Re: Lottery Probabilities

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Originally Posted by

**milliondollarsocks** hi all,

new here and looking for some help with this lotto problem!

**PART 1) (easy)** do you have better odds buying multiple tickets in the same lotto, or the same number of tickets in different lottos? eg

imagine fixed probability of winning is 1/100. does buying 2 tickets give you 2/100?

Yes. If we call your tickets "A" and "B", then the probability of A winning is 1/100. If A does not win, there are 99 tickets left of which B is 1. The probability that A does NOT win and B does is (99/100)(1/99)= 1/100.

So the probability of **either** A or B winning is 1/00+ 1/100= 2/100.

However, that does NOT mean you are likely to wind up with more money than if you bought only one ticket. Lotteries are set up to **bring** in money, not give it away! Suppose each of those 100 tickets cost $1. Then the prize will have to be worth less than $100. Let's say it is worth $80. If you buy one ticket your probability of winning is 1/100 so your "expected value" is $80/100= $0.80. That means you will have a net $1- 0.80= -$.20. On the average you would lose 20 cents. If you buy two tickets, your probability of winning is 2/100 so your expected value is $80(2/100)= $1.60. But you had to spend $2 for those two tickets so have a net $2- 1.60= -$0.40 cents. On the average you would lose 40 cents rather than 20 cents. That should be no surprise. On the average you lose 20 cents per ticket no matter how many tickets you buy. (If you bought all 100 tickets, in order to **guarantee** that you win, you would win $80 but the tickets cost you $100 so your net is **negative** 20. You have still lost 20 cents per ticket.

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what if you bought 1 this week, and 1 next week? i want to say again it's 2/100, but there's also the possibility of BOTH tickets winning, which is not possible in first scenario. can anybody please explain the maths to me!? (Happy)

The probability of "win this week, lose next week" is (1/100)(99/100)= 99/10000. The probability of "lose this week, win next week" is (99/100)(1/100)=99/10000. The probability of "win both weeks" is (1/100)(1/100)= 1/10000. The probability of winning "at least once in two weeks" is 99/10000+ 99/10000+ 1/10000= 199/10000= .0199 which is just less than 2/100.

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**PART 2) (medium)** in a lotto where you are eligible to win multiple prizes (eg if you win 1st division you also win a certain amount of 2nd division, 3rd division etc) do you need to include the probability of winning higher prizes when trying to calculate return?

eg:

take a standard lotto with 6 balls chosen from 38. the formula for each number of correct picks is (k,b)(n-k,k-b)/(n,k)

here is the expectancy table:

correct | | probability | | odds | | | prize | return |

6 | | 3.6223E-07 | | 2,760,681 | | | 20,000 | 0.007245 |

5 | | 6.9548E-05 | | 14,379 | | | 1,000 | 0.069548 |

4 | | 0.00269499 | | 371 | | | 100 | 0.269499 |

3 | | 0.03593316 | | 28 | | | - | 0 |

2 | | 0.19538657 | | 5 | | | - | 0 |

1 | | 0.43766592 | | 2 | | | - | 0 |

| | | | | | | | 0.346291 |

the question is for eg calculating the return for the '5 correct' scenario, do i need to also add the probability that i could get all 6 correct? or is that already considered in the formula =COMBIN(6,5)*COMBIN(38-6,6-5)/COMBIN(38,6)

That depends, of course, upon the rules of the specific lottery. But I don't believe any lottery gives "prize for 5 in a row" to a person who has "6 in a row".

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thanks so much for any answers to **Part 1) **or **Part 2)**

(Rofl)(Happy)(Clapping)

Re: Lottery Probabilities

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Originally Posted by

**HallsofIvy** Yes. If we call your tickets "A" and "B", then the probability of A winning is 1/100. If A does not win, there are 99 tickets left of which B is 1. The probability that A does NOT win and B does is (99/100)(1/99)= 1/100.

So the probability of **either** A or B winning is 1/00+ 1/100= 2/100.

The probability of "win this week, lose next week" is (1/100)(99/100)= 99/10000. The probability of "lose this week, win next week" is (99/100)(1/100)=99/10000. The probability of "win both weeks" is (1/100)(1/100)= 1/10000. The probability of winning "at least once in two weeks" is 99/10000+ 99/10000+ 1/10000= 199/10000= .0199 which is just less than 2/100.

thanks for your help! so the probability is less if you enter consecutive lotteries rather than putting 2 tickets in the same lottery. do you know if this loss in probability is compensated for by the opportunity to win both times (in expectancy terms)? ie is the expected return the same?

thanks!

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Originally Posted by

**HallsofIvy** That depends, of course, upon the rules of the specific lottery. But I don't believe any lottery gives "prize for 5 in a row" to a person who has "6 in a row".

let's just pretend for a moment that this lottery does allow you to win multiple prizes. then do i need to include the higher division wins with the probability of the lower division wins?

thansk again!

Re: Lottery Probabilities

Quote:

Originally Posted by

**milliondollarsocks** thanks for your help! so the probability is less if you enter consecutive lotteries rather than putting 2 tickets in the same lottery. do you know if this loss in probability is compensated for by the opportunity to win both times (in expectancy terms)? ie is the expected return the same?

thanks!

Expected return for two tickets in different weeks (assuming probability of winning is $\displaystyle \dfrac{1}{100}$ and the prize is $80):

$\displaystyle \begin{align*}\$80\dfrac{99}{10000} + \$80\dfrac{99}{10000} + (\$80+\$80)\dfrac{1}{10000} - \$2 & = \$160\cdot \left( \dfrac{99}{10000} + \dfrac{1}{10000} \right) - \$2 \\ & = \$160\cdot \dfrac{1}{100} - \$2 \\ & = \$1.60 - \$2 = -\$0.40\end{align*}$.

In other words, you have the same losses as if you bought two tickets in the same lottery.

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Originally Posted by

**milliondollarsocks** let's just pretend for a moment that this lottery does allow you to win multiple prizes. then do i need to include the higher division wins with the probability of the lower division wins?

thansk again!

Anyone who gets six numbers correct also has both four and five numbers correct. Just adjust the winnings for the higher "division" prizes to include the winnings for the lower "division" prizes. Then you use the method you already showed.

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Re: Lottery Probabilities

thanks very much for your proof! everyone here has been so helpful

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Originally Posted by

**SlipEternal** Anyone who gets six numbers correct also has both four and five numbers correct. Just adjust the winnings for the higher "division" prizes to include the winnings for the lower "division" prizes. Then you use the method you already showed.

so just to make sure i'm understanding you correctly, i attached a spreadsheet with both the original table and what i expect is the table including the lower division prizes. could someone please take a look to double-check if i understand the concept correctly? ie if you get 6 numbers correct, you'd win the '5 correct' prize a total of combin(6,5) times etc

thanks again!

Re: Lottery Probabilities

bumping for some assistance!

this may be simple for some of you out there but unfortunately i'm not such an intuitive mathematician so need help from you genii :)

cheers

Re: Lottery Probabilities

More help with **what**? If you have a specific question, ask it.