Probability in a dice game

[anuttarasammyak]

Following up the post ##32 and #33 I show here counting relevant cases of N(10), N(11), N(12), N(13) and N(14). N(n) is introduced in post #32.

*********************************************************
Is the number of cases we want is written as

14∑n=7 14Cn∗N(n)∑n=714 14Cn∗N(n)​

\sum_{n=7}^{14} \ _{14}C_n*N(n)
where N(n) is the number of cases in game of n dices of five faces,
satisfying 7-of-a-kind or (n-7)-of-a-kind with no redundant count ?
*********************************************************

*************************************
N(10)

$Three\ 3-of-a-kind$

For brevity I introduce < > notation for number of multiset multiset permutation of the case AAABBBCCCD
$$<3+3+3+1> = \frac{10!}{3!3!1!}$$
$<3+3+3+1>*5*4$... choice D,E

$Two\ 3-of-a-kind$

AAABBBCCCC
$<3+3+4>*5*\ 4C_2$ ...choice C,(AB)

AAABBBCCDD
$<3+3+2+2>*5*\ 4C_2$ ...choice E,(AB)

AAABBBCCDE
$<3+3+2+1+1>*5*\ 4C_2$ ...choice C,(AB)

$One\ 3-of-a-kind$

AAABBBBBBB (*)
$<3+7>*5*4$ ...choice A,B

AAABBBBBBC
$<3+6+1>*5*4*3$ ...choice A,B,C

AAABBBBBCC
$<3+5+2>*5*4*3$... choice A,B,C

AAABBBBBCD
$<3+5+1+1>*5*4*3$ ...choice A,B,E

AAABBBBCCD
$<3+4+2+1>*5*4*3*2$... choice A,B,C,D

AAABBBBCDE
$<3+4+1+1+1>*5*4$... choice A,B

AAABBBBCDE
$<3+2+2+2+1>*5*4$... choice A,B

$7-of-a-kind$
In addition to (*)

AAAAAAABBC
$<7+2+1>*5*4*3$ ...choice A,B,C

AAAAAAABCD
$<7+1+1+1>*5*4$ ...choice A,E********************************************
N(11)

$Two\ 4-of-a-kind$

AAAABBBBCCC
$<4+4+3>*5*\ _4C_2$... choice C,(AB)

AAAABBBBCCD
$<4+4+2+1>*5*4*3$... choice C,D,E

AAAABBBBCDE
$<4+4+1+1+1>*\ _5C_2$... choice (AB)

$One\ 4-of-a-kind$

AAAABBBBBBB (*)
$<4+7>*5*4$ choice A,B

AAAABBBBBBC
$<4+6+1>*5*4*3$ choice A,B,C

AAAABBBBBCC
$<4+5+2>*5*4*3$ choice A,B,C

AAAABBBBBCD
$<4+5+1+1>*5*4*3$ choice A,B,E

AAABBBCCCD
$<4+3+3+1>*5*4*3$ choice A,D,E

AAAABBBCCDD
$<4+3+2+2>*5*4*3$ choice A,B,E

AAAABBBCCDE
$<4+3+2+1+1>*5*4$ choice A,B

$7-of-a-kind$
In addition to (*)

AAAAAAABBBC
$<7+3+1>*5*4*3$ choice A,B,C

AAAAAAABBCC
$<7+2+2>*5*\ _4C_2$ choice A,(BC)

AAAAAAABBCD
$<7+2+1+1>*5*4*3$ choice A,B,E

AAAAAAABCDE
$<7+1+1+1+1>*5$ choice A

*****************************************
N(12)

$Two\ 5-of-a-kind$

AAAAABBBBBCC
$<5+5+2>*5*\ _4C_2$... choice C,(AB)

AAAAABBBBBCD
$<5+5+1+1>*5*\ _4C_2$... choice E,(AB)

$One\ 5-of-a-kind$

AAAAABBBBBBB (*)
$<5+7>*5*4$ choice A,B

AAAAABBBBBBC
$<5+6+1>*5*4*3$ choice A,B,C

AAAAABBBBCCC
$<5+4+3>*5*4*3$ choice A,B,C

AAAAABBBBCCD
$<5+4+2+1>*5*4*3*2$ choice A,B,C,D

AAAAABBBCCCD
$<5+3+3+1>*5*4*3$ choice A,D,E

AAAAABBBCCDD
$<5+3+2+2>*5*4*3$ choice A,B,E

AAAAABBBBCDE
$<5+4+1+1+1>*5*4$ choice A,B

AAAAABBBCCDE
$<5+3+2+1+1>*5*4*3$ choice A,B,C

AAAAABBCCDDE
$<5+2+2+2+1>*5*4$ choice A,E

$7-of-a-kind$
In addition to (*)

