The above image is of the Andromeda spiral galaxy. What has it to do with winning the National Lottery? Answer: the numbers concerning both are astronomic.
The other day, we were in the Co-Op and a lady was having a discussion with the check out lady about the different types of voluntary taxation offered by the operator Camelot. The assistant explained to that lady that she understood very little about the lottery and asked for her colleague to come over to help the lady whilst she got on with dealing with other customers. When eventually we were able to pay for our goods we commented about the variety of “games” available to those punters who presumably “feel lucky.” The lady responded that she never played the lottery but both her two adult sons did each of whom spent £40 every week! £20 on Wednesdays and £20 on Saturdays.
Thus, on the basis of pro bono publico, the British Gazette offers this explanation as to how to calculate the odds of winning the Wednesday or the Saturday lottery.
As you may or may not know the draws for the Wednesday and Saturday Lotto comprise a random draw of six numbers in the form of balls. There are 49 balls, numbered 1 to 49.
Thus when the first ball is drawn it is 1 ball drawn from 49 balls. When the second ball is drawn it is from 48 remaining balls. When the third ball is drawn it is from 47 remaining balls. When the fourth ball is drawn it is from 46 remaining balls. When the fifth ball is drawn it is from 45 remaining balls, and when the sixth ball is drawn it is from 44 remaining balls.
The formula therefore is:
(49 x 48 x 47 x 46 x 45 x 44) divided by (6 x 5 x 4 x 3 x 2 x 1)
or: 10,068,347,520 divided by 720
Which equals 13,983,816.
Thus the odds of winning are: 1 in 13,983,816
Where two or more people either select of have a random selection presented to them in the form of a “Lucky Dip” ticket, the odds of these identical tickets are exactly the same as they comprise in effect the same entry. The difference being that should these tickets become the winning tickets the prize is shared equally between the multiple holders.
At this point many consider that they can increase their chances of winning the lottery by the simple expedient of buying more tickets! This of course is correct. However, where they go wrong is to make this erroneous calculation:
If I buy two tickets, I double my chances. Thus, odds of 1 in nearly 14 million, become 2 in 14 million, in other words, 1 in 7 million!
Err…. No!
Each lottery ticket must be viewed as a discrete random draw of six numbers that have been entered into the draw. Thus each ticket whatever the numerical selection of numbers has exactly the same chance of corresponding to the numbers drawn by the lottery machine. Thus the odds of one ticket winning are 1 in 13,983,816. The odds that one of the two lottery tickets purchased are 1 in 13,983,816 minus 1 or 1 in 13,983,815! So for a doubling of your stake you have increased your chances of winning the lottery by approximately 0.000007152 percent!
If you thought odds of nearly 14 million to 1 were long, these are nothing when compared to the odds of winning the Euromillions draw!
Up to Friday 6th May 2011 the draw for this exercise in futile hope was once a week late on a Friday night and comprised two random draws, one of five numbers out of a selection of 50 and a second of two numbers out of a selection of 9.
The formula for the two draws was slightly more complex, being in two parts:
Firstly:
(50 x 49 x 48 x 47 x 46) divided by (5 x 4 x 3 x 2 x 1)
or: 254,251,200 divided by 120
Which equals 2,118,816.
Secondly:
(9 x 8) divided by (2 x 1)
Or: 72 divided by 2
Which equals 36.
The first part is then multiplied by the second part:
2,118,816 x 36
Which equals 76,275,360
Thus the odds of winning are: 1 in 76,275,360!!!!
From Tuesday 10th May, 2011, a second weekly draw was launched together with revisons includiung the addition of two extra numbers (10 & 11) in the second set of numbers, to make winning the jackpot even less likely:
Firstly (unchanged):
(50 x 49 x 48 x 47 x 46) divided by (5 x 4 x 3 x 2 x 1)
or: 254,251,200 divided by 120
Which equals 2,118,816.
Secondly:
(11 x 10) divided by (2 x 1)
Or: 110 divided by 2
Which equals 55.
The first part is then multiplied by the second part:
2,118,816 x 55
Which equals 116,534,880
Thus the odds of winning are: 1 in 116,534,880!!!!
These are hardly prudent investments! It is because so many disregard or are ignorant of this that many regard lotteries as a tax on fools!
Speaking the Truth unto the Nation
This article is wrong.
If there were 10 possible combinations and you had 1 ticket, the chances of you winning would be 1/10, 1 in 10, 10%.
If you had 2 tickets, the chances of you winning would be 2/10, 2 in 10, 20%. You have doubled your chances.
