HP 15C and TI-60: The Lottery Probability Function

HP 15C and TI-60:  The Lottery Probability Function

The Lottery Probability Function

The expression, of what I call the Lottery Probability Function:

prob =  nCr(k, b) × nCr(n-k, k-b) ÷ nCr(n, k)

where:

*  nCr is the combination function.  nCr(x,y) = x! ÷ (y! × (y - x)!)

*  n = the number of total balls in the lottery

*  k = the number of balls drawn

*  b = the number of balls your ticket matches

*  prob = probability 

Lottery odds are often stated as "1 in number" instead of decimal form.  The number is the statement "1 in number" is 1/p.   For example, "1 in 5" means that the probability is 1/5 or 0.20.  (20%)

HP 15C (Emulator) Program Code

Program Size:

19 steps, 30 bytes

Store before running label A:

R1 = n

R2 = k

R3 = b

Step : Step Code : Key

001 : 42,21,11 : LBL A

002 : 45, 2 : RCL 2

003 : 45, 3 : RCL 3

004 : 43,40 : Cy,x

005 : 44, 0 : STO 0

006 : 45, 1 : RCL 1

007 : 45,30, 2 : RCL- 2

008 : 45, 2 : RCL 2

009 : 45,30, 3 : RCL- 3

010 : 43,40 : Cy,x

011 : 44,20, 0 : STO× 0

012 : 45, 1 : RCL 1

013 : 45, 2 : RCL 2

014 : 43,40 : Cy,x

015 : 44,10, 0 : STO÷ 0

016 : 45, 0 : RCL 0

017 : 15 : 1/x

018 : 44, 0 : STO 0

019 : 43, 32 : RTN

Caution:  HP 15C Limited Edition.  It is a known bug when the combination (Cy,x) and permutation (Py,x) are run multiple times, the display flashes as if a overflow error occurs.  The program otherwise runs fine (with the examples I tested - 2011/04/15 firmware).   

To eliminate unnecessary flashing, use the following key sequence:

[ GTO ] ( A ) [ R/S ]

Full a list of known bugs, as compiled by Katie Wasserman, click here:

https://www.hpmuseum.org/cgi-sys/cgiwrap/hpmuseum/articles.cgi?read=1089

TI-60 (Programmable Scientific/Advanced Scientific) Program Code

57 steps.   No commands are merged.  

For the combination (and permutation) functions, the number must be formatted as the following: nnn.rrr

The command sequence INV EE cancels scientific notation display.  

Step : Step Code : Key

01 : 71 : RCL

02 : 02 :  2

03 : 85 : +

04 : 03 : 3

05 : 55 : ÷

06 : 01 : 1

07 : 42 : EE

08 : 03 : 3

09 : 95 : =

10 : 80 : nCr

11 : 61 : STO 

12 : 00 : 0

13 : 71 : RCL

14 : 01 : 1

15 : 85 : +

16 : 71 : RCL

17 : 02 : 2

18 : 55 : ÷

19 : 01 : 1

20 : 42 : EE

21 : 03 : 3

22 : 95 : =

23 : 80 : nCr

24 : 61 : STO

25 : 55 : ÷

26 : 00 : 0

27 : 71 : RCL

28 : 01 : 1

29 : 75 : -

30 : 71 : RCL

31 : 01 : 1

32 : 85 : +

33 : 53 : (

34 : 71 : RCL

35 : 02 : 2

36 : 75 : -

37 : 71 : RCL

38 : 03 : 3

39 : 54 : )

40 : 55 : ÷

41 : 01 : 1

42 : 42 : EE

43 : 03 : 3

44 : 95 : =

45 : 80 : nCr

46 : 61 : STO

47 : 65 : ×

48 : 00 : 0

49 : 71 : RCL

50 : 00 : 0

51 : 76 : 1/x

52 : 61 : STO

53 : 00 : 0

54 : 12 : INV

55 : 42 : EE

56 : 13 : R/S

57 : 22 : RST

Examples

1.  49 balls and 6 are drawn.    Set up variables:  R1 = n =  49 and R2 = k = 6.

R3 = b = 6:  1 in 13,983,816

R3 = b = 5:  1 in 54,200.83721

R3 = b = 4:  1 in 1,032.3396899

2.  52 balls and 6 are drawn.  Set up variables:  R1 = n = 52 and R2 = k = 6.

