What are the odds of winning Toto?

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:

$Crn=n!r!(n−r)!\displaystyle{{C}_{{r}}^{{n}}}=\frac{{{n}!}}{{{r}!{\left({n}-{r}\right)}!}}$

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 $1\displaystyle{1}$ in C(49,6). That is:

$1C(49,6)=113,983,816\displaystyle\frac{1}{{{C}{\left({49},{6}\right)}}}=\frac{1}{{{13},{983},{816}}}$

$=7.15×10−8\displaystyle={7.15}\times{10}^{ -{{8}}}$

That is, there are $13,983,816\displaystyle{13},{983},{816}$ 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:

$C(6,5)×C(43,1)C(49,6)×143\displaystyle\frac{{{C}{\left({6},{5}\right)}\times{C}{\left({43},{1}\right)}}}{{{C}{\left({49},{6}\right)}}}\times\frac{1}{{43}}$

$=25813,983,816×143\displaystyle=\frac{258}{{{13},{983},{816}}}\times\frac{1}{{43}}$

$=12,330,636\displaystyle=\frac{1}{{{2},{330},{636}}}$

$=4.29×10−7\displaystyle={4.29}\times{10}^{ -{{7}}}$

Explanation: We chose 5 of the 6 winning numbers [C(6,5)], and chose the correct “additional” number from the $43\displaystyle{43}$ remaining numbers that did not win anything [C(43,1)].

There is $1\displaystyle{1}$ chance in $43\displaystyle{43}$ that we chose the additional number, so multiply by $143\displaystyle\frac{1}{{43}}$ .

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

Group 3 (5 correct)

Odds:

$C(6,5)×C(43,1)C(49,6)×4243\displaystyle\frac{{{C}{\left({6},{5}\right)}\times{C}{\left({43},{1}\right)}}}{{{C}{\left({49},{6}\right)}}}\times\frac{42}{{43}}$

$=25813,983,816×4243\displaystyle=\frac{258}{{{13},{983},{816}}}\times\frac{42}{{43}}$

$=155,491.3\displaystyle=\frac{1}{{{55},{491.3}}}$

$=1.80×10−5\displaystyle={1.80}\times{10}^{ -{{5}}}$

We chose 5 of the 6 winning numbers and chose $1\displaystyle{1}$ number from the $43\displaystyle{43}$ 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 $43\displaystyle{43}$ numbers not containing the additional number, which is $1−143=4243\displaystyle{1}-\frac{1}{{43}}=\frac{42}{{43}}$ .

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

Group 4 (4 + additional)

Odds:

$C(6,4)×C(43,2)C(49,6)×243\displaystyle\frac{{{C}{\left({6},{4}\right)}\times{C}{\left({43},{2}\right)}}}{{{C}{\left({49},{6}\right)}}}\times\frac{2}{{43}}$

$=13,54513,983,816×243\displaystyle=\frac{{{13},{545}}}{{{13},{983},{816}}}\times\frac{2}{{43}}$

$=122,196.53\displaystyle=\frac{1}{{{22},{196.53}}}$

$=4.505×10−5\displaystyle={4.505}\times{10}^{ -{{5}}}$

We chose 4 of the 6 winning numbers [C(6,4)], and chose $2\displaystyle{2}$ numbers from the $43\displaystyle{43}$ remaining numbers that did not win anything [C(43,2)]. But we chose 6 numbers originally, so there are $2\displaystyle{2}$ chances in $43\displaystyle{43}$ that we chose the additional number, so multiply by $243\displaystyle\frac{2}{{43}}$ .

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

Group 5 (4 correct)

Odds:

$C(6,4)×C(43,2)C(49,6)×4143\displaystyle\frac{{{C}{\left({6},{4}\right)}\times{C}{\left({43},{2}\right)}}}{{{C}{\left({49},{6}\right)}}}\times\frac{41}{{43}}$

$=13,54513,983,816×4143\displaystyle=\frac{{{13},{545}}}{{{13},{983},{816}}}\times\frac{41}{{43}}$

$=11082.7577\displaystyle=\frac{1}{{1082.7577}}$

$=9.236×10−4\displaystyle={9.236}\times{10}^{ -{{4}}}$

We chose $4\displaystyle{4}$ of the $6\displaystyle{6}$ winning numbers and chose $2\displaystyle{2}$ numbers from the $43\displaystyle{43}$ remaining numbers that did not win. Once again, we need to consider the probability of the additional number not being one of our $2\displaystyle{2}$ remaining (non-winning) numbers. This probability is $1−243=4143\displaystyle{1}-\frac{2}{{43}}=\frac{41}{{43}}$ . So we multiply by $4143\displaystyle\frac{41}{{43}}$ .

