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Friday, April 2, 2010 - 5:18amSanction this postReply
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The "Dinner Party Problem" (see link source in quote) is relatively more complicated than basic financial accounting. Yet I meet too many people who ought to know better who either cannot or will not do the latter. On election day, their votes count just as much as ours. No wonder we have the government we have.

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Friday, April 2, 2010 - 2:46pmSanction this postReply
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This is essentially like the Monty Hall problem.

Very few people answer correctly the first time. (Luke didn't.) It's harder than it first appears. The article explains why you should switch.


Post 2

Friday, April 2, 2010 - 3:26pmSanction this postReply
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Yes, indeed, the problem "busted" me, and I am decent at mathematics. That was my point. People who are not decent at mathematics and not good at economics get to vote on issues with profound economic implications. Here in Florida, "the people" approved a referendum to institute light rail to connect key cities in the state. The issue as presented on the ballot included no economic information. Now our legislature finds itself struggling to finance this boondoggle. AAARRRRRGGGGGHHHHH!!!!!

Post 3

Friday, April 2, 2010 - 4:38pmSanction this postReply
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I first heard the problem maybe about 20 years ago. My first thought was switching or not didn't matter, but I was wary that somebody would be posing the problem to me if it were that easy. So I thought about it a few minutes before giving my final and correct answer.

Post 4

Saturday, April 3, 2010 - 7:04amSanction this postReply
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My first answer is to switch because, after I said "Door 1" -- Monty Hall opened Door 3 (as if it were "Door 1" in his mind).

If I stick with Door 1 (i.e., "Door 3" -- in Monty Hall's mind), then I'm stuck with the goat -- and it'd be no fun riding a goat to work.

Ed

Post 5

Saturday, April 3, 2010 - 9:22amSanction this postReply
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Only depending on whether riding it backwards... ;-)

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Post 6

Saturday, April 3, 2010 - 6:43pmSanction this postReply
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Merlin,

The Monty problem really gets my goat.

The probability (after seeing a goat) probably should be updated to one chance in two that each of the remaining doors has a car behind it. Let's say you first chose the door with the car behind it. Now, the first other door gets opened and lo'-and-behold there's a goat behind it (Monty knew this). Then, if you switch your choice, you lose the car!

Now, when you made your choice, you had one chance in three of being right. There were ("were" is the key word) 2 chances in 3 that the car was behind one of the other 2 doors and ... wait a minute ... oh crap ... what the ... holy ... I got it now.

Nevermind.

Ed


Post 7

Saturday, April 3, 2010 - 8:29pmSanction this postReply
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The Monty Hall problem is a bunch of fun.  But in the end I was able to wrap my mind around it.  Parrondo's Paradox  is another very counter-intuitive math problem, one that I can't "get".

It states:  If given two coin tossing games, each having odds clearly against your winning, you might be able to win long term, merely by switching from one game to the other in some pattern.

WTF!!!

When this problem was published and verified repeatedly, academics got really excited to try this in evolutionary biology, economics, and other game-theory fields.  I'm not aware of any discovered practical applications.

If anyone here could explain this is a elementary fashion, I'd be much obliged.


Post 8

Sunday, April 4, 2010 - 6:31amSanction this postReply
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Mr. Doug Fischer,

The game you linked to that has two coin games, the first has a probability of winning of (1/2 - 0.005) = 0.495, the second has a probability of winning of (((1/10) - 0.005) * .3836) + (((3/4) - 0.005) * .6164) = 0.49566.

But in the second coin game, if you play it when your sum of money is divisible by 3 with no remainder, then your probability of winning is (3/4) - 0.005 = 0.745.

So play the first coin game where its practically 50/50 whether you win or loose, until your sum is divisible by three, then win big playing the second coin game.

Post 9

Wednesday, April 7, 2010 - 11:22pmSanction this postReply
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Thanks very much Mr Gores. Sorry for the late reply.

That is remarkably simple.

Post 10

Thursday, April 8, 2010 - 5:07amSanction this postReply
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The switch logic only applies when information is added about the set originally not chosen.

Initial guess is 1:3, meaning other set has winner 2:3. If information is added about other set, eliminating all but one of the choices, then the remaining single choice will be winner 2:3 times.

For the 1:50 and 49:50 example to payoff at 49:50, information would need to be added about 48 of the 49 choices. All but one box would need to be eliminated. (Deal or no Deal...)

It is clear that information needs to be added about the other choices for the switch to payoff. Otherwise, everyone would do this:

Go to buy a lottery ticket, and pick a non repeating set of two digit numbers. Your chances of winning are astronomically low, against astronomically high that the winning number is in the set not chosen. Change your mind, and pick a different non repeating set of two digit numbers. Your chances of winning are not now 'astronomically high.' Without information about the other set, there is no advantage to switching.

Go back to the 3 choice problem. It is true that the unchosen set has a 2:3 chance of having the Gold, but when you second chance pick before the additional information, you only have a 1:2 chance of picking from that 2:3 set. Without the added information before your second chance pick, the arbitrary second choice has the same 1:3 chance as the original 1:3 choice. (Otherwise, all but one of us would win the lottery every day, simply by changing our mind when we picked a lottery number.)

With the additional information, your second chance pick is the better choice.

Even with the 1:50 example, your second choice is marginally better if even just one box is eliminated from the 49:50.

Your chances of the 1st pick are 1:50.

Your chnaces of the second pick are 1/48 x 49/50

or 49/48 of 1/50

If you eliminate 48 of the 49 before making your second puck, your seconf pick is

1/1 x 49/50

Just as, if you eliminate 1 of 2 before making your second pick of 3, your second pick is

1/1 x 2/3

(Edited by Fred Bartlett on 4/08, 5:19am)


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