R3 = b = 6:  1 in 20,358,520

R3 = b = 5:  1 in 73,762.75362

R3 = b = 4:  1 in 1,311.337842

Note that this function does not take the power ball, which only 1 is drawn in addition to the lottery balls, into account.   If you have a power ball of p balls, we can calculate the odds as follows:

You match the power ball:

prob = 1 ÷ p × nCr(k, b) × nCr(n-k, k-b) ÷ nCr(n, k)

1 in odds:  1/prob

You don't match the powerball:

prob = (p - 1) ÷ p × nCr(k, b) × nCr(n-k, k-b) ÷ nCr(n, k)

1 in odds:  1/prob

Source:

"Lottery mathematics".  Wikipedia.  Last edited July 10, 2023.  Retrieved July 23, 2023.  https://en.wikipedia.org/wiki/Lottery_mathematics

Eddie 

All original content copyright, © 2011-2023.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

Casio fx-9750GIII and Swiss Micros DM42: Bijections

Casio fx-9750GIII and Swiss Micros DM42:  Bijections

Bijection - Three Problems 

I have a subscription to Brilliant (brillant.org, this is not a paid ad), where I am taking courses of probability, quantitative finance, and quantum mechanics.  

Today's blog is dealing with bijections, a technique to solve a class of counting problems.   A counting problem calculates the number of ways an event, such as distributing pieces of candy among a group of people, can happen.  

Here, I am working with three types of problems:

Problem 1:  How many positive integer solutions are possible of the equation:

A + B + C + ... = M

M = sum of the integers A, B, C, etc.

M > A, M > B, M > C.  

Problem 2:  How many positive integer solutions are possible of the equation:

A + B + C + ... ≤ M

M = sum of the integers A, B, C, etc.

M > A, M > B, M > C.  

Problem 3:  How many ways can distribute a set amount of balls into a set of urns?  Assume that the urns do not have to have an equal amount of balls.  

Similar problem would be:

How many ways to distribute candy among a population of people?  

How many ways to give a deck of cards from a deck to a table full of players, given that each card is randomly distributed?  Assume that the players do not have equal amount of cards.  

The programs on this blog uses two different approaches.   The program for the fx-9750GIII uses a prompt approach while the program for the DM42 uses a direct approach.  

Casio fx-9750GIII Program:  BIJECT

Code:

"EWS 2023-06-12"

0 → S

ClrText

Menu "BIJECTIONS","A+B+C+...=M",1,"A+B+C+...≤M",2,"BALLS IN URNS",3

Lbl 2

1 → S

Lbl 1

"# OF VARIABLES"? → B

V + S - 1 → V

"SUM (POS. INTEGER)"? → M

Goto 4

Lbl 3

"# OF URNS"? → V

V - 1 → V

"# OF BALLS"? → M

Lbl 4

(M + V) nCr V → S

"# SOLUTIONS = "

S

Notes:

The character # is from the character map (CHAR) of the default program menu in program editing mode (Mode B).

The combination function is displayed as a bold C.  I have nCr in the code listing for clarity.

Swiss Micros DM42 Program:  BIJECT

Code also works for HP 42S, Plus42, Free42

Code:

00  {14-Byte Prgm}

01  LBL "BIJECT"

02  +

03  LAST x

04  COMB

05  END

Stack Set Up:

Y:  number of dividers

X:  number of balls

Set up the stack for the three types of problems.

Problem 1:

Y:   number of variables - 1

X:  sum

Problem 2:

Y:  number of variables

X:  sum

Problem 3:

Y:  number of urns - 1

X: number of balls

Examples

Problem 1:  How many positive integer solutions are possible of the equation:

A + B + C = 20

Number of variables:  3

Sum:  20

Number of dividers:  2   (the number of plus signs in the equation)

Number of solutions:  nCr(20 + 2, 20) = 231

Problem 2:  How many positive integer solutions are possible of the equation:

A + B + C ≤ 20

Number of variables:  3

Sum:  20

Number of dividers:  3   

Since we are allowing for A, B, and C to add up to less than 20, add a slack variable, S, to "catch" any deficit to make the sum 20.   Therefore, we are really solving:

A + B + C + S = 20   

In listing the solutions, S is ignored. 

Number of solutions:  nCr(20 + 3, 20) = 1,771

Problem 3:  How many ways can distribute 40 balls in to 6 urns?

Set the urns in a line, and imagine a divider in between each urn.

urn 1 | urn 2 | urn 3 | urn 4 | urn 5 | urn 6

Number of dividers: 5

Number of balls:  40

Number of distributions:  nCr(40 + 5, 40) = 1,221,759

Try listing all possible distributions.