AAAAAAABBBBC
$<7+4+1>*5*4*3$ choice A,B,C

AAAAAAABBBCC
$<7+3+2>*5*4$ choice A,B,C

AAAAAAABBBCD
$<7+3+1+1>*5*4*3$ choice A,B,E

AAAAAAABBCCD
$<7+2+2+1>*5*4*3$ choice A,D,E

AAAAAAABBCDE
$<7+2+1+1+1>*5*4$ choice A,B

*****************************************
N(13)

$Two\ 6-of-a-kind$

AAAAAABBBBBC
$<6+6+1>*5*\ _4C_2$... choice C,(AB)

$One\ 6-of-a-kind$

AAAAAABBBBBBB (*)
$<6+7>*5*4$ choice A,B

AAAAAABBBBBCC
$<6+5+2>*5*4*3$ choice A,B,C

AAAAAABBBBCCC
$<6+4+3>*5*4*3$ choice A,B,C

AAAAAABBBBBCD
$<6+5+1+1>*5*4*3$ choice A,B,E

AAAAAABBBBCCD
$<6+4+2+1>*5*4*3*2$ choice A,B,C,D

AAAAAABBBCCCD
$<6+3+3+1>*5*4*3$ choice A,D,E

AAAAAABBBCCDD
$<6+3+2+2>*5*4*3$ choice A,B,E

$7-of-a-kind$
In addition to (*)

AAAAAAABBBBBC
$<7+5+1>*5*4*3$ choice A,B,C

AAAAAAABBBBCC
$<7+4+2>*5*4*3$ choice A,B,C

AAAAAAABBBCCC
$<7+3+3>*5*\ _4C_2$ choice A,(BC)

AAAAAAABBBBCC
$<7+4+1+1>*5*4*3$ choice A,(BC)

AAAAAAABBBCCD
$<7+3+2+1>*5*4*3*2$ choice A,B,C,D

AAAAAAABBCCDD
$<7+2+2+2>*5*4$ choice A,E

AAAAAAABBBCDE
$<7+3+1+1+1>*5*4$ choice A,B

AAAAAAABBCCDE
$<7+2+2+1+1>*5*\ _4C_2$ choice A,(BC)

****************************
N(14)

$Two\ 7-of-a-kind$

AAAAAAABBBBBBB
$<7+7>*5*4$... choice A,B

$One\ 7-of-a-kind$

AAAAAAABBBBBBC
$<7+6+1>*5*4*3$ choice A,B,C

AAAAAAABBBBBCC
$<7+5+2>*5*4*3$ choice A,B,C

AAAAAAABBBBCCC
$<7+4+3>*5*4*3$ choice A,B,C

AAAAAAABBBBBCD
$<7+5+1+1>*5*4*3$ choice A,B,E

AAAAAAABBBBCCD
$<7+4+2+1>*5*4*3*2$ choice A,B,C,D

AAAAAAABBBCCCD
$<7+3+3+1>*5*4*3$ choice A,D,E

AAAAAAABBBCCDD
$<7+3+2+2>*5*4*3$ choice A,B,E

AAAAAAABBBBCDE
$<7+4+1+1+1>*5*4*3$ choice A,B,E

AAAAAAABBBCCDE
$<7+3+2+1+1>*5*4*3$ choice A,B,C

AAAAAAABBCCDDE
$<7+2+2+2+1>*5*4$ choice A,E

**********************************

 

[PeroK]

I spotted a short cut of sorts. We can simply convert the successful patterns from one value of $n$ to the next. Every pattern for seven dice with at least one 0 can be changed to a pattern for eight dice with at least one 1 and vice versa. There are eleven successful patterns for every $n$ from $0$ to $7$. E.g., from above, the successful patterns for $n = 0$ (must have a $7$):

7, 7, 0, 0, 0
7, 6, 1, 0, 0
7, 5, 2, 0, 0
7, 5, 1, 1, 0
7, 4, 3, 0, 0
7, 4, 2, 1, 0
7, 4, 1, 1, 1
7, 3, 3, 1, 0
7, 3, 2, 2, 0
7, 3, 2, 1, 1
7, 2, 2, 2, 1

These can immediately be converted to the successful patterns for $n = 1$ (must have a $6$), which are:

7, 6, 0, 0, 0 [20]
6, 6, 1, 0, 0 [30]
6, 5, 2, 0, 0 [60]
6, 5, 1, 1, 0 [60]
6, 4, 3, 0, 0 [60]
6, 4, 2, 1, 0 [120]
6, 4, 1, 1, 1 [20]
6, 3, 3, 1, 0 [60]
6, 3, 2, 2, 0 [60]
6, 3, 2, 1, 1 [60]
6, 2, 2, 2, 1 [20]

And, to get the patterns for $n = 2$ just take these and replace a $6$ with a $5$ etc.

