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[GendoIkari]

What are the odds of this?

Code: [Select]

(GendoIkari's first hand: 2 Estates and 3 Coppers.)
(ElGuapo's first hand: 2 Estates and 3 Coppers.)


--- GendoIkari's turn 1 ---
GendoIkari plays 3 Coppers.
GendoIkari buys an Oasis.
(GendoIkari draws: an Estate and 4 Coppers.)
   

   --- ElGuapo's turn 1 ---
   ElGuapo plays 3 Coppers.
   ElGuapo buys an Oasis.
   (ElGuapo draws: an Estate and 4 Coppers.)


--- GendoIkari's turn 2 ---
GendoIkari plays 4 Coppers.
GendoIkari buys a Nomad Camp.
... putting it on the deck.
(GendoIkari reshuffles.)
(GendoIkari draws: an Estate, a Nomad Camp, and 3 Coppers.)
   

   --- ElGuapo's turn 2 ---
   ElGuapo plays 4 Coppers.
   ElGuapo buys a Nomad Camp.
   ... putting it on the deck.
   (ElGuapo reshuffles.)
   (ElGuapo draws: an Estate, a Nomad Camp, and 3 Coppers.)


--- GendoIkari's turn 3 ---
GendoIkari plays a Nomad Camp.
... getting +1 buy and +$2.
GendoIkari plays 3 Coppers.
GendoIkari buys a Merchant Ship.
(GendoIkari draws: an Oasis, an Estate, and 3 Coppers.)
   

   --- ElGuapo's turn 3 ---
   ElGuapo plays a Nomad Camp.
   ... getting +1 buy and +$2.
   ElGuapo plays 3 Coppers.
   ElGuapo buys a Merchant Ship.
   (ElGuapo draws: an Oasis, an Estate, and 3 Coppers.)


--- GendoIkari's turn 4 ---
GendoIkari plays an Oasis.
... drawing 1 card and getting +1 action and +$1.
... discarding a card.
GendoIkari plays 4 Coppers.
GendoIkari buys a Merchant Ship.
(GendoIkari reshuffles.)
(GendoIkari draws: a Merchant Ship, 2 Estates, and 2 Coppers.)
   

   --- ElGuapo's turn 4 ---
   ElGuapo plays an Oasis.
   ... drawing 1 card and getting +1 action and +$1.
   ... discarding a card.
   ElGuapo plays 4 Coppers.
   ElGuapo buys a Merchant Ship.
   (ElGuapo reshuffles.)
   (ElGuapo draws: 3 Estates and 2 Coppers.)



[WanderingWinder]

A LOT better if you're playing with identical starting hands.

[GendoIkari]

A LOT better if you're playing with identical starting hands.

We weren't. But it would seem that the odds of starting with identical starting hands even if you didn't select that option are pretty decent... should be 1 in 4, right? (And really 1 in 2, if you aren't thinking about opening things like Nomad or Inn where 4/3 is different than 3/4).

[WanderingWinder]

A LOT better if you're playing with identical starting hands.

We weren't. But it would seem that the odds of starting with identical starting hands even if you didn't select that option are pretty decent... should be 1 in 4, right? (And really 1 in 2, if you aren't thinking about opening things like Nomad or Inn where 4/3 is different than 3/4).

But this DOES have Nomad camp, so that's important.
Also, your numbers are off, probably because you're making the common fallacy of assuming 50-50 splits (i.e., there are two possibilities and you don't know which will happen, the probability is 50-50, right? Actually no, or at least, not usually; for example, you drive to work tomorrow; you don't know whether or not you will get in a car accident on your way; but unless you are a really terrible driver, the chance of you getting into a wreck is well below 50%). There is a 5/12 chance of 3/4, 5/12 chance of 4/3, 1/12 chance of 5/2, 1/12 chance of 2/5. Which gets us to a 52/144 = 13/36 or just over a 36% chance that you have the same split. If you count 3/4 and 4/3 as the same, etc., you get a bit over a 72% chance of the same split.

[GendoIkari]

Well I was wrong, but not because of the 50/50 fallacy. I wasn't assuming it was 50/50 because it was either "same" or "not same", but I was thinking "if person A got 4/3, then it's higher than 50% that they will have the same split. If person A got 5/2, then it's lower than 50% that they will have the same split. These will balance out, leading to a net of 50%." I see why I was mistaken. Even so, the odds of drawing the exact same hand on turns 3 and 4 though.... (as well as the odds of us choosing the exact same buys; though that just suggests that either the best strategy was obvious, or he was just copying me).

[qmech]

Intuition pump: a bag contains 100 black balls and one white ball.  Two people each draw a ball then put it back.  How likely is it that the balls were the same colour?

Ignoring which card was drawn by Oasis, the probability of your first 4 turns going as they did (if that's what you're aiming for) is 5/99.  Square that for it to go the same way for both players, so around 0.255%.  So it's unlikely, but not impossibly so.

[WanderingWinder]

The oasis is going to make it somewhat more likely of course.
If you allow for a t1/t2 flip and/or a t3/t4 flip, it's not that unlikely at all.

[GendoIkari]

The oasis is going to make it somewhat more likely of course.
If you allow for a t1/t2 flip and/or a t3/t4 flip, it's not that unlikely at all.

