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[blueblimp]

It makes all sorts of expected value calculations unexpectedly easy.

I wouldn't be mad, doesn't look like an intended pun.

Wasn't intended originally, but I chose to leave it in, so you can legit be mad about that.

[skip wooznum]

If I have 2 different colors of socks in my drawer (100 red, 100 blue), the chance of getting a pair by picking 2 random socks is different than first picking a sock, looking at its color and determining the chances of matching it with the second sock.

Is this true even when the number of socks of each color is equal? I'd imagine that both before and after I pick my first sock, the chances of matching it are 99/199.

[ConMan]

If I have 2 different colors of socks in my drawer (100 red, 100 blue), the chance of getting a pair by picking 2 random socks is different than first picking a sock, looking at its color and determining the chances of matching it with the second sock.

Is this true even when the number of socks of each color is equal? I'd imagine that both before and after I pick my first sock, the chances of matching it are 99/199.

This, I'm pretty sure. Probability of finding a match given the first sock is, say, blue, is 99/199 (number of matching socks out of total number of socks). Same value given the first sock is red. Then the unconditional probability is then P(match|first sock blue)×P(first sock blue) + P(match|first sock red)×P(first sock red), and since the chance of getting either colour first is 0.5 that all cancels out. However, if you had an imbalance of sock colours (say 50 red and 150 blue), then you'd have Pr(match|first sock red) = 49/199, Pr(match|first sock blue) = 149/199, Pr(match) = 49/199×50/200 + 149/199×150/200 = 24800/39800 = 124/199 = about 62.3%

[qmech]

I think I might be missing the point of Davio's example, as I can't find any interpretation that doesn't make the statement false.  If two different methods of calculating a quantity give different answers then one of them has to be wrong.

[Asper]

I think I might be missing the point of Davio's example, as I can't find any interpretation that doesn't make the statement false.  If two different methods of calculating a quantity give different answers then one of them has to be wrong.

I don't think we calculate the same quality. In one case we allready drew a sock. Let's say it's red. We can only get a pair by drawing another red one. In the other case, we can either draw two blue or two red socks to get a pair. The probability of things can change depending on what happened allready.

Let's see. I'll break it down to two of each color. I have one sock from the drawer, and it's red. Left are a red and two blue socks. Probability to draw the remaining red sock is 1/3. In the other case, i am supposed to simply draw two identical socks, there are six combinations of socks i can draw, two of which are pairs. So, 1/3.

Now let's imagine one red sock and two blue ones (which is far more realistic as everybody who does their own laundry knows). The probability to get a pair by random draw is 1/3. The probabilty to draw a second red when you drew the first, though, is zero. The probability to draw a second blue one after you drew the first is 1/2.

So i guess in my examples there's a difference only when the groups are not equally likely to be drawn, but at least it shows that we are, in fact, not computing the same thing.

[qmech]

Right, if you get some information (like the colour of the first sock) then it affects the distribution of future observations.  But I found Davio's original phrasing a little misleading as these effects of course wash out if you just want the the probability of drawing a pair.  In the 1 and 2 example you have a 1/3 chance of hitting the 0 case and a 2/3 chance of hitting the 1/2 case, so you recover the original 1/3.

This takes me back to the discussion about whether there is a strategy that can beat 50% success in the following game: you turn over as many cards as you like from the top of the deck, then guess that the next card is red.  Rather embarrassingly that was the proximate cause for my registering an account on the forums.  (The idea was that it has something to do with Venture, although reshuffles make those questions much harder.)

[Asper]

Right, if you get some information (like the colour of the first sock) then it affects the distribution of future observations.  But I found Davio's original phrasing a little misleading as these effects of course wash out if you just want the the probability of drawing a pair.  In the 1 and 2 example you have a 1/3 chance of hitting the 0 case and a 2/3 chance of hitting the 1/2 case, so you recover the original 1/3.

This takes me back to the discussion about whether there is a strategy that can beat 50% success in the following game: you turn over as many cards as you like from the top of the deck, then guess that the next card is red.  Rather embarrassingly that was the proximate cause for my registering an account on the forums.  (The idea was that it has something to do with Venture, although reshuffles make those questions much harder.)

Of course. If you draw a sock and don't update the question to say "and how likely is it NOW", you are in fact still asking for the same thing (and kind of drew a sock for nothing), and of course the old question would not have changed in its probability. I was just trying to point out that when some people here said that after picking/drawing/rolling something "the probability changes", they actually meant that the problem we look at changes, which may or may not have a different probability.

[Ozle]

I'm starting this thread as a place to talk about the question in probability that arose during the recently-finished game of Harry Potter Mafia: http://forum.dominionstrategy.com/index.php?topic=14209.0

The setup is the following.  7 letters are selected at random.  For each letter individually, there is a 1/2 chance of selecting a "T", 1/10 of selecting a "P", 1/10 of selecting a "K" and 3/10 of choosing some other letter. 

At some point in the game it became known that:
1) Exactly 3 Ts had been chosen.
2) At least 2 Ps had been chosen.
3) At least 1 K had been chose.

The question: given this information, what is the probability that in fact 3 Ps had been chosen?  (ie. the letters were some arrangement of TTTPPPK?)
The "obvious" answer: 20%, since the chance of getting a P out of a single letter which is known not to be a T is 1/5.

I argue that this is wrong.

The argument is based on the following fact. 

I flip two (fair) coins and tell you that at least one of them came up heads.  Then the probability of both being heads, given that information is 1/3. 
This is because there are 4 equally likely possibilities: HH, HT, TH and TT.  Of these, exactly 3 contain one or more heads, and exactly 1 contains 2 heads, the P(2 heads | 1 or more heads)=1/3.

By a similar argument (which I will present in the next post), I argue that the answer to the larger question is 2/23, as unintuitive as this seems.

Let's discuss.

The answer is 6

[Titandrake]

The answer is 6

I'm sorry Ozle, we have a new Town Jester, but if you'd like to re-apply submit your credentials and we'll get back to you soon.

[Asper]

The answer is 6

I'm sorry Ozle, we have a new Town Jester, but if you'd like to re-apply submit your credentials and we'll get back to you soon.

Huh, who is that new jester? I need to know so i can excessively complain how the old implementation was better and for free.

[Ozle]

The answer is 6

I'm sorry Ozle, we have a new Town Jester, but if you'd like to re-apply submit your credentials and we'll get back to you soon.

Huh, who is that new jester? I need to know so i can excessively complain how the old implementation was better and for free.

Somebody just whisphered to me that it was you...

Also, town Jester pffft, I've applied for the job of Town Chancellor.
Well, i would have done if i hadnt accidently eaten the application form, thought it was rice paper.