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I have just a question about probability/odds and combinatorics here.

Context:

  • In Singapore, we have a lottery called Toto.
  • Buyers can pick at least six numbers, from 1 to 49.
  • Group 1 prize odds is 1 in 13,983,816 (49c6), where the buyer must win all 6 numbers bought

Question:

  • Why is Group 2 odds 1 in 2,330,636, where the buyer must win 5 numbers plus an additional number?
  • I figured it was from (49c6)/6 but have no idea why)
  • Would be great if anyojne could give insights to Group 3 to 7 as well
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2 Answers 2

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According to the definition given in the wiki page, we play six numbers and an additional one (so, 6+1). Group 2 win is achieved when 5 of your 6 chosen numbers and the separated number matches the given 5+1. Number of total cases of this 5+1 is simply C(49,6). But, out of your 6+1 play, you can generate six (5+1) sets by choosing any 5 of the first 6 numbers. That makes your odds 6 in C(49,6), i.e. 1 in 2330636.

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For Group 1, 3, 5 and 7, that is, with "$k$ numbers" match, the odds of winning (denoted as $W(k)$) is

$W(k)=\frac{C_{49}^{6}}{C_6^k \times C_{42}^{6-k}}$, where $k$ is integers from 3 to 6.

For instance, when $k=4$, $W(4) = \frac{C_{49}^{6}}{C_6^4 \times C_{42}^2}=1082.76$, same as given by the link of Toto.

For Group 2, 4 and 6, that is, with "$k$ numbers plus the additional number" match, the odds of winning (denoted as $W_+(k)$) is

$W_+(k) = \frac{C_{49}^{6}}{C_6^k \times C_{42}^{5-k}}$, where $k$ is integers from 3 to 5.

For instance, when $k=4$, $W_+(4) = \frac{C_{49}^{6}}{C_6^4 \times C_{42}^1}=22196.53$, same as given by the link of Toto.

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