If you had 5 tickets you would be 5 times more likely to win than having 1 ticket, and the odds would be 1/2, or 50%.
The same rule applies if you have 14,000,000 combinations. If you have two different tickets, you have doubled your chances to 1 in 7,000,000. It is still extremely unlikely you will win.
If you have 3 tickets, the odds are 3 in 14,000,000, or 4,666,667 to 1. You are three times more likely to win than if you had 1 ticket. Again, the odds are still huge.
If you had 7,000,000 tickets the chances of winning would be 1/2 and you would be 7,000,000 times more likely to win than if you had 1 ticket.
It’s not a hard concept to grasp.
Derek,
It is people like you who make millions, no, billions, for lotteries the world over. Just like the muppets who are to be seen standing by roulette tables counting the number of times the ball lands on red or black. It is not the British Gazette that is wrong. It is YOU. When you buy a £1 Lotto ticket, the chances of that ticket winning are 13,983,816 to 1. If you buy another ticket its chances are EXACTLY the same. Thus buying two tickets reduce your odds from 13,983,8161 to 1 to 13,983,815 to 1! If you can’t see that, then carry on wasting your money!
Jack,
I am not wrong. I am sorry you can’t do maths.
If you bought 2 tickets for two separate lotteries then yes, the odds would be the same.
I’m not saying the lottery isn’t a waste of time as buying two tickets still results in odds of 1 in 7 million. However, your chances have still doubled!
Dear Oh Dear Oh Dear!!!!!!!!!
Derek, My Son,
You just don’t get it do you?
OK then, let’s explain it another way. Suppose you sallied forth to your local newsagent yesterday, your £2 coin clasped in your little hand to purchase your Euromillions lottery ticket.
This means that you had ONE chance in 116,534,880 of waking up this morning the deliriously happy winner of £161 million quid. By logical implication this also meant that your chances of NOT waking up as rich as Croesus was 116,534,879! If, perchance, you’d had found another £2 coin and bought two lines on your ticket, you would have had TWO chances in 116,534,880 of winning and 116,534,878 chances of NOT winning!
So your chances of NOT winning from a near certain 99.999999141887819337866911606208% TO a slightly less near certain 99.999998283775638675733823212415%
Comprende Amigo?
Jack.
I get the percentages, thank you. I’m with you on that one. The percentages are so small because of the large number of combinations.
You don’t seem to understand that your chances have STILL doubled.
Imagine there were only 100 possible combinations.
If you bought 2 tickets then you would have 2 chances of winning out of 100. That equates to 2 in 100, or 1 in 50. That is double the chances of having just 1 ticket (1 in 100). Why can’t you see that it works the same if there were 14,000,000 or 14,000,000,000,000,000 different combinations?
By your logic, if you bought HALF the 14,000,000 different combinations (7,000,000) then you would only have a 1 in 7,000,000 chance of winning.
In fact, you would have a 50% chance of winning, as you have HALF the potential winning tickets.
“If, perchance, you’d had found another £2 coin and bought two lines on your ticket, you would have had TWO chances in 116,534,880 ”
2 in 116,534,880 is the same as 1 in 58,267,440. It is double 1 in 116,534,880.
2 in 100 is the same as 1 in 50. It is double 1 in 100.
Don’t be so smug about something which you are wrong.
TWO chances in 116,534,880 = ONE chance in 58,267,440.
YOU HAVE DOUBLED YOUR CHANCES.
The total number of combinations does not change.
The amount of combinations you have bought has doubled.
Are you thick?
Derek’s correct.
This article is embarrassing. Was it posted as a joke?
Sorry for raising this thread from the dead just to have a good laugh, but Bwahahahahahahahahahahahahahahahaha!
Well done Jack Ketch for keeping your cool in the face of blank ignorance.
Just found this thread and can’t believe this article was published!!
IT IS WRONG!!!
Derek is 100% correct in what he is saying.
The odds of winning the National Lottery are just under 14 million to 1.
That is because there are 14 million different possible combinations, so if you buy 1 ticket you have covered 1 combination out of 14 million giving you a 1 in 14 million chance that your ticket has the winning combination.
If you buy 2 tickets you have covered 2 combinations out of 14 million giving you a 2 in 14 million chance of winning the jackpot.
2 in 14 million equates to 1 in 7 million
So by buying 2 tickets you have halved the odds.
Just like if you buy 14 tickets you have covered 14 combinations out of 14 million, giving you a 14 in 14million chance of winning.
