# Singapore TOTO

### Later, on this page

Odds of getting any prize

System odds

### Related Sections

Don't miss...

Combinations (for background on this section)

Probability and Poker

In the Singapore game of TOTO, 6 numbers plus one "additional" number are drawn at random from the numbers 1 to 49. In the Ordinary game, players spend $1 and they choose 6 numbers in the hope of becoming instant millionaires.

A prize pool is established at 54% 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 5.5 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 | 38% of prize pool (min $1 M) | 6 numbers |

2 | 8% of prize pool | 5 numbers + additional number |

3 | 5.5% of prize pool | 5 numbers |

4 | 3% of prize pool | 4 numbers + additional number |

5 | $50 per winning combination | 4 numbers |

6 | $25 per winning combination | 3 numbers + additional number |

7 | $10 per winning combination | 3 numbers |

## 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:

`C_r^n=(n!)/(r!(n-r)!`

We can write (and type) 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:

`1/(C(49,6))=1/(13,983,816)=7.15xx10^-8`

That is, there are `13,983,816` ways of choosing 6 numbers from 49 numbers but there is only one correct combination.

So there is **1 chance in 13,983,816** of getting the Group 1 prize.

This means we have to buy almost 14 million tickets (at a cost of $14 million) before we can confidently say we will probably win the top prize...

### Group 2 (5 + additional)

Odds:

`(C(6,5)xxC(43,1))/(C(49,6))xx1/43`

`=258/(13,983,816)xx1/43`

`=1/(2,330,636)`

`=4.29xx10^-7`

**Explanation: **We chose 5 of the 6 winning numbers [*C*(6,5)], and chose the correct "additional" number from the `43` remaining numbers that did not win anything [*C*(43,1)].

There is `1` chance in `43` that we chose the additional number, so multiply by `1/43`.

So there is **1 chance in 2,330,636** of getting the Group 2 prize.

### Group 3 (5 correct)

Odds:

`(C(6,5)xxC(43,1))/(C(49,6))xx42/43`

`=258/(13,983,816)xx42/43`

`=1/(55,491.3)`

`=1.80xx10^-5`

We chose 5 of the 6 winning numbers and chose `1` number from the `43` 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 `43` numbers **not** containing the additional number, which is `1 − 1/43 = 42/43`.

So there is **1 chance in 55,491** of getting the Group 3 prize.

### Group 4 (4 + additional)

Odds:

`(C(6,4)xxC(43,2))/(C(49,6))xx2/43`

`=(27,090)/(13,983,816)xx2/43`

`=1/(22,196.53)`

`=4.505xx10^-5`

We chose 4 of the 6 winning numbers [*C*(6,4)], and chose `2` numbers from the `43` remaining numbers that did not win anything [*C*(43,2)]. But we chose 6 numbers originally, so there are `2` chances in `43` that we chose the additional number, so multiply by `2/43`.

So there is **1 chance in 22,197** of getting the Group 4 prize.

### Group 5 (4 correct)

Odds:

`(C(6,4)xxC(43,2))/(C(49,6))xx41/43`

`=(13,545)/(13,983,816)xx41/43`

`=1/1082.7577`

`=9.236xx10^-4`

We chose `4` of the `6` winning numbers and chose `2` numbers from the `43` 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 − 2/43 = 41/43`. So we multiply by `41/43`.

So there is **1 chance in 1,083** of getting the Group 5 prize.

### Group 6 (3 + additional)

Odds:

`(C(6,3)xxC(43,3))/(C(49,6))xx3/43`

`=(185,115)/(13,983,816)xx3/43`

`=1/812.068`

`=1.23142xx10^-3`

We chose `3` of the `6` winning numbers [*C*(6,3)], and choose `3` numbers from the `43` remaining numbers that did not win anything [*C*(43,3)]. But we chose `6` numbers originally so there are `3` chances in `43` that we chose the additional number, so multiply by `3/43`.

So there is **1 chance in 812** of getting the Group 6 prize.

### Group 7 (3 correct)

Odds:

`(C(6,3)xxC(43,3))/(C(49,6))xx40/43`

`=(185,115)/(13,983,816)xx40/43`

`=1/60.905`

`=1.642xx10^-2`

We chose `3` of the `6` winning numbers and chose `3` numbers from the `43` remaining numbers that did not win. Again, we need to consider the probability of the additional number not being one of our `3` remaining (non-winning) numbers. This probability is `1 − 3/43 = 40/43`. So we multiply by `40/43`.

So there is **1 chance in 61** of getting the Group 7 prize.

### Reader's Question

A reader wrote in to ask:

"In that group 6 (3 + additional number), there are `812` combinations, please help me to know all the possible `812` combinations."

## 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:

`= 7.15 × 10^-8 + 4.29 × 10^-7 ` `+ 1.80 × 10^-5 + 4.505 × 10^-5 ` `+ 9.236 × 10^-4 + 1.23142 × 10^-3 + 1.642 xx 10^-2`

`=1.864 × 10^-2`

So the odds of getting any prize is `1` in `1/(1.864 × 10^-2) = 1\ "in"\ 53.6`.

## System Entries

In most Lotto and Toto games, you can buy a "System". Your chances of winning increase, but of course, you pay more as well. For example:

**System 7** means you choose 7 numbers (instead of the usual 6). This gives you 7 times the chance of winning (so it costs 7 times as much), since it is equivalent to buying 7 different 6-number games, or *C*(7,6). Say you chose 1, 3, 5, 7, 9, 11, 13 as your 7 numbers. You have the following 7 ways of winning if the 6 winning numbers happened to be:

1 3 5 7 9 11

3 5 7 9 11 13

1 5 7 9 11 13

1 3 7 9 11 13

1 3 5 9 11 13

1 3 5 7 11 13

1 3 5 7 9 13

**System 8** means you choose 8 numbers and it gives you the equivalent of 28 ordinary bet combinations, so costs 28 times as much, or `C(8,6)`.

Similarly, **System 9** gives you `C(9,6) = 84` ordinary bet combinations, **System 10** gives `C(10,6) = 210` ordinary combinations, **System 11** gives `C(11,6) = 462` combinations and **System 12** (the maximum in the Singapore game) gives `C(12,6) = 924` combinations.

The probability of winning with a System 12 is `924` times the probability of winning when you buy 1 game, that is:

`924/(13,983,816)` or `1` in `15,134`.

### Reader's Question

Reader David wrote in to ask about the probability of winning a Group 6 prize when you buy a System 12. He quite rightly asked what happens when the probability appears to be greater than 1.

What he's getting at is, for a Group 6 prize, the probability is 1 in 61, so a System 12 will give you 924 chances in 61 or 15.15 chances - greater than 1!

It just means you can be "almost certain" to get a Group 6 prize when you buy a System 12.

In fact, you would expect to win `924/321 = 2.88` (almost 3) prizes of some sort.

But of course, if those prizes were (say) one Group 4, one Group 5 and one Group 6 prizes, you would be out of pocket, since your System 12 ticket would have cost you $924 and your (typical) prize winnings for those 3 prizes could be around $410.

Don't believe anyone who says any strategy is a "sure thing". The only sure thing is the "house" (or in many cases, the government) are the real winners.

### 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 occur most frequently.

### 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.

### See also...

Should we teach gambling in math classes? (in the math blog.)

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?

### Online Algebra Solver

This algebra solver can solve a wide range of math problems. (Please be patient while it loads.)

Go to: Online algebra solver

### Math Lessons on DVD

Easy to understand math lessons on DVD. See samples before you commit.

More info: Math videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!