Zilch is a parlour game introduced to my family by Camilla who learnt it on her honeymoon with David from a man called Bali Bill. There are variations on the rules which we will not further discuss.
Each player takes it in turn to play and it is decided who starts by each player throwing just one dice and seeing who gets the highest number. The highest number starts. If two or more players get the same highest number, those players throw again until a starter is found. The game proceeds with the starter making a play. Once they have finished the next player on the left plays and so it goes found and round. It is important to note that every player gets the same number of plays.
In order to start scoring points a player must make 500 points in their first scoring play.
⇒ Whenever a player throws (any number of dice) there are two possible outcomes:
1) The thrown dice get no points. That is Zilch. Their play ends, Z for Zilch is written against their score. They lose any score they made that play. The next player plays.
2) The dice thrown score some points. The player may then stop and their accumulated total for that play is added to their score; the next player plays. Or the player keeps some of the scoring dice and throws the remainder again. Often there is no choice of how many dice to keep. The total scored so far is accumulated for that play. Now go back to ⇒ (with less dice to throw).
On step 2 above their may be no dice left. This is excellent. The player may start again with all six dice and continue to accumulate points for that play. (Go back to →).
Since a player must get a score of 500 or more in one play to get into the game they must throw relentlessly until that happens. This can cause numerous zilches at the start of the game.
If a player gets four zilches in a row 500 is deducted off their score. Subsequent consecutive zilches incur the same penalty of 500. With bad luck, it is quite possible to go seriously negative at the start of the game. I have seen 5000 and the player involved ended the whole game on a record breaking zero.
Clearly throws including the last three are very desirable as are three 6s, three 5s and two sets of three of a kind.
Some of the examples are deliberately tricky, but it is easy to make an error in the excitement of the game. For example a throw like 255552 might easily be scored as 550 (for three 5s and the singleton 5). It is also quite easy to not see three pairs or a run. The last example, 131113, is interesting because one could also take 1000 points for the three pairs and then throw all six dice again. This is normally the better strategy.
When less than six dice are thrown, a run or three pairs are impossible. When only one or two dice are thrown, three of a kind or three 1s are also impossible.
The dice that were previously thrown, whose score is 'in the bank', have no effect on the score of the thrown dice.
Here is an example of Alice's first play: She threw the six dice and got 146523. She kept the 1 and the 5 (worth 150) and threw the other four dice again. With those she got four 2s. She kept three of those bringing her total on that play to 350. She has one dice left and needs 150 points to get into the game with 500. She throws it and it's a 1! She now has 450 and can throw all six again. But then she threw 364632 which is no points so she lost all her score that play and got Zilch. On Alice's fifth play she got zilch again so another 500 was deducted.
Alice must throw more than 500 in one play to get going on the right direction. If she got exactly 1000 in her sixth play 0 not Z would be put in her score. Z would be wrong and confusing.
Doris threw three 1's as her first throw and wisely stopped. 1000 went in her score and she was in the game. Next time round (after all the others got zilch) she threw 153562 and kept the score of 200. She probably would have been better to keep the 1 and throw five dice again.
Notes:
1) Does not help much with calculating the odds of getting over 2000 in a play.
2) Not guaranteed correct.
3) These are all calculated in zilch (Click Read more) except for the first three zilch odds which were done with a simulator. All the others were checked by said simulator which gave the same answer.
By George, November 2019 with thanks to David and Camilla who checked this for me (and also told me that the rules I had been using were wrong).
With two dice we have a total of ##6^2=6\times6=36## possible results. One can think of these laid out on a ##6\times6## grid with each cell containing one possible result:
This time, counting the number of results that contain a 1 or a 6 is more difficult. It's not just the results in the 1 row + the results in the 1 column + the results in the 5 row and column because there are the awkward results containing (1,1) (1,5) (5,1) and (5,5) which need to be subtracted out of the count. The situation only gets worse with three dice when we need a ##6\times6\times6## grid with ##6^3## possible results. Think about it. We need another way of counting.
Look again at the table. If we cut out all the calls containing 1 or 5 we are left with
That's a ##4\times4## grid and much easier to count. So the ways of not getting 1 or 5 are ##4^2##, the total number of all outcomes is ##6^2## and the total number of ways of getting a 1 or a 5 is what's left. That's ##6^24^2=3616=20##. You can count those in the bigger table to check. Finally, the probability of getting a 1 or 5 is that divided by the total number of all outcomes which know, so$$
\frac{20}{36}=0.56=56\%
$$
That will work for any number of dice. With ## n## dice, there are ##6^n## total possible outcomes (think of an ## n##dimensional hypercube if you will), there are ##4^n## total outcomes without a 1 or a 5, ##6^n4^n## outcomes with a 1 or a 5. So the probability of getting a 1 or a 5 with ## n## dice is $$
\frac{6^n4^n}{6^n}=1\left(\frac{4}{6}\right)^n
$$
and we can fill in the first column of the Zilch Odds Table. Note that this includes getting two 1's with two dice (##n=2##) or even five 1's with five dice.
When throwing one or two dice the probability of getting a zilch is just the probability of all the dice being 2,3,4,5 or 6. We know what those are so we can fill in the bottom two entries of the zilch column in the Zilch Odds Table.
When throwing three dice, zilch is avoided by throwing a 1 or 5 and also avoided by throwing 222, 333, 444 or 666. (we don't count 111 or 555 because they are in the 1 or 5 results total). There are no other ways of avoiding a zilch. There is only one way to get 222 with three dice and only four ways of getting 222, 333, 444 or 666. (Think of a ##6\times6\times6## cube if you will). So the total ways of avoiding zilch with three dice are those four and the ways of getting a 1 or a 5 which we luckily calculated above. If we divide that by ##6^3## we get the probability of avoiding zilch with three dice and subtract that from 1 to get the probability of getting zilch with three dice. So that is$$
1\frac{6^34^3+4}{6^3}
$$We can either work that out or simplify it first to$$
1\frac{6^34^3+4}{6^3}=11+\left(\frac{2}{3}\right)^3\frac{4}{6^3}=\frac{8}{27}\frac{4}{216}=0.31=27\%
$$So we can put that into the Zilch Odds Table.
It's more complicated with four dice: Imagine a ##6\times6\times6\times6## hypercube. Then we must count the ways of getting 222 with four dice and add that times four (for 333,444,666) to the formula at (1) for ## n=4##. It's even more complicated with five dice. But say we had a formula for the number of ways of getting 3 of a particular number with ## n## dice and it was ## Q\left(n\right)## then the ways of scoring with ## n=4\ or\ 5## dice would be
$$
6^n4^n+4\times\ Q\left(n\right)
$$
With six dice we must add in the ways of getting a run or three pairs. It is quite easy to calculate the odds of a run, you may want to try it yourself. The answers are coming below.
The picture shows the dice thrown one after another under the numbers 1,2,3,4,5,6 at the top. At each stage we show a 1 being thrown by an upward line and a 2,3,4,5 or 6 being thrown by a downward line. So after two throws there are four outcomes for the number of 1's. From the top 2,1,1 and 0. Each number is 1+ the number to the left on an up branch and the same as the number on the left on a down branch. I have not written all of them in.