What do you mean by the Oasis making it more likely?

[WanderingWinder]

The oasis is going to make it somewhat more likely of course.
If you allow for a t1/t2 flip and/or a t3/t4 flip, it's not that unlikely at all.

What do you mean by the Oasis making it more likely?

Well, if I have CCCCO and draw an estate, this is the same as having CCCEO and drawing a copper. Perhaps more important, CCCCO drawing an estate is the same as CCCCO drawing a copper.
Well, I guess it largely depends on what you're comparing TO, which I guess goes to what exactly you mean by "Ignoring which card was drawn by Oasis"

[GendoIkari]

The oasis is going to make it somewhat more likely of course.
If you allow for a t1/t2 flip and/or a t3/t4 flip, it's not that unlikely at all.

What do you mean by the Oasis making it more likely?

Well, if I have CCCCO and draw an estate, this is the same as having CCCEO and drawing a copper. Perhaps more important, CCCCO drawing an estate is the same as CCCCO drawing a copper.
Well, I guess it largely depends on what you're comparing TO, which I guess goes to what exactly you mean by "Ignoring which card was drawn by Oasis"

Ah, got it. (It wasn't me who said "ignoring which card was drawn by Oasis"). I wouldn't ignore that, because we actually did both draw the same card from Oasis. We both started with an Oasis, an Estate, and 3 Coppers, and we both drew a 4th Copper with Oasis.

[^_^_^_^]

There are times when I learn more math here in 5 minutes than I ever learned at school in 40. This is one of those times.

[qmech]

I ignored Oasis for a couple of reasons: you can't tell what it was from the log, and even if you do know what it is there's a question about whether we're measuring "draw the same" or "both draw an Estate" (say).  More generally, as WW suggested, equivalent turns are far more likely than identical turns, which bumps it up quite a lot if you find those equally interesting.

[GendoIkari]

I ignored Oasis for a couple of reasons: you can't tell what it was from the log, and even if you do know what it is there's a question about whether we're measuring "draw the same" or "both draw an Estate" (say).  More generally, as WW suggested, equivalent turns are far more likely than identical turns, which bumps it up quite a lot if you find those equally interesting.

How about "identical logs"? In this case of the Oasis turn, you can tell that we both drew a Copper from the Oasis, because you can see that our starting hands only had 3 coppers, but we played 4. It's the identical logs that I noticed... Of course, the logs (or the game) doesn't tell you anything about the order that you drew your 5 card hand in, but what it comes down to is that we drew, played, and bought all the same cards, in the same order, across our first 4 turns.

[qmech]

Identical logs is trickier because there are many more cases to examine.  As a rough estimate you could take the 5/99 figure, which is the probability that the second player has the posted log given that the first player does.  If this is a typical log then that will be close to the probability that the logs match. 

In bagland this corresponds to the case when there are lots of different colours of ball (corresponding to different logs) and there are the same number of each type of ball: if there are b types of ball then the probability of a match is 1/b.  If instead there are different numbers of each type then the probability of a match is higher, for the same reason as in the 100 to 1 case above.

A similar phenomenon is that on average most people's friends have more friends than they do.

[GendoIkari]

Identical logs is trickier because there are many more cases to examine.  As a rough estimate you could take the 5/99 figure, which is the probability that the second player has the posted log given that the first player does.  If this is a typical log then that will be close to the probability that the logs match. 

In bagland this corresponds to the case when there are lots of different colours of ball (corresponding to different logs) and there are the same number of each type of ball: if there are b types of ball then the probability of a match is 1/b.  If instead there are different numbers of each type then the probability of a match is higher, for the same reason as in the 100 to 1 case above.

For the record, I love the fact that here at this board, someone can post a common idiom such as "wow, what are the odds of that?" and get a whole bunch of replies with analysis and discussion aimed at actually getting an accurate answer to that question!

A similar phenomenon is that on average most people's friends have more friends than they do.

You just blew my mind.

[^_^_^_^]

Wait so I have more friends than my friends from their perspective but they have more friends than me from my perspective?

[eHalcyon]

Wait so I have more friends than my friends from their perspective but they have more friends than me from my perspective?

No, it's on average.

So, suppose there are three groups of friends:

A1, A2, A3

B1, B2

C1, C2, C3, C4

A1 is friends with A2 and A3, A2 is friends with A1 and A3.  They have no other friends.  Likewise for the other groups.

Now introduce D0, who is friends with EVERYONE.

A1 has 2 friends, but his friends (A2, A3 and D0) have more friends on average than he does, because of D0.  The same goes for everyone else -- their friends have more friends on average than them, because of D0.  The only exception is D0, because he is D0.

[zahlman]

A similar phenomenon is that on average most people's friends have more friends than they do.

You just blew my mind.

Friendship is distributed unevenly in the population; some people are far more social than others.

Consider the extreme example where one person is friends with 99 others, all of whom are strangers to each other.

The average number of friends is 1.98: 99 people have 1 friend, and 1 person has 99 friends.

The average number of friends of friends is 98.02: 99 people have only the social butterfly as a friend, so they count 99 friends of friends. The social butterfly has 99 friends, each of which has only one friend (the social butterfly), thus counting an average of 1 friend of friends.