29-12-2004 22:25									#91
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Ok what about this :
Suppose we have a coin or whatever with certain odds, but with don't know what the odds are.
Say it has P chance of landing heads, and 1-P chance of landing tails.
(like .75 and .25 or something)

Suppose me and you want to make an even bet.
Can you find a procedure using our coin we can make an even bet (.5 chance of winning each) without knowing what the odds for the coin actually are?


No spoilers please from those who know...
BTW this is usually attributed to von Neumann, one of the fathers of Computer Science, and Game Theory also.
And the trick is still used in many areas of Computer Science...

	30-12-2004 01:31									#92
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Heads I win, tails you lose.

spoiler(highlight to read): At a quick glance I would say you could set it so that we both toss the coin until only one of us gets a heads. In a kind of penalty shoot out fashion. Not sure if that is what you had in mind though.

	30-12-2004 01:57									#93
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If you restart in case of equalities, that could work.

You didnt state it like the solution I was looking for, but after some thinking about it, they are both almost equivalent.

	30-12-2004 02:27									#94
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Hmmm...

Toss the coin. Then toss the coin again. If the coin lands on the opposite side the second time it's a win (for the person who bet on the first toss correctly). If it lands on the same side you start over.

	30-12-2004 05:38									#95
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Here's a puzzle that's similar to the original one (since it involves a decision whether or not to switch from your original choice).

A wealthy friend places two envelopes in front of you. Your friend explains that one envelope contains twice as much money as the other one (but you don't know which is which). You may select one envelope and keep whatever money you find inside.

You select one of the envelopes at random, open it, and find $100. Your friend now tells you that you may, if you wish, change your mind and select the other envelope. Should you do so?

Common sense tells you that, since your original choice was made at random, there is no advantage to switching. However, what if you reason as follows: The envelope that you didn't choose either contains $50 or $200. If you choose the second envelope, your expected payout is one-half of $200 + $50, or $125. Based on that reasoning, it seems that you should switch.

The reasoning is absurd, since it implies that you improve your expectation by always switching to the second envelope (no matter which one you chose first). The puzzle is to explain the flaw in the reason given above for switching.

	30-12-2004 05:46									#96
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I'll stick with a simple double-blind study with four pages of statistics, thank you.

	30-12-2004 11:43									#97
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quote: Originally posted by Mercator Hmmm... Toss the coin. Then toss the coin again. If the coin lands on the opposite side the second time it's a win (for the person who bet on the first toss correctly). If it lands on the same side you start over.

Yep thats what I was looking, again in a slightly different form, but Dauphins idea is equivalent I think.

	30-12-2004 11:48									#98
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quote: Originally posted by Petek Here's a puzzle that's similar to the original one (since it involves a decision whether or not to switch from your original choice). A wealthy friend places two envelopes in front of you. Your friend explains that one envelope contains twice as much money as the other one (but you don't know which is which). You may select one envelope and keep whatever money you find inside. You select one of the envelopes at random, open it, and find $100. Your friend now tells you that you may, if you wish, change your mind and select the other envelope. Should you do so? Common sense tells you that, since your original choice was made at random, there is no advantage to switching. However, what if you reason as follows: The envelope that you didn't choose either contains $50 or $200. If you choose the second envelope, your expected payout is one-half of $200 + $50, or $125. Based on that reasoning, it seems that you should switch. The reasoning is absurd, since it implies that you improve your expectation by always switching to the second envelope (no matter which one you chose first). The puzzle is to explain the flaw in the reason given above for switching.

This is a tricky one.The fallacy I view this way
There are 2 different possible situations :
The enveloppes had 50 and 100 and you chose the 100 one.
or The enveloppes had 200 and 100 and you chose the 100 one.

But by computing the averages, you assumed that these were somehow equiprobable, but in fact we have no knowledge of the probability distribution of the situations so we are not able to compute an average.

	30-12-2004 12:10									#99
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What LulThyme said.

If you are guaranteed the situations are equiprobable then you should switch.

Computing probabilities with 0 information as to the distribution of events is impossible.

	30-12-2004 12:11									#100
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Probabilities are what you do with incomplete, not 0 information.

	01-01-2005 00:24									#101
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But don't you know what the distribution could be. Its either $100, $50 or $200, $100. Isn't the fallacy that the two distributions you use in your expectation calc have different means and variances.

I may be oversimplifying, but couldn't you take a geometric mean, as opposed to an arithmetic mean, to get around this.
i,e ($200 * $50)^1/2 = $100 expectation if you switch. Therefore no gain or loss if you switch.

The same then works if you added in a third envelope of money with either 4 or .25 times as much money as you first pick.
i.e. ($400*$200*$50*$25)^1/4 = $100. Therefore no gain or loss if you switch.