Life is simple after throwing three dice: There is only one outcome with three 1s. The outcome at the top. We know that there are ##6^3=216## possible outcomes in all so the odds of throwing three 1s with three dice are ##\frac{1}{216}=0.5\%##. Pretty remote!
It's more complicated after throwing four dice. There is one outcome with four 1s and four with three 1s. There is only one way of getting to the four 1s outcome, but each three 1s outcome has a downward branch in it and there a five ways of doing a downward branch, because one can get 2,3,4,5 or 6 on a downward branch. So there are ##4\times5=20## ways of getting three 1s and 21 ways of getting three or more. So the odds of getting at least three 1s with four dice are ##\frac{21}{6^4}=1.6\%##. A slight improvement on the odds with three dice.
The first time I did it for four dice I forgot to count the lowest three 1s outcome in the tree  it's an outlier  and so I got the wrong answer. So I turned to Excel to help me with counting and calculating. There is a tab in the spread sheet called '3 of a kind tree' which has the answer. The top part just contains the tree, which is done fairly automatically as indicated above.
Below the 'tree' we have a table showing the number of 1s in the tree for that number of throws. It is reproduced here:
So after one throw there is one 0 in the tree and one 1 in the tree.After two throws there are one 0, two 1s and one 2.
It only gets interesting after three throws when there is one 3 in the tree. Which means three ones were thrown. The interesting part of the table are the results in bold where we have more than three 3s.
With five dice we have 10 outcomes with exactly three 1s. Each of them comes with two downward branches and there are ##5^2## ways of doing those. So there are ##10\times5^2=250## ways of getting exactly three 1s. Similarly there are ##5\times5## ways of getting exactly four 1s and one way of getting exactly five 1s. So the total number of ways of getting at least three 1s with five dice is ##250+25+1=276##. Divide that by the total ways of throwing five dice ##6^5=7,776## and we have odds of ##276/7,776=3.5\%## of getting at least three 1s with five dice. These numbers are calculated in the spreadsheet.
The same is done for six dice and the odds of getting three 1s with six dice are 6.2%. These numbers go into the 'three 1s' of the Zilch Odds Table.
You might remember that that is Pascals triangle. Each number is the sum of the numbers to left and right of it in the row above it. The numbers in each row are also the dreaded binomial coefficients. In row ## n## (0 is the top row) they are the coefficients in the expansion of ##\left(a+b\right)^n## . For example$$\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3
$$I haven't worked out why the numbers in the table are binomial coefficients but I do know the formula for the ## k##th coefficient in row ## n## (because I looked it up). It is
\begin{align}
\frac{n!}{k!\left(nk\right)!}&\phantom {10000}(3)\nonumber
\end{align}The ##!## is called 'factorial' and it just means that you take the number and multiply it by all the numbers below it. So$$
6!=6\times5\times4\times3\times2\times1=720
$$These numbers get very big very fast, ##22!\approx1,124,000,000,000,000,000,000##. Luckily we won't be going anywhere near there.
So the numbers in the table are given by that formula (3) where ## n## is the number of dice and ## k## is the exact number of 1s. So we need to add up all the ways of getting exactly three times, exactly four, exactly five … and divide by the all the ways of throwing the dice. That's
\begin{align}
\frac{1}{6^n}\times\sum_{k=3}^{k=n}{\frac{n!}{k!\left(nk\right)!}\times5^{nk}}&\phantom {10000}(4)\nonumber
\end{align}Well done us.
[Note:
##\Sigma## means take the sum of something for values of k (between k=3 and k=n in this case).]
In the spreadsheet the odds of getting three 1s are calculated up to 10 dice and after throwing 10 dice the odds of getting three 1s are nearly 23% which sounds fair.
In another tab on the spreadsheet (with a bit of code, see below) we can calculate the odds of getting three 1s with huge numbers of dice.
If you threw over 40 dice you are almost bound to get three 1s but however many dices you throw you can never be absolutely certain that you will get three dice. The odds approach 100% but never quite reach it. All is as it should be.
The first thing we notice is that with 13 dice one is certain to get some three of a kind, so there is no nice smooth graph like we had for three 1s. With 13 or more dice the probability of getting some 3 of a kind is always 100%.
With six dice all the ways of getting exactly three 1s also contain ways of getting three 1s and three of a 2,3,4,5 or 6. We have double counted those ways. There is no problem with three, four or five dice so we can safely multiply by six for those cases. With six dice we need to subtract those ways of getting three 1s and three of a 2,3,4,5 or 6. Referring back to the table there were 20 ways of getting exactly three 1s and each of them will have 5 ways of getting three of another number so that's ##20\times5=100## results we need to subtract out of the previous total which was 2,906 and we end up with 2,806 then divide that by ##6^6## and the odds of getting three of a kind are 36.1% which is a little less than ##6.23\%\times6=37.4\%## which was the wrong answer we might have got by the simplistic method. However, in a game of zilch, who cares whether the odds are 36% or 37%? They are both about 1 in 3.
We have now completed the three of a kind column in the Zilch Odds Table.
With seven dice it gets worse because there are 35 ways of getting exactly three 1s and also 35 ways of getting exactly 4 1s. They contain ##5^2## and ##5## ways of getting three of another kind respectively. Subtracting these out and turning the handle we get the right answer 46% plays the wrong answer 58%  a much more substantial correction. For eight dice it just gets trickier but for nine doubly trickier. Luckily we are not concerned with any of these in this game.
\frac{6\times5\times4\times3\times2\times1}{6^6}=\frac{20}{6^4}=1.5\%
$$That was easy and we add it to the Zilch Odds Table.
The first dice can be anything. There are 6 outcomes all of which may lead to success. That was easy.
3a) From 2a, the same as the first two ##6\times1\times1=6## (three of a kind)
3b) From 2a, different from the first two ##6\times1\times5=30## (a pair and 1 singleton)
3c) From 2b, the same as one of the first two which were different ##6\times5\times2=60## (a pair and 1 singleton)
3d) From 2b, different from both the first two ##6\times5\times4=120## (three singletons)
The numbers are the numbers of ways to get to that outcome so they should add up to ##6^3=216## and indeed they do. For each outcome we have just multiplied the number of ways of getting to the previous scenario by the number of ways of getting to that outcome.
Scenario 3d is quite simple because to get three pairs, the fourth dice must be the same as one of the first three: 3 ways to do that; the fifth must be the same as one of the two other first three: 2 ways to do that; and the sixth dice must be the same as the remaining one of the first three: 1 way to do that. So, using the same procedure of multiplying the number of these outcomes into the number for the scenario, the probability of getting three pairs on scenario 3d is$$
\frac{6\times5\times4\times3\times2\times1}{6^6}=\frac{6!}{6^6}=\frac{720}{46,656}=0.015=1.5\%
$$To check our sums we calculate the probability of not getting three pairs on this scenario. It must be$$
\frac{6\times5\times4}{6^3}\times\left(1\frac{3\times2\times1}{6^3}\right)=\frac{120\times\left(6^36\right)}{6^6}=\frac{120\times210}{6^6}=\frac{25,200}{46,656}=0.54=54\%
$$
We have now eliminated 3d which covers more than half the half the possibilities. We will be more interested in the number of ways of having successful (720) and unsuccessful (25,200) outcomes than in the probabilities.
We will investigate 3a,3b,3c.
So the fourth dice could be
