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SixHorse
SixHorse
Joined: Nov 3, 2011
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November 3rd, 2011 at 6:55:34 AM permalink
First time at this site. I was hoping someone could help me derive a classical process to calculate the expected number of dice throws with two dice to obtain all of the eleven sums. I understand the same problem with one die and the six possible outcomes, but I'm having difficulty when the outcomes are not equally likely. Thanks for any help.
FleaStiff
FleaStiff
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November 3rd, 2011 at 7:00:38 AM permalink
The faces are equally likely, they just add up differently.
dm
dm
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November 3rd, 2011 at 10:29:44 AM permalink
Quote: FleaStiff

The faces are equally likely, they just add up differently.



Really? You think that is the answer?
dwheatley
dwheatley
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November 3rd, 2011 at 11:17:06 AM permalink
Let X be a random variable (r.v.) that represents the sum of both dice.
Let X1 be r.v. for outcome of die 1
Let X2 be r.v. for outcome of die 2

P( X = 2 ) = P( X1 = 1 ) * P( X2 = 1 ) = 1/36
P( X = 3 ) = P( X1 = 1 ) * P( X2 = 2 ) + P( X1 = 2 ) * P( X2 = 1 ) = 2/36

and so forth.

E[X] = 7

Edit: I clearly misunderstood the OP
Wisdom is the quality that keeps you out of situations where you would otherwise need it
mustangsally
mustangsally
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November 3rd, 2011 at 1:45:28 PM permalink
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silly
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SixHorse
SixHorse
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November 6th, 2011 at 11:39:11 AM permalink
I tried to find BrunoZ and could not find the post. I did arrive at the same results in different simulation. Thanks for the info, but I would like to find BrunoZ.
Wizard
Administrator
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November 6th, 2011 at 5:38:56 PM permalink
Quote: mustangsally

This was solved this past summer over at 2+2 forum by the handsome math guru BruceZ.



I used the BruceZ method and agree with his answer.

For now, I don't think I can explain his answer any better than he did, but will try for my next Ask the Wizard column. Meanwhile, the following table shows the expected number of rolls to get a 2, 2-3, 2-4, 2-5, ...

Highest Number Prob Exp wait Prob rolled before last Prob rolled last Exp rolls
2 0.027778 36.0 0.000000 1.000000 36.000000
3 0.055556 18.0 0.666667 0.333333 42.000000
4 0.083333 12.0 0.850000 0.150000 43.800000
5 0.111111 9.0 0.922222 0.077778 44.500000
6 0.138889 7.2 0.956044 0.043956 44.816484
7 0.166667 6.0 0.973646 0.026354 44.974607
8 0.138889 7.2 0.962994 0.037006 45.241049
9 0.111111 9.0 0.944827 0.055173 45.737607
10 0.083333 12.0 0.911570 0.088430 46.798765
11 0.055556 18.0 0.843824 0.156176 49.609939
12 0.027778 36.0 0.677571 0.322429 61.217385
It's not whether you win or lose; it's whether or not you had a good bet.
SixHorse
SixHorse
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November 6th, 2011 at 8:50:59 PM permalink
Thanks heaps Sally - I knew it was something simple. SixHorse
MathExtremist
MathExtremist
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November 6th, 2011 at 10:40:56 PM permalink
Quote: mustangsally

You want to know the average wait time for all 11 numbers to roll from 2 dice. Excellent math question.

This is not even a normal distribution.
The mean of 61.2 is not even close to a mode of around 35-38 and they are not close to the median of 52.

I would like to see someone here at WoV create a formula that would solve for the distribution.


It's not normal, it's Poisson. See this post for the formula (from a textbook by Sheldon Ross).
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
mustangsally
mustangsally
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November 8th, 2011 at 4:52:07 PM permalink
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silly
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