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

Group 6 (3 + additional)

Odds:

$C(6,3)×C(43,3)C(49,6)×343\displaystyle\frac{{{C}{\left({6},{3}\right)}\times{C}{\left({43},{3}\right)}}}{{{C}{\left({49},{6}\right)}}}\times\frac{3}{{43}}$

$=246,82013,983,816×343\displaystyle=\frac{{{246},{820}}}{{{13},{983},{816}}}\times\frac{3}{{43}}$

$=1812.068\displaystyle=\frac{1}{{812.068}}$

$=1.23142×10−3\displaystyle={1.23142}\times{10}^{ -{{3}}}$

We chose $3\displaystyle{3}$ of the $6\displaystyle{6}$ winning numbers [C(6,3)], and choose $3\displaystyle{3}$ numbers from the $43\displaystyle{43}$ remaining numbers that did not win anything [C(43,3)]. But we chose $6\displaystyle{6}$ numbers originally so there are $3\displaystyle{3}$ chances in $43\displaystyle{43}$ that we chose the additional number, so multiply by $343\displaystyle\frac{3}{{43}}$ .

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

Group 7 (3 correct)

Odds:

$C(6,3)×C(43,3)C(49,6)×4043\displaystyle\frac{{{C}{\left({6},{3}\right)}\times{C}{\left({43},{3}\right)}}}{{{C}{\left({49},{6}\right)}}}\times\frac{40}{{43}}$

$=246,82013,983,816×4043\displaystyle=\frac{{{246},{820}}}{{{13},{983},{816}}}\times\frac{40}{{43}}$

$=160.905\displaystyle=\frac{1}{{60.905}}$

$=1.642×10−2\displaystyle={1.642}\times{10}^{ -{{2}}}$

We chose $3\displaystyle{3}$ of the $6\displaystyle{6}$ winning numbers and chose $3\displaystyle{3}$ numbers from the $43\displaystyle{43}$ remaining numbers that did not win. Again, we need to consider the probability of the additional number not being one of our $3\displaystyle{3}$ remaining (non-winning) numbers. This probability is $1−343=4043\displaystyle{1}-\frac{3}{{43}}=\frac{40}{{43}}$ . So we multiply by $4043\displaystyle\frac{40}{{43}}$ .

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 $C(8,6)\displaystyle{C}{\left({8},{6}\right)}$ .

Similarly, System 9 gives you $C(9,6)=84\displaystyle{C}{\left({9},{6}\right)}={84}$ ordinary bet combinations, System 10 gives $C(10,6)=210\displaystyle{C}{\left({10},{6}\right)}={210}$ ordinary combinations, System 11 gives $C(11,6)=462\displaystyle{C}{\left({11},{6}\right)}={462}$ combinations and System 12 (the maximum in the Singapore game) gives $C(12,6)=924\displaystyle{C}{\left({12},{6}\right)}={924}$ combinations.

The probability of winning with a System 12 is $924\displaystyle{924}$ times the probability of winning when you buy 1 game, that is:

$92413,983,816\displaystyle\frac{924}{{{13},{983},{816}}}$ or $1\displaystyle{1}$ in $15,134\displaystyle{15},{134}$ .

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

Author: Gilbert Tan TS

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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 (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:

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

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

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

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

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

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 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).

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

The odds of winning the top Group 1 prize are $1\displaystyle{1}$ in C(49,6). That is:

That is, there are $13,983,816\displaystyle{13},{983},{816}$ ways of choosing 6 numbers from 49 numbers but there is only one correct combination.

