What is the probability that the most frequent outcome is not a seven?
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June 2nd, 2019 at 11:24:25 PM
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If I roll two fair six-sided dice 72 times (on a Craps table), what is the probability that the most frequent outcome (between 2 and 12 inclusive), is not a seven? (expectation is it will be the 7)
one may solve this by simulation and or calculation
the most frequent outcome means we track my rolls and table the data (an example)
this sample shows the 7 was the most frequent outcome
10 more samplessample[6,] 7&8 Tied for the most frequent outcome.
The 7 was still the most frequent outcome (It shared the title) so it was not counted as 'the most frequent outcome is not a seven.
one may solve this by simulation and or calculation
the most frequent outcome means we track my rolls and table the data (an example)
| roll (sum) | frequency |
|---|---|
| 2 | 3 |
| 3 | 4 |
| 4 | 9 |
| 5 | 7 |
| 6 | 8 |
| 7 | 19 |
| 8 | 8 |
| 9 | 3 |
| 10 | 6 |
| 11 | 3 |
| 12 | 2 |
| total | 72 |
10 more samples
2 3 4 5 6 7 8 9 10 11 12
[1,] 0 2 6 10 7 12 13 7 9 3 3
[2,] 1 0 4 20 8 10 10 9 6 3 1
[3,] 3 3 11 3 8 16 9 8 5 4 2
[4,] 1 4 2 10 11 12 14 8 6 3 1
[5,] 2 5 7 9 14 13 4 7 7 3 1
[6,] 2 4 7 8 11 14 14 5 3 3 1
[7,] 3 3 4 8 14 12 7 9 5 4 3
[8,] 5 3 6 5 12 10 9 13 6 2 1
[9,] 4 4 6 9 5 12 8 13 7 3 1
[10,] 2 1 4 7 15 13 11 7 6 5 1
The 7 was still the most frequent outcome (It shared the title) so it was not counted as 'the most frequent outcome is not a seven.
Last edited by: 7craps on Jun 3, 2019
winsome johnny (not Win some johnny)
June 3rd, 2019 at 4:23:37 AM
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I just did it 60 times and got exactly 50% of the time it was a 7 and 50% of the time it was either a non-7 or shared with 7.
June 3rd, 2019 at 4:53:28 AM
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the chances that it will be a 7 each time, or never a 7, are easy to calculate
I can't calculate it for your question but there are plenty here who could, I have to think
I can't calculate it for your question but there are plenty here who could, I have to think
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
June 3rd, 2019 at 8:34:00 AM
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me too. I count that the 7 was NOT the most frequent sum. If it was tied with others for the most frequent I do not count that as NOT the most frequent sum.Quote: RSI just did it 60 times and got exactly 50% of the time it was a 7 and 50% of the time it was either a non-7 or shared with 7.
> print(sims)
[1] 60
> print(success)
[1] 30
> print(success/sims)
[1] 0.5
> print(craps[(sims-9):sims, ])#last 10 - sum frequencies
2 3 4 5 6 7 8 9 10 11 12
51 1 4 5 6 13 14 8 11 3 6 1
52 2 4 6 11 11 12 12 4 8 1 1 <<<<<
53 3 5 3 12 13 10 8 7 8 1 2
54 1 4 9 5 10 10 10 10 7 5 1 <<<<<
55 4 7 7 4 13 9 13 3 6 4 2
56 2 5 4 10 9 12 10 4 6 7 3
57 2 4 6 2 9 15 13 11 3 1 6
58 4 1 11 11 5 10 16 8 4 1 1
59 1 1 13 6 15 10 10 7 4 2 3
60 2 2 3 12 10 16 11 8 3 4 1
sumRoll <- 7
x <- tabulate(rollDice())#1:12
R code: (x[sumRoll] < max(x))
winsome johnny (not Win some johnny)
June 23rd, 2019 at 7:34:07 AM
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Is there an answer by calculation? Yes, but it has to be numerical, not analytical.
The overall setting is MultiNomial.
N=72 trials, m=11 outcomes with probabilities <1,2,3,4,5,6,5,4,3,2,1>/36.
The result is the random vector (x2, x3,... ,x12) in the subspace SUM(x’s)=72.
The probability desired is P(x2<x7,x3<x7, etc.,x12<x7).
Let us call a the value of x7. Decomposing the Multinomial we write this as
P(x7=a)*P(x2<a, etc.)
Let’s rewrite this as f(a)*G_a(a), where f(.) is the Binomial (N, 1/6 ) and g_a(.) is the multinomial (N-a, <1,...,5,5,...1>/30 ).
G is the cumulative of g. Lots of sums, as this has no analytical form.
The solution is the SUM f(a)G_a(a) over all a’s
This has no analytical solution. You must tediously compute each possible vector.
Is an approximation available?
When N is sufficiently large, MN is approximated by a multivariate Normal, and sums by integrals.
But the Gaussian integral also is only computable numerically.
.
The overall setting is MultiNomial.
N=72 trials, m=11 outcomes with probabilities <1,2,3,4,5,6,5,4,3,2,1>/36.
The result is the random vector (x2, x3,... ,x12) in the subspace SUM(x’s)=72.
The probability desired is P(x2<x7,x3<x7, etc.,x12<x7).
Let us call a the value of x7. Decomposing the Multinomial we write this as
P(x7=a)*P(x2<a, etc.)
Let’s rewrite this as f(a)*G_a(a), where f(.) is the Binomial (N, 1/6 ) and g_a(.) is the multinomial (N-a, <1,...,5,5,...1>/30 ).
G is the cumulative of g. Lots of sums, as this has no analytical form.
The solution is the SUM f(a)G_a(a) over all a’s
This has no analytical solution. You must tediously compute each possible vector.
Is an approximation available?
When N is sufficiently large, MN is approximated by a multivariate Normal, and sums by integrals.
But the Gaussian integral also is only computable numerically.
.
Reperiet qui quaesiverit
June 24th, 2019 at 8:44:37 AM
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Over 80,000 samples, I get a proportion of 40.02% samples where ‘’7’’ is the lonely maximum.
Reperiet qui quaesiverit
June 24th, 2019 at 5:44:38 PM
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I ran 50 million runs of 72 rolls, and got at least one number being rolled more times than a 7 50.38% of the time
Also, 9.44% of the time, 7 tied with at least one other number for the most
Also, 9.44% of the time, 7 tied with at least one other number for the most
June 24th, 2019 at 8:22:42 PM
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I got 0.5034092 after 10 million runsQuote: ThatDonGuyI ran 50 million runs of 72 rolls, and got at least one number being rolled more times than a 7 50.38% of the time
looks like 75 rolls is the 1st value greater than 50% (0.5012322)
I may get back to this later than sooner
winsome johnny (not Win some johnny)
