# 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*(49,6). 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`

`=(13,545)/(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`

`=(246,820)/(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`

`=(246,820)/(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."

My reply

We can re-write the fraction as: `(185,115)/(13,983,816)xx3/43 = (12,915)/(13,983,816)`.

So actually there are 12,915 Group 6 winning combinations (the numerator of the final fraction), not 812 (which was obtained by cancelling).

For example, in the 19 Mar 2007 draw, the Winning Numbers were:

2 15 16 19 25 32

The Additional Number was 37

I won't list out all 12,915 combinations :-), but I will show how to get started.

Choosing 3 of the 6 winning numbers:

2 15 16 2 15 19 2 15 25 2 15 32etc

There are 20 such combinations: `C(6,3)`

Now with each of these, we would have chosen 3 non-winning numbers, and there are C(43,3) = 12,341 such combinations for each of the 20 winning triples. For example, with the first 3 winning numbers, the non-winning combinations could include:

2 15 16 1 3 4 2 15 16 1 3 5 2 15 16 1 3 6 2 15 16 1 3 7 2 15 16 1 3 8 etc

There are `C(6,3) × C(43,3) = 246,820` such combinations.

Only some of those (3 in every 43 of them, in fact) will include the additional number 37, eg

2 15 16 1 3 37 2 15 16 1 37 38 2 15 16 37 38 39 etc

Other possible Group 6 combinations, using other winning numbers, are:

15 16 19 37 38 39 19 25 32 37 38 39 etc

There are `C(6,3) × C(43,3) × 3/43 = 17,220` of these.

There are `13,983,816` possible combinations, so the fraction

`(17,220)/(13,983,816) = 1\ "in"\ 812`

is the chance of getting the Group 6 prize. The "812" does not refer to the number of possible combinations.

I hope that makes sense.

Please support IntMath!

## 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 asked what happens when the probability appears to be greater than 1.

What he means is this: For a Group 6 prize, the probability is 1 in 812, so (it appears) a System 12 will give you 924 chances in 812 or 1.138 chances - greater than 1! However, this refers to the high chance in relation to your 12 chosen numbers - not to the total 49 possible numbers.

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.

If you win a Group 6 prize with a System 12, you actually get 28 Group 6 shares (28 × $25 = $700) and because you have 3 correct numbers as well, you get 56 Group 7 shares (56 × $10 = $560) so your total prize amount is $1,260 (or a "profit" of $1260 - $924 = $336).

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.

### See also...

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

### 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

NOTE:At the time, there were only 45 numbers to choose from in Singapore Toto, not 49 as there is now.

Let us start with a similar but simpler game. Say we have 9 numbers (1,2,3,4,5,6,7,8,9) and we draw 4 numbers each time. Four numbers have been drawn (say 1,2,3,4) We want to find the probability that two of these four numbers come out again in the next draw.

We divide the numbers into 2 groups:

Category A: 1,2,3,4 (four numbers came out in the first draw)

Category B: 5,6,7,8,9 (five remaining numbers)

Suppose 1 and 2 repeat:

1,2,5,6 1,2,5,7 1,2,5,8 1,2,5,9 1,2,6,7 |
1,2,6,8 1,2,6,9 1,2,7,8 1,2,7,9 1,2,8,9 |

There are *C*(5,2) = 10 numbers in this group. We choose 2 from the 5 numbers from Category B.

Similarly there is another group in which 1,3 repeat, and another where 1,4 repeat, then 2,3 repeat etc.

Altogether there are *C*(4,2) = 6 groups (4 is the number of numbers in group A and 2 of them are drawn).

So the probability of the 2 numbers repeating in 2 different draws is:

`(C(5,2)xxC(4,2))/(C(9,4))=(10xx6)/126=10/21`

The TOTO case is similar: In each game, we draw 7 numbers (6 plus one additional) from a pool of 45. We want to find the probability of having 5 numbers repeating.

There are 7 numbers in Category A and 38 numbers in Category B.

The number of ways to have the first five numbers repeating:

We choose 2 from 38 numbers in group B: *C*(38,2) = 703.

There are *C*(7,5) = 21 groups, and there are 7 numbers in group A and 5 of them are drawn.

The denominator is *C*(45,7) = 45 379 620.

The probability is

`(703xx21)/(45\ 379\ 620)=(4\ 921)/(15\ 126\ 540)`

`=3.2532xx10^-4` or `1` in `3073.9`.

Of course, this is only the probability of having EXACTLY 5 numbers repeating.

The probability of 6 numbers repeating can be found using similar working:

`(38xx7)/(45\ 379\ 620)=133/(22\ 689\ 810)`

`=5.8617xx10^-6` or `1` in `170\ 600`.

Also the probability of having all the same 7 numbers coming out again is `1/(45\ 379\ 620)`.

The answer for the probability of having AT LEAST 5 repeating numbers in the next draw is

`(14\ 763)/(45\ 379\ 620)` `+(266)/(45\ 379\ 620)` `+1/(45\ 379\ 620)`

`=167/(504\ 218)`

`=0.00033121`

or 1 in 3019.2.

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