14 in 14 million equates to 1 in 1 million.
Still massive odds (you wouldn’t back a horse if it was a million to 1(would you??))
I’ll try and explain how it works!!
If you enter a raffle that has 100 tickets (100 different combinations just like the lottery has 14 million combinations) and you buy 1 ticket you have a 1 in 100 chance of winning.
If you buy 2 tickets you have a 2 in 100 chance of winning which equates to 1 in 50
Let’s just say you have “number 1 ticket” and “number 2 ticket”
When the raffle is drawn the organiser announces that the winning raffle ticket is…………………….An odd number
So that rules out my chances of winning with my “number 2” ticket
But im still in with a chance of winning with my “number 1” ticket
And as there are only odd number winning combinations left (50 odd numbers between 1 and 100) I still have a 1 in 50 chance of winning the raffle.
This rule of thought can work for 10 tickets
I buy all the tickets that end in 5 ( 5,15,25,35,45,55,65,75,85,95)
The organiser announces that the winning raffle ticket is………………………..in the Fifties
As I have number 55 I still have a 1 in 10 chance of winning
If I’d bought numbers 1, 2,3,4,5,6,7,8,9,10
The organiser announces that the winning raffle ticket is………………………Ending in a 7
The are only 10 possible tickets that it can be
( 7,17,27,37,47,57,67,77,87,97)
I have a 1 in 10 chance that it will be me
This is exactly the same principle as the lottery but on a much larger scale
I know this is really old but I can’t help myself….
Just imagine there are 14 million playing cards face down, only one of which is the ace of spades, picking the ace wins you the jackpot. So you pay your £2 for a chance to choose a card..1/14 million we all agree, now if you pay for two chances to pick the card..you still have 14 million cards face down…in what world do 7 million of those cards disappear? you have 2 chances in 14 million to win but that does not reduce down to 1 in 7 million. You pick your first card and its not the ace……now on your second pick you have a 1/13,999,999 chance. I agree it’s the fact people don’t understand this is how lotteries make their money.
I’m not sure that Stephen’s playing card example isn’t a bit misleading though, as you’re re-assessing the odds during the draw. Which is also what the article does in fact, and not what happens in Lotto draws.
For instance, tweak the scenario a bit – you turn over the first card and it IS the ace of spades. The odds of you having the ace in your hand at the end of the draw have now increased to 14million/14million. The odds of the next card being the ace of spades have become zero though. So if you measure your odds halfway through a draw based on your initial bet they will naturally differ from the odds at the outset.
I think the point is that each Lotto draw is a distinct event. Prior to it, buying 2 unique tickets for the same draw will halve the odds.
Think about a simple scenario of 3 playing cards face down, J,Q,K. You buy 2 unique tickets, one for a Jack and one for a Queen. One of the 3 face down cards will be turned over, you need to match it to win.
Consider the Article’s working:
“…the odds of one ticket winning are 1 in 13,983,816. The odds that one of the two lottery tickets purchased are 1 in 13,983,816 minus 1 or 1 in 13,983,815!”
Transpose this working to our new playing card example:
“…the odds of one ticket winning are 1 in 3. The odds that one of the two lottery tickets purchased are 1 in 3 minus 1, or 1 in 2!”
Before the draw, would you agree with that assessment of your chances?
It’s a shame that some of the comments on here were so snide. Well done Derek for spotting the error.
Well done to Derek for challenging this mistake. The author really should withdraw this mistaken article. For all those who are still confused, let me try this simplification for you…
1. We agree that there are 13,983,816 possible combinations? Which for simplicity sake we shall call 14 million
2. So, we can agree that if I buy 14 million tickets and put a different combination on each ticket, I am guaranteed to have 1 ticket with the winning combination. So my chances of winning are 100%, which could also be called 14 million in 14 million, or 1 in 1.
3. If I only bought 7 million tickets, each with a unique combination on it, I would have a 50% (1 in 2) chance of winning agreed? Which could also be called 7 million in 14 million.
4. If I bought 2 tickets, with different numbers on them, I would have 2 chances in 14 million, or 1 chance in 7 million. This would be double the (very small) chance of 1 in 14 million that I had with 1 ticket.
Here’s another simple concrete illustration…
Let’s imagine a simplified lottery. There are 4 balls and you have to pick 2 of them.
Using the (correct) maths in this article, the possible combinations are (4 x 3) divided by (2 x 1) = 12 / 2 = 6. Which (for sake of completeness) are…
1,2
1,3
1,4
2,3
2,4
3,4
Now, if you buy 1 ticket you have a 1 in 6 chance of winning.
