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11 on 11 is the last number where defender has advantage

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Re: 11 on 11 is the last number where defender has advantage

Postby Viceroy63 on Sat Nov 23, 2013 7:31 am

Koganosi wrote:
Viceroy63 wrote:
bonzifan wrote:The odds of getting heads and tails never changes, regardless of previous tosses.

If you have tossed 5 tails, you have already beaten odds of 1/32 (3.1%) To toss another tails would only halve your odds to 1.55%

Another way of looking at it is that, although when you started tossing you had a 3.1% chance of getting 5 tails. Having tossed them though, the chance that you did toss 5 tails is now 100%. Because it's in the past, we know the result 100%. Therefore, the next toss is still only 50/50.


In the test, 13 people were asked to flip a coin 300 times, trying to get as many heads as possible. All 13 participants got more heads than tails. Seven out of the thirteen had statistically significant margins of heads over tails (meaning almost certainly not a matter of chance). The highest was one individual had 68% of the coin flips land heads. In other words, a coin toss isn't particularly random.
http://www.techdirt.com/articles/200912 ... 7292.shtml

I know from personal eperience that I can make a coin come out ether heads or tails...
Most of the time. :)


Its indeed true, that you can make a head or tails appear more wich certain techiniques, but then lets go with somethign else. I have 2 balls a red and a white one, I put them both under a cup at random, you cant see through the cup, neither did you see me put the balls under the cup!

Now you can remove one of the caps, youll have to find the white ball to win!

Your odds to find the white one are 50%.

You fail the first time, second time you try now still 50% on finding it. THough in the sequence you failing twice has odds off 1/2^2, aka 1/4=0,25 aka 25%.

Third one still 50%

Sequence 1/2^3, 1/8 = 0,125. so 12,5%.....

Urs

Koganosi


Of course you are correct. It is a very specific technique to that coin trick. =)

But instead of the upside down cups and a ball, let us use two right side up see through glass beer mugs instead!
You with me so far? ;)

Now one of the glass mugs is filled with your favorite brand of Beer and the other is just some other kind of identical looking beer. And you did not see me pour, OK.

What are the odds then of you picking your favourite beer with out smelling or tasting first?

Let me see you calculate those odds? :lol:
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Re: 11 on 11 is the last number where defender has advantage

Postby Koganosi on Wed Dec 18, 2013 11:08 am

Viceroy63 wrote:
Koganosi wrote:
Viceroy63 wrote:
bonzifan wrote:The odds of getting heads and tails never changes, regardless of previous tosses.

If you have tossed 5 tails, you have already beaten odds of 1/32 (3.1%) To toss another tails would only halve your odds to 1.55%

Another way of looking at it is that, although when you started tossing you had a 3.1% chance of getting 5 tails. Having tossed them though, the chance that you did toss 5 tails is now 100%. Because it's in the past, we know the result 100%. Therefore, the next toss is still only 50/50.


In the test, 13 people were asked to flip a coin 300 times, trying to get as many heads as possible. All 13 participants got more heads than tails. Seven out of the thirteen had statistically significant margins of heads over tails (meaning almost certainly not a matter of chance). The highest was one individual had 68% of the coin flips land heads. In other words, a coin toss isn't particularly random.
http://www.techdirt.com/articles/200912 ... 7292.shtml

I know from personal eperience that I can make a coin come out ether heads or tails...
Most of the time. :)


Its indeed true, that you can make a head or tails appear more wich certain techiniques, but then lets go with somethign else. I have 2 balls a red and a white one, I put them both under a cup at random, you cant see through the cup, neither did you see me put the balls under the cup!

Now you can remove one of the caps, youll have to find the white ball to win!

Your odds to find the white one are 50%.

You fail the first time, second time you try now still 50% on finding it. THough in the sequence you failing twice has odds off 1/2^2, aka 1/4=0,25 aka 25%.

Third one still 50%

Sequence 1/2^3, 1/8 = 0,125. so 12,5%.....

Urs

Koganosi


Of course you are correct. It is a very specific technique to that coin trick. =)

But instead of the upside down cups and a ball, let us use two right side up see through glass beer mugs instead!
You with me so far? ;)

Now one of the glass mugs is filled with your favorite brand of Beer and the other is just some other kind of identical looking beer. And you did not see me pour, OK.

What are the odds then of you picking your favourite beer with out smelling or tasting first?

Let me see you calculate those odds? :lol:


Pretty high, caus I would pick both at the same time and just drink em both. God gave me 2 hands after all!

So where did you leave those beers?

Urs

Koganosi
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Re: 11 on 11 is the last number where defender has advantage

Postby Viceroy63 on Wed Dec 18, 2013 5:26 pm

OH; I forget to mention that you're blind folded with one arm tied behind your back. It's a blind taste test. :lol:

No sooner do you grab one mug, that I go for the other. LOL.
:lol:

Yup; Just dive right into it. :lol:

See, I don't care about the odds or the beer. Cheers Mate. LOL.
:lol:
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