4a) from 3a, the same as the first three: ##6\times1\times1\times1## (four of a kind)
4b) from 3a, different from the first three: ##6\times1\times1\times5## (1 triplet and 1 singleton)
3b and 3c were both a pair and one other dice (a singleton) adding the number outcomes for them that was 90. We can deal with them at the same time.
4c) from 3b/c, same as one of the pair ##90\times1## (1 triplet and 1 singleton)
4d) from 3b/c, same as the singleton ##90\times1## (2 pairs)
4e) from 3b/c, different from both of the first two kinds: ##90\times4## (1 pair and 2 singletons)
If we were also worrying about following 3d, there would have been ##120\times6## outcomes from that at the fourth throw, because we don't care what was thrown. We'll use this in the bookkeeping and call it 4f.
After having thrown four dice we should have a total ##6^4=1,296## possible outcomes. Let's check:
Hurrah! (I got 1,032 the first time because I used a ##+## instead of a ##\times## with the 90s. Good thing to check). We're now almost home and dry we can throw the fifth and sixth dice together.
4b and 4c resulted in the same outcome: 1 triplet and 1 singleton. So we can deal with them at the same time. There were ##90+30=120## outcomes like this all of which could lead to success.
6b) from 4b/4c, for success one of the dice must be the same as the singleton and the other the same as the triplet. The fifth dice can be any 2 of 6 and the sixth must be the other 1 of 6. Successful outcomes: ##120\times2\times1=240##, failures ##120\times4\times6+120\times2\times5=4,080##. We can fail on this branch by getting any 1 of the wrong 4 on the fifth dice, then it doesn't matter what we get on the last dice ##4\times6##; or by getting a good one on the fifth and the wrong one on the sixth ##2\times5##. So the total number of outcomes on this endbranch is ##2\times1+4\times6+2\times5=36## as it should be.
6c) from 4d, which was 2 pairs: Our last two must be the same for 3 pairs in all. We either get three different pairs or a pair and four of a kind. That's like 6a but we started with 90 outcomes so there are ##90\times6## successful outcomes and ##90\times30## failures.
6d) from 4e which 360 outcomes of 1 pair and 2 singletons . The fifth dice must be the same as one of the singletons: 2 outcomes like this; and the sixth dice must be the same as the other: 1 outcome like this. So successful outcomes were ##360\times2\times1## against ##360\times34## failures. It is easy to check the latter properly as we did for 6b.
6e) we must also remember 3d/4f where we had 720 successful and 25,200 failures.
Now we are done and can make a table of the outcomes we have found for each possibility.
And adding the totals we get 46,656 which is exactly ##6^6##. Hallelujah!
So now we can say that the odds of getting three pairs is $$
\frac{2,256}{\ 46,656}=0.048=4.8\%
$$and we can add that to the Zilch Odds Table.
How long do you think that took? In percentage terms, estimating the ##\pm## part by eye, throwing 100 times gave odds of about ##4.4\pm1.5\%## and 1000 times gave ##4.6\pm0.5\%##. The uncertainty reduced by a factor of 3 as the number of trials increased by factor of 10. Both results tend to verify our calculated, 'exact', result of ##4.6\%##. There's a more refined branch of probability theory for telling how many trials give what certainty and whether streams of random numbers (like the digits of ## \pi##) are truly random. It's called 'random sampling'. The new Google 54 qubit quantum computer was able to do this in three minutes against the best supercomputer by IBM at 10,000 years (Google's estimate). IBM dispute this.
As for the time it took me: the ten 100 throw trials took about five minutes and the ten 1000s about two days because that's how long it took me to create the spreadsheet that does it all and counts other results. Excel with its random number function and other formulas sufficed.
You can see all these results and a summary in the remaining tabs in the spreadsheet. If you look at the tab '1000 throws 6 dice' you will see the 1000 random throws down the left hand side and the results for each throw calculated on the right. At the top the total is added up for each result. One just has to use the Calculate Now (F9) to throw another 1000 dice and then copy and paste the result into the summary tab. Once it's all set up it takes less than a second to throw 6 dice a 1000 times.
There is also a little table at the top left which shows the number of times each result occurs in the individual dice result columns and the total thereof. This is just to check that the dice throwing is reasonably random and adds up to the right amount.
We can now reproduce the Zilch Odds Table using the results from the spreadsheet.
They are in very good agreement with the calculated zilch odds table. :)
Sub CalculateProbabilities()
' Calculate probabilities of getting three 1s with N dice, N= 1 to 100
Dim N As Integer
Dim MyCells As Range
Set MyCells = Worksheets("3 of a kind graph").Cells
For N = 1 To 100
MyCells(N + 3, 1).Value = N
MyCells(N + 3, 2) = Probability(N, 3)
DoEvents
Next
End Sub
Function Probability(ByVal N As Integer, ByVal K As Integer) As Double
Dim Result As Double
Result = 0
For K = K To N
Result = Result + Factorial(N) / Factorial(K) / Factorial(N  K) * Power(5, N  K)
Next
Probability = Result / Power(6, N)
End Function
Function Factorial(ByVal N As Integer) As Double
Dim Result As Double
Result = 1
While N > 1
Result = Result * N
N = N  1
Wend
Factorial = Result
End Function
Function Power(ByVal N As Integer, ByVal Exponent As Integer) As Double
Dim Result As Double
Result = 1
While Exponent <> 0
Result = Result * N
Exponent = Exponent  1
Wend
Power = Result
End Function
The spreadsheet is at: zilch probabilities.xlsx
How to play
Start
The game is played with six ordinary dice by two to six people. Six is quite a lot and it usually gets out of hand with more. One of the players must be chosen as the scorer.Each player takes it in turn to play and it is decided who starts by each player throwing just one dice and seeing who gets the highest number. The highest number starts. If two or more players get the same highest number, those players throw again until a starter is found. The game proceeds with the starter making a play. Once they have finished the next player on the left plays and so it goes found and round. It is important to note that every player gets the same number of plays.
In order to start scoring points a player must make 500 points in their first scoring play.
Middle: Plays, scoring
→ When a player plays they start by throwing all six dice.⇒ Whenever a player throws (any number of dice) there are two possible outcomes:
1) The thrown dice get no points. That is Zilch. Their play ends, Z for Zilch is written against their score. They lose any score they made that play. The next player plays.
2) The dice thrown score some points. The player may then stop and their accumulated total for that play is added to their score; the next player plays. Or the player keeps some of the scoring dice and throws the remainder again. Often there is no choice of how many dice to keep. The total scored so far is accumulated for that play. Now go back to ⇒ (with less dice to throw).
On step 2 above their may be no dice left. This is excellent. The player may start again with all six dice and continue to accumulate points for that play. (Go back to →).
Since a player must get a score of 500 or more in one play to get into the game they must throw relentlessly until that happens. This can cause numerous zilches at the start of the game.
If a player gets four zilches in a row 500 is deducted off their score. Subsequent consecutive zilches incur the same penalty of 500. With bad luck, it is quite possible to go seriously negative at the start of the game. I have seen 5000 and the player involved ended the whole game on a record breaking zero.
Scoring
You might be wondering how you do score points in this game. Here's the answer:Clearly throws including the last three are very desirable as are three 6s, three 5s and two sets of three of a kind.
Examples
You may want to cover up the answers to test yourself! They are written quite faintly on the right.
Thrown dice

