Need a math person to answer a Lotto/probability/game theory question

Factual Questions
1

I hope someone can help me with a probability problem I’ve been wondering about for several years.

Say you have a Lotto of 42 numbers, the Lotto people pick the winning ticket of 6 at random
(without replacement) A person can win with 6 matching numbers, or 5 or 4.

My question is, if you could buy a limited number of tickets, say 7 total, would it be to your advantage
to hit all 42 numbers when you buy the tickets (instead of picking them randomly)?
In other words, you would dump 42 numbered slips into a hat and chose seven tickets of six numbers
without replacement. By advantage, I mean do you increase your probability of winning any prize?

Put another way, if you picked randomly with replacement and by chance picked the same 6 numbers
all 7 times, your chance of winning any prize would be reduced for that set of numbers. Would it also
be reduced with only partially unique sets of numbers? For example, if you picked 1,2,3,4,5,6, and then
also 1,4,5,7,9,10, etc.

How would one go about figuring out the solution to this problem?

I tried it with a simpler version, using 6 numbers, chosing three, and making a match on 2 or 3, I figured the probability of winning any game but then I got stuck.

I feel that if you have an advantage with this method, it only would work with a lottery that allows
a range of matches for a winning ticket.

I hope my question makes sense.

2

You still have the same effective amount of money won whether you pick some same numbers or only different numbers. If you repeat numbers, you increase the variance; that is, you have a higher chance of getting more money, but you also have a higher chance of winning nothing at all.

The fact that you have a range of winning right numbers to win matters naught. It’s just a superposition of the separate chances of winning with an exact number of correct numbers.

3

I don’t care about the “amount of money”, the only thing I cared about was the probability of winning a prize, any prize.

I’m more interested in: if I have a “higher chance of winning nothing at all” if I repeat numbers, does that mean I have a higher chance of winning outright if I don’t?

4

By reducing the variance you have a lower chance of winning more money, but a higher chance of winning something.

5

IANAMG (I am not a math geek)

Intuitively, it would seem to me that if you had coverage of all numbers in your set of tickets, you’d have better chance of winning something at all, because if you can have duplicate numbers across tickets, you could, conceivably have duplicate tickets and the duplicate would add no additional probability of winning any prize.

6

I haven’t thought too hard about this problem yet, but my intuition is that the best strategy, i.e. the one that will maximize the probability of winning some prize, is to maximize the number of distinct four-element subsets of your tickets.

So, begin by choosing six numbers. That’s your first ticket. Then choose another six numbers for your second ticket in such a way that no more than three of the numbers that were on the first ticket are on the second ticket. E.g., if the first ticket was {1,2,3,4,5,6}, the second ticket shouldn’t be {3,4,5,6,7,8}, because then {3,4,5,6} would be on both tickets. But the second ticket can be {4,5,6,7,8,9}, since that only shares {4,5,6} with the first ticket.

Of course, you can also achieve this by having each ticket share no numbers with any other ticket. So you could use just as well use

{1,2,3,4,5,6},
{7,8,9,10,11,12},
{13,14,15,16,17,18},
etc.

Note that if the tickets share no four-element subsets, then they share no five- or six-element subsets either, so you’ve maximized your chances for winning one of those prizes also.

So, I conclude that you can do no better than choosing your tickets as above. Now that I’ve thought about this more while writing, I’m pretty sure that what I have here constitutes a proof.

7

Puno, you are wrong about something: it does make a difference if there is a range of winning numbers and here’s why:

If there were only one 6 number winning ticket, then it wouldn’t matter at all if you had duplicate numbers between sets, as long as the tickets you bought weren’t identical. In that scenario, it wouldn’t matter how you picked your numbers, your probability of winning would be exactly the same.

8

RuskalaIf you picked the numbers randomly, it’s possible two tickets would be identical. But it’s a good point that if only 6 matches win, it doesn’t matter if two tickets differ only by one number.

9

Tyrrell McAllister’s analysis strikes me as correct, and clearly explained, too.

10

This would be true if the prize for 6 matched numbers were always the same. In real lotteries, as I understand it, the winner(s) divide the jackpot. So if there’s $5million up for grabs, and 5 winning tickets, each ticket gets $1million. So if you buy 5 identical tickets, your average payoff is the same as if you buy one; if you buy 5 different tickets, your expected payoff is 5 times higher.

11

If you don’t mind a simplifying assumption, suppose that there is one and only one prize and it is the jackpot for matching all six numbers. There will be one distinct combination that can win. Overlap is not an issue, as long as each combo is unique.

Suppose that combos that match 4 or 5 can win, then you’ll have three distince classes of combinations, each combination being unique, with a greater number of possible winners for each set of numbers drawn. So for each ticket you’ll have 6 Choose 4 “tickets” for the “four number lottery” and similarly for the “five number lottery”.

To maximize the number of “tickets” for the four and five number sub-lotteries, you’ll want to minimize the overlap, right? If so, then I guess I’m with Tyrrell McAllister as well.

