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Posted: **Sat Jul 31, 2010 4:24 pm**

Let's talk about Bin or Win. There can be 6 killer balls in all: 4 in round 1, 1 in round 2 and 1 in Bin or Win round. There are 6 balls to bin and 5 to win. That means, there's a probability that all killer balls are binned. But first, please go here:

http://en.wikipedia.org/wiki/Binomial%20coefficient before I start explaining it
OK, are you back?

I'm here.

The probability of non-killer win is C(n, 5)/C(11, 5) (n - number of non-killer balls).
But if you don't want to calculate, here's the chart

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Number of killer balls  Probability
0                       Not possible because of required killer ball in Bin or Win round
1                       252/462=6/11
2                       126/462=3/11
3                       56/462=4/33
4                       21/462=1/22
5                       6/462=1/77
6                       1/462

After one binned and one won ball the probability is C(n, 4)/C(9, 4). Here's the chart.

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Number of killer balls  Probability
0                       126/126=1 (in all cases) LOL
1                       70/126=5/9
2                       35/126=7/18
3                       15/126=5/42
4                       5/126
5                       1/126
6                       Not possible because only 5 binned balls remain

After two binned and two won balls the probability is C(n, 3)/C(7, 3). Here's the chart.

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Number of killer balls  Probability
0                       35/35=1 (in all cases) LOL
1                       20/35=4/7
2                       10/35=2/7
3                       4/35
4                       1/35
5                       Not possible because only 4 binned balls remain
6                       See above

After 3 binned and 3 won balls the probability is C(n, 2)/C(5, 2). Here's the chart.

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Number of killer balls  Probability
0                       10/10=1 (in all cases) LOL
1                       6/10=3/5
2                       3/10
3                       1/10
4                       Not possible because only 3 binned balls remain
5                       THOSE BALLS AREN'T LOVING ME!! :x 
6                       6 from 5?! LOLOL

Here goes the final decision: which from remaining 3 balls to win. The probability is C(n, 1)/C(3, 1)=n/3. But if you don't want to calculate this simple formula, I still give you a chart.

Code: Select all

Number of killer balls  Probability
0                       3/3=1 (in all cases) LOLOLOL
1                       2/3
2                       1/3
3                       THOSE BALLS AREN'T LOVING ME!! :x

Posted: **Sat Jul 31, 2010 4:36 pm**

Why did you put this up?
And why'd you make a pointless poll for it?

Oh, and why are the LOL's back?

Posted: **Sat Jul 31, 2010 4:43 pm**

Eoin Monaghan wrote:Oh, and why are the LOL's back?

Just because I copy-pasted Golden Balls game show probability theory topic in Other game shows

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Golden Balls game show probability theory

Golden Balls game show probability theory

Re: Golden Balls game show probability theory

Re: Golden Balls game show probability theory