Maximum score

Reason


235643

50

One 5


143575

200

One 1 and two 5s


321211

1000

Three 1s


542444

450

Three 4s =400 + one 5


255552

1000

Three pairs


231

100

One 1


364632

0=Z=zilch

Nothing scores


51

150

One 1 and one 5


532641

1000

A run


434334

700

Three 4s + three 3s


23226

200

Three 2s


131113

1100

Three 1s +one 1

Some of the examples are deliberately tricky, but it is easy to make an error in the excitement of the game. For example a throw like 255552 might easily be scored as 550 (for three 5s and the singleton 5). It is also quite easy to not see three pairs or a run. The last example, 131113, is interesting because one could also take 1000 points for the three pairs and then throw all six dice again. This is normally the better strategy.
When less than six dice are thrown, a run or three pairs are impossible. When only one or two dice are thrown, three of a kind or three 1s are also impossible.
The dice that were previously thrown, whose score is 'in the bank', have no effect on the score of the thrown dice.
The
scores should be laid out as shown to the right. There are four players
Alice, Bob, Doris and Chris. Alice was the starter, chosen as described
above, so her score is in the first column. The players were not sitting in
alphabetical order round the table. She is not doing well, she has scored
zilch five times in a row. Bob was next. He scored 700 in his first play,
zilch in his second then 200, 800. Doris 1000, 200, Z, 500. Scores for each
play are not recorded. It is easy to tell who must play last because the
scores are laid out so neatly.


