Maths Problem

[KrazyHorse]

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?

[KrazyHorse]

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.

[civman2000]

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.

[Lul Thyme]

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.

[Lul Thyme]

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

[Urban Ranger]

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.

[Snowflake]

Ok I've kind of forgotten how to do Permulation and Combination. How do you get 105 and 120?

[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?

[Urban Ranger]

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.

[Lul Thyme]

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...

[Snowflake]

That is so very confusing ...

[Urban Ranger]

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.

[Snowflake]

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.

[Ramo]

Just wondering, what does 3d6 mean?

[KrazyHorse]

I think it means that it's 3 six-sided dice

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