Also, it would work if you changed the ratio.

Though I must admit, I feel as if I am using circular logic in thinking the maths of this one through.

	01-01-2005 01:10									#102
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Here's a link to an explanation of the problem:

http://www.maa.org/devlin/devlin_0708_04.html

For lots more information, Google on "two envelope paradox".

	01-01-2005 09:13									#103
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What KH and LT said. It's a mis-application of the Principle of Indifference.

	01-01-2005 23:27									#104
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quote: Originally posted by Dauphin But don't you know what the distribution could be. Its either $100, $50 or $200, $100. Isn't the fallacy that the two distributions you use in your expectation calc have different means and variances. I may be oversimplifying, but couldn't you take a geometric mean, as opposed to an arithmetic mean, to get around this. i,e ($200 * $50)^1/2 = $100 expectation if you switch. Therefore no gain or loss if you switch. The same then works if you added in a third envelope of money with either 4 or .25 times as much money as you first pick. i.e. ($400*$200*$50*$25)^1/4 = $100. Therefore no gain or loss if you switch. Also, it would work if you changed the ratio. Though I must admit, I feel as if I am using circular logic in thinking the maths of this one through.

but your calculations still assume that somehow you know something about how likely it was to be 100-200 vs 50-100....

	07-01-2005 23:56									#105
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Ok this is not a trick question. Just one that I didn't know how to do.

If you do a 3d6 12 times, and then throw away 6 lowest scores, what would be the expected value of total score?

	08-01-2005 22:36									#106
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quote: Originally posted by Dauphin But don't you know what the distribution could be. Its either $100, $50 or $200, $100.

Yes, it could be. With what probabilities?

	08-01-2005 22:39									#107
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Since you don't know anything about the original probability distribution you can't tell me anything after you eliminate all but two of the choices.

	09-01-2005 00:37									#108
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quote: Well if if makes you feel good to think that its warped then be my guest. but the fact is that if you know that there are two children in a household, and a boy answers the door (and we suppose the one who answers the door is random), then the chances are 1/3 that the other is a boy...

Not so. What makes it 1/3 and not 1/2 is that you only know that at least one of the two is a boy, not that a specific one is a boy. Once you know the one answering the door is a boy, the other one is completely independent:

Case 1: The boy at the door is the firstborn
Subcase 1: The two kids are M F. The other is a girl.
Subcase 2: The two kids are M M. The other is a boy.
Case 2: The boy at the door is the secondborn.
Subcase 1: The two kids are F M. The other is a girl.
Subcase 2: The two kids are M M. The other is a boy.

Each of the cases has a 50% probability. Given that one of the cases is true, each of the subcases has a 50% probability. Hence each subcase overall has a 25% probability and the other is a boy with 25%+25%=50% probability.

This is different from the red/black card trick since instead of a single two-colored card there are two: MF and FM. Thus while in the card trick 1 of the 3 reds is paired with a black and the other 2 with reds, 2 of every 4 boys are paired with a girl.

Or looking at it another way:
You go up to a house.
Case 1 (1/4): The children are MM
Then one that comes to the door will be a boy, and so is the other.
Case 2 (1/2): The children are MF or FM.
Subcase 1 (1/4): The girl comes to the door. We ignore this case.
Subcase 2 (1/4): THe boy comes to the door. Then the other is a girl.
Case 3 (1/4): The children are FF.
Then a girl will come to the door, so we ignore this case.

Of the cases we don't ignore, then, in half the second kid is a boy and the other half it's a girl.

	09-01-2005 10:59									#109
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quote: Originally posted by Snowflake Ok this is not a trick question. Just one that I didn't know how to do. If you do a 3d6 12 times, and then throw away 6 lowest scores, what would be the expected value of total score?

I dont see a way to easily compute this...
We could approximate with a Normal distribution and get a pretty close , specially since there are 3 dices...
A semi brute force, dealing of cases would get an exact answer but would be long.