Here is an example of Alice's first play: She threw the six dice and got 146523. She kept the 1 and the 5 (worth 150) and threw the other four dice again. With those she got four 2s. She kept three of those bringing her total on that play to 350. She has one dice left and needs 150 points to get into the game with 500. She throws it and it's a 1! She now has 450 and can throw all six again. But then she threw 364632 which is no points so she lost all her score that play and got Zilch. On Alice's fifth play she got zilch again so another 500 was deducted.
Alice must throw more than 500 in one play to get going on the right direction. If she got exactly 1000 in her sixth play 0 not Z would be put in her score. Z would be wrong and confusing.
Doris threw three 1's as her first throw and wisely stopped. 1000 went in her score and she was in the game. Next time round (after all the others got zilch) she threw 153562 and kept the score of 200. She probably would have been better to keep the 1 and throw five dice again.
End
As stated above every player gets the same number of plays. The game ends when a player's score gets to 10,000 or more. We'll call that person Bob. When that happens any other players who have not had as many plays as Bob get one last chance to equal or overtake Bob. So that's Doris and Chris in our example. If Doris succeeded she would be declared the winner (unless Chris overtook her on the last play of the game.)Tactics and Etiquette
 If in a throw any dice fall off the table or are cocked (leaning at an angle due to other dice or some other obstacle) all the thrown dice must be thrown again.
 It's important not to get zilch in a play but also important to get a decent score. Therein lies the tension in each play and the judgement required.
 After any throw is made, nobody should touch the dice until a few players (particularly the scorer) have seen them all. Moving them around may be considered cheating. Peter ✠ used to wrap his arms around his thrown dice so only he could see them. This was banned.
 It may be best not to comment after a throw is made. The player may fail to spot a good score. There is no need to tell them (until it's too late). On the other hand you may innocently call a run a 100 and see if the player falls for your trap. Camilla says this is unsporting.
 Consider the scorer. They not only have to play but they have to add up everybody else's score. In particular do not start a new play until the scorer has written down the score from the last play.
 If you’re in Bali and Chris wins, for the next game someone will make him go and get the next round of drinks. While he’s away from the table they’ll move to sit in his place because they believe the seat is affecting his luck – and they want it!
 Cheating: It is surprisingly easy to cheat if all the players are discussing the latest gossip or otherwise entertained. A cheat may slyly turn over a dice as they are 'rearranging' a throw. If all the other players are really being so inattentive it serves them right. Be warned, be attentive and devise a punishment if a cheater persists.
The Zilch Odds Table
We all know that the odds of getting a 1 with one dice is 1 in 6 = 1/6 = 17%. What are the odds (the probability) of getting a 1 with six dice? They aren't six times the odds of getting a 1 with one dice. That would be 100%  a dead cert. Here is a little table with some useful probabilities.
Zilch Odds Table
Throwing dice

Probability of
1 or 5

zilch

three 1s

three of a kind

three pairs

run

6

91%

2.5%

6%

36%

5%

1.5%

5

87%

16%

4%

21%

0

0

4

80%

10%

2%

10%

0

0

3

70%

27%

.5%

3%

0

0

2

56%

44%

0

0

0

0

1

33%

67%

0

0

0

0

Notes:
1) Does not help much with calculating the odds of getting over 2000 in a play.
2) Not guaranteed correct.
3) These are all calculated in zilch (Click Read more) except for the first three zilch odds which were done with a simulator. All the others were checked by said simulator which gave the same answer.
By George, November 2019 with thanks to David and Camilla who checked this for me (and also told me that the rules I had been using were wrong).
Calculation of odds
This is somewhat mathematical (but it only involves counting and dividing!) and it may not be of much interest to some people. But if you want to check that my odds table above is correct you need to read it. I have never been very good at calculating probabilities, which is similar to counting. My methods may be flawed and I take it very slowly.Probability of throwing 1 or 5 with ##{n}## dice
If you throw one dice you can get six possible results. Of those two are a 1 or a 5. That's the counting part. So the probability of getting a 1 or a 5 is ##\frac{2}{6}=33\%##. That's the probability, or dividing, part. One counts the total ways of getting the desired result then counts the total number of results and then divides one by the other.With two dice we have a total of ##6^2=6\times6=36## possible results. One can think of these laid out on a ##6\times6## grid with each cell containing one possible result:
1,1

1,2

1,3

1,4

1,5

1,6

2,1

2,2

2,3

2,4

2,5

2,6

3,1

3,2

3,3

3,4

3,5

3,6

4,1

4,2

4,3

4,4

4,5

4,6

5,1

5,2

5,3

5,4

5,5

5,6

6,1

6,2

6,3

6,4

6,5

6,6

This time, counting the number of results that contain a 1 or a 6 is more difficult. It's not just the results in the 1 row + the results in the 1 column + the results in the 5 row and column because there are the awkward results containing (1,1) (1,5) (5,1) and (5,5) which need to be subtracted out of the count. The situation only gets worse with three dice when we need a ##6\times6\times6## grid with ##6^3## possible results. Think about it. We need another way of counting.
Look again at the table. If we cut out all the calls containing 1 or 5 we are left with
2,2

