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Please help me explain to FIL why his lottery picking strategy is no better than random.
Djeet
Registered User regular
Registered User regular
Not that I think any strategy is better than random selection. Also, this is not homework. I haven't had to do homework in more than a decade. FIL has been bugging the shit out of me about this and I just don't want to hear it anymore.
The lottery game is to pick 5 (non-replacement) out of 36. Odds of 5 of 5 are 1 out of 376992.
His strategy is to make sure to buy a card that has each number represented, so 8 tickets, e.g.:
1,2,3,4,5
6,7,8,9,10
11,12,13,14,15
16,17,18,19,20
21,22,23,24,25
26,27,28,29,30
31,32,33,34,35
36,1,2,3,4
The order of the numbers don't matter, you need 8 to get all 36 represented. I told him that he has guaranteed 100% to have on his card a pick that will have the 1st number drawn, but that's it. His odds are still 8 out of 376992 or 1 out of 47124.
Problem is when I try to math it out I get even lower odds: 1 out of 35!/4! or 1 out of 52360. So all I have managed to math out is if you get the 1st number in pick 5 of 36, then the resultant chances are better than a pick 4 (non-replacement) of 35.
Weird thing is 52360 - 47124 = 5236, or exactly 10% of 52360, which is a real head-scratcher for me.
The lottery game is to pick 5 (non-replacement) out of 36. Odds of 5 of 5 are 1 out of 376992.
His strategy is to make sure to buy a card that has each number represented, so 8 tickets, e.g.:
1,2,3,4,5
6,7,8,9,10
11,12,13,14,15
16,17,18,19,20
21,22,23,24,25
26,27,28,29,30
31,32,33,34,35
36,1,2,3,4
The order of the numbers don't matter, you need 8 to get all 36 represented. I told him that he has guaranteed 100% to have on his card a pick that will have the 1st number drawn, but that's it. His odds are still 8 out of 376992 or 1 out of 47124.
Problem is when I try to math it out I get even lower odds: 1 out of 35!/4! or 1 out of 52360. So all I have managed to math out is if you get the 1st number in pick 5 of 36, then the resultant chances are better than a pick 4 (non-replacement) of 35.
Weird thing is 52360 - 47124 = 5236, or exactly 10% of 52360, which is a real head-scratcher for me.
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@Paladin
@Phyphor
Your friend's odds are going to be low, no matter how you slice it, unless he buy so many tickets that the payout from the lottery probably won't be worthwhile. Buying 7 extra tickets is simply a waste of money.
You have a 5/36 chance of picking any number that is chosen. Assuming you do that, you then have a 4/35 chance of picking the next, and so on (3/34, 2/32, 1/31). Multiply those together and you happen to get 1/376992
To figure out how much you actually expect to win you need the expected value - you calculate the probability that any random ticket gives each possible award, times the award amount, them sum them all up and compare it to the cost of the initial ticket. There are a few cases where the EV is positive (you can actually expect to make money) - but it's rare. It's also risky, because while you are guaranteed to hit the jackpot and win if you buy every ticket, because tickets are not limited (presumably) you run the risk of spending a ton of money covering everything and have some random person guess it and then have to share the jackpot
If the ball selection is truly random, then no single ticket has any higher chance of winning than any other ticket.
{1,2,3,4,5} has exactly the same probability as {13,6,27,29,2}.
Humans instinctively perceive numbers with no discernible pattern as more random, but that perception can be wrong.
Another way to look at it: FIL's strategy is to cover as many possibilities as possible. There are 376,992 possibilities in this game, which means he needs to buy 376,992 unique tickets to break even. If the jackpot is less than $376,992, then the most likely outcome from any strategy will be to lose money.
This does not take into account secondary prizes (for example: win $5 if you match 3 balls) but in most real-world lottery games the payouts from such secondary prizes are so small as to be negligible.
the "no true scotch man" fallacy.
Where we diverge is he thinks his 8 ticket strategy (where across the tickets he has all 36 numbers included) gives him the some significant edge (cause now "you only need to be lucky 4 times" which to me sounds absurd) while I'm thinking it would be the same as picking 8 sets of 5 numbers at random* (1/47124). I just cannot figure out the math to show him.
