What are the odds of winning Toto?

4D-stats-768x402

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!(nr)!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}}1431=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}}1432=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}}1433=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

Author: Gilbert Tan TS

IT expert with more than 20 years experience in Apple, Andriod and Windows PC. Interests include hardware and software, Internet and multimedia. An experienced Real Estate agent, Insurance agent, and a Futures trader.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s