The only tedious bit is to check each of these patterns for the first calculation. If the pattern is of the form $xyyyy$, then we have $5$ sub-patterns; $xyzzz$ has $20$ etc. I've put these in square brackets above.

The second calculation can be done by setting up a rule, where $k = 14 - n$ and the five numbers in each row are $r_1$ to $r_5$. This is the number of ways to get each sub-pattern:
$$\binom k {r_1} \binom {k - r_1}{r_2} \binom{k - r_1 - r_2}{r_3} \binom {k-r_1 -r_2 - r_3}{r_4} \binom{k - r_1 -r_2 - r_3 - r_4}{r_5}$$
We just multiply this by the number from the first calculation (in square brackets).

This can all just be put in a spreadsheet and cut and pasted. E.g. for $n = 1$ we have:

7​

6​

0​

0​

0​

20​

1716​

1​

1​

1​

1​

34320​

6​

6​

0​

0​

0​

30​

1716​

7​

1​

1​

1​

360360​

6​

5​

2​

0​

0​

60​

1716​

21​

1​

1​

1​

2162160​

6​

5​

1​

1​

0​

60​

1716​

21​

2​

1​

1​

4324320​

6​

4​

3​

0​

0​

60​

1716​

35​

1​

1​

1​

3603600​

6​

4​

2​

1​

0​

120​

1716​

35​

3​

1​

1​

21621600​

6​

4​

1​

1​

1​

20​

1716​

35​

3​

2​

1​

7207200​

6​

3​

3​

1​

0​

60​

1716​

35​

4​

1​

1​

14414400​

6​

3​

2​

2​

0​

60​

1716​

35​

6​

1​

1​

21621600​

6​

3​

2​

1​

1​

60​

1716​

35​

6​

2​

1​

43243200​

6​

2​

2​

2​

1​

20​

1716​

21​

10​

3​

1​

21621600​

140214360​

0.114864​

The probability of success, given one $1$ is $0.114864$. Then, we just put all this together and we get:

n	p(n)		q(n)	p(n)q(n)
0	0.077887​		0.046058​	0.003587​
1	0.218082​		0.114864​	0.02505​
2	0.283507​		0.25962​	0.073604​
3	0.226806​		0.489781​	0.111085​
4	0.124743​		0.692675​	0.086406​
5	0.049897​		0.820961​	0.040964​
6	0.014969​		0.886374​	0.013268​
7	0.003422​		0.78496​	0.002686​
	0.999313​			0.35665​

That's a final answer of $p = 0.35665$.

 

[pbuk]

That's a final answer of $p = 0.35665$.

That looks a bit high compared with @FactChecker's Monte Carlo simulation in #6. I wonder why?

 

[FactChecker]

That looks a bit high compared with @FactChecker's Monte Carlo simulation in #6. I wonder why?

0.35665 is close to what I ended up with when the rules were all decided on and I fixed a bug in my code. I made 20 runs just now of 1 million tests each (1 test=14 dice tossed). The 20 results ranged from 0.3561 to 0.3574 and averaged 0.35660.

 

[pbuk]

0.35665 is close to what I ended up with when the rules were all decided on and I fixed a bug in my code. I made 20 runs just now of 1 million tests each (1 test=14 dice tossed). The 20 results ranged from 0.3561 to 0.3574 and averaged 0.35660.

I translated your code in post #6 into JavaScript almost exactly (I used zero based arrays) and got similar results to those you posted in #6 so I guess this must be before you fixed the bug. My code runs in the browser on jsfiddle.

 

[FactChecker]

I translated your code in post #6 into JavaScript almost exactly (I used zero based arrays) and got similar results to those you posted in #6 so I guess this must be before you fixed the bug. My code runs in the browser on jsfiddle.

I don't remember posting code in a post #6. Do you mean post #26? I posted a couple of versions. That first version was in post #19. It had an early bug in the random number generation call which I fixed as mentioned in post #21. The second version was in post #26. The problem in that was that it did not count a success if there were 7 1's and other numbers with 0. It was decided that those should count as a success and that the results of #26 were too low. After fixing that my results agreed closely to the 0.35665.

 

[pbuk]

I don't remember posting code in a post #6. Do you mean post #26?

Yes I linked to post #26 but labelled it as #6 by mistake.

The second version was in post #26. The problem in that was that it did not count a success if there were 7 1's and other numbers with 0. It was decided that those should count as a success and that the results of #26 were too low. After fixing that my results agreed closely to the 0.35665.

Ah OK, I made the same change and agree with this result.