It is not that he is playing sequential numbers (not that it would matter if he did). But that amongst his 40 numbers (8 picks of 5) he has 1-36 represented.
To be clear, he is not planning on buying every permutation, just 8, so he has 1-36 somewhere in his picks. EV is going to be hard to calculate cause payout is not fixed (estimated $200K, but it depends on number of tickets sold, and there is a roll-down payout if no one hits 5/5), there are secondary prizes 2-4/5, and there is possibility of grand prize being split in case there are multiple 5/5 winners. So eventhough i think he understands the basic gist of EV I'm not even going to try to calculate it. I would however like to be able to show him that his 8 pick strategy is no better than 8 random picks.
*Random: I am making the assumption that quick pick will not pick the same 5 numbers several times but in a different sequence should he buy 8 quick picks.
If your friend's magical lottery system is supposed to win him lottery jackpots, one might ask why it hasn't actually won him any jackpots yet. There can be deep maths involved in the exploration of that question, but there doesn't need to be any deep maths involved in having the question asked (assuming the premise - that it hasn't won him any jackpots - is true). If you have a proposed system that works in theory but doesn't work when applied to real situations, that's a pretty good indicator that something may be wrong with your theory.
"You only need to be lucky four times" is gibberish.
He seems to think that his 8 picks are special in some way. They are not.
All combinations are equally likely. This means that the combinations
are no more likely the combinations
or any other unique set of 8 combinations.
Another way to look at it: if you map out all 376992 combinations... how many times do his combinations appear?
His combinations appear exactly once each. Just like any other combination.
the "no true scotch man" fallacy.
it's an interesting brainteaser - please someone correct me if this is way off!
Then let him cover the spread by making two bets (H? and T?) - he now wins 50% of the time but has an EV of:
3*(1/2) (50% chance of winning 3) - 1*(1/2) (50% chance of losing 1) - 1 (the branch we didn't take which he always loses) = 0
*Technically the last ticket repeats a few of the first ticket's numbers, but it's not really relevant overall (or intended, from what I gather of his scheme).
Now, there is a possibility that the balls aren't chosen at random. The drum may not truly randomize them, and allow the initial order to impact which balls are chosen. The balls may not be uniform in weight or size, causing certain numbers to be more or less likely to be drawn. Unfortunately, it's HIGHLY unlikely that - without active tampering - any of those factors will be sufficient to significantly change the odds of any particular sequence. Additionally, it's unlikely that any lottery will do enough draws under the same conditions that someone could discern a pattern to capitalize on.
That leaves betting strategy. Any game with fixed prizes is going to be a loss - unless the lottery fucks up bad in calculating odds and prizes, you're never going to be able to beat the odds on a Pick-3 or Pick-4, etc. You may win, but it's just gambling - in the long term, the odds are always against you.
The only place where you can POTENTIALLY beat the system is a game like the Powerball or Mega Millions, where the jackpot is not fixed and is based on the buy-in. In theory, the jackpot could get high enough that - even with taxes taken into consideration - the overall odds could be in favor of the gambler. This would be a situation where buying 'every ticket' would cost less than you win in the grand prize.
However, because jackpots are split, there is a significant risk that you buy 'every ticket' and the winning ticket is split two, three, or more ways. Realistically, with that taken into consideration, no betting strategy is likely to push the odds in your favor.
I also want to note that while it doesn't matter on a 'fixed prize' lottery, your FIL's choice of tickets is exceptionally bad for something like the Powerball or Mega Millions. While the odds of any particular set of numbers - even an ordered sequence like 1-2-3-4-5-6 (7 Mega) are the same, a common sequence like 1-2-3-4-5-6 (7 Mega) is far more likely to be chosen by other people. That means even if you hit the big jackpot, you could end up splitting it with dozens or hundreds of people, and your $400,000,000 jackpot could 'just' be a few hundred thousand.
TL,DR: The system is rigged, you can't beat it. There's a reason people call the lottery a tax on people who are bad at math.