Eddie 

All original content copyright, © 2011-2023.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

HP Prime: Graphs of Permutation and Combination

HP Prime:  Graphs of Permutation and Combination

Definitions

Permutation:

perm(n, x) = n! ÷ (n - x)!

Combination:

comb(n, x) = n! ÷ (x! × (n - x)!)

Combination with Repeated Choices allowed:

comb(n + x - 1, x) = (n + x - 1)! ÷ (x! × (n - 1)!)

For the following graphs, I use screen shots with the HP Prime Virtual Calculator.

Discrete Graphs  

Let n and x be positive integers.

Continuous Graphs

N is still a positive integer while x is a positive real number.   The factorial can be represent as the Gamma function as  x! = Γ(x+1).  

3D-Graphs

The variables  x and y take real values.

Eddie 

All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

7000G Retro Month - June 19 Edition

7000G Retro Month - June 19 Edition

Introduction

Welcome to the 7000G Retro Month, which features programming for the classic Casio calculators from the mid/late 1980s:  primarily fx-7000G and fx-7500G.  Since the programming language stays similar throughout the years, programs can be translated to the fx-6300G and later graphing calculators with little to no adjustments.  Non graphic programs should be ported to the fx-4000P, fx-4500P (A), fx-3650p (II), fx-50F Plus (II), and fx-5800P with little to no adjustments.  

7000G Retro Month takes place every Saturday during June 2021.

To make text easier to type, I can going to use the following text friendly symbols for the following:

->  for →

/I for ⊿

=> for ⇒

What do you think?   Unicode or simple text equivalents?  

- - - - - - -- - -- - -

Today's subject revolves around Probability and Random Numbers.  Enjoy!

- - - -- - - -- - -- -- -

Random Integers: Repeats Allowed 

This program allows the user to generate a number of integers between A and B, repeats are allowed.   Each integer is displayed one at a time.

"A"? -> A

"B"? -> B

"N"? -> N

Lbl 1

Int ((B-A+1) Rnd#) /I

Dsz N

Goto 1

Combinatorics

The program allows the user to choose between three options:

1.  PERM:  permutation:  nPr

2.  COMB:  combination:  nCr

3.  COMB REPLACE:  combination with replacements allowed (n+r-1)Cr

"N"? -> N

"R"? -> R

"1. PERM"

"2. COMB"

"3. COMB REPLACE"

? -> K

K=1 => N!÷(N-R)! -> X

K=2 => N!÷(R!(N-R)!) -> X

K=3 => (N+R-1)!÷(R!(N-1)!) -> X

X

Binomial Distribution

This program calculates the sum of probabilities for a binomial distribution:

total probability = ∑( nCr(N,K) p^K (1-p)^(N-K), K=A to B)

Probability:  0 < p < 1

"A"? -> A : "B"? -> B

"N"? -> N : "P"? -> P

0 -> M : Lbl 1

M+(N!×P^A×(1-P)^(N-A))÷(A!(N-A)!) -> M

A+1 -> A

A>B => Goto 2

Goto 1

Lbl 2

M

Confidence Interval

This program generates a confidence interval using one of four probabilities are assigned to the following variables:

F: 99%  (z* ≈ 2.576)

G: 98% (z* ≈ 2.326)

H: 95% (z* ≈ 1.96)

I: 90% (z* ≈ 1.645)

interval = mean ± z* × variance / √n

"CONF INTERVAL"

2.576 -> F

2.326 -> G

1.96 -> H

1.645 -> I

"MEAN"? -> A

"VAR"? -> B

"N"? -> N

"F=.99, G=.98"

"H=.95, I=.90"

? -> J

A-JB÷√N -> E /I

A+JB÷√N -> F

E:  low, F: high

Eddie

All original content copyright, © 2011-2021.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author. 

fx-260 Solar Algorithms Part II

fx-260 Solar Algorithms Part II

Decimal to Binary Conversions

This is probably best demonstrated by example. 

Algorithm:

To the decimal integer D to binary integer B:

1.  Determine the number of digits (zeroes or ones) that the binary integer is going to have.  Also, we'll store D in memory.

n = int(log D/log 2)

Each digit will represent the powers 2^(n) to 2^0.

Keystrokes: D [SHIFT] (Min) [ log ] [ ÷ ] 2 [ log ] [ = ]  // ignore the decimal part

2.  Starting with n and going to 0, calculate 2^n.  Compare 2^n to the number in memory. 

If 2^n ≤ Memory, then write a 1.  Subtract 2^n from memory:  2^n [ +/- ] [M+].  Decrease n by 1 and continue.