Okay, let me ask this: Suppose you’re given the opportunity to pick two numbers from one to six, not necessarily different so that you may pick three twice, for example. I then roll a die and pay you $1 if the die matches your number.

If you choose two seperate numbers, your expectation is:

(1/6)*1 + (1/6)*1 = (1/6) + (1/6) = (2/6) = (1/3) or about $0.33.

If you choose one number twice, your expectation is:

(1/6)*2 = (2/6) = (1/3) or about $0.33.

It doesn’t matter whether you choose one number on the die twice, or two numbers one time each. The expectation is the same either way.

It seems analogous to the four and five match tickets since each ticket pays a set amount no matter how many you bring in; e.g. a ticket that matches four numbers pays $2,500 no matter how many people match four numbers. So if my little thought experiment is right, and if it is an appropriate analog, then maximizing the number of four and five number subsets won’t improve your odds. Having, say, ten different four number subsets gives you the same odds as having ten copies of one four number subset. So the subset issue seems moot.

This brings us back, to my suprise quite frankly, to my original assumption where it doesn’t matter as long as each six number set you choose is unique to you. This is not analogous to the die experiment since the pot is split between the number of tickets that match. So the numbers can overlap as much as you like so long as each ticket you buy has a six number set different from every other ticket you buy.

Final note: I have an acquaintance who insists that you’d be better off purchasing, say, seven tickets with the same six number combination because the probability of somebody else matching is “relatively” high. His reasoning is that if you do so and you and one other person win, then you’ll get seven-eighths of the pot instead of one-half of the pot. He figures that buying seven different tickets won’t improve your odds much since each one has such a small probability of winning; whereas increasing your share of the pot will have a bigger relative effect on expectation. I’ll let somebody else do the math on that one.

12

I just re-read the thread and the OP doesn’t care about expectation, only about whether a win is logged in the books. In that case I’m back with Tyrrell McAllister. Sorry about missing that!

Punoqllads, I see what you were saying about the variance. Thanks for pointing it out!

13

Well, no; your expected payoff is always X/(Y+X), where X is the number of winning jackpot numbers you bought and Y is the total number sold to other people.

If the jackpot was $5 million and you bought one winning ticket and another person bought another, you would win $2.5 million. but if you bought NINE winners, you’d win $4.5 million… not nine times more.

No way. Consider our big lotto, the Lotto 6/49. The odds of a single $1 ticket winning are 1 in 13,983,816. Purchasing (to make the math easier) ten DIFFERENT tickets therefore gives you a 1 in 1,398,381 shot at winning the jackpot.

Purchasing ten IDENTICAL tickets means you still have a 1 in 13,983,816 chance, but as your friend points out you’ll win a bigger share of the jackpot. However, the expected payout is only greater than ten different tickets if the average number of winners per draw is above 10. In the Different Ticket Strategy your expected payout on, say, a $2 million jackpot is about 14 cents on the dollar. In the Same Ticket scenario, it’s not that high, no matter how many people win. For instance, if 1 other person picked the same number, you’d win $1.8 million instead of $1 million - a marginal payoff of $800,000 for wasting your other nine numbers on the saem number you’d already picked. It’s not worth the cost of giving up nine other chances to win $1 million. Would you rather have one chance of winning $1.8 million, or ten chances of winning $1 million? Obvious choice, right?

If TEN other people picked the same number, you would win $1 million instead of $181,818.18… better than if only one other person hit the jackpot, but it’s still not ten times better. Still a bad bet.

If TWENTY other people picked the number, ten identical tickets deliver $666,666.66 instead of $95,238 - again, not a good exchange. The expected payoff is better for $1 to win $95K than $10 to win $666K.

The best you can ever do is more or less even, and that’s only if a really large number of people pick the winning number. In every Lotto I’ve ever heard of that is a very rare thing; most Lottos have zero to three/four winners per draw. Picking 10 identical numbers is a dumb bet.

14

<sigh> Y’all are gonna make me break out the graphite? Well, in the interest of fighting ignorance, here’s a simplified example.

Suppose there are only two tickets, not seven. Suppose also that a winning ticket has five or six winning numbers on it.

First, let’s see what our odds are of picking a winning ticket when we don’t repeat any numbers. Then, our odds of picking a winning ticket are:

choose(6,6)*choose(36,0)/choose(42,6) + choose(6,5)*choose(36,1)/choose(42,6) + choose(6,1)*choose(36,5)/choose(42,6) * choose(5,5)*choose(31,1)/choose(36,6) + choose(6,0)*choose(36,6)/choose(42,6) * choose(6,6)*choose(30,0)/choose(36,6) + choose(6,0)*choose(36,6)/choose(42,6) * choose(6,5)*choose(30,1)/choose(36,6)

which simplifies to 31/374699.