If you buy 6 tickets – with all 6 possible combinations – you are guaranteed to win with a chance of 6 in 6 One of your tickets MUST have the correct combination.
If you buy 2 tickets – *with 2 different combinations* (e.g. 1,2 and 1,3), you now have a 2 in 6 chance, or a 1 in 3. You can try this at home with a die. It also has 6 possible outcomes from which you can pick 2. Throw it 30 times. Each time it falls on a 1 or a 2 record a yes. Each time it doesn’t record a no. You will find (approimately) that 1/3 of the outcomes (10 of them) are a yes and 2/3 are a no.
Don’t make the mistake of thinking the 2 tickets are completely independent of one another – they are not – the numbers on the second ticket you picked are influenced by the first ticket – as you wont pick the same numbers a second time.
The mistake that many people make (including Derek, Stephen, John, Craig and Simon) is in believing that 2 chances in 14 million is the same as 1 chance in 7 million. It’s not. That’s just plain wrong maths.
There seem to be many fools here who are far too easily parted from their money.
Mike,
2/14 is the same as 1/7.
2/140 is the same as 1/70.
2/1,400 is the same as 1/700
2/14,000 is the same as 1/7,000
2/140,000 is the same as 1/70,000
2/1,400,000 is the same as 1/700,000
Therefore 2/14,000,000 is the same as 1/7,000,000.
If there were 14 potential winning combinations and you bought 1 ticket then there would be a 13/14 chance that you won’t win and a 1/14 chance that you will win.
If there were 14 potential winning combinations and you bought 2 different tickets then there would be a 12/14 chance that you won’t win and a 2/14 chance that you will win. 2/14 = 1/7 (see above).
If there were 14,000,000 potential winning combinations and you bought 1 ticket then there would be a 13,999,999/14,000,000 chance that you won’t win and a 1/14,000,000 chance that you will win.
If there were 14,000,000 potential winning combinations and you bought 2 different tickets then there would be a 13,999,998/14,000,000 chance that you won’t win and a 2/14,000,000 chance that you will win. 2/14,000,00 = 1/7,000,000 (see above).
Let’s take this further because I’m sure you still don’t get it:
If there were 14,000,000 potential winning combinations and you bought 32,926 different tickets (somehow) then there would be a 13,967,074/14,000,000 chance that you won’t win and a 32,926/14,000,000 chance that you will win. 32,926/14,000,00 = 1/425.2
If there were 14,000,000 potential winning combinations and you bought 459,777 different tickets (you store them in a warehouse) then there would be a 13,540,223/14,000,000 chance that you won’t win and a 459,777/14,000,000 chance that you will win. 459,777/14,000,00 = 1/30.4
If there were 14,000,000 potential winning combinations and you bought 2,298,302 different tickets (you hide them in an abandoned mineshaft) then there would be a 11,701,698/14,000,000 chance that you won’t win and a 2,298,302/14,000,000 chance that you will win. 2,298,302/14,000,000 = 1/6.1
If there were 14,000,000 potential winning combinations and you bought 7,000,000 different tickets (you hide them in 2 abandoned mineshafts) then there would be a 7,000,000/14,000,000 chance that you won’t win and a 7,000,000/14,000,000 chance that you will win. 7,000,000/14,000,000 = 1/2.
That means you would have to spend £7million just to get a coin-flip chance of winning the jackpot.
That is maths, my friend.
This is all very funny to read. You are wrong Derek. By your rational if you bought 20 tickets your odds of wining would be 1/13. If 13 of you bought 20 different tickets each you would not be guaranteed to win.
You are a fool and that is fact which will no doubt be disputed by you too.
Merry Christmas.
Clive, you are the fool.
Merry Christmas.
If you bought 20 different combinations out of a possible 14,000,000 then there would be a 20/14,000,000 of winning, which is actually 1 in 700,000 not 1/13. What are you talking about?
If you meant 20 different combinations out of a possible 14 then that is a mathematical impossibility. And you are still wrong.
Derek:
When you buy your second ticket, you double your chance of winning (to 2 in 14 million) but you don’t halve the odds against it happening.
Err, yes you have. Doubling the chance of winning = halving the odds of winning.
I’m not saying that the chances of winning are now a half; you’d have to buy half the possible winning tickets for that to be true.
If you buy 2 tickets you have doubled your chance of winning and halved your odds no matter what the starting number of possible winning combinations is.
