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:
Crn=n!r!(n−r)!\displaystyle{{C}_{{r}}^{{n}}}=\frac{{{n}!}}{{{r}!{\left({n}-{r}\right)}!}}Crn=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\displaystyle{1}1 in C(49,6). That is:
1C(49,6)=113,983,816\displaystyle\frac{1}{{{C}{\left({49},{6}\right)}}}=\frac{1}{{{13},{983},{816}}}C(49,6)1=13,983,8161
=7.15×10−8\displaystyle={7.15}\times{10}^{ -{{8}}}=7.15×10−8
That is, there are 13,983,816\displaystyle{13},{983},{816}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)×C(43,1)C(49,6)×143\displaystyle\frac{{{C}{\left({6},{5}\right)}\times{C}{\left({43},{1}\right)}}}{{{C}{\left({49},{6}\right)}}}\times\frac{1}{{43}}C(49,6)C(6,5)×C(43,1)×431
=25813,983,816×143\displaystyle=\frac{258}{{{13},{983},{816}}}\times\frac{1}{{43}}=13,983,816258×431
=12,330,636\displaystyle=\frac{1}{{{2},{330},{636}}}=2,330,6361
=4.29×10−7\displaystyle={4.29}\times{10}^{ -{{7}}}=4.29×10−7
Explanation: We chose 5 of the 6 winning numbers [C(6,5)], and chose the correct “additional” number from the 43\displaystyle{43}43 remaining numbers that did not win anything [C(43,1)].
There is 1\displaystyle{1}1 chance in 43\displaystyle{43}43 that we chose the additional number, so multiply by 143\displaystyle\frac{1}{{43}}431.
So there is 1 chance in 2,330,636 of getting the Group 2 prize.
Group 3 (5 correct)
Odds:
C(6,5)×C(43,1)C(49,6)×4243\displaystyle\frac{{{C}{\left({6},{5}\right)}\times{C}{\left({43},{1}\right)}}}{{{C}{\left({49},{6}\right)}}}\times\frac{42}{{43}}C(49,6)C(6,5)×C(43,1)×4342
=25813,983,816×4243\displaystyle=\frac{258}{{{13},{983},{816}}}\times\frac{42}{{43}}=13,983,816258×4342
=155,491.3\displaystyle=\frac{1}{{{55},{491.3}}}=55,491.31
=1.80×10−5\displaystyle={1.80}\times{10}^{ -{{5}}}=1.80×10−5
We chose 5 of the 6 winning numbers and chose 1\displaystyle{1}1 number from the 43\displaystyle{43}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\displaystyle{43}43 numbers not containing the additional number, which is 1−143=4243\displaystyle{1}-\frac{1}{{43}}=\frac{42}{{43}}1−431=4342.
So there is 1 chance in 55,491 of getting the Group 3 prize.
Group 4 (4 + additional)
Odds:
C(6,4)×C(43,2)C(49,6)×243\displaystyle\frac{{{C}{\left({6},{4}\right)}\times{C}{\left({43},{2}\right)}}}{{{C}{\left({49},{6}\right)}}}\times\frac{2}{{43}}C(49,6)C(6,4)×C(43,2)×432
=13,54513,983,816×243\displaystyle=\frac{{{13},{545}}}{{{13},{983},{816}}}\times\frac{2}{{43}}=13,983,81613,545×432
=122,196.53\displaystyle=\frac{1}{{{22},{196.53}}}=22,196.531
=4.505×10−5\displaystyle={4.505}\times{10}^{ -{{5}}}=4.505×10−5
We chose 4 of the 6 winning numbers [C(6,4)], and chose 2\displaystyle{2}2 numbers from the 43\displaystyle{43}43 remaining numbers that did not win anything [C(43,2)]. But we chose 6 numbers originally, so there are 2\displaystyle{2}2 chances in 43\displaystyle{43}43 that we chose the additional number, so multiply by 243\displaystyle\frac{2}{{43}}432.
So there is 1 chance in 22,197 of getting the Group 4 prize.