This phenomenon is easier to see if we look at the lottery where the player has to pick a single number from 1 to 4, and wins if his number is drawn. If you bought the four tickets 1,2,3,4, you would always win exactly once. If you bought 4 tickets at random, you would win once on average. However, you would have a (3/4)^4 ~ 31% chance of not winning at all, as well as about a 21% chance of winning exactly twice, a 5% chance of winning exactly3 times and a 0.4% chance of winning all 4 times. The expected values are the same, but the distributions are quite different. Your FIL might want to maximize his chance of winning at least once, and his strategy does exactly that (as do many other strategies).
Its worth pointing out that the differences in my made-up example are much more extreme than in your original example - practically, the difference is tiny.
When I try to say something to the effect of "you have 8 non-repeating picks, your chances are still 8/376992 or the same as 8 quickpicks" but my math doesn't support it.
Now lets say on draw night 10 is the first ball picked. My intuition is that the maths should then be the same as a pick 4 (non-replacement) of 35, or (4/35)*(3/34)*(2/33)*(1/32), or 1/52360. Which is actually less likely than 8 random (1/47124). Shouldn't it be the same? I'm sure I'm missing something very basic, but why doesn't it add up?
Or another way to state the question, assuming I have a ticket and the first number drawn is represented on my ticket, what are the odds that the 4 remaining numbers on my ticket will be picked from the 35 balls?
Edit: Now that I think about it the point CycloneRanger and mumbly_pie bring up makes sense, and I can see how it could be the reason my math is off. If it were 8 quickpicks there is some probability that the first ball will be represented more than once, and thus he has more than one shot. While his strategy almost ensures the 1st ball will only be represented once across his picks. I will see if this explanation goes anywhere with him.
Imagine if instead of having any ties to numbers, the lottery has 376,992 different colored marbles. One particular color variation will be selected, and if that's the one you chose, you win.
By making a special attachment to the first number, your father-in-law is arbitrarily separating the 376,992 marbles into 36 piles, and then choosing one marble from each pile. When the lottery guy says "the winning marble is in this pile!" it doesn't mean that you somehow have a better chance of winning than if all the marbles had been kept in one big pile.
Or wait, here's another way to look at it.
Let's say you have a lottery with 376,992 possible choices, but instead of the first digit always being a choice of numbers 1-36, it varies day by day. One Monday the first digit can only be the numbers 1-5, and Tuesday the first digit can be numbers 1-50. There are always 376,992 choices, though, it's just that the variation of the rest of the digits also goes up in down depending on what they did with the 1st digit that day.
With your FIL's strategy, on Monday he would feel he only had to buy 5 tickets to be "covered" on his first digit strategy, but on Tuesday he would have to buy 50 tickets to be "covered." But really, he is just giving himself a 5 in 376,992 chance to win on Monday, and a 50 in 376,992 chance to win on Tuesday, just as with his real-life strategy he is giving himself an 8 in 376,992 chance to win every single time, same as if he just chose 8 tickets at random.
The best way to explain stats or probability to someone without a deep understanding of them is to increase or decrease the scale a ton. For instance, the old problem of choosing one of three doors, and it being beneficial to switch after is more easily explained when you increase the number of doors from 3 to a million.
In this case a good way to explain it would be, instead of the numbers being from 1 to 32, what if they were from 1-4 and you only picked 2? Your tickets would be
1,2
3,4
The total number of combinations is 6, and they are:
1,2 1,3 1,4
2,3 2,4
3,4
You have a 2/6, or 33% chance to pick a correct one by randomly choosing 2 of any combinations.
If we ignore the first pick, because "now you only have to get lucky once," that means now we only have one number left, and there are only three choices. One in 3 is 33%.
Pick 4 of 35 is less likely than pick 5 of 36 times 8. Your math isn't wrong but the probability of each is different.
