# Probability in a dice game

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• Norway
In summary, we are playing a game with regular, fair, six-sided dice and we are throwing 14 of them. The "1" is a "joker", meaning it can count as any other number. We want to know the probability of getting exactly seven of any face value. After attempting various calculations, it seems that the probability is around 35.7%. However, it is not a simple calculation as we must consider the possibility of getting seven of more than one face value simultaneously. A Monte Carlo simulation may be a more practical approach to finding an approximate solution.
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.

anuttarasammyak said:
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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 ?
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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##

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##

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##

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##

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

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

Last edited:
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##.

Last edited:
FactChecker, sysprog and anuttarasammyak
PeroK said:
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?

pbuk said:
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.

FactChecker said:
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.

pbuk said:
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
FactChecker said:
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.

FactChecker said:
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.

FactChecker

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