Expected Number

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.

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

Quote: FleaStiff

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

Really? You think that is the answer?

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

removed
silly

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.

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

Thanks heaps Sally - I knew it was something simple. SixHorse

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).

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silly

Quote: mustangsally

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

Quote: MathExtremist

It's not normal, it's Poisson. See this post for the formula (from a textbook by Sheldon Ross).

ok.
there are a few solutions for calculating the distribution (including simulation that was easiest) but no data from any of them could I find.

I used the Markov chain solution from question 29 (page 38)
in this pdf after a first/second attempt in Excel and R stalled:
http://www.madandmoonly.com/doctormatt/mathematics/dice1.pdf

mean: 61.217384763957.... (769,767,316,159/12,574,325,400)
mode: 36
median: 52

the views

pmf



cumulative



the distribution from 11 to 300 rolls can be found here
https://sites.google.com/view/krapstuff/dice/2d6-stuff

it is in a Google sheet
direct link: https://goo.gl/SKfUZB
after using the PariGp code in a Dell i7 layptop
*****
a 2048x2048 transition matrix was used with
no harm to any math

that dice pdf is fun to read over (helps if you really like dice too)
he (Matthew M. Conroy)
did (and still is doing) a super job on it! Well done!!

Sally