1) Each lotto ticket consists of 5 numbers
2) The 5 numbers each range from between 1 to 36
3) Numbers never repeat - each number can only be used once
4) The order of the numbers do not matter (e.g., "1,2,3,4,5" and "5,4,3,2,1" are the exact same ticket)
5) The "winning" lotto ticket is a single set of 5 numbers
Is this correct? I feel like some of these aren't actually lotto rules but are the FIL's rules for picking numbers?
Anyways this can be closed.
I've gotten as far with him as I think I can. He thinks that having the 1st ball being represented in his 8 ticket card gives him some kind of advantage. And he thinks anyone who would do 8 quick picks and just get a probability of 0-8 of the tickets having the the 1st ball is somehow gambling more than he is. He's got this attachment to the 1st ball that I cannot disabuse him of.
Personally I wouldn't play this particular game anyways. I sometimes throw a fiver or something into one of the big multi-state progressive jackpot lottos when it gets big, but I don't play a strategy or anything.
Anyway, don't spend more than a few minutes trying to explain to someone why their gambling system is wrong / won't work. If you can't get some traction in unseating their beliefs in a few minutes, it's probably a lost cause and will only build animosity.
If nothing else, his odds with those sets of numbers are no different than the odds of any other numbers - so his irrational belief those are the 'best' numbers is no different than anyone else who gambles a particular set of numbers.
Basically, as long as they aren't going to blow their life savings / mortgage their house / empty their 401k (which is intervention point regardless of their system or not) let them have their fun. Just keep an eye on them to make sure they don't sink into a gambling addiction. $10 in lotto tickets every few days isn't like someone who CAN'T LOSE using the Martingale System at the local casino and starts betting more and more.
It's probably the combination of the idea that "the order doesn't matter" and "I'm putting all the numbers across all the tickets" that is causing the gap in your FIL's reasoning. Since he doesn't seem to have a big math background, short of literally showing him, in a giant spreadsheet, all the possible winning numbers (organized into "equivalent sets"), comparing the numbers he chooses against the other possibilities, and then showing that the chances are exactly the same, you will probably never disavow him of this belief. And that assumes that he is open to the idea that he might be wrong, which most people aren't (it's a fact of human cognition, unfortunately). He could be pretending to listen to you the entire time while not actually listening, and that will only serve to antagonize the issue.
Since he's your FIL I think you have the right idea in just letting it go and moving on. There are plenty of things that I'm sure both you and he are "wrong" about, and if your relationship consists of just proving how irrational the other person is, that's a flame war that neither of you will win.
This is not correct. Absence of evidence is not conclusive proof of absence, but it is indeed one piece of evidence that points to absence - and if that's the only evidence kicking around, chances are that it's pointing you in the right direction.
As a side note, I really do not recommend tossing that phrase around. It's mostly used by people that espouse belief in crackpot theories (as well as the occasional salty old Worthless Degree of Choice professor).
Anyways I actually like the statement. It turns it back on whomever it is directed to use more powerful evidence/reasoning, which is only a good thing.
And, as @The Ender alluded to, generally this statement is only used as a last resort by proponents who cannot in good faith provide affirmative evidence as to their propositions (that's a really fancy way of saying it's a hand-wavy cop-out to absolve yourself of having to actually prove what you are saying). As in, "I believe that the flying spaghetti monster exists!", "No one has ever seen the FSM", "Well just because you can't prove it isn't there doesn't mean it's not there!"
My 5th Grade teacher used to tell this horribad joke:
"What's an elephant's favorite place to hide?"
"I don't know."
"A tree!"
"What do you mean, a tree?"
"Have you ever seen an elephant in a tree?"
1000000 drawings with 8 tickets for each strategy per drawing, for a total of 16000000 total tickets.
