# Let's discuss Mathematics

Discussion in 'Science & Technology' started by ParadigmShifter, Mar 16, 2009.

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Meh, applied maths, too useful

2. ### dutchfireDeityRetired Moderator

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Well, being applied is the entire point of Calculus.

3. ### TruronianQuite unfamiliarRetired Moderator

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I like similarity solutions... they're great for turning unsolvable PDEs into unsolvable ODEs.

4. ### AtticusDeityRetired Moderator

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Are there any probability guys here?

Friend of mine was asked what is the expected value when you cast three dice, disregard the smallest result and add the two other together (if there's two or three dice with the smallest result, you disregard only one).

For example if you cast 1-4-6, the result is 10.
If you cast 2-2-4, the result is 6.
If you cast 3-3-3, the result is 6.

This is of course very easy to calculate if you go through all the possibilities of the smallest die, and I need no help for that. Instead I want to know if there's easier way to do this, something involving less calculating. Does anybody know?

5. ### dutchfireDeityRetired Moderator

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Hmm, I don't see any obvious simplification here. So I'd probably just calculate all the possibilities in excel.

6. ### AtticusDeityRetired Moderator

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No sportsmanship in that solution, and that's why I'm asking. It's probably easier to use brute force, but much much more boring.

And, GOOD GOD, EXCEL? I can see you're physicist...

7. ### dutchfireDeityRetired Moderator

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Sportsmanship? I can see you're not a physicist
I'll think about it some more, but it'd only take 5 minutes in Excel.

8. ### AtticusDeityRetired Moderator

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Well no need to think it more, I didn't bother either....

It just feels like there must be some simpler way, and that people who know stochastics better know it.

9. ### dutchfireDeityRetired Moderator

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Okay, so I made tables with the first throw fixed.
Results:
Spoiler :

Sum of all options with first throw:
252
263
283
310
342
377

Difference of these sums:
11
20
27
32
35

Difference of difference
9
7
5
3

Difference of difference of difference ))
-2
-2
-2
Constant
I believe that could help to make a formula, but I forgot how it works.

With the first throw is 1, the expected value is obviously 7. If you increase the first throw from 1 to 2, you change 6+6-1=6+5=11 fields, adding 1 to them, giving an expected value of 7+11/36. Then from 2 to 3, you change 6+6+6+6-4=20 fields...
So you change 12n-n^2 fields when you increase the first throw from n to n+1.
So the expected values are 7, 7+12-1, 7+12-1+24-2^2 etc. Take the average off these 6 values, and you get 8,46.

So the expected value if the first throw is n: En= 7 + (sum(i=1 -> n) 12n-n^2)/36
And then your total expected value is the average of these values, might be a nice formula for that too.

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I'd enumerate the possibilities as well.

But here's a (geometric argument) sketch:

I'll do the 2 dice case first while I work out what's going on...

P[1 is lowest] = P[A=1 or B=1] = 11/36
P[2 is lowest] = P[A=2| B>=2 or B=2|A>=2] = 9/36
P[3 is lowest] = 7/36
P[4 is lowest] = 5/36
P[5 is lowest] = 3/36
P[6 is lowest] = 1/36

This is just the area of an L-shaped slice in a 6x6 grid, easy to visualise.

For the 3D case we have a corner shape for the probabilities, so

P[1 is lowest] = 36 + 30 + 25 / 216 = 91/216
P[2 is lowest] = 25 + 20 + 16 / 216 = 61/216
P[3 is lowest] = 16 + 12 + 9 / 216 = 37/216
P[4 is lowest] = 9 + 6 + 4/216 = 19/216
P[5 is lowest] = 4 + 2 + 1 = 7/216
P[6 is lowest] = 1/216

Then you need to use these weights to multiply the expected roll on the other dice, which is 7 when 1 is the lowest, 8 when 2 is lowest, 9 when 3 is lowest, 10 when 4 is lowest, 11 when 5 is lowest and 12 when 6 is lowest.

91*7/216 + 61*8/216 + 37*9/216 + 19*10/216 + 7*11/216 + 1*12/216

= (91*7 +61*8 + 37*9 + 19*10 + 7*11 + 1*12)/216
= (637 + 488 + 333 + 190 + 77 + 12)/216
= 1737/216 = 8.0416666.....

I think

EDIT: Looks like I disagree with dutchfire... I used windows calculator though...

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12. ### Miseisle of lucy

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I get 8.46 too. It's a pretty curve. EDIT: I also used Excel.

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Can you see a faw in my argument though... maybe I just got some of the sums wrong...

14. ### dutchfireDeityRetired Moderator

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I think the weighing you do might be fishy. I thought about that first too, but deemed it too complicated.

The sum of your Ps is 1, so that part is right.

15. ### dutchfireDeityRetired Moderator

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For example, when 5 is the lowest, these are the possibilities:
10,11,11,12,11,12,12,12
Giving an average of 11,38, not 11.

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I'll try and recalc the P[x is lowest] working backwards:

P[6 is lowest] = 1/216. Obvious.
P[5 is lowest] = P[5 or 6 is lowest] - P[6 is lowest] = 2*2*2/216 - 1/216 = 7/216
P[4 is lowest] = 3*3*3/216 - 2*2*2/216 = 19/216
P[3 is lowest] = 4*4*4/216 - 3*3*3/216 = 37/216
P[2 is lowest] = 5*5*5/216 - 4*4*4/216 = 61/216
P[1 is lowest] = 216/216 - 5*5*5/216 = 91/216

So those weights look right to me???

EDIT: Ah OK it was the expected value on the remaining dice I got wrong.

17. ### dutchfireDeityRetired Moderator

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Sorry, I meant these

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Hold on, when 5 is lowest we get these possibilities:

{5,5,5} so 10
{5,5,6} so 11
{5,6,5} so 11
{5,6,6} so 12
{6,5,5} so 11
{6,5,6} so 12
{6.6.5} so 12
{6,6,6} so 12

so EV is (10 + 3*11 + 4*12) / 8 = 91/8 = 11.375

Yep, I wasn't taking the conditional probability into account in the final step.

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At least I had a go without cheating

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