$C(6,5)×C(43,1)C(49,6)×143\displaystyle\frac{{{C}{\left({6},{5}\right)}\times{C}{\left({43},{1}\right)}}}{{{C}{\left({49},{6}\right)}}}\times\frac{1}{{43}}$

$=25813,983,816×143\displaystyle=\frac{258}{{{13},{983},{816}}}\times\frac{1}{{43}}$

$=12,330,636\displaystyle=\frac{1}{{{2},{330},{636}}}$

$=4.29×10−7\displaystyle={4.29}\times{10}^{ -{{7}}}$

Explanation: We chose 5 of the 6 winning numbers [C(6,5)], and chose the correct “additional” number from the $43\displaystyle{43}$ remaining numbers that did not win anything [C(43,1)].

There is $1\displaystyle{1}$ chance in $43\displaystyle{43}$ that we chose the additional number, so multiply by $143\displaystyle\frac{1}{{43}}$ .

$C(6,5)×C(43,1)C(49,6)×4243\displaystyle\frac{{{C}{\left({6},{5}\right)}\times{C}{\left({43},{1}\right)}}}{{{C}{\left({49},{6}\right)}}}\times\frac{42}{{43}}$

$=25813,983,816×4243\displaystyle=\frac{258}{{{13},{983},{816}}}\times\frac{42}{{43}}$

$=155,491.3\displaystyle=\frac{1}{{{55},{491.3}}}$

$=1.80×10−5\displaystyle={1.80}\times{10}^{ -{{5}}}$

$C(6,4)×C(43,2)C(49,6)×243\displaystyle\frac{{{C}{\left({6},{4}\right)}\times{C}{\left({43},{2}\right)}}}{{{C}{\left({49},{6}\right)}}}\times\frac{2}{{43}}$

$=13,54513,983,816×243\displaystyle=\frac{{{13},{545}}}{{{13},{983},{816}}}\times\frac{2}{{43}}$

$=122,196.53\displaystyle=\frac{1}{{{22},{196.53}}}$

$=4.505×10−5\displaystyle={4.505}\times{10}^{ -{{5}}}$

$C(6,4)×C(43,2)C(49,6)×4143\displaystyle\frac{{{C}{\left({6},{4}\right)}\times{C}{\left({43},{2}\right)}}}{{{C}{\left({49},{6}\right)}}}\times\frac{41}{{43}}$

$=13,54513,983,816×4143\displaystyle=\frac{{{13},{545}}}{{{13},{983},{816}}}\times\frac{41}{{43}}$

$=11082.7577\displaystyle=\frac{1}{{1082.7577}}$

$=9.236×10−4\displaystyle={9.236}\times{10}^{ -{{4}}}$

$C(6,3)×C(43,3)C(49,6)×343\displaystyle\frac{{{C}{\left({6},{3}\right)}\times{C}{\left({43},{3}\right)}}}{{{C}{\left({49},{6}\right)}}}\times\frac{3}{{43}}$

$=246,82013,983,816×343\displaystyle=\frac{{{246},{820}}}{{{13},{983},{816}}}\times\frac{3}{{43}}$

$=1812.068\displaystyle=\frac{1}{{812.068}}$

$=1.23142×10−3\displaystyle={1.23142}\times{10}^{ -{{3}}}$

$C(6,3)×C(43,3)C(49,6)×4043\displaystyle\frac{{{C}{\left({6},{3}\right)}\times{C}{\left({43},{3}\right)}}}{{{C}{\left({49},{6}\right)}}}\times\frac{40}{{43}}$

$=246,82013,983,816×4043\displaystyle=\frac{{{246},{820}}}{{{13},{983},{816}}}\times\frac{40}{{43}}$

$=160.905\displaystyle=\frac{1}{{60.905}}$

$=1.642×10−2\displaystyle={1.642}\times{10}^{ -{{2}}}$

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 $C(8,6)\displaystyle{C}{\left({8},{6}\right)}$ .

The probability of winning with a System 12 is $924\displaystyle{924}$ times the probability of winning when you buy 1 game, that is:

$92413,983,816\displaystyle\frac{924}{{{13},{983},{816}}}$ or $1\displaystyle{1}$ in $15,134\displaystyle{15},{134}$ .