2,3

2,4

2,6

3,2

3,3

3,4

3,6

4,2

4,3

4,4

4,6

6,2

6,3

6,4

6,6

That's a ##4\times4## grid and much easier to count. So the ways of not getting 1 or 5 are ##4^2##, the total number of all outcomes is ##6^2## and the total number of ways of getting a 1 or a 5 is what's left. That's ##6^24^2=3616=20##. You can count those in the bigger table to check. Finally, the probability of getting a 1 or 5 is that divided by the total number of all outcomes which know, so$$
\frac{20}{36}=0.56=56\%
$$
That will work for any number of dice. With ## n## dice, there are ##6^n## total possible outcomes (think of an ## n##dimensional hypercube if you will), there are ##4^n## total outcomes without a 1 or a 5, ##6^n4^n## outcomes with a 1 or a 5. So the probability of getting a 1 or a 5 with ## n## dice is $$
\frac{6^n4^n}{6^n}=1\left(\frac{4}{6}\right)^n
$$
and we can fill in the first column of the Zilch Odds Table. Note that this includes getting two 1's with two dice (##n=2##) or even five 1's with five dice.
Probability of getting zilch with one, two or three dice
It is very important to know the probability of getting zilch, because, if you do, you lose all the score build up in a turn. This fate should be avoided.When throwing one or two dice the probability of getting a zilch is just the probability of all the dice being 2,3,4,5 or 6. We know what those are so we can fill in the bottom two entries of the zilch column in the Zilch Odds Table.
When throwing three dice, zilch is avoided by throwing a 1 or 5 and also avoided by throwing 222, 333, 444 or 666. (we don't count 111 or 555 because they are in the 1 or 5 results total). There are no other ways of avoiding a zilch. There is only one way to get 222 with three dice and only four ways of getting 222, 333, 444 or 666. (Think of a ##6\times6\times6## cube if you will). So the total ways of avoiding zilch with three dice are those four and the ways of getting a 1 or a 5 which we luckily calculated above. If we divide that by ##6^3## we get the probability of avoiding zilch with three dice and subtract that from 1 to get the probability of getting zilch with three dice. So that is$$
1\frac{6^34^3+4}{6^3}
$$We can either work that out or simplify it first to$$
1\frac{6^34^3+4}{6^3}=11+\left(\frac{2}{3}\right)^3\frac{4}{6^3}=\frac{8}{27}\frac{4}{216}=0.31=27\%
$$So we can put that into the Zilch Odds Table.
It's more complicated with four dice: Imagine a ##6\times6\times6\times6## hypercube. Then we must count the ways of getting 222 with four dice and add that times four (for 333,444,666) to the formula at (1) for ## n=4##. It's even more complicated with five dice. But say we had a formula for the number of ways of getting 3 of a particular number with ## n## dice and it was ## Q\left(n\right)## then the ways of scoring with ## n=4\ or\ 5## dice would be
$$
6^n4^n+4\times\ Q\left(n\right)
$$
With six dice we must add in the ways of getting a run or three pairs. It is quite easy to calculate the odds of a run, you may want to try it yourself. The answers are coming below.
Probability of getting three of a kind
So what are the odds of getting three of a kind? We'll think about getting three 1's to make it a bit more concrete. After struggling with this for a bit and getting quite a few hopeless answers, I came up with this method of counting the ways of getting three 1's. We need this picture.The picture shows the dice thrown one after another under the numbers 1,2,3,4,5,6 at the top. At each stage we show a 1 being thrown by an upward line and a 2,3,4,5 or 6 being thrown by a downward line. So after two throws there are four outcomes for the number of 1's. From the top 2,1,1 and 0. Each number is 1+ the number to the left on an up branch and the same as the number on the left on a down branch. I have not written all of them in.
Life is simple after throwing three dice: There is only one outcome with three 1s. The outcome at the top. We know that there are ##6^3=216## possible outcomes in all so the odds of throwing three 1s with three dice are ##\frac{1}{216}=0.5\%##. Pretty remote!
It's more complicated after throwing four dice. There is one outcome with four 1s and four with three 1s. There is only one way of getting to the four 1s outcome, but each three 1s outcome has a downward branch in it and there a five ways of doing a downward branch, because one can get 2,3,4,5 or 6 on a downward branch. So there are ##4\times5=20## ways of getting three 1s and 21 ways of getting three or more. So the odds of getting at least three 1s with four dice are ##\frac{21}{6^4}=1.6\%##. A slight improvement on the odds with three dice.
The first time I did it for four dice I forgot to count the lowest three 1s outcome in the tree  it's an outlier  and so I got the wrong answer. So I turned to Excel to help me with counting and calculating. There is a tab in the spread sheet called '3 of a kind tree' which has the answer. The top part just contains the tree, which is done fairly automatically as indicated above.
Below the 'tree' we have a table showing the number of 1s in the tree for that number of throws. It is reproduced here:
n Nr throws →

1

2

3

4

5

6

k nr of 1s ↓


0

1

1

1

1

1

1

1

1

2

3

4

5

6

2

0

1

3

6

10

15

3

0

0

1

4

10

20

4

0

0

0

1

5

15

5

0

0

0

0

1

6

6

0

0

0

0

0

1

So after one throw there is one 0 in the tree and one 1 in the tree.After two throws there are one 0, two 1s and one 2.
It only gets interesting after three throws when there is one 3 in the tree. Which means three ones were thrown. The interesting part of the table are the results in bold where we have more than three 3s.
With five dice we have 10 outcomes with exactly three 1s. Each of them comes with two downward branches and there are ##5^2## ways of doing those. So there are ##10\times5^2=250## ways of getting exactly three 1s. Similarly there are ##5\times5## ways of getting exactly four 1s and one way of getting exactly five 1s. So the total number of ways of getting at least three 1s with five dice is ##250+25+1=276##. Divide that by the total ways of throwing five dice ##6^5=7,776## and we have odds of ##276/7,776=3.5\%## of getting at least three 1s with five dice. These numbers are calculated in the spreadsheet.
The same is done for six dice and the odds of getting three 1s with six dice are 6.2%. These numbers go into the 'three 1s' of the Zilch Odds Table.
A brief excursion with Pascal and 60 dice
It would be nice to be able to calculate the numbers in the table without bothering to count nodes in the tree  imagine in an exotic game of zilch with ten dice! You may have noticed something about the numbers. They can be rewritten thus
n=

