PopCan
Joined: Feb 15, 2012
• Posts: 178
February 20th, 2012 at 10:46:00 PM permalink
Quote: WongBo

It doesn't really have an answer does it?
It just has a range of probabilities?

I'm sure the Wizard could use on of his iterative techniques to figure out the exact number of average rolls. That's well beyond my highschool math education, however.
andysif
Joined: Aug 8, 2011
• Posts: 430
February 21st, 2012 at 1:24:02 AM permalink
Quote: MathExtremist

I don't think he should be able to turn that in.

Question: does a roll that puts you below zero not count, equivalent to Chutes and Ladders? Or is the ending criteria zero or lower, equivalent to Candy Land?

(It's a safe bet I have kids, huh?)

lets say is zero or below
andysif
Joined: Aug 8, 2011
• Posts: 430
February 21st, 2012 at 1:37:07 AM permalink
Quote: WongBo

It doesn't really have an answer does it?
It just has a range of probabilities?

actually, getting an "expected number of roll" is the easier part of the question. "a range of probabilities" would be more difficult as you would have to figure out the standard deviation (in order to find a 95% confidence interval etc).
CrystalMath
Joined: May 10, 2011
• Posts: 1668
February 21st, 2012 at 7:41:02 AM permalink
I calculate 33.37217893, which comes very close to another poster's simulation.

So, how many times did it take your son?
I heart Crystal Math.
odiousgambit
Joined: Nov 9, 2009
• Posts: 7915
February 21st, 2012 at 9:22:13 AM permalink
something tells me what the teacher thinks the answer is, is actually incorrect
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
andysif
Joined: Aug 8, 2011
• Posts: 430
February 21st, 2012 at 5:47:33 PM permalink
Quote: CrystalMath

I calculate 33.37217893, which comes very close to another poster's simulation.

So, how many times did it take your son?

before i started i think it is going to take FOREVER, but luckily i reached 0 in about 20 rolls
andysif
Joined: Aug 8, 2011
• Posts: 430
February 21st, 2012 at 6:18:04 PM permalink
i tried calculating using excel assuming it takes 10 rolls to reach 0 (without hitting the 1), and in average you hit the 1 in every 6 rolls.
it shows you need about 40.2 rolls.
boymimbo
Joined: Nov 12, 2009
• Posts: 5985
February 21st, 2012 at 8:14:24 PM permalink
Yeah, I'm having a hard time getting this calculated.

I know that if you just discarded 1s, the probabilities are:
# of RollsProbability
70.0004608
80.0291072
90.1779712
100.336336384
110.28534636544
120.129685766144
130.034703417344
140.00574724915200001
150.000600678760448001
160.0000393485418496
170.00000155612348416
180.00000003414556672
190.000000000347602944
200.000000000001048576

The problem is what happens when a one is rolled?
----- You want the truth! You can't handle the truth!
boymimbo
Joined: Nov 12, 2009
• Posts: 5985
February 21st, 2012 at 8:14:25 PM permalink
Doubled
----- You want the truth! You can't handle the truth!
CrystalMath
Joined: May 10, 2011
• Posts: 1668
February 22nd, 2012 at 11:21:04 AM permalink
This, I calculated through the use of a Markov chain. Yesterday, I used a stochastic matrix with an absorbing state to calculate only the average number of rolls to end. There is almost exactly a 40% chance of finishing in 20 rolls or fewer. About 1 in 20 kids would have to go to 85 or more rolls.

# of Rolls Probability
7 0.000128601
8 0.006790838
9 0.035641678
10 0.06121854
11 0.054355896
12 0.036897524
13 0.028020105
14 0.025764809
15 0.025430765
16 0.025400113
17 0.025389245
18 0.025307461
19 0.024970031
20 0.024189534
21 0.023035933
22 0.021817628
23 0.020776793
24 0.019942031
25 0.019230398
26 0.018567709
27 0.017919202
28 0.017274663
29 0.016634267
30 0.016003349
31 0.015389895
32 0.01480094
33 0.01423942
34 0.013703981
35 0.013191251
36 0.012698135
37 0.012222728
38 0.011764137
39 0.011322
40 0.010896106
I heart Crystal Math.