
Trump winning majorities in 5 states is overvalued on PredictIt - ioannes
http://www.flightfromperfection.com/trump-wins-50-in-5.html
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gus_massa
> _It 's safe to assume independence in this case because the five primaries
> (Connecticut, Delaware, Maryland, Pennsylvania, Rhode Island) will all occur
> at roughly the same time, in the same time zone. It would be unlikely for
> the result of one of these primaries to influence the result of another, so
> we will assume that there's no inter-primary influence._

No way they are independent. If the previous day he appears in all the mayor
TV networks killing a kitty and drinking the blood, the probability of winning
will decrease in all the states, so there is a lot of correlation.

(You can come up with a more realistic scenario.)

~~~
ioannes
Doesn't independence mean that, for two events, the occurrence of one event
doesn't effect the probability of the other event? You can observe
correlations between two independent events, they just can't be causally
connected.

In your example, the press conference where Trump drinks kitty blood is
driving the change in probability for all primaries, so it's functioning as a
confounding variable.

You wouldn't expect the results of one primary to affect the results of
another (because they're happening near simultaneously), so I think it's safe
to assume independence.

~~~
gus_massa
Independence is a more strict condition. The probabilities must follow the
rule

P(A and B) = P(A) * P(B)

that is the rule that is used in the article.

Let's pick an easier example. You buy two equal decks of cards while traveling
abroad. When you return home, you realize that it may be a deck with 48 cards
(123456789JQK)or a deck with only 40 cards (1234567JQK) without the "8" and
"9".

To make it simple, suppose that you magically know that you have a 50% chance
of having bought a 48 cards deck and a 50% chance of having bought a 40 cards
deck.

One of your friend come and start to shuffle both decks, one in each hand,
without mixing them.

If your friend picks a card from the right deck, what is the probability that
it's a "8"? (spoiler alert: There is a 50% that the deck has only 40 cards
(without the "8" and "9") and you can't pick the "8", and there is a 50% that
the deck has 48 cards and there is a 1/12 to pick the "8", so the probability
is: 1/2 * 0 + 1/2 * 1/12 = 1/24)

If your friend picks a card from the left deck, what is the probability that
it's a "8"? (again 1/24)

Are they independent? Whatever he does with the left decks doesn't affect the
right deck. To get less relation, you can use two friends, and put them in a
different room, and make them pick the card exactly simultaneously.

If the events were independent, the probability that both have picked an "8"
would be 1/24 * 1/24 = 1/576

But the correct answer is more complicated. There is a 50% that the deck has
only 40 cards (without the "8" and "9") and you can't pick the "8", and there
is a 50% that the deck has 48 cards and there is a 1/12 to pick the "8" in
each one, so the probability is: 1/2 * 0 + 1/2 * 1/12 * 1/12 = 1/288\. That is
2x bigger than if they were independent.

~~~
ioannes
Thanks for the example! I didn't understand it, but then I asked about it on
cross-validated and the cross-validated hive mind helped me get it:
[http://stats.stackexchange.com/questions/209057/independence...](http://stats.stackexchange.com/questions/209057/independence-
of-two-decks-of-cards-each-with-a-different-numbers-of-cards/209062#209062)

