Does It Matter?

[DJA]

This has been posted before... as you can see by the posts you should switch...

The best explanation I have read that is very intuitive is this... Say there are 100 cards 99 red and 1 black and you have to pick the black card. You pick 1 and he does you a favor and turns over 98 red cards... would you switch your pick?

[karlson]

I think you guys are full of it.

You should switch whether he flipped a random card or not.
Think of it this way: you are given the option of sticking with your card or picking both of the other ones as a set. By flipping a red card, he is giving you the second option.

I am anxious to hear why I am wrong.

[lorinda]

Three people have a bet...
Player A picks a card a random
Player B picks a card a random (and flips it over if you like)
Player C picks the other card
You are telling me player C is favourite now to win the game?

[heihojin]

Whether or not the propositioner could have turned over a black card is irrelevant, so long as the probability of that event occurring is less than 1.

Let the following events be defined:

A: You select the black card on your first attempt.
B: The propositioner turns over a red card.
C: You hold the black card at the end of the game.

The question is, given that the propositioner turns over a red card, do you increase the probability of event C occurring by switching or not switching?

The probability of event C occurring given that event B occurs is expressed as P(C|B).

If you do not switch:

P(A) = 1/3
P(C|A*B) = 1

P(!A) = 1 - P(A) = 2/3
P(C|!A*B) = 0

P(C|B) = P(A) * P(C|A*B) + P(!A) * P(C|!A*B) = 1/3 * 1 + 2/3 * 0 = 1/3

If you do switch:

P(A) = 1/3
P(C|A*B) = 0

P(!A) = 2/3
P(C|!A*B) = 1

P(C|B) = P(A) * P(C|A*B) + P(!A) * P(C|!A*B) = 1/3 * 0 + 2/3 * 1 = 2/3

Because P(C|B) is greater for switching than not, you should switch. Notice that this is regardless of the probability of event B occurring, so long as it is non-zero.


heihojin

[irchans]

I loved your very clear argument. I am worried that your formula

P(C|B)=P(A)*P(C|A*B)+P(!A)*P(C|!A*B)

might not always be correct.

I think the formula is correct if A and B are independent events because of the following reasoning (using ! for not and * for "and"):

Spitting the set B*C into two parts gives:

P(B*C)=P(A*B*C)+P(!A*B*C)

From the definition of | :
P(C|B)*P(B)= P(B*C),
P(A*B*C) = P(A*B)*P(C|A*B), and
P(!A*B*C) = P(!A*B)*P(C | !A*B).

Substituting gives

P(C|B)*P(B) = P(A*B)*P(C|A*B) + P(!A*B)*P(C | !A*B).

IF A AND B ARE INDEPENDENT, THEN THIS REDUCES TO

P(C|B)=P(A)*P(C|A*B)+P(!A)*P(C|!A*B).

A and B are independent if P(B)=1, but would they are not be independent if the "propositioner" has no knowledge of the location of the black card. Does the formula hold if A and B are not independent?

[irchans]

[lorinda]

Okay, Im gonna lose this argument even though I understand the concepts [img]/forums/images/icons/frown.gif[/img]

My notation is non-existent Im afraid having been self-taught only in probability so even though I can understand arguments, I cant follow notation [img]/forums/images/icons/frown.gif[/img]

In the original question , we have cards ABC where we always pick card A,
The propositioner is considered to know where the black card is and so removes a non-black card.

ABC A is black, propositioner removes B switch loses
ABC B is black, propositioner removes C switch wins
ABC C is black, propositioner removes B switch wins

This is elementary

Now to the random case

ABC A is black, prop removes B, and it is red switch loses
ABC A is black, prop removes C, and it is red switch loses
ABC B is black, prop removes B, game cancelled
ABC B is black, prop removes C, switch wins
ABC C is black, prop removes B, switch wins
ABC C is black, prop removes C, game cancelled

six equally likely events, two draws, two wins, two losses, what am I missing?

[lorinda]

The longhand version of the non-random case reads

ABC A is black, prop removes B, switch loses
ABC A is black, prop removes C, switch loses
ABC B is black, prop removes B, changes mind, removes C,win
ABC B is black, prop removes C, switch wins
ABC C is black, prop removes B, switch wins
ABC C is black, prop removes C, changes mind, removes B, win

4 wins, 2 losses, as we know

[karlson]

Of course you are right.
I was somehow incorporating the events where a black card is turned over into wins.
By the way, this is the same thing that heihojin's formula does, as far as I can tell.

[irchans]

I think lorinda's reasoning correct .

In the case where the propositioner always picks a red card, it is best to switch. (We can use lorinda's enumeration or heihojin's formula to show that this works.)

In the "randomly choosing propositioner" case, it does not matter whether you switch or not. (We can use case lorinda's enumeration or her three gambler post on 9/24 to understand this. Heihojin's formula seems to fail in this case due to lack of independence. Baggins explanation for this case also works.)

We can modify Heihojin's formula to make it work for the "random" case. If we don't assume independence, Heiholjin's formula becomes:

P(B) * P(C|B) = P(A*B) * P(C|A*B) + P(!A*B) * P(C|!A*B)

As Heiholjin points out,

P(C|A*B) = 1 and P(C| !A *B) =0 if you don't switch, so

P(B) * P(C|B) = P(A*B)
P(C|B) = P(A*B)/P(B) = P(A|B).

Recall that in Heiholjin's notation C = winning. In the "random" case, P(B) = 2/3 and P(A*B) = 1/3, so P(C|B) = 1/2. So you win 1/2 the time when you don't switch. You win the other half the time if you switch. So it does not matter which strategy you choose in the random case.