional probability, but the "possibilities" that you're accounting for in your calculation are not the actual possibilities for the frog scenario. the actual possibilities for the frog scenario are as follows: DM, MD, DF, FD; whereby D represents the given male.
as to whether or not there is a female frog on one side and or the other, so that you would be able to survive:
(in the clearing) DM, MD, DF, FD - 2/4 50%
(on the stump) F, M - 1/2 50%
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Leo Smith you're wrong. flipping a coin once does not change the probability if you were to flip the coin again. if the condition was to flip two consecutive heads or two consecutive tails, only then does the probability lessen. let's use dice as an example. if i were to use only one die, and i rolled a 6, re-rolling this die does not increase the likelihood that i will roll a 1, 2, 3, 4, or 5, on the next roll. this is known as the gambler's fallacy. if we had two die, however, and the condition for our survival was rolling a 6 with each die, this would be more unlikely. BUT, we already know that one of the die has been rolled and it's a 6, therefore the roll of the next die becoming any of the six numbers is equally likely. it is not all of a sudden more likely to be a 1, 2, 3, 4, or 5.
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i live in canada and always use the oxford comma if practicing good grammar. from my experience, it usually avoids more confusion than it creates. the oxford comma makes sense, period.
i.e.
a) 'he had stolen, bought and sold, and squeezed lemons.'
b) 'he had stolen, bought, sold, and squeezed lemons.'
c) 'he had stolen, bought, sold and squeezed lemons.'
personally, i believe a) and b) express two different notions, if only slightly; whereas c) is just confusing and borderline nonsensical.
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icemd24 In short, the method they use treats effectively identical outcomes as different possibilities, and it uses these additional possibilities, which are irrelevant to the outcome, in its calculations of the probability. Likewise, they disregard relevant information, which in turn has generated an inclination to consider irrelevant information as relevant information in their calculations. In other words, it would be no different from declaring:
(1) buses travel to only two different destinations, Destination A (DA) and Destination Z (DZ);
(2) there are as many buses travelling to DA as there are travelling to DZ;
(3) in order to survive, you must get to DZ;
(4) you notice that there is 1 bus at the bus stop on the right, and 2 buses at the bus stop on the left;
(5) you know that 1 of the 2 buses at the bus stop on the left is travelling to DA;
and, after asking 'What are your odds of survival if you go right?' and 'What are your odds of survival if you go left?', then claiming:
(1) that because you do not know which bus at the bus stop on the left is travelling to DA, even though 1 of the 2 buses is indeed travelling to DA, there are more unique possibilities, and therefore more effectively unique outcomes, and for this reason your odds of getting to DZ in order to survive are 33.33% better than if you were to go to the bus stop on right whereby there is also only 1 bus with an unknown destination.
I hope that this is a clear description. Let me know if you have any more questions!
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