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Offline teffy

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Re: Let's talk about how non-random shuffling is one more time. https://elementscommunity.org/forum/index.php?topic=7601.msg85639#msg85639
« Reply #12 on: June 08, 2010, 09:40:11 am »
Quote from: PhuzzY LogiK
   
Quote from: teffy
The chances that the 3 Quintessences are on position x, y and z is   1 / Binomial[30,3]
    How many positive cases do we have, with 2 Quintessences in a row?

    I count 28+27+26...+1+1+2+...+27=1/2*28*(28+1)+1/2*27*(27+1)=784

    Binomial[30,3]=4060

    784/4060~0,193=19,3%, you had it twice in 10 games ,that´s 20%

I don't follow what you're doing here:
1.  I don't understand how you're getting 784 for your "positive cases" (I don't see what method you're using).
2.  In your binomial, you're finding the number of ways to choose 3 cards from 30, but you only have positive cases for drawing 2 Quints.  How can you compare these if they are for a different number of draws?
The chances that the Quintessences are on position x,y,z are equal to the chances that they are on position a,b,c
Then I made an intuitive counting, how many ways do I have to get a row of 2

1= Quintessence
0= the Rest

111000....
11010000...
1100100000...
...
01110000...
01101000.....

Then, I forgot the pairs with the 3rd Quintessence above the pair e.g.
1011000000
That lets me come to that number
The chances are the same for all cases, exactly one of this cases must be, if you draw cards, so I come to 19,3%.


Quote from: PhuzzY LogiK
I came up with my numbers like this:
Drawing 3 Shards in a row:
Probability of drawing 1 Shard (# of Shards/Total number of cards) * Probability of drawing a second Shard (-1 potential Shard, -1 cards total) *
Probability of drawing a third Shard (-1 potential Shard, -1 cards total)

This works out to:
(4/30) * (3/29) * (2/28) = 9.85x10^-4 = 0.0985% ~ 0.1%

 
In my understanding, this are the chances, that 3 of the 4 shards are on position x, y and z. e.g. 1,2 and 3.
But thats not the same as the chances that you can find a row of 3 in your deck of 30 cards.

A case in which your way of calculation would produce results which are obviously wrong:
Have more than 15 Quantum Towers in your deck and calculate the chances of 2 in a row.
If you have more than 15 Towers, the chances for this are 100% but you calculate a value <100%
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Re: Let's talk about how non-random shuffling is one more time. https://elementscommunity.org/forum/index.php?topic=7601.msg85668#msg85668
« Reply #13 on: June 08, 2010, 11:58:37 am »
So phuzzy, are you trying to do what demagog was doing? Seeing if certain cards have a better chance of being close together, or are you just simply trying to see about the randomness of shuffling.
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PuppyChow

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Re: Let's talk about how non-random shuffling is one more time. https://elementscommunity.org/forum/index.php?topic=7601.msg85790#msg85790
« Reply #14 on: June 08, 2010, 04:24:44 pm »
Quote
I see what you're saying here, but I just don't find it intuitive.  It's like saying "There are nearly infinite ways you could die each day, so it is inevitable that at least one will happen and you will die today."  If the probabilities of every one of these potential combinations is so low, why am I seeing the same ones over and over?  Shouldn't the fact that they are so rare mean I should see several of them before one repeats?

You did see several of them.
1) In one game, you drew a pulvy followed by a quint.
2) In another, you drew a permafrost followed by a bone wall.
3) The list goes on and on...

Just because one of those possibilities happens to be repeated a few times doesn't mean it isn't random. In fact, I would say it actually proves it IS random. If one didn't repeat, I'd be skeptical because then it seems like the game is intentionally not repeating certain combinations (like if you flipped a coin and it was heads, you would rig it next time to make it tails).

 

blarg: