My name: Jamie Level: Secondary Who: Other (I'm writing a book on lotteries) Please help me in confirming the following deduction of the probabilities for a 1 from 35 Lottery: A number is randomly drawn from an urn containing 35 balls numbered 00 to 34. How to win Division 1 Match the number drawn Division 2 Match the last digit of the number drawn Calculation of probabilities for this lottery: The total number of possible outcomes is 35. The probability of matching any number is 1/35: If the winning number is 14, then P(14) = 1/35 If the winning number is 15, then P(15) = 1/35 And a particular situation is observed when working out the probabilities for the second digit: If the winning number is 14, then P(24 or 34 or 44) = P(24) + P(34) + P(44) = 1/35 + 1/35 +1/35 = 3/35 If the winning number is 15, then P(25 or 35 ) = P(25) + P(35) = 1/35 + 1/35 = 2/35 Because the digits 5 to 9 are not repeated by the same number of times that the digits 0 to 4, the probabilities are different. My question is: Is it correct to find two different probabilities for the same event?? A typical lottery in which this situation occurs was Lucky 6, played in New South Wales since 1977 to 2000. Thank you very much Hi Jamie, These are not the same event. In fact there are 10 events here. The last dight can be any of the 10 digits from 0 to 9. If the last digit is between 0 and 4, as in the first situation that you describe (the winning number is 14) the Division 2 winning tickets would be 04, 24, 34, and P(04 or 24 or 34) = 3/35 If the last digit is between 5 and 9 as is true when the winning ticket is 15, then the Division 2 winning tickets would be 05, 25, and P(05 or 25) = 2/35 The events "winning with the number 19" and "winning with the number 23" are not the same, so it is normal that they get different probabilities. If the amount of money that you win in each division is fixed, then the probabilities show that it is always better to pick a number that ends in 0, 1, 2, 3 or 4, even if your lucky number is 17. But if you have to share the jackpot with other winners, then the situation is much more complex: It could be appropriate to shift your choice towards a number like 17, in the hope of sharing the jackpot with fewer people, but this depends on the proportion of people aware of the bias, and on the number of people that have the same strategy. Cheers, Claude and Penny Go to Math Central