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}=r!(n−r)!n! $
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:
$C(,)1 =,,1 $
$=7.15×_{−8}$
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(,)C(,)×C(,) ×431 $
$=,,258 ×431 $
$=,,1 $
$=4.29×_{−7}$
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 $431 $.
So there is 1 chance in 2,330,636 of getting the Group 2 prize.
Group 3 (5 correct)
Odds:
$C(,)C(,)×C(,) ×4342 $
$=,,258 ×4342 $
$=,1 $
$=1.80×_{−5}$
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 $1−431 =4342 $.
So there is 1 chance in 55,491 of getting the Group 3 prize.
Group 4 (4 + additional)
Odds:
$C(,)C(,)×C(,) ×432 $
$=,,, ×432 $
$=,1 $
$=4.505×_{−5}$
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 $432 $.
So there is 1 chance in 22,197 of getting the Group 4 prize.
Group 5 (4 correct)
Odds:
$C(,)C(,)×C(,) ×4341 $
$=,,, ×4341 $
$=1082.75771 $
$=9.236×_{−4}$
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 $1−432 =4341 $. So we multiply by $4341 $.
So there is 1 chance in 1,083 of getting the Group 5 prize.
Group 6 (3 + additional)
Odds:
$C(,)C(,)×C(,) ×433 $
$=,,, ×433 $
$=812.0681 $
$=1.23142×_{−3}$
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 $433 $.
So there is 1 chance in 812 of getting the Group 6 prize.
Group 7 (3 correct)
Odds:
$C(,)C(,)×C(,) ×4340 $
$=,,, ×4340 $
$=60.9051 $
$=1.642×_{−2}$
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 $1−433 =4340 $. So we multiply by $4340 $.
So there is 1 chance in 61 of getting the Group 7 prize.
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 $C(,)$.
Similarly, System 9 gives you $C(,)=84$ ordinary bet combinations, System 10 gives $C(,)=210$ ordinary combinations, System 11 gives $C(,)=462$ combinations and System 12 (the maximum in the Singapore game) gives $C(,)=924$ combinations.
The probability of winning with a System 12 is $924$ times the probability of winning when you buy 1 game, that is:
$,,924 $ or $1$ in $15,134$.
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 |