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 ofeitherA 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 tobringin 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 toguaranteethat you win, you would win $80 but the tickets cost you $100 so your net isnegative20. You have still lost 20 cents per ticket.

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.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!?

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".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)

thanks so much for any answers toPart 1)orPart 2)