B Probability in a dice game

[PeroK]

Although the idea of using the expected number works, that was only necessary because I thought the rules were more complicated. As it stands, here is a way to do it that could be programmed:

For $n = 0$ to $7$ ($n$ is the number of 1's).

Calculate $p(n)$ (it's a binomial distribution - I'll not put too many details in this thread).

The criterion for success is to have at least one number occurring $7-n$ times, from $14 - n$ dice. We need all the patterns that meet this. For example, for $n = 0$, the successful patterns are:

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

You could write a subroutine that would do this for any number. In this case, input $14, 7, 5$ and it would generate all patterns of $5$ numbers that have at least one $7$ and sum to $14$.

Or, do a subroutine to do them all and then select any with a $7$.

That's programming challenge #1!

Next, you need to calculate the probability of each of these patterns. In this case with five random possibilities - as we are assuming we are given the number of 1's. For example, for 7, 6, 1, 0, 0, you need the probability of getting 7 of one number, six of another and 1 of a third number. The way I would calculate this is:

60 sub-patterns (where a sub-pattern would be 7 x 2, 6 x 3, 1 x 4) - i.e. 60 is just 5 x 4 x 3.

For each of these the number of possibilities is $\binom{14} {7} \binom {7} {6}$.

I.e. there are $60 \times \binom{14} {7} \binom {7} {6}$. ways of getting the pattern 7, 6, 1, 0, 0.

You can add them all up and divide by $5^{14}$, which is in general $5^{14-n}$. We could call this $q(n)$ - the probability of success given $n$ 1's.

That's programming challenge #2!

Once you've done this, the total probability is just the sum of $p(n)q(n)$.

Obviously an alternative is to do something similar for all 6 dice and work the addition of the 1's into the logic.

I can't see any short-cuts.

 

[anuttarasammyak]

I still wonder whether I understand the rule rightly so I rewrite the problem.*********************************************************
Is the number of cases we want is written as
\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 ?
*********************************************************

Here 14-n is number of Jokers and $\ _{14}C_{14-n}=\ _{14}C_n$ is number of their appearance.
Some N(n) is easy to get,
N(7)=5
N(8)=5*\ _8C_1 * 4^7
N(14)=\ _{14}C_7(5*(4^7 - 4)+\ _5C_2) if not mistaken.

I should appreciate it if you confirm/correct my understanding.

 

[anuttarasammyak]

EDIT to post #32
Revision of N(8)

I found N(8) in #32 is wrong. The following is the revision.

Say five face of dice is ABCDE. We think 1-of-a-kind cases by its numbers.

FOUR 1-of-a-kind

Type of occurence is expressed as
ABCDEEEE or 1+1+1+1+4
\frac{8!}{1!1!1!1!4!}*5
coefficient 5 : choice of ETHREE 1-of-a-kind

ABCDDDDD or 1+1+1+5
\frac{8!}{1!1!1!5!}*5*4
coefficient 5*4 : choice of D,E

ABCDDEEE or 1+1+1+2+3
\frac{8!}{1!1!1!2!3!}*5*4
coefficient 5*4 : choice of D,ETWO 1-of-a-kind

ABCCCCCC or 1+1+6
\frac{8!}{1!1!6!}*5*\ _4C_2
coefficient $5*\ _4C_2$ : choice of C,(AB)

ABCCCCDD or 1+1+4+2
\frac{8!}{1!1!4!2!}*5*4*3
coefficient 5*4*3 : choice of C,D,E

ABCCCDDD or 1+1+3+3
\frac{8!}{1!1!3!3!}*5*\ _4C_2
coefficient $5*\ _4C_2$ : choice of E,(AB)

ABCCDDEE or 1+1+2+2+2
\frac{8!}{1!1!2!2!2!}*_5C_2
coefficient $\ _5C_2$ : choice of (AB)

ONE 1-of-a-kind

ABBBBBBB or 1+7
\frac{8!}{1!7!}*5*4
coefficient 5*4 : choice of A,B

7-of-a-kind count is included here. No new counting is necessary.

ABBBBBCC or 1+5+2
\frac{8!}{1!5!2!}*5*4*3
coefficient 5*4*3 : choice of A,B,C

ABBBBCCC or 1+4+3
\frac{8!}{1!4!3!}*5*4*3
coefficient 5*4*3 : choice of A,B,C

ABBBCCDD or 1+3+2+2
\frac{8!}{1!3!2!}*5*4*3
coefficient 5*4*3 : choice of A,B.ESum of these cases gives N(8)=3,118,240

Similarly for N(9)

FOUR 2-of-a-kind
AABBCCDDE 2+2+2+2+1
\frac{9!}{2!2!2!2!1!}*5
multiplied by choice of E.

THREE 2-of-a-kind
AABBCCDDD 2+2+2+3
\frac{9!}{2!2!2!3!}*5*4
multiplied by choice of D,E.

TWO 2-of-a-kind
AABBCCCCC 2+2+5
\frac{9!}{2!2!5!}*5*\ _4C_2
multiplied by choice of C,(AB).

