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.
