Brainteaser for TeamLiquid! - Page 19

The unedited OP didn't say "other game as well". It just said "other game". The poll itself STILL says "other game". People are going to continue to say 1/3 because, regardless of the explanation at the beginning, the poll itself still asking the probability of drawing Zerg in a single game.

On June 11 2011 00:41 iStarKraft wrote:
...since we know that one of the games was zerg, all we need to do is find the probability that the other game was zerg. This gets affected in no way by the game we KNOW was zerg. Therefore, there are three possibilities for the 'other' game played:

He played as zerg.
He played as terran.
He played as protoss.

Zerg will occur 1/3 of the time. Therefore the answer is 1/3. There is absolutely no purpose to writing out the different combinations of games (ZZ PP TT PT TP... etc.), as we only need to find the probability he played one game as zerg, since we already know one of his games was zerg.

I will happily respond to any counter-arguments, but I would most likely be rewording / repeating myself. (^_^)

iSK

On June 11 2011 02:11 foxmeep wrote:
Here's an easier way to understand the problem. Say I flip two coins, we can all agree that the possible outcomes are:

H-T 1/4
H-H 1/4
T-T 1/4
T-H 1/4

Now I tell you that if, and only if, a heads comes up, if the other coin is tails, I win. If the other coin is heads, you win. Simply making this proposition does not change the probability of each result coming up.

H-T I win
H-H I lose
T-T Draw
T-H I win

We already know each of the outcomes has an equal chance of occuring, so in 2/4 results, I win, 1/4 you win. Therefore the odds are 2:1, or 2/3 in my favour, and 1/3 in yours. NOT 1/2.

On June 11 2011 02:21 Cyba wrote:


That's because you represent the problem differently, the question is what are the odds for TT to drop if we flip it twice in a row. It's only understandable in the way you say it because the OP was a bit bad ;p

On June 11 2011 02:49 JKira wrote:
I learned conditional probability in grade 11, and I don't even think that's early or anything. I was 16 at the time.

Are the majority of people on TL really under 16 years old? I was under the impression that the majority is over 20 years old.

On June 11 2011 03:00 Millitron wrote:
OP, In your solution, you claim that there are 5 possible games, ZZ, ZP, ZT, PZ, TZ. This is nonsense, because ZvT is the same as TvZ; meaning you count two of the matchups twice for no reason.

Its 1/3.

On June 11 2011 02:11 foxmeep wrote:
Here's an easier way to understand the problem. Say I flip two coins, we can all agree that the possible outcomes are:

H-T 1/4
H-H 1/4
T-T 1/4
T-H 1/4

Now I tell you that if, and only if, a heads comes up, if the other coin is tails, I win. If the other coin is heads, you win. Simply making this proposition does not change the probability of each result coming up.

H-T I win
H-H I lose
T-T Draw
T-H I win

We already know each of the outcomes has an equal chance of occuring, so in 2/4 results, I win, 1/4 you win. Therefore the odds are 2:1, or 2/3 in my favour, and 1/3 in yours. NOT 1/2.

On June 11 2011 02:59 L3g3nd_ wrote:
i read the solution, then voted 1/3. youre making the mistake of counting ZT and TZ as two seperate cases, when they are actually exactly the same, given that ordering doesnt matter.

On June 11 2011 02:59 SolC361 wrote:
Condescending much? Even the great mathematician Erdos initially got the Monty Hall problem wrong, which is considered a basic conditional probability problem. And most of the fuss is over the wording ambiguity of the problem, not the problem solving itself.