AABBCCCCD 2+2+4+1
\frac{9!}{2!2!4!1!}*5*4
multiplied by choice of C,D.

AABBCCCDE 2+2+3+1+1
\frac{9!}{2!2!3!1!1!}*5*\ _4C_2
multiplied by choice of C,(AB).

ONE 2-of-a-kind

AABBBBBBB 2+7 (*)
\frac{9!}{2!7!}*5*4
multiplied by choice of A,B.

AABBBBBBC 2+6+1
\frac{9!}{2!6!1!}*5*4*3
multiplied by choice of A,B,C.

AABBBBCCC 2+4+3
\frac{9!}{2!4!3!}*5*4*3
multiplied by choice of A,B,C.

AABBBBBCD 2+5+1+1
\frac{9!}{2!5!1!1!}*5*4*3
multiplied by choice of A,B,E.

AABBBCCCD 2+3+3+1
\frac{9!}{2!3!3!1!}*5*4*3
multiplied by choice of A,D,E.

AABBBBCDE 2+4+1+1+1
\frac{9!}{2!4!1!1!1!}*5*4
multiplied by choice of A,B

7-of-a-kind
In addition to above (*)
AAAAAAABC 7+1+1
\frac{9!}{7!1!1!}*5*\ _4C_2
multiplied by choice of A,(BC)

N(9) is given by summing up the above.

 

[Norway]

@Norway: In my opinion, you exhibit a (not especially unusual) propensity for adding new conditions subsequently to the statement of the problem.

I agree with your observation. As you suggest, I am indeed interesting in getting things right. English is not my first language, and so I'm struggling to formulate and phrase my thoughts and intentions clearly. If I could, I would want the problem to be clearly and unambiguously stated in the first post. Sadly, as mentioned, many of the great questions asked, I hadn't even considered myself, so some elaboration had to be done subsequently. Hopefully, these "new" conditions did not turn the problem upside-down, but rather clear up some ambiguity. The one notable exception being that the jokers must be counted, which surprised even me. I was almost about to send a long post I had written about how the jokers should be optional, just as the game master called me and told me they are mandatory. I do apologize for all uncertainty, and the extra time and effort it has cost the contributors of this thread. I am very grateful for all participation here.

Here I'd like to re-state what I see as your current version of the rules:

14 6-sided fair dice $-$
exactly 7 dice matching wins [if 8 of the dice match, no win is possible] $-$
1s:

may be any value other than 1 $-$​

all must be the same value $-$​

all must be used in any solution $\cdots$​

[All 3 of those restrictions are significant departures from standard 'jokers wild' rules.]

Is that in your view an accurate and adequate summary of the rules?

Eight of the same kind doesn't immediately disqualify the throw, for instance if the throw includes: {1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4, ...} then we have eight "2"s (which is not a success), but at the same time we have seven "3"s, which makes the throw successful.
Otherwise, yes, I do think that's accurate. Every time you count a value, you must also count all the "1"s as that value. For instance, counting the number of "2"s in a throw, you would have to also add the number of "1"s. @PeroK: Thank you for laying it out in such a clear and organized fashion! I manage to follow along, and I agree with everything you wrote. And a programming challenge it is indeed! It's a little unsatisfying that there are no easier solution, but at the same time it's nice to know that I wasn't a total buffoon when I was called up in the middle of the night by some friends playing a dice game, querying me for a probability calculation. Usually when they do that, I can answer on the fly, or at the very least within a minute or two, but this time I was completely stumped, and I spent hours and hours on this before I just decided to simulate it. At least I can know that I didn't really "fail" the way it felt like then.

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 ?

I am sorry, but I'm not completely able to take in this statement. I'm not very good at maths, so I don't completely understand your formula either, but I can't seem to relate that to my problem. As I said, English is not my first language, so I have struggles both formulating myself and understanding others, which may be an explanation to why I still haven't been able to properly formulate the problem to you. It seems that both @FactChecker and @PeroK have understood the problem correctly, and they are way better at presenting than I am, so if you're still interested in calculating this further (I think @PeroK has a spot-on solution, and @FactChecker and myself have already simulated it), then I hope you will consider reading their posts. They explain it better than I could. Thank you so much for your participation and interest in my thread!

 

[sysprog]

exactly 7 dice matching wins [if 8 of the dice match, no win is possible]

Eight of the same kind doesn't immediately disqualify the throw, for instance if the throw includes: {1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4, ...} then we have eight "2"s (which is not a success), but at the same time we have seven "3"s, which makes the throw successful.

Evidently I didn't elaborate on the bracketed remark sufficiently. By my interpretation, in the case of exactly 4 1s, 3 3s, and 4 2s, 1 4, and 2 others, you have either 7 or 8 matching. If you assign the 1s such that you have 8 2s matching, the 8 don't win, and the remaining 6 can't win either; if you assign the 1s to make exactly 7 matching 3s, then you don't have 8 matching. Without that clarification, my bracketed remark is open to the interpretation that you presented -- my intention in making that remark was to briefly illustrate the 'exactly 7 matching' rule.

 

[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
&lt;3+3+3+1&gt; = \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.