Chapter Contents

• Counting and Probability - Introduction
• 1. Factorial Notation
• 2. Basic Principles of Counting
• 3. Permutations
• 4. Combinations
• 5. Introduction to Probability Theory
• 6. Probability of an Event
• Singapore TOTO
• Probability and Poker
• 7. Conditional Probability
• 8. Independent and Dependent Events
• 9. Mutually Exclusive Events
• 10. Bayes’ Theorem
• 11. Probability Distributions - Concepts
• 12. Binomial Probability Distributions
• 13. Poisson Probability Distribution
• 14. Normal Probability Distribution
• The z-Table
• Counting and Probability Problem Solver

• Comments, Questions?

Feelings about Math

Recommendation

Easy to understand math lessons on DVD. Try before you commit. More info:
MathTutorDVD.com

Online Algebra Solver

Solve your algebra problem step by step!

Online Algebra Solver »

Share

Share this page.

From the math blog...

Singapore TOTO

In the Singapore game of TOTO, 6 numbers plus one "additional" number are drawn at random from the numbers 1 to 45. In the Ordinary game, players spend 50c per game (minimum 2 games) and they choose 6 numbers in the hope of becoming instant millionaires.

A prize pool is established at 55% of sales for a draw. Typically, $2.8 million dollars is "invested" in each game - and games are offered twice per week. This is quite a lot for a country of 4.5 million people...

Plenty of other countries have similar Toto games, usually called Lotto. The more numbers in a game, the worse your chances become.

Summary of the Prizes (Singapore Toto)

Grp	Prize Amount	Winning Numbers Matched
1	33% of prize pool (min $500K)	6 numbers
2	13% of prize pool	5 numbers + additional number
3	13% of prize pool	5 numbers
4	13% of prize pool	4 numbers + additional number
5	$30 per winning combination	4 numbers
6	$20 per winning combination	3 numbers + additional number

TOTO Odds

Recall (from the Combinations section) that the number of ways in which r objects can be selected from a set of n objects, where repetition is not allowed, is given by:

We can write the left hand side more conveniently as C(n,r).

Now let's look at the probabilities for each prize.

Group 1 (Choose all 6)

The odds of winning the top Group 1 prize are 1 in C(45,6). That is:

That is, there are 8,145,060 ways of choosing 6 numbers from 45 numbers but there is only one correct combination.

So there is 1 chance in 8,145,060 of getting the Group 1 prize.

This means we have to buy 8 million tickets before we can confidently say we will probably win the top prize...

Group 2 (5 + additional)

Odds:

Explanation: We chose 5 of the 6 winning numbers [C(6,5)], and chose the correct "additional" number from the 39 remaining numbers that did not win anything [C(39,1)].

There is 1 chance in 39 that we chose the additional number, so multiply by 1/39.

So there is 1 chance in 1,357,510 of getting the Group 2 prize.

Group 3 (5 correct)

Odds:

We chose 5 of the 6 winning numbers and chose 1 number from the 39 remaining numbers that did not win. In the Group 3 prize, we cannot include the "additional" number, so we need to multiply by the probability of the remaining 39 numbers not containing the additional number, which is 1 − 1/39 = 38/39.

So there is 1 chance in 35724 of getting the Group 3 prize.

Group 4 (4 + additional)

Odds:

We chose 4 of the 6 winning numbers [C(6,4)], and chose 2 numbers from the 39 remaining numbers that did not win anything [C(39,2)]. But we chose 6 numbers originally, so there are 2 chances in 39 that we chose the additional number, so multiply by 2/39.

So there is 1 chance in 14,290 of getting the Group 4 prize.

Group 5 (4 correct)

Odds:

We chose 4 of the 6 winning numbers and chose 2 numbers from the 39 remaining numbers that did not win. Once again, we need to consider the probability of the additional number not being one of our 2 remaining (non-winning) numbers. This probability is (1 − 2/39 = 37/39). So we multiply by 37/39.

So there is 1 chance in 772 of getting the Group 5 prize.

Group 6 (3 + additional)

Odds:

We chose 3 of the 6 winning numbers [C(6,3)], and choose 3 numbers from the 39 remaining numbers that did not win anything [C(39,3)]. But we chose 6 numbers originally so there are 3 chances in 39 that we chose the additional number, so multiply by 3/39.

So there is 1 chance in 579 of getting the Group 6 prize.

Reader's Question

A reader wrote in to ask:

"In that group 6 (3 + additional number), there are 579 combinations, please help me to know all the possible 579 combinations."

My reply

Odds of Getting Any Prize:

Total probability of getting any prize is simply the sum of all the probabilities for Group 1 to Group 6 prizes:

= 1.2277 × 10^-7 + 7.3664 × 10^-7 + 2.7992 × 10^-5 + 6.9979 × 10^-5 + 1.2947 × 10^-3 + 1.7262 × 10^-3

= 3.1197 × 10^-3

So the odds of getting any prize is 1/(3.1197 × 10^-3) = 1 in 320.546.

Which are the Best Toto Numbers?

Click here for the Toto number frequency chart (at Singapore Pools site.)

Of course, there are no "best" numbers, but it is interesting to see which ones are most frequent.

An Interesting Problem...

On Monday 11 June 2001, the winning numbers for Singapore Toto were 1,10,19,23,29,45 with additional number 34.

The next consecutive draw on Thursday 14 June 2001 had five of the same numbers from the Monday draw. The winning numbers for this draw were 1,10,14,19,23,34 with additional number 33.

Thus the numbers 1,10,19,23,34 were repeated.

This is very rare and quite amazing. What is the probability of this occurring?

Answer