Example 1:
If there is a lottery with only 10 winning combinations and you buy 1 ticket then your chances of winning are 1 in 10. You have a 10% chance of winning.
If, before the lottery is drawn, you buy another ticket (with a different combination) then you have doubled your chances because you now own 20% of the possible combinations.
The odds have been halved because 2 over 10 is the same as 1 over 5. This is how fractions work.
Example 2:
If there is a lottery with 100 winning combinations and you buy 1 ticket then your chances of winning are 1 in 100. You have a 1% chance of winning.
If, before the lottery is drawn, you buy another ticket (with a different combination) then you have doubled your chances because you now own 2% of the possible combinations.
The odds have been halved because 2 over 100 is the same as 1 over 50. Fractions, fractions, fractions.
Example 3:
If there is a lottery with 14,000,000 winning combinations and you buy 1 ticket then your chances of winning are 1 in 14,000,000. You have a 0.000007142857% chance of winning.
If, before the lottery is drawn, you buy another ticket (with a different combination) then you have doubled your chances because you now own 0.000014285714% of the possible combinations.
The odds have been halved because 2 over 14,000,000 is the same as 1 over 7,000,000.
Thanks.
Both schools of thought are correct here.
Derek is correct because he is discussing the odds of winning based on a single draw. where the chance of winning goes with the fractions
1 ticket is 1 in 14,000,000
2 tickets is 1 in 7,000,00
4 tickets is 1 in 3.500,000
8 tickets is 1 in 1.750.000
etc etc
BUT this is only based on buying more tickets for the SAME DRAW.
this is were the other people are correct…if someone buys 1 ticket each draw, then they are playing the 1 in 14,000,000 EVERY TIME,,
i haven’t done maths since 15 years ago but i believe this is where .. week on week of buying just 1 ticket your overall odds of getting a win are something like…
week 1 1/14.000.000
week 2 1/13.999.999
week 3 1/13.999.998
which is much worse odds than or am I wrong ??
I originally found this article because I want to know if it makes sense to buy more tickets at specific times than playing weekly- I presume buying more tickets at once makes more sense.
I say this with the utmost respect,, can a “nerd” help me out with this one please.
Hi Matt,
The article implies that buying 2 tickets for the same draw will not double your chances of winning. This is simply wrong and I have been trying to point that out. I am glad you understand.
You are right that if you buy 1 ticket each week then the odds would be 14,000,000 each time.
If, however, before a succession of lotteries, you committed to buying a ticket for each lottery, then yes, the odds do change ever so slightly – but not to the same extent as buying 2 tickets for the same lottery.
For example, if you flip a coin then the chances of it being heads is 50%. If you then flip the coin again then then the chances are still 50%.
However, before flipping the coin twice, the odds of flipping at least one heads is 75% because the possible outcomes are:
Heads/Heads – 25%
Heads/Tails – 25%
Tails/Heads – 25%
Tails/Tails – 25%
Is that sort of what you are talking about?
Thanks Derek,
I have done some guess work calculations based on simple maths I know to be true to come to this conclusion.(and Yes I do think in coin tosses;) >>>>> I will only play the national lottery a few times in my life and I will play a relatively large amount of tickets.
lets just say someone plays the lotto once per week for 19 and a half years… based on my rusty numbers they finally win the lottery on the last game, the odds of which are ~approximately
13,999,999/14,000,000 to the power 1999 x 1/13,999,999 = 0.00000007142
0.00000007142 + 198*0.00000007143 = .000017143 which is 1 in 70,000 chance of winning.
terrible odds.
now lets consider this same person was to buy all their tickets on the same day of a quadruple roll over, in which case, the second tier of winners share the jackpot if no one has all 6 numbers.
2000/14,000,000= 0.00014286 which is 1 in 7000 of 6 matching numbers and approximately 1 in 1,166 for getting 5 numbers plus the bonus ball. much better odds indeed, but still I wonder is the lotto really a potential investment ?
in an attempt to do a cost risk analysis which is half guess work to be honest but still…
taking the 1/1166 — 1166* £4000 = £4.664.000. meaning you need more than £4.6 million return to make economic sense
taking the 1/7000 (all 6 numbers) means you would need more than £28 million return to have a half chance of making economic sense.
overall I think the lottery is a good Charity and the rest of the money goes to the tax man,, not such a bad thing, far better than a casino in my opinion.
when the time is right I will buy as many tickets as I can afford to loose hoping that for every 2 pounds I spend statically I might get about 90p back which isn’t so bad. I was thinking about it, the people who play the lottery are helping other people’s dreams come true, which is nice, and maybe just maybe 🙂
You guys are so sick! You all are wasting your time. I am sure this article is not misleading.
