Math riddle / philosophy problem
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Math riddle / philosophy problem
by k3nt » Mon Aug 21, 2006 12:10 pm
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k3nt - Enthusiast (Online)
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by WildBillHickok » Mon Aug 21, 2006 12:50 pm
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WildBillHickok - BTP Tourny #6 Winner
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by NorthView » Mon Aug 21, 2006 1:22 pm
Mon May 12, 2008 1:46 am
When I play a patient and relaxed game I win - that simple.
Mon May 12, 2008 10:55 pm
Seriously, fuck poker.
==================================================================
[21:03] NorthViewBTP: mac is a fellow mexican
[21:03] Mekosking: yup
[21:03] NorthViewBTP: you should support your bro
[21:03] Mekosking: therefore hes a fat worthless tsr obv
When I play a patient and relaxed game I win - that simple.
Mon May 12, 2008 10:55 pm
Seriously, fuck poker.
==================================================================
[21:03] NorthViewBTP: mac is a fellow mexican
[21:03] Mekosking: yup
[21:03] NorthViewBTP: you should support your bro
[21:03] Mekosking: therefore hes a fat worthless tsr obv
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NorthView - Posts: 5998
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by Nortonesque » Mon Aug 21, 2006 2:28 pm
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Nortonesque - Enthusiast (B&M & Online)
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by black_knight6 » Mon Aug 21, 2006 4:42 pm
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black_knight6 - Semi Pro (Online)
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by GodlikeRoy » Tue Aug 22, 2006 3:24 am
Poker is silly.
It is not enough to be good at chess, you must also play well.
Somewhere in the world someone is training when you are not. When you race him, he will win.
It is not enough to be good at chess, you must also play well.
Somewhere in the world someone is training when you are not. When you race him, he will win.
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GodlikeRoy - Posts: 7430
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by Xaston » Tue Aug 22, 2006 8:19 am
The envelope you choose has X dollars in it.
The other envelope can have two possible amounts of money.
2x or .5x.
By switching we get 2x-x = x more than first envelope
or
By switching we get x-.5x = .5x less than the first envelope.
Each outcome is equally likely. For all non zero amounts of money (which is stated in the problem, we shall neglect negatives as I don't think it's even possible to put negative amounts of money into an evelope (bills?)), it is +EV to switch.
Can someone please show me the flaw in my arithmatic and/or logic?
The other envelope can have two possible amounts of money.
2x or .5x.
By switching we get 2x-x = x more than first envelope
or
By switching we get x-.5x = .5x less than the first envelope.
Each outcome is equally likely. For all non zero amounts of money (which is stated in the problem, we shall neglect negatives as I don't think it's even possible to put negative amounts of money into an evelope (bills?)), it is +EV to switch.
Can someone please show me the flaw in my arithmatic and/or logic?
Boy, you got me confused with a man who repeats himself.
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Xaston - Posts: 8833
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