Number of times a ticket had N matching values while using his strategy:
0 matching numbers: 3605766
1 matching numbers: 3337379
2 matching numbers: 954800
3 matching numbers: 98831
4 matching numbers: 3211
5 matching numbers: 13
Number of times a ticket had N matching values while picking tickets randomly:
0 matching numbers: 3605016
1 matching numbers: 3339158
2 matching numbers: 954182
3 matching numbers: 98283
4 matching numbers: 3337
5 matching numbers: 24
The ruby code:
require 'rubygems' require "backports/1.9.2/array/rotate" num_drawings = 1_000_000 possible_numbers = 36 numbers_per_ticket = 5 tickets_per_drawing = (possible_numbers.to_f/numbers_per_ticket).ceil def get_random_numbers_no_replacement(num_values, range) picked_numbers = [] unpicked_numbers = *range 1.upto(num_values) do |x| picked_numbers.push(unpicked_numbers.rotate!(rand(unpicked_numbers.size))[0]) unpicked_numbers = unpicked_numbers.drop(1) end picked_numbers end strategy_tickets = [] 0.upto(tickets_per_drawing-1) do |ticket_num| ticket = [] 0.upto(numbers_per_ticket - 1) do |offset| ticket.push(1 + (numbers_per_ticket*ticket_num + offset)%possible_numbers) # "1 +" to shift tickets from 0-35 to 1-36 end strategy_tickets.push(ticket) end drawing_histogram = Array.new(possible_numbers, 0) random_histogram = Array.new(possible_numbers, 0) strategy_winning_tickets = Array.new(numbers_per_ticket + 1, 0) random_winning_tickets = Array.new(numbers_per_ticket + 1, 0) num_drawings.times do drawing = get_random_numbers_no_replacement(numbers_per_ticket, 1..possible_numbers) drawing.each do |x| drawing_histogram[x-1] += 1 end 1.upto(tickets_per_drawing) do |x| strategy_winning_tickets[(drawing & strategy_tickets[x-1]).size] += 1 random_ticket = get_random_numbers_no_replacement(numbers_per_ticket, 1..possible_numbers) random_winning_tickets[(drawing & random_ticket).size] += 1 random_ticket.each do |x| random_histogram[x-1] += 1 end end end puts "#{num_drawings} drawings with #{tickets_per_drawing} tickets for each strategy per drawing, \ for a total of #{num_drawings*tickets_per_drawing*2} total tickets." puts "Number of times a ticket had N matching values while using the strategy:" strategy_winning_tickets.each_with_index do |x, index| puts "#{index} matching numbers: #{x}" end puts "Number of times a ticket had N matching values while picking tickets randomly:" random_winning_tickets.each_with_index do |x, index| puts "#{index} matching numbers: #{x}" endIf basic probability concepts could allow you to develop a strategy that gave great chances to win the lottery, lottery companies would have been driven out of business by maths majors decades ago.
Be the one selling the tickets
Ask him to consider a pick 2 in 10, and apply his strategy:
(1, 2)
(3, 4)
(5, 6)
(7, 8)
(9,10)
Call the two drawn two numbers X and Y. He will definitely have a set that contains X, but then there's an 89% (8 out of 9) chance that Y is going to be in a different set.
So if X is 1, there's an 11% (1:9) chance that Y is 2. If X = 3, 11% that Y is 4, and so on. The win rate for all possible X values is 11%.
X : Win%
1-10 : 11%
That's an 11% to win a pick 2 in 10 by covering all the numbers. The chance to win a 2 in 10 with one ticket is only 2%, so he must be doing something right, right?
But what about these tickets:
(1, 2)
(2, 3)
(1, 3)
(2, 4)
(1, 4)
They certainly look riskier; you're missing out on over half the numbers.
But despite the fact that there's a 60% chance you'll lose on the first number, there is still an 11% chance to win.
X : Win %
1 : 33% (2, 3, 4)
2 : 33% (1, 3, 4)
3 : 22% (1, 2)
4 : 22% (1, 2)
5-10 : 0%
(33+33+22+22+0+0+0+0+0+0)/10 = 11%
So if picking all the numbers gives him an 11% chance, and picking less than half the numbers gives you an 11% chance of winning, then why does the chance of winning increase at all? Because the odds of winning off of one ticket is 1/45, or 2.2%. 2.2% * 5 tickets gives you an 11% chance of winning, regardless of what the numbers on that ticket are.
PS: I'm not explaining this to YOU like you're five, I'm suggesting a dumbed down proof for him. Maybe even try 2 in 8.