[HN Gopher] Perplexing the Web, One Probability Puzzle at a Time
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Perplexing the Web, One Probability Puzzle at a Time
Author : isaacfrond
Score : 21 points
Date : 2024-09-02 12:17 UTC (3 days ago)
(HTM) web link (www.quantamagazine.org)
(TXT) w3m dump (www.quantamagazine.org)
| gosub100 wrote:
| my favorite problem, which I originally read here on hn:
|
| > I have two children. One is a boy born on a Tuesday. What is
| the probability I have two boys?
|
| Apparently the answer is 13/27
|
| https://math.stackexchange.com/questions/4400/boy-born-on-a-...
| lupire wrote:
| This is an example of how probability problems are ambiguously
| or misleadingly stated, to make them seem harder than they are.
|
| The top answer on SE explains why the wording you chose is
| nonsensical.
| Terretta wrote:
| SPOILER
|
| (Note that the article gives the answer, and discusses arriving
| at the answer. So go read it first.)
|
| Article opens with a puzzle. This is commenting on the article's
| explanation .. avoiding some words in the comment just so this
| spoiler doesn't spoil at a glance, you'd have to think about the
| following before reading:
|
| Curiously, this article doesn't touch on an instantaneous and no-
| maths intuition about what can (or what cannot) come up as the
| set after all draws including your first. There are a finite set
| of possible sets in the container. One interesting set is ruled
| out by the first draw: there is a set it _cannot_ be. With that
| set ruled out, you 're at least (sets/(sets-1)) more likely the
| next draw isn't from the missing and interesting set that is
| intuitively known to you.
|
| Further, if one did do this in the real world, it's intuitively
| evident that the missing set and its inverse would be more likely
| chosen for the container due to their interestingness. (The
| article alludes to this when it points out the sets are not coin
| flips, they are deliberately selected by a human.) That tips the
| odds even more in your intuition's favor of how the probabilities
| must balance.
|
| So I'd argue this one is solvable by psychology as an alternative
| to math.
| lupire wrote:
| Ypur first argument is interstellar but doesn't help solving,
| because the set ruled out is exactly balanced (cancelled) by
| the first red ball being removed from al the other sets.
|
| The second paragraph just says that guesses can sometimes be
| right, which of course is true, but doesn't tell us which
| guesses are likely to be correct. Wha y you claim as
| intuitively obvious at all, and I think most people would
| _avoid_ those sets. And it 's not the same problem anyway,
| because in the problem the sets are chosen randomly according
| to a stated rule, not based on a personal bias.
| Terretta wrote:
| Exactly right, I'd missed: _" number of red balls by picking
| a number between zero and 100 from a hat."_
|
| Or, rather, when they talked about this being different from
| a coin flip, I misremembered that the set was a human choice.
| tzs wrote:
| > [...] imagine that instead of starting with 100 balls, you
| start with 101 balls in a row. Pick a ball at random. Then color
| the balls to the left of it green and the ones to the right of it
| red. Throw that ball away, leaving 100 balls.
|
| > Then pick a second ball at random. That ball corresponds to the
| first ball in the original problem. The problem tells you that
| you picked a red ball, so it was to the right of the ball you
| threw away. Now pick a third ball. This ball is either to the
| left of the first ball, between the first ball and the second, or
| to the right of the second. In two of the three possibilities,
| the third ball is red. So the probability that the ball is red is
| 2/3.
|
| I think that explanation will confuse a lot of people. Let's say
| the first ball picked is ball 94 (numbering from 1 to 101 left to
| right). So 1-93 end up green and 95-101 are red. Let's say the
| second ball is 98.
|
| The third ball then must be from one of these three intervals:
| green 1-93, red 95-97, red 99-101. Those intervals have lengths
| 93, 3, and 3. OK, so now we've got three intervals which sounds
| like the situation the explanation is talking about, and two of
| those three give a red ball, so the probability the second ball
| is red is 2/3.
|
| But that is clearly wrong, since the probability of the ball
| being picked from an interval is proportional to the length of
| the interval. There is a 93/99 chance that ball 3 is green and
| only a 6/99 chance it is red for these particular intervals.
|
| It needs to be made clear that what you have to look at is all
| possible ways to pick the three balls.
|
| Maybe something like this. I'm going to call the position of the
| ball that divides red and green D, which is an integer in
| [1,101], P1 the position of the first colored ball picked, also
| an integer in [1,101] not equal to D, and P2 the position of the
| second colored ball, also an integer in [1,101] not equal to D or
| P1.
|
| Imagine a list of every possible legal (D, P1, P2) 3-tuple. Note
| that each consists of 3 distinct integers in [1,101]. Partition
| that list into several sublists, where two of the 3-tuples, (D,
| P1, P2) and (D', P1', P2') are in the same sublist if and only if
| they contain the same numbers. E.g., (5, 2, 12) and (2, 5, 12)
| would be in the same sublist.
|
| Each sublist will have 6 of the 3-tuples, one for each
| permutation of the three distinct numbers shared by all the
| 3-tuples in that sublist.
|
| Each of those 6 permutations corresponds to a different ordering
| of D, P1, and P2. For example (5, 2, 12) corresponds to the
| ordering P1 < D < P2 and (2, 5, 12) corresponds to D < P1 < P2.
|
| Here are the 6 possible orderings, and what the outcome is when
| the balls are drawn: D < P1 < P2 P1 red, P2 red
| D < P2 < P1 P1 red, P2 red P1 < D < P2 P1 green, P2 red
| P1 < P2 < D P1 green, P2 green P2 < D < P1 P1 red, P2
| green P2 < P1 < D P1 green, P2 green
|
| In the 3 cases where P1 is red, P2 is red in 2 of them and green
| in 1. That gives a 2/3 chance of P2 being red if P1 was red.
|
| This is true in all of the partitions of our list of all legal(D,
| P1, P2), so is true in all cases.
| smusamashah wrote:
| What if there were just 4 balls instead of hundred. Second
| probability of 2nd ball being of same color will be higher only
| if the distribution of first was higher.
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