	09-01-2005 11:01									#110
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quote: Originally posted by civman2000 Not so. What makes it 1/3 and not 1/2 is that you only know that at least one of the two is a boy, not that a specific one is a boy. Once you know the one answering the door is a boy, the other one is completely independent: Case 1: The boy at the door is the firstborn Subcase 1: The two kids are M F. The other is a girl. Subcase 2: The two kids are M M. The other is a boy. Case 2: The boy at the door is the secondborn. Subcase 1: The two kids are F M. The other is a girl. Subcase 2: The two kids are M M. The other is a boy. Each of the cases has a 50% probability. Given that one of the cases is true, each of the subcases has a 50% probability. Hence each subcase overall has a 25% probability and the other is a boy with 25%+25%=50% probability. This is different from the red/black card trick since instead of a single two-colored card there are two: MF and FM. Thus while in the card trick 1 of the 3 reds is paired with a black and the other 2 with reds, 2 of every 4 boys are paired with a girl. Or looking at it another way: You go up to a house. Case 1 (1/4): The children are MM Then one that comes to the door will be a boy, and so is the other. Case 2 (1/2): The children are MF or FM. Subcase 1 (1/4): The girl comes to the door. We ignore this case. Subcase 2 (1/4): THe boy comes to the door. Then the other is a girl. Case 3 (1/4): The children are FF. Then a girl will come to the door, so we ignore this case. Of the cases we don't ignore, then, in half the second kid is a boy and the other half it's a girl.

You are of course right I didnt state the problem properly, like the first time around.
I needed to add that boys always answer door if possible

	09-01-2005 11:23									#111
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quote: Originally posted by Snowflake If you do a 3d6 12 times, and then throw away 6 lowest scores, what would be the expected value of total score?

Impossible to tell, except that it is higher than 105. In essence, you are squashing the distribution curve to the right, but the numerical amount can't be computed in general terms.

Throwing 3D6 12 times is the same as throwing 36 dice all at once. Suppose this is an ideal case, that means you are just going to get rid of all the 1's, all 6 of them. The expected value in this case is 120.

	09-01-2005 11:51									#112
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Ok I've kind of forgotten how to do Permulation and Combination. How do you get 105 and 120?

	09-01-2005 11:56									#113
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If we do 3d6 for six times, I got the expected value is

[(1+2+..+6)*6*6/(6*6*6)]*6=63 only. Is that not right?

	09-01-2005 12:19									#114
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quote: Originally posted by Snowflake If we do 3d6 for six times, I got the expected value is [(1+2+..+6)*6*6/(6*6*6)]*6=63 only. Is that not right?

You're right. I read the question as removing the lowest 6 dice.

	09-01-2005 22:42									#115
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quote: Originally posted by Urban Ranger Impossible to tell, except that it is higher than 105. In essence, you are squashing the distribution curve to the right, but the numerical amount can't be computed in general terms. Throwing 3D6 12 times is the same as throwing 36 dice all at once. Suppose this is an ideal case, that means you are just going to get rid of all the 1's, all 6 of them. The expected value in this case is 120.

Well since its finite, of course it can be computed.
We could draw the distribution for results for 3D6, thats not too bad.
1/216 : 3/18
3/216 : 4/17
6/216 : 5/16
10/216 : 6/15
15/216 : 7/15
21/216 : 8/13
25/216 : 9/12
27/216 : 10/11
(This took about 10 minutes, had to write out the possibilites)

Now the possibilities for the result of 6 times 3d6 which is what were going to be left with , are from 18 to 108.
We can actually compute the possibilites for each by hand although its very lengthy, but a semi automated process with a computer could do it.

to get 18, you need to get 3 with each of the 12 3D6
to get 19, you need to get 3 with 11 of the 3D6 and 4 with another
to get 20 you can either get 11 "3" and 1 "5" or 10 "3" and 2 "4"s
etc...
so we get :
18 : (1/216)^12
19 : (12)(3/216)(1/216)^11
20 : (12)(6/216)(1/216)^11+(12*11/2)(3/216)^2(1/216)^10
etc...

Once you get to 108, just compute the expected value.
Im sure we could write a program that computes those exactly, or we could approximate with a normal, or just write a program to do random tries...

	10-01-2005 03:52									#116
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That is so very confusing ...

	10-01-2005 07:07									#117
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quote: Originally posted by Lul Thyme Once you get to 108, just compute the expected value.

You're not done yet, you still need to throw away the 6 lowest values.

	10-01-2005 12:20									#118
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What, do we really need to complicate things this much?

The way I look at it is like this:

The expected value of score from a six side dice is:
(1+2+...+6)/6=3.5
So expected score of 3d6 would be 10.5.

If it was thrown six times, the expected score is 6*10.5=63.

If it was thrown 12 times, and the six lower ones are thrown away, then I suppose we could assume six scores are lower than the average and six scores are higher than average. So the expected value of the six remain scores would be perhaps the average of 10.5 and 18 (maximum), which is 14.25. Therefore the total score would be 6*14.25=85.5

Would that be a reasonable estimate? Hopefully a computer program would be able to confirm this.

	10-01-2005 12:23									#119
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Just wondering, what does 3d6 mean?

	10-01-2005 12:38									#120
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I think it means that it's 3 six-sided dice

At least that's what I've gleaned from what's been written