This is true for any game but here I will present results for pick 3. For one ticket and one drawing, it does not matter what numbers you pick. However, if you buy multiple tickets, you can adjust the probability of winning or losing based on whether you buy n tickets for one drawing or buy one ticket for each of the next n drawings. Let's say you bet $1 per ticket, bet on a permutation, and pick 3 digits that are different. The house edge is 50%, so the prize is $83.33.
Of the 1000 possible choices for pick 3, 720 of them have 3 different digits. You wish to buy all the possible combinations that have 3 different digits. You don't need to buy 720 tickets. You only need to buy 120 tickets, because 4-5-2 is the same win as 4-2-5, 5-4-2, 5-2-4, 2-4-5, and 2-5-4. 720 divided by 6 is 120.
Scenario 1: you buy 120 tickets with all the possible permutations of 3 distinct digits for one drawing. Hand the cashier $120. The probability of winning is 72%, and if you win, you win $83.33. You can't win less than that and you can't win more than once because all of your tickets are different. The probability of losing is 28%.
Scenario 2: buy 4-5-2 as a permutation for the next 120 drawings. Now you can win more than once. I ran a simulation (1 million trials) and produced the following:
win |
win amt |
probability |
none |
$0.00 |
48.55% |
once |
$83.33 |
35.21% |
twice |
$166.66 |
12.64% |
3 times |
$249.99 |
2.99% |
4 times |
$333.32 |
0.53% |
5 times |
$416.65 |
0.0684% |
6 times |
$499.98 |
0.0074% |
7 times |
$583.31 |
0.0003% |
8 times |
$666.64 |
0.0001% |
As you can see, the probability of losing went way up, from 28% to 48.55%. The probability of winning $83.33 dropped significantly, from 72% to 35.21%. However, the probability of winning twice or more is 16.24% whereas in scenario 1 that is a zero.
In scenario 1, the expected value of winnings is 72% times $83.33 which is $60, half of your $120 (50% house edge, makes sense).
In scenario 2, the expected value of winnings is the sum of the product of win amt and probability for each row in the above table. That value is also $60. So you don't change the expected value; you can't. It's always 50%.
However, by buying 120 tickets over 120 drawings, you change the probability distribution of winnings, increasing the chance of losing and decreasing the chance of winning $83.33 in exchange for a 16% chance of winning $83.33 more than once. Even just twice will result in a profit of $46.66. You have a 3% chance of netting $130.
Scenario 1 is kinda silly because you are guaranteed to lose money. The only question is, do you lose all $120, or do you win $83.33 and only lose $36.67?
However, you can modify this and say "I'll buy just half of all the permutations for $60 and try to win $83.33". Similarly, if you buy one ticket for each of the next 60 drawings, your probability of losing will go up in exchange for getting a chance to win $83.33 more than once. You're simply increasing the stakes.
It takes about half an hour to run 1 million trials. I will post the results for buying one ticket for the next 60 drawings and see how that compares to buying 60 tickets for one drawing (i.e., if you win, you profit $83.33-$60=$23.33)