Group 5 (4 correct)
Odds:
C(6,4)×C(43,2)C(49,6)×4143\displaystyle\frac{{{C}{\left({6},{4}\right)}\times{C}{\left({43},{2}\right)}}}{{{C}{\left({49},{6}\right)}}}\times\frac{41}{{43}}C(49,6)C(6,4)×C(43,2)×4341
=13,54513,983,816×4143\displaystyle=\frac{{{13},{545}}}{{{13},{983},{816}}}\times\frac{41}{{43}}=13,983,81613,545×4341
=11082.7577\displaystyle=\frac{1}{{1082.7577}}=1082.75771
=9.236×10−4\displaystyle={9.236}\times{10}^{ -{{4}}}=9.236×10−4
We chose 4\displaystyle{4}4 of the 6\displaystyle{6}6 winning numbers and chose 2\displaystyle{2}2 numbers from the 43\displaystyle{43}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\displaystyle{2}2 remaining (non-winning) numbers. This probability is 1−243=4143\displaystyle{1}-\frac{2}{{43}}=\frac{41}{{43}}1−432=4341. So we multiply by 4143\displaystyle\frac{41}{{43}}4341.
So there is 1 chance in 1,083 of getting the Group 5 prize.
Group 6 (3 + additional)
Odds:
C(6,3)×C(43,3)C(49,6)×343\displaystyle\frac{{{C}{\left({6},{3}\right)}\times{C}{\left({43},{3}\right)}}}{{{C}{\left({49},{6}\right)}}}\times\frac{3}{{43}}C(49,6)C(6,3)×C(43,3)×433
=246,82013,983,816×343\displaystyle=\frac{{{246},{820}}}{{{13},{983},{816}}}\times\frac{3}{{43}}=13,983,816246,820×433
=1812.068\displaystyle=\frac{1}{{812.068}}=812.0681
=1.23142×10−3\displaystyle={1.23142}\times{10}^{ -{{3}}}=1.23142×10−3
We chose 3\displaystyle{3}3 of the 6\displaystyle{6}6 winning numbers [C(6,3)], and choose 3\displaystyle{3}3 numbers from the 43\displaystyle{43}43 remaining numbers that did not win anything [C(43,3)]. But we chose 6\displaystyle{6}6 numbers originally so there are 3\displaystyle{3}3 chances in 43\displaystyle{43}43 that we chose the additional number, so multiply by 343\displaystyle\frac{3}{{43}}433.
So there is 1 chance in 812 of getting the Group 6 prize.
Group 7 (3 correct)
Odds:
C(6,3)×C(43,3)C(49,6)×4043\displaystyle\frac{{{C}{\left({6},{3}\right)}\times{C}{\left({43},{3}\right)}}}{{{C}{\left({49},{6}\right)}}}\times\frac{40}{{43}}C(49,6)C(6,3)×C(43,3)×4340
=246,82013,983,816×4043\displaystyle=\frac{{{246},{820}}}{{{13},{983},{816}}}\times\frac{40}{{43}}=13,983,816246,820×4340
=160.905\displaystyle=\frac{1}{{60.905}}=60.9051
=1.642×10−2\displaystyle={1.642}\times{10}^{ -{{2}}}=1.642×10−2
We chose 3\displaystyle{3}3 of the 6\displaystyle{6}6 winning numbers and chose 3\displaystyle{3}3 numbers from the 43\displaystyle{43}43 remaining numbers that did not win. Again, we need to consider the probability of the additional number not being one of our 3\displaystyle{3}3 remaining (non-winning) numbers. This probability is 1−343=4043\displaystyle{1}-\frac{3}{{43}}=\frac{40}{{43}}1−433=4340. So we multiply by 4043\displaystyle\frac{40}{{43}}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(8,6)\displaystyle{C}{\left({8},{6}\right)}C(8,6).
Similarly, System 9 gives you C(9,6)=84\displaystyle{C}{\left({9},{6}\right)}={84}C(9,6)=84 ordinary bet combinations, System 10 gives C(10,6)=210\displaystyle{C}{\left({10},{6}\right)}={210}C(10,6)=210 ordinary combinations, System 11 gives C(11,6)=462\displaystyle{C}{\left({11},{6}\right)}={462}C(11,6)=462 combinations and System 12 (the maximum in the Singapore game) gives C(12,6)=924\displaystyle{C}{\left({12},{6}\right)}={924}C(12,6)=924 combinations.
The probability of winning with a System 12 is 924\displaystyle{924}924 times the probability of winning when you buy 1 game, that is:
92413,983,816\displaystyle\frac{924}{{{13},{983},{816}}}13,983,816924 or 1\displaystyle{1}1 in 15,134\displaystyle{15},{134}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 |