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The forums' hardest game


baltoist

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(edited)

Time for some MATH:

 

I have gone over the previous 10 games and recorded the value at which the count was stopped. The numbers are:

 

20, 12, 5, 8, 3, 8, 3, 5, 13, 5

 

Let’s make the assumption that every game has a 50% chance to be over by the time the game reaches a number of posts equal to the mean (average) number of posts the last 10 games reached (rounded to the nearest integer for simplicity, which is 8) Let’s call this event D.

 

i.e. P(D)=0.5

 

Now let’s calculate the probability for the game to continue after each individual post. Let’s call this event A.

 

Since event B occurs when A does occurs exactly 8 times, we have this relationship between P(D) and P(A):

P(D) = (P(A))^8

Rearranging:

P(A) = (P(D))^(1/8)

Substituting values:

P(A) = (0.5)^(1/8)

P(A) = 0.917 (rounded to 3 significant figures)

 

Now we know that the probability of the game continuing after each post is 91.7%. With this information, we can find out what the probability is of the game reaching 50 posts (Let this be event E):

P(E) = (P(A))^50

P(E) = 0.917^50

P(E) = 0.013139 (rounded to 5 significant figures)

 

So there you have it. If I’ve done my math right and my assumption is reasonable, we have a 1.3139% chance of winning this game. This means we should win once for every 76 games we play.

 

Why did I do this, you ask? For fun! Why else?

 

Oh, and 2!

 

Edit: I couldn't use 'event B' or 'event C' because apparently 'B' followed by ')' gives you B) and '(' followed by 'C' followed by ')' gives you ©

Edited by FortyTwo42

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Thanks to Pink for the lovely avatar and W.G.A. for the amazing signature!

My OCs: Aero Wind, Shadowhide, Ebony (WIP)

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