1


1

1

1


2

1

2

1


3

1

3

3

1


4

1

4

6

4

1


5

1

6

10

10

5

1


6

1

6

15

20

15

6

1

$$I haven't worked out why the numbers in the table are binomial coefficients but I do know the formula for the ## k##th coefficient in row ## n## (because I looked it up). It is
\begin{align}
\frac{n!}{k!\left(nk\right)!}&\phantom {10000}(3)\nonumber
\end{align}The ##!## is called 'factorial' and it just means that you take the number and multiply it by all the numbers below it. So$$
6!=6\times5\times4\times3\times2\times1=720
$$These numbers get very big very fast, ##22!\approx1,124,000,000,000,000,000,000##. Luckily we won't be going anywhere near there.
So the numbers in the table are given by that formula (3) where ## n## is the number of dice and ## k## is the exact number of 1s. So we need to add up all the ways of getting exactly three times, exactly four, exactly five … and divide by the all the ways of throwing the dice. That's
\begin{align}
\frac{1}{6^n}\times\sum_{k=3}^{k=n}{\frac{n!}{k!\left(nk\right)!}\times5^{nk}}&\phantom {10000}(4)\nonumber
\end{align}Well done us.
[Note:
##\Sigma## means take the sum of something for values of k (between k=3 and k=n in this case).]
In the spreadsheet the odds of getting three 1s are calculated up to 10 dice and after throwing 10 dice the odds of getting three 1s are nearly 23% which sounds fair.
In another tab on the spreadsheet (with a bit of code, see below) we can calculate the odds of getting three 1s with huge numbers of dice.
If you threw over 40 dice you are almost bound to get three 1s but however many dices you throw you can never be absolutely certain that you will get three dice. The odds approach 100% but never quite reach it. All is as it should be.
Probability of getting (any) three of a kind
We're also interested in calculating the odds of getting three of a kind  that is three 1s, three 2s, three 3s …. or three 6s because they all score points when one has three or more dice. Naively one might think that these odds are just six times the odds of getting three 1s because there are five more ways to success. But that would obviously be wrong for throwing ten dice because we would calculate the odds of three of any kind at ##22.5\%\times6=135\%## which is better than absolutely certain and therefore impossible. What has gone wrong?The first thing we notice is that with 13 dice one is certain to get some three of a kind, so there is no nice smooth graph like we had for three 1s. With 13 or more dice the probability of getting some 3 of a kind is always 100%.
With six dice all the ways of getting exactly three 1s also contain ways of getting three 1s and three of a 2,3,4,5 or 6. We have double counted those ways. There is no problem with three, four or five dice so we can safely multiply by six for those cases. With six dice we need to subtract those ways of getting three 1s and three of a 2,3,4,5 or 6. Referring back to the table there were 20 ways of getting exactly three 1s and each of them will have 5 ways of getting three of another number so that's ##20\times5=100## results we need to subtract out of the previous total which was 2,906 and we end up with 2,806 then divide that by ##6^6## and the odds of getting three of a kind are 36.1% which is a little less than ##6.23\%\times6=37.4\%## which was the wrong answer we might have got by the simplistic method. However, in a game of zilch, who cares whether the odds are 36% or 37%? They are both about 1 in 3.
We have now completed the three of a kind column in the Zilch Odds Table.
With seven dice it gets worse because there are 35 ways of getting exactly three 1s and also 35 ways of getting exactly 4 1s. They contain ##5^2## and ##5## ways of getting three of another kind respectively. Subtracting these out and turning the handle we get the right answer 46% plays the wrong answer 58%  a much more substantial correction. For eight dice it just gets trickier but for nine doubly trickier. Luckily we are not concerned with any of these in this game.
Probability of getting a run
What is the probability of getting 123456 not necessarily in that order. 436521 would do. Once again we think of throwing the dice one at a time and we are throwing all six. The first can be anything  there are 6 ways of doing that, the second can be any 5 of 6  5 ways, the third can be any 4 of 6 and so on until the sixth which can only be 1 of 6. Multiply those for the total number of ways of getting a run and, as usual, divide by all possible results:$$\frac{6\times5\times4\times3\times2\times1}{6^6}=\frac{20}{6^4}=1.5\%
$$That was easy and we add it to the Zilch Odds Table.
Probability of getting three pairs
Again we think of throwing the six dice one at a time. This time we sort of get a tree but it turns in on itself into clumps, so we don't need a diagram but we calculate the number of outcomes that lead to success and failure as we weave through the tree.The first dice can be anything. There are 6 outcomes all of which may lead to success. That was easy.
Second dice
Second dice could be 2a) same as the first or 2b) different from first. Of the ##6^2## outcomes ##6\times1## are two pairs (2a) and ##6\times5## are two different dice (2b). The total of those is 36, as it should be. We can still get thee pairs with all of them.Third dice
Continuing the scenarios, the third dice could be3a) From 2a, the same as the first two ##6\times1\times1=6## (three of a kind)
3b) From 2a, different from the first two ##6\times1\times5=30## (a pair and 1 singleton)
3c) From 2b, the same as one of the first two which were different ##6\times5\times2=60## (a pair and 1 singleton)
3d) From 2b, different from both the first two ##6\times5\times4=120## (three singletons)
The numbers are the numbers of ways to get to that outcome so they should add up to ##6^3=216## and indeed they do. For each outcome we have just multiplied the number of ways of getting to the previous scenario by the number of ways of getting to that outcome.
Fourth dice
We are now going to follow on from the four scenarios for three dice.Scenario 3d is quite simple because to get three pairs, the fourth dice must be the same as one of the first three: 3 ways to do that; the fifth must be the same as one of the two other first three: 2 ways to do that; and the sixth dice must be the same as the remaining one of the first three: 1 way to do that. So, using the same procedure of multiplying the number of these outcomes into the number for the scenario, the probability of getting three pairs on scenario 3d is$$
\frac{6\times5\times4\times3\times2\times1}{6^6}=\frac{6!}{6^6}=\frac{720}{46,656}=0.015=1.5\%
$$To check our sums we calculate the probability of not getting three pairs on this scenario. It must be$$
\frac{6\times5\times4}{6^3}\times\left(1\frac{3\times2\times1}{6^3}\right)=\frac{120\times\left(6^36\right)}{6^6}=\frac{120\times210}{6^6}=\frac{25,200}{46,656}=0.54=54\%
$$
We have now eliminated 3d which covers more than half the half the possibilities. We will be more interested in the number of ways of having successful (720) and unsuccessful (25,200) outcomes than in the probabilities.
We will investigate 3a,3b,3c.
So the fourth dice could be
4a) from 3a, the same as the first three: ##6\times1\times1\times1## (four of a kind)
4b) from 3a, different from the first three: ##6\times1\times1\times5## (1 triplet and 1 singleton)
3b and 3c were both a pair and one other dice (a singleton) adding the number outcomes for them that was 90. We can deal with them at the same time.
4c) from 3b/c, same as one of the pair ##90\times1## (1 triplet and 1 singleton)
4d) from 3b/c, same as the singleton ##90\times1## (2 pairs)
4e) from 3b/c, different from both of the first two kinds: ##90\times4## (1 pair and 2 singletons)
If we were also worrying about following 3d, there would have been ##120\times6## outcomes from that at the fourth throw, because we don't care what was thrown. We'll use this in the bookkeeping and call it 4f.
After having thrown four dice we should have a total ##6^4=1,296## possible outcomes. Let's check:
Outcome

