User:JumpDiscont

likely errors in Phyti's explanation:

"All actions are independent, ..."

Prob(card #2 is the A card) = Prob(card #3 is the A card)

Prob(card #2 is one of the 50 blanks that gets removed) = Prob(card #3 is one of the 50 blanks that gets removed)

Prob([card #2 is the A card] and [card #2 is one of the 50 blanks that gets removed])

=

Prob([card #3 is the A card] and [card #3 is one of the 50 blanks that gets removed])

=

0

<

1/(51*2*52)

<=

Prob([card #2 is the A card] and [card #3 is one of the 50 blanks that gets removed])

=

Prob([card #3 is the A card] and [card #2 is one of the 50 blanks that gets removed])

Thus, which card is the A card is not independent of the host's action.

If you just mean [the actions are independent of each other] and

you're not regarding [which card is the A card] as an action, then:

What about [which card is p1]? That is also not independent of the host's action.

If you're regarding that as also not an action, then it seems to

me that there's only one thing you're regarding as an action.

This would make "All actions are independent" trivial and not sufficient, since

[Does p1 win?] and [Does r win?] both depend on things that you'd be regarding as not actions.

With "All actions are independent" gone, "there is no conditional probability."

[doesn't follow from anything else is your explanation] and [is a false assertion].

If the "player makes a random 2nd guess" , then the win ratio certainly will be 1/2.

However, that applies no matter what the probabilities are for each of the two:

If the prize has [a 0.1% chance of being in Z and not in Y] and [a 99.9% chance of

being in Y and not in Z], a random guess still has only a 50% chance of being correct.

For the game Rick Block most recently described, Rick Block chooses switch every time.

"having no knowledge of A location."

Do you regard [a 0.1% chance of the prize being in Z and not in Y]

and [a 99.9% chance of the prize being in Y and not in Z] as

[no knowledge of the prize location], because it still could be in either?

If no, then "no knowledge of A location" is what you're trying to show.

"The host randomly reveals the 0-cards, ..."

Mathematically speaking, this is correct, but it's possible that your interpretation of it is wrong.

When either [fixing at most one of [[which card is the A card], [which card is p1]]]

or [fixing the same card as both],

[which cards the host reveals] indeed is a non-degenerate random variable.

However, when [which card is the A card] and [which card is p1] are fixed

as different cards, [which cards the host reveals] becomes degenerate:

Fixing [which is the A card] and [which is p1] as different cards leaves

a set of 50 that has probability 1 of being the set the host reveals.

"Comparing p1 and r will" only "show there is no advantage to switch"

if one does it incorrectly.

I could expand here, but I've pointed out several other things too, so for this I'll wait on

whether your comparison of p1 and r is roughly "51 = 51" or has something significantly more.