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Author Topic: Maths Question  (Read 7033 times)
kinboshi
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« Reply #30 on: January 16, 2012, 12:34:10 AM »

Use the coin to buy another one, or a dice, or a deck of cards.
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« Reply #31 on: January 16, 2012, 12:41:59 AM »

Nope. You're making an assumption that neither person knows about the bias, and this might not always be the case.

no I'm not, you told us that


Has this one been done on here before?

I have a standard coin. But... it is biased to one side. We don't know which side, nor do we know the size of the bias. How can we use this coin to get a genuine 50/50 chance?

my answer is correct unless you want to move the goalposts
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Jon MW
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« Reply #32 on: January 16, 2012, 06:08:32 AM »

Has this one been done on here before?

I have a standard coin. But... it is biased to one side. We don't know which side, nor do we know the size of the bias. How can we use this coin to get a genuine 50/50 chance?

Do the thing where you hide it in one hand and get someone else to guess which hand it is in.

Is also correct.

When CF says:

Nope. Too many other factors for that to be 50/50.

What he actually means is:
Yes. But that's not the way I was thinking of.
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« Reply #33 on: January 16, 2012, 06:41:20 AM »

Has this one been done on here before?

I have a standard coin. But... it is biased to one side. We don't know which side, nor do we know the size of the bias. How can we use this coin to get a genuine 50/50 chance?

Do the thing where you hide it in one hand and get someone else to guess which hand it is in.

Is also correct.

When CF says:

Nope. Too many other factors for that to be 50/50.

What he actually means is:
Yes. But that's not the way I was thinking of.

This method only works if the person doing the hand holding is 100% unreadable.

My solution can be mathematically proven. It's done by tossing the coin and observing the result. (hint: results). No trickery etc required.
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gatso
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« Reply #34 on: January 16, 2012, 12:01:30 PM »

you could just accept that there's more than one way to answer the question the way you worded it
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« Reply #35 on: January 16, 2012, 02:24:29 PM »

Ask punter to call heads or tails (lets say he calls tails)

Toss coin twice.  If it is a tails and then a tails then start again.
If it is a heads and then a heads then start again
If it is a heads and then a tails he loses
If it is a tails and then a heads he wins

The answer works for whatever bias the coin has, but lets assume it is 10% heads and 90% tails. 

Chances of = 9%
Chances of TH = 9%

By removing all other possibilities we have a total "chance population" of 18% and we now have a 50/50 proposition.

When the guy calls the toss Tails he is in effect picking the Tails/Heads combo without knowing it and the other guy has the Heads/Tails combo.

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« Reply #36 on: January 16, 2012, 02:25:32 PM »

Ah - it pulled the old Ten of Hearts when I typed TH.

"Chances of Heads/Tails = 9%
Chances of Tails/Heads = 9%"
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« Reply #37 on: January 16, 2012, 03:32:45 PM »


in b4 cf gives a reason why this is wrong
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« Reply #38 on: January 16, 2012, 03:41:50 PM »

I think it's mathematically sound, but I'm not sure if I am allowed to "kick out" toss observations like I have suggested.
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« Reply #39 on: January 16, 2012, 03:54:55 PM »

good work. this is kinda like taking the C boundary but without all the expensive maths Wink
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« Reply #40 on: January 16, 2012, 07:11:00 PM »

That is indeed the answer Smiley
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George2Loose
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« Reply #41 on: January 17, 2012, 12:47:54 AM »

worst thread ever
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« Reply #42 on: January 17, 2012, 12:58:06 AM »

worst thread ever

this
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« Reply #43 on: January 17, 2012, 06:50:57 AM »


just cos you didn't get the right answer Wink
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kinboshi
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« Reply #44 on: January 17, 2012, 07:23:22 AM »


just cos you didn't get the answer I wanted, and instead gave another correct answer Wink


FYP
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