If 2^n > Memory, then write a 0.  Decrease n by 1 and continue.

Each digit will be written to the right of the preceding digit.

Example:  Convert 462 to binary.

Determine n:
462 [SHIFT] (Min) [ log ] [ ÷ ] 2 [ log ] [ = ]
Result:  8.851749041

Start with n = 8.  462 is stored in Memory.

In M:  462  (n = 8)
2 [ x^y ] 8 [ = ] 256,  256 ≤ 462,  [+/-] [ M+ ]   // first digit is 1
Binary:   1________

In M:  206 (n = 7)
2 [ x^y ] 7 [ = ] 128,  128 ≤ 206,  [+/-] [ M+ ]  // next digit is 1
Binary:   11_______

In M:  78 (n = 6)
2 [ x^y ] 6 [ = ] 64,  64 ≤ 78,  [+/-] [ M+ ]  // next digit is 1
Binary:   111______

In M:  14  (n = 5)
2 [ x^y ] 5 [ = ] 32,  32 > 14  // next digit is 0
Binary:   1110_____

In M:  14  (n = 4)
2 [ x^y ] 4 [ = ] 16,  16 > 14  // next digit is 0
Binary:   11100____

In M:  14  (n = 3)
2 [ x^y ] 3 [ = ] 8,  8 ≤ 14,  [+/-] [ M+ ]  // next digit is 1
Binary:   1110001___

In M:  6  (n = 2)
2 [ x^y ] 2 [ = ] 4,  4 ≤ 6,  [+/-] [ M+ ]  // next digit is 1
Binary:   11100011__

In M:  2  (n = 1)
2 [ x^y ] 1 [ = ] 2,  2 ≤ 2,  [+/-] [ M+ ]  // next digit is 1
Binary:   111000111_

In M:  2  (n = 0)
2 [ x^y ] 01 [ = ] 1,  1 > 0  // last digit is 0
Binary:   1110001110

Result:  462_10 = 1110001110_2

Combinations that Allow for Repeated Picks

Sometimes when we are choosing r objects out of a group of n objects, repeated picks are allowed.  That is, any object that is picked is put back in the pool and has a chance to be picked again.  The formula to calculate such calculations is:

nHr = (n + r - 1)! / (r! * (n - 1)!)

We can state nHr in terms of nCr (number of combinations where no repeats are allowed).

aCb = a! / (b! * (a - b)!)
Let a = n + r - 1 and b = n - 1.
Then a - b = n + r - 1 - (r - 1) = r

Then:
nHr = (n + r -1)C(n - 1)

Algorithm:
[ ( ] n [ + ] r [ - ] 1 [ ) ] [SHIFT] (nCr) [ ( ] n [ - ] 1 [ ) ] [ = ]

Example:
Find the number of combinations of picking 10 objects out of the pool of 38, where repeats are allowed.

n = 38, r = 10

[ ( ] 38 [ + ] 10 [ - ] 1 [ ) ] [SHIFT] (nCr) [ ( ] 38 [ - ] 1 [ ) ] [ = ]

Result: 5,178,066,751

Harmonic Mean of Numbers

The harmonic mean of a set of n numbers is calculated by:
HM = n / Σ(1 / x_i)

We can use the Statistics mode to calculate the harmonic mean.

Algorithm:
[ON]  // clear everything and reset calculator to COMP mode
[MODE] 0  // Mode 0 is SD mode (single data, standard deviation)
x_i [SHIFT] (1/x) [M+](DATA)
....
[SHIFT] (n) [ × ] [SHIFT] (Σx) [ = ]

Example:
Data:  3.8, 4.6, 5.9, 7.1, 7.6, 9.0  (n = 6)

[ON]
[MODE] 0 
3.8 [SHIFT] (1/x) [M+](DATA)
4.6 [SHIFT] (1/x) [M+](DATA)
5.9 [SHIFT] (1/x) [M+](DATA)
7.1 [SHIFT] (1/x) [M+](DATA)
7.6 [SHIFT] (1/x) [M+](DATA)
9.0 [SHIFT] (1/x) [M+](DATA)
[SHIFT] (n) [ × ] [SHIFT] (Σx) [ = ]

Result:  6.20145512

Atwood Machine

Given the masses of two weights (in kg) on an Atwood Machine, the following system describes the relationship between the masses, tension, and acceleration of the system:

T + M1 * a = M1 * g
T - M2 * a = M2 * g

where:
T = tension of the system (N)
a = acceleration, positive means the pulley rotates counter-clockwise; negative means the pulley rotates clockwise (m/s^2)
g = Earth's gravity constant, 9.80665 m/s^2

Assumptions:
1.  The mass of both the pulley and the string are negligent
2.  Mass 1 is on the left side of the pulley while Mass 2 is on the right. 