Now let’s see what the odds are of picking a winning ticket when we repeat one number:

choose(6,6)*choose(36,0)/choose(42,6) + choose(6,5)*choose(36,1)/choose(42,6) + choose(6,2)*choose(36,4)/choose(42,6) * 2/6 * choose(4,4)*choose(32,1)/choose(36,5) + choose(6,1)*choose(36,5)/choose(42,6) * 1/6 * choose(5,5)*choose(31,0)/choose(36,5) + choose(6,1)*choose(36,5)/choose(42,6) * 1/6 * choose(5,4)*choose(31,1)/choose(36,5) + choose(6,1)*choose(36,5)/choose(42,6) * 5/6 * choose(5,5)*choose(31,0)/choose(36,5) + choose(6,0)*choose(36,6)/choose(42,6) * choose(6,5)*choose(30,0)/choose(36,5)

which simplifies to 31/374699.

As both are the same number, then, it shows that the odds of picking a winning ticket are the same whether you pick distinct numbers for each ticket, or reuse one number from the first ticket.

You can extend the above to accomidate a winning ticket on four winning numbers, as well as to seven tickets, but the mathematics gets ugly very rapidly. You will find, however, that all picking algorithms will have the same expected number of winning tickets.

15

I may be saying the same thing as others, but if the goal is to match 4 or more numbers on the same ticket, then just make sure you have no two tickets having the same four numbers on them.

Subject to that, how you select them, whether randomly, out your back side or numerically will not improve or reduce your odds of winning.

So if you want, have all seven tickets with 1-2-3, but no other number repeated across two or more tickets. Or all different numbers or whatever. Every combination of four numbers has an equal chance, so don’t repeat them.

16

Why is the chance of losing 0% under the following similar but
simpler scenario?

Out of six numbers, 1,2,3,4,5,6 a match on 2 or 3
numbers will win. 1,2,3, and 4,5,6 are the two tickets
I chose. THERE IS A CERTAINTY I will win no
matter what the winning ticket is.
There is NO CHANCE I COULD LOSE. If my neighbor
Punoqllads picks two tickets randomly, (i.e.1,2,3, and 2,3,4)
There is a chance he could lose. WHY IS THERE A
DIFFERENCE?

For the above example I would NEVER lose, but my neighbor would lose if the winning lotto ticket is 1,5,6, or 2,5,6, or 3,5,6, or 4,5,6 or 1,4,5. Of course there is a high probability he would win, but it wouldn’t be guaranteed. My win would be guaranteed.

Thanks

17

Because there is a chance that both of his tickets will win. The expected number of winning tickets will be the same over the long haul for both of you, however. It’s what I was talking about earlier with the higher and lower variance.

18

It couldn’t “be the same” if there was a chance he could lose and I would not. Would you pick random numbers if you could only buy two tickets and my method was a sure bet and the random method was not? My point is there IS a difference, however it evens out over many tries. Lottery tickets cost money.

I thought the probability of winning + the probability of losing must = 1. So I am puzzled.

19

Your loss is guaranteed as well. You will always win on one ticket and always lose on one ticket. What your scheme seems to be is essentially making two bets on the flip of a coin, one bet for heads and one bet for tails. You are guaranteed a win, yes, but you are guaranteed a loss as well. Your neighbor is making two bets on the coin as well, but choosing randomly, i.e. HH, TT, or HT with equal probability. If the coin comes up T, then your neighbor loses everything on the HH bets, wins everything on the TT bets, and breaks even on the HT bets. You break even with certainty. In essence, you aren’t gambling. If you play only once, then you stand a chance to gain nothing and lose nothing; if you play repeatedly, you still gain and lose nothing. Your neighbor can win 2 on a single play, lose 2, or break even. Over many plays, the ratio of wins to losses for your neighbor will come to 50-50. (Since it’s the ratio and not the absolute number your neighbor may actually be behind or ahead by quite a bit after enough plays. After 10,000 plays with a win-loss ratio of 4,900 to 6,100, your neighbor is down by 200, but you are still exactly even with certainty.)

20

Ya’ll are talking about two different things, Rusalka and ** Punoqllads**.

We have a pick 6 out of 42 lottery, for a total of 5245786 different possible tickets (for simplicity, let’s assume the only winners are the ones who match all 6; the point is still the same). Say person A and person B each buy 5245786 lottery tickets; A’s tickets are all the same, while B buys up all the different tickets.

Rusalka wants to know what’s the probability of winning, period (what’s the probability of having a winning ticket at all, without consideration of how many of your tickets are winners). Here, clearly A has a 1/5245786 chance of winning, while B has a 100% chance of winning.

Punoqllads wants to know, on average, how many of your tickets can be expected to be winners. Here, in a given lottery pick, we’re not interested just in whether we have a winning ticket or not, but how many winning tickets we have:

A: 5245785 out of 5245786 times, A will have no winning ticket, but that 1 time out of 5245786 where A does win, he wins big (in terms of the number of winning tickets he has)–he has 5245786 winning tickets. So A’s expected number of winning tickets is:

0 * (5245785/5245786) + 5245786 * (1/5245786) = 1.

B: For each of the 5245786 different tickets, B has exactly one winner, so B’s expected number of winning tickets is:

1 * (1/5245786) + 1 * (1/5245786) + … + 1 * (1/5245786) (5245786 terms)

= 5245786 * (1/5245786) = 1.