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I run a Lottery syndicate for the UK lottery, and we play 30 lines per draw.

The odds of winning £10 (3 matching numbers) is deemed to be 1 in 56.7 (or 1/57 for the purposes of this question).

I'd loosely determined that the odds of getting 3 numbers from 30 lines was approximated as 1 in 2 (or 30/57) - firstly, is this correct?

Secondly, we recently played 30 lines and matched 3 numbers on two of those lines. I'd like to tell my syndicate how likely that was expressed as a fraction, but I can't work it out (or be sure it is correct).

1 / 56.7 = 0.0176366843

I'd always learned that the occurance of something happening AND something else happening is the multiple of the probabilities. So, 0.0176366843 x 0.0176366843 = 0.00031105263.

Is that correct, and if so, what is the nearest representation of this number as a fraction? The fact this calculation doesn't seem to take into account that 30 lines were played makes me think it is wrong.

monkeymatrix
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  • no and no, the reason is the following. By your logic, the odd of getting 3 numbers from 57 lines is probability 1. Do you think that will be the case?

    I'd always learned that the occurance of something happening AND something else happening is the multiple of the probabilities.

    No only if they are independent. The chance 3 numbers came up out of 1,2,3,4,5,6 is likely to effect the chance whether 3 numbers comes up out of 1,2,3,4,9,10, for example. If 3 numbers came up out of 1,2,3,4,5,6, the chance 3 numbers come up out of 7,8,9,10,11,12 is greatly decreased.

     – Lost1 Mar 30 '13 at 23:45

3 Answers3

1

I assume that the chance for getting $3$ matching numbers is $1/57$, and you do your bets on the different lines independently.

Independence means that you draw your $30$ bets randomly and uniformly from the set of all possible bets each time you are playing. If instead, you use a fixed system of $30$ bets, the below computations still should give reasonable results within the calculation accuracy. If you want exact results, you have to provide us with details on how you choose your 30 bets.

No 3 matches on 30 bets: The chance for getting $3$ matching numbers is $1/57$, so the chance for not getting three matching numbers is $1 - 1/57 = 56/57$. Now not getting three matching numbers on $30$ independent tries is $$(56/57)^{30} \approx 58.8\%.$$

3 Matches on 30 bets at least once: The chance to get three matches at least once is $$1 - (56/57)^{30} \approx 41.2\%.$$

3 Matches on 30 bets exactly once: Using the Bernoulli distribution, we find the chance of getting three matches exactly once as $$30 \cdot (56/57)^{29} \cdot (1/57) \approx 31.5\%.$$

3 Matches on 30 bets at least twice: Now the chance to get three matches at least twice is $$ 1 - \text{no three matches} - \text{three matches exactly once} \\ = 1 - (56/57)^{30} - 30 \cdot (56/57)^{29} \cdot (1/57) \approx 9.7\%.$$ If you want to write it as a fraction, $$\frac{1}{10}$$ is a reasonable approximation.

3 Matches on 30 bets exactly twice: Again, the Bernoulli distribution gives us the chance as $$ \binom{30}{2}\cdot (56/57)^{28} (1/57)^2 \approx 8.2\%$$

A reasonable fractional approximation is $$\frac{1}{12}.$$

azimut
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  • The lines are not drawn independently. – zyx Apr 03 '13 at 12:13
  • @zyx: How do you know? Furthermore, as long as the selection is not extremely biased (say you use the same bet $30$ times), assuming independence should not do any harm (there is already some rounding taking place). – azimut Apr 03 '13 at 12:28
  • The question stated this is a gambling syndicate. Syndicates are usually organized to make multiple bets that are as extremely biased as possible. – zyx Apr 03 '13 at 12:34
  • @zyx: I see. Still, assuming independence should give reasonable approximations: If the system of 30 bets is something "extreme", then it won't be like using the same bet 30 times (this would not make any sense), but some quite homogeneous way to cover the numbers. – azimut Apr 03 '13 at 13:33
  • That's possible. The question is whether the range of numbers used in the lottery is large enough. – zyx Apr 04 '13 at 01:42
  • The range of numbers used in the UK lottery is 1-49. You choose 6 numbers from that range. All of the 30 lines are selected using a 'lucky dip' which is essentially a randomised selection. This is the best answer so far though, thank you :) – monkeymatrix Apr 04 '13 at 12:51
  • To clarify further, that means 30 lines per single draw, rather than 30 times in 30 independent draws. – monkeymatrix Apr 04 '13 at 12:55
  • Ok. Since you write "essentially randomised", the values in my answer should be quite accurate. Now that I know that the system is "6 out of 49", I see that the approximation $1/57$ is not bad, but not completely exact (compare the answer of Charles). If you want me to use the exact value, just tell me. Then I will update my answer accordingly. – azimut Apr 05 '13 at 16:56
  • The exact value would be interesting, but unless it radically changes the calculated odds I'm not too bothered. I think you've done enough to earn the bounty here, with my thanks :) – monkeymatrix Apr 06 '13 at 22:06
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The odds of winning the £10 prize are 8815/499422 and so the odds that at least one of the 30 independent draws will win is $1-(1-8815/499422)^{30}\approx41.4\%$ (about 5/12). I'm not sure how your syndicate places bets but your odds are probably pretty close to this.

Charles
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The odds of winning £10 (3 matching numbers) is ... 1/57 for the purposes of this question). ... the odds of getting 3 numbers from 30 lines was approximated as 1 in 2 (or 30/57) - firstly, is this correct?

30 times the probability for one line is an upper bound (an overestimate) on the probability, is a reasonable estimate for low-probability events, and is an exactly correct calculation of a related probability quantity, the expected number of matching triples.

To get the exact answer you need to specify how the lottery works and how the 30 lines are constructed.

zyx
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