Solving for T and a give:
a = (M1 - M2) * g / (M1 + M2)
T = M1 * (g - a) = M2 (g + a)

Algorithm:
[ ( ] M1 [ - ] M2 [ ) ] [ × ] 9.80665 [ ÷ ] [ ( ] M1 [ + ] M2 [ ) ] [ = ]   
// acceleration is displayed

[ +/- ] [ + ] 9.80665 [ = ] [ × ] M1 [ = ]
// tension is displayed

Example:
M1 = 18 kg, M2 = 12 kg

[ ( ] 18 [ - ] 12 [ ) ] [ × ] 9.80665 [ ÷ ] [ ( ] 18 [ + ] 12 [ ) ] [ = ]   
Acceleration: 1.96133 m/s^2  (pulley is rotating counter-clockwise)

[ +/- ] [ + ] 9.80665 [ = ] [ × ] 18 [ = ]
Tension:  141.21576 N


Eddie

All original content copyright, © 2011-2019.  Edward Shore.   Unauthorized use and/or unauthorized distribution for commercial purposes without express and written permission from the author is strictly prohibited.  This blog entry may be distributed for noncommercial purposes, provided that full credit is given to the author.

Adventures in Python: Combinations and Permutations

Adventures in Python: Combinations and Permutations

This program offers the user a choice of one of three calculations:

1.  Combinations, no repeats

2.  Permutations

3.  Combinations, with repeats

The program demonstrates how a menu is made.  Equality is tested with two equal signs (==).  A single equal sign (=) represents assignment for the Python.

The math.factorial function uses integers.

#  Program 005: Combinations, Permutations, Repeated Combination

#  Demonstration of the If structure

import math

# factorial is only for integers

# choices

# we will need to format the choice variable as an integer (number)

print("Make your choice:")

print("1.  Combination")

print("2.  Permutation")

print("3.  Repeated Combination")

choice = int(input("Choice: "))

# error condition, also input routine

if choice < 1 or choice > 3:

    print("Not a valid choice")

else:

    n = int(input("n = "))

    r = int(input("r = "))

    nf = math.factorial(n)

    rf = math.factorial(r)

    df = math.factorial(n-r)

# calculation test.  Equality is tested using 2 equal signs

if choice == 1:

    calc = nf/(df*rf)

    print("Combinations = ",calc)

if choice == 2:

    calc = nf/df

    print("Permutations = ",calc)

if choice == 3:

    calc = math.factorial(n+r-1)/(rf*math.factorial(n-1))

    print("Combinations = ",calc)

Example: n = 52, r = 5

Make your choice:

1.  Combination

2.  Permutation

3.  Repeated Combination

Choice: 1

n = 52

r = 5

Combinations =  2598960.0

>>>

Make your choice:

1.  Combination

2.  Permutation

3.  Repeated Combination

Choice: 2

n = 52

r = 5

Permutations =  311875200.0

>>>

Make your choice:

1.  Combination

2.  Permutation

3.  Repeated Combination

Choice: 3

n = 52

r = 5

Combinations =  3819816.0

Eddie

This blog is property of Edward Shore, 2017.

High Rollers: Game Show and Possible Sums with the TI-84 Plus

High Rollers:  Game Show and Possible Sums with the TI-84 Plus

If you have a group of numbers, what are the possible sums you can make?  We’re talking about combinations of 1, 2, to as many numbers you have.  This is one of the key elements of the classic game show High Rollers (1970s, 1987-1988), a dice rolling game based on Shut The Box. 

To whet your appetite, click on the links below to see classic episodes (the links should be good as of 3/15/2017).  The first two links are from episodes from the 1978-1980 series with Alec Trebek (with facial hair) and the last two links are from episodes from the 1987-1988 series with Wink Martindale. We’ll get to the game show later in this blog entry. (no ownership implied, these are from other YouTube accounts)

https://www.youtube.com/watch?v=nsxQoDM0rd0&t=1172s

https://www.youtube.com/watch?v=urzJINgMu3E

https://www.youtube.com/watch?v=j951e0r0EB8

https://www.youtube.com/watch?v=OeD__GYvZZk

The Program POSSUMS

The program POSSUMS calculates all possible sums created by a set of numbers (2, 3, or 4).  With n numbers available, there are 2^n – 1 possible sums.