Occurrences

4a

6

4b

30

4c

90

4d

90

4e


4f

720

Total

1,296

Hurrah! (I got 1,032 the first time because I used a ##+## instead of a ##\times## with the 90s. Good thing to check). We're now almost home and dry we can throw the fifth and sixth dice together.
Fifth and Sixth dice
6a) from 4a, we need to throw both dice the same and we'll either have six of a kind or 4 of one and two of another. The fifth can be anything and the sixth must be the same as the fifth, 6 outcomes success, 30 failure. In total ##6\times6## and ##6\times30## when we multiply these by 6 to get the total number of outcomes on this branch.4b and 4c resulted in the same outcome: 1 triplet and 1 singleton. So we can deal with them at the same time. There were ##90+30=120## outcomes like this all of which could lead to success.
6b) from 4b/4c, for success one of the dice must be the same as the singleton and the other the same as the triplet. The fifth dice can be any 2 of 6 and the sixth must be the other 1 of 6. Successful outcomes: ##120\times2\times1=240##, failures ##120\times4\times6+120\times2\times5=4,080##. We can fail on this branch by getting any 1 of the wrong 4 on the fifth dice, then it doesn't matter what we get on the last dice ##4\times6##; or by getting a good one on the fifth and the wrong one on the sixth ##2\times5##. So the total number of outcomes on this endbranch is ##2\times1+4\times6+2\times5=36## as it should be.
6c) from 4d, which was 2 pairs: Our last two must be the same for 3 pairs in all. We either get three different pairs or a pair and four of a kind. That's like 6a but we started with 90 outcomes so there are ##90\times6## successful outcomes and ##90\times30## failures.
6d) from 4e which 360 outcomes of 1 pair and 2 singletons . The fifth dice must be the same as one of the singletons: 2 outcomes like this; and the sixth dice must be the same as the other: 1 outcome like this. So successful outcomes were ##360\times2\times1## against ##360\times34## failures. It is easy to check the latter properly as we did for 6b.
6e) we must also remember 3d/4f where we had 720 successful and 25,200 failures.
Now we are done and can make a table of the outcomes we have found for each possibility.
Success
with

Successful
Outcomes

Failed
Outcomes


6a

6 of a kind or
4 of a kind and a pair

36

180

6b

4 of a kind
and a pair

240

4,080

6c

4 of a kind
and a pair or 3 pairs

540

2,700

6d

3 pairs

720

12,240

6e

3 pairs

720

25,200

Totals

44,400

And adding the totals we get 46,656 which is exactly ##6^6##. Hallelujah!
So now we can say that the odds of getting three pairs is $$
\frac{2,256}{\ 46,656}=0.048=4.8\%
$$and we can add that to the Zilch Odds Table.
Probability of getting zilch with for five or six dice
I might come back to this problem a later but there's another way …Checking the results  the Zilch simulator
I often make terrible blunders in maths and I wanted to check these calculated results. The only way is by playing zilch or throwing dice and recording the result each time. There's a thing about lady luck, she doesn't obey the laws of averages until she's played many times. How many times would you have to throw six dice to get a good estimate of the odds of getting three pairs of any kind? To find out I threw six dice 100 times counting the times that I got three pairs, I then did that again and again ten times over in all. I then did that all over again throwing six dice a 1000 times. Here are the results.How long do you think that took? In percentage terms, estimating the ##\pm## part by eye, throwing 100 times gave odds of about ##4.4\pm1.5\%## and 1000 times gave ##4.6\pm0.5\%##. The uncertainty reduced by a factor of 3 as the number of trials increased by factor of 10. Both results tend to verify our calculated, 'exact', result of ##4.6\%##. There's a more refined branch of probability theory for telling how many trials give what certainty and whether streams of random numbers (like the digits of ## \pi##) are truly random. It's called 'random sampling'. The new Google 54 qubit quantum computer was able to do this in three minutes against the best supercomputer by IBM at 10,000 years (Google's estimate). IBM dispute this.
As for the time it took me: the ten 100 throw trials took about five minutes and the ten 1000s about two days because that's how long it took me to create the spreadsheet that does it all and counts other results. Excel with its random number function and other formulas sufficed.
You can see all these results and a summary in the remaining tabs in the spreadsheet. If you look at the tab '1000 throws 6 dice' you will see the 1000 random throws down the left hand side and the results for each throw calculated on the right. At the top the total is added up for each result. One just has to use the Calculate Now (F9) to throw another 1000 dice and then copy and paste the result into the summary tab. Once it's all set up it takes less than a second to throw 6 dice a 1000 times.
There is also a little table at the top left which shows the number of times each result occurs in the individual dice result columns and the total thereof. This is just to check that the dice throwing is reasonably random and adds up to the right amount.
We can now reproduce the Zilch Odds Table using the results from the spreadsheet.
The Zilch Odds Table by the simulator
These odds are taken from the 'Summary of 1000 throws' tab of the spreadsheet.
Zilch Odds Table by simulation
Throwing dice

Probability of
1 or 5

zilch

three 1s

three of a kind

three pairs

run

6

91%

2.5%

6%

37%

5%

1.6%

5

86%

16%

3%

21%

0

0

4

80%

10%

2%

10%

0

0

3

70%

29%

0.4%

3%

0

0

Code for calculating the odds of getting three 1s with up 100 dice
Option ExplicitSub CalculateProbabilities()
' Calculate probabilities of getting three 1s with N dice, N= 1 to 100
Dim N As Integer
Dim MyCells As Range
Set MyCells = Worksheets("3 of a kind graph").Cells
For N = 1 To 100
MyCells(N + 3, 1).Value = N
MyCells(N + 3, 2) = Probability(N, 3)
DoEvents
Next
End Sub
Function Probability(ByVal N As Integer, ByVal K As Integer) As Double
Dim Result As Double
Result = 0
For K = K To N
Result = Result + Factorial(N) / Factorial(K) / Factorial(N  K) * Power(5, N  K)
Next
Probability = Result / Power(6, N)
End Function
Function Factorial(ByVal N As Integer) As Double
Dim Result As Double
Result = 1
While N > 1
Result = Result * N
N = N  1
Wend
Factorial = Result
End Function
Function Power(ByVal N As Integer, ByVal Exponent As Integer) As Double
Dim Result As Double
Result = 1
While Exponent <> 0
Result = Result * N
Exponent = Exponent  1
Wend
Power = Result
End Function
Links to files pdfs
All the above is contained in these pdf files. The latter contains links to other useful files.
Zilch probabilities.pdfThe spreadsheet is at: zilch probabilities.xlsx
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