For n = 2, with let A and B represent the numbers.   There are 2^2 -1  = 3 possible sums:

A, B, A + B

For n = 3 (A, B, C), there are 2^3 – 1 = 7 possible sums:

A, B, C, A + B, A + C, B + C, A + B + C

For n = 4 (A, B, C, D) there are 2^4 – 1 = 15 possible sums:

A, B, C, D, A + B, A + C, A + D, B + C, B + D, B + C, A + B + C, A + B + D, A + C + D, B + C + D, A + B + C + D

The program POSSUMS allows all possible sums, including repeats.

TI-84 Plus Program POSSUMS

Menu("SUMS FROM AVAIL. NUMBERS","2",2,"3",3,"4",4)

Lbl 2

Prompt A,B

{A,B,A+B}→L₁

Goto 5

Lbl 3

Prompt A,B,C

{A,B,C,A+B,A+C,B+C,A+B+C}→L₁

Goto 5

Lbl 4

Prompt A,B,C,D

{A,B,C,D,A+B,A+C,A+D,B+C,B+D,C+D,A+B+C,A+C+D,A+B+D,B+C+D,A+B+C+D}→L₁

Goto 5

Lbl 5

SortA(L₁)

Pause L₁

Examples

3 Numbers:  A = 2, B = 3, C = 5

Result:  {2, 3, 5, 5, 7, 8, 10}

Notes:  There are two 5s, meaning 5 can be made by two combinations (5, 2 + 3)

4 Numbers:  A = 1, B = 6, C = 7, D = 9

Result:  {1, 6, 7, 7, 8, 9, 10, 13, 14, 15, 16, 16, 17, 22, 23}

Notes:  7 can be made two ways (7, 6 + 1); 16 can be made two ways (7 + 9, 1 + 6 + 9)

The Game Show High Rollers

 

High Rollers was a game show that aired in three series: 1974-1976, 1978-1980, and 1987-1988.  The premise of the game was to clear as many numbers, ranging from 1 to 9.  In the original 1974-1976 series, each number was attached to a prize.  In the more famous 1978-1980 and 1987-1988 series, the numbers were aligned (seemingly at random) on a 3 x 3 grid.  Each column represented a prize or a group of prizes. 

In the main game there are two contestants.  You would win by either rolling the last number off the board or most likely, force your opponent to roll a number that can’t be cleared.  Obviously the total on the dice is used to clear numbers.  For example, a roll of a 6 (the total counts, not the pips on the individual dies), can clear any of the following combinations: 6 itself, 1 and 5, 2 and 4, or 1 and 2 and 3.  Starting with the 1978 series, rolling doubles earned the contestant an insurance marker, basically an extra life.

Winning two games entitled the champion to play the Big Numbers.  The object remained the same, get rid of the numbers 1 to 9 for a major cash prize or car.

One thing to note:  unlike Shut the Box, High Rollers offered no provision should the last number remaining on the board be a 1.

Below are all the possible combinations that can be cleared with each roll.  There are 61 combinations.  Statistically, rolling a 7 is the most likely event, followed by 6 or 8.  However, the most powerful rolls are 12, followed by 11, then 10. 

All the Possible Combos in High Rollers

Total	Combinations that can be Cleared
2	2
3	3, 1-2
4	4, 1-3
5	5, 1-4, 2-3
6	6, 1-5, 2-4, 1-2-3
7	7, 1-6, 2-5, 3-4, 1-2-4
8	8, 1-7, 2-6, 3-5, 1-2-5, 1-3-4
9	9, 1-8, 2-7, 3-6, 4-5, 1-2-6, 1-3-5, 2-3-4
10	1-9, 2-8, 3-7, 4-6, 1-2-7, 1-3-6, 1-4-5, 2-3-5, 1-2-3-4
11	2-9, 3-8, 4-7, 5-6, 1-2-8, 1-3-7, 1-4-6, 2-3-6, 2-4-5, 1-2-3-5
12	3-9, 4-8, 5-7, 1-2-9, 1-3-8, 1-4-7, 1-5-6, 2-3-7, 2-4-6, 3-4-5, 1-2-3-6

Let’s have some fun,

Eddie

This blog is property of Edward Shore, 2017