Buying 2 tickets doubles your chance of winning. Anyone arguing otherwise has no grasp of probability or reality. The chance of winning is still very small, on the basis that two times a minute number is still a minute number
So if the first ball drawn is an even number, does this mean the odds of drawing an odd number next are increased, because there are more odds than even? And to take this further if the first 3 balls were all even, does this greately improve the chances of an odd ball being the next 3 drawn.
If so then either pick odd or even numbers in sequence to improve you chances of winning.
Whoever wrote this article, and whoever disagrees with Derek, basically doesn’t understand probability! I think it’s the big numbers that are make it difficult for some to understand.
If you break it down to a draw with 10 numbers, then clearly if you have 2 tickets, you have a 2 in 10 or 1 in 5 chance of winning. This principle doesn’t change just because the numbers are larger. Do people really not understand this!
Essentially, what the laws of probability say is that if you did 14,000,000 draws, you’d expect each combination to come out once. If you have two tickets, then you’d expect one of your tickets to come out in the first 7,000,000 draws.
By the way, Witsend – What are you talking about? To start with, of course if an even number comes out first, then as you look at the next ball being drawn, there’s a slightly higher chance that it’s odd. However, your logic about picking even or odds in sequence makes no sense. Ultimately, it doesn’t matter what you pick, you end up with a ticket that has one of the 14,000,000 combinations. You’ve as much chance of winning with 1,2,3,4,5,6 as anything else.
I am genuinely thankful to the holder of this web page who has shared this impressive article at at this time.
By Derek logic that means if u were to play 14 million tickets in lotto at 1 in 14million odds you are guaranteed the jackpot….
I wonder why people who have 28mill to spend don’t do this coz 1 ticket is £2 x 14milltickets is £28million but by his theory if the jackpot is 40 mill you just made like 70% profit. Repeat til billionaire right?
The odds of winning the lottery were recently changed when extra balls were added (roughly 45 million to 1).
The jackpot is very rarely, if ever, larger than the odds of winning. And even if it is, good luck buying 10,000 lottery tickets let alone 45 million.
Also, when someone wins, the prize money is heavily reduced in the next draw.
The maths is still the same though. Buy 2 tickets for 1 draw in the new format and the odds of winning will be 22.5 million to 1 (i.e. you’re still not going to win).
Derek mate I’m afraid your wrong here bro.
Based on your logic
1 Ticket = 1 in 14 Million
2 Tickets = 1 in 7 Million
3 Tickets = 1 in 3,500,000
4 Tickets = 1 in 1,750,000
5 Tickets = 1 in 875,000
6 Tickets = 1 in 437,500
..
..
17 Tickets = 1 in 106
18 Tickets = 1 in 53
19 Tickets = 1 in 26
20 Tickets = 1 in 13
21 Tickets = 1 in 6
22 Tickets = 1 in 3
23 Tickets = 1 in 1 (1.6689 to be exact)
So by your maths you only need 23 tickets to win the lottery. That’s clearly absurd.
My point is you don’t just remove 7 million possible number sequences (or 50% of the total) with each extra ticket. Use your logic here.
Think about it like this. The 14 million combinations are written on to playing cards and spread out over a (humongous) area. The winning combination is drawn and you are allowed to turn one card over per go to find the winner. If you decide to pay for a second go it is not logical to presume that 7 million cards will simply dissappear. It’s really as simple as that.
No, you are wrong. You’re halving the odds each time. That’s not correct.
1 Ticket = 1 in 14 Million
2 Tickets = 1 in 7 Million
3 Tickets = 1 in 4,666,667
4 Tickets = 1 in 3,500,000
5 Tickets = 1 in 2,800,000
6 Tickets = 1 in 2,333,333
…
23 Tickets = 1 in 608,695
Sorry.
7,000,000 Tickets = 1 in 2.
DO YOU GET IT YET?
Read my earlier posts.
Remember that one winning ticket doesn’t mean that there can be no other winning tickets. It’s not a raffle.
Buying 2 tickets with different numbers means that you have one ticket that def can’t win and one that can. So you have only taken one combination out of the game. Therefore doesnt that mean that your chances are now one in (all possible combinations) minus one. Ie one in 4499999 rather than in in 45m?
There are a few insulting comments on here but i find it a very interesting debate. I also wish That I had stuck in a t maths now!
What I’m saying is that I don’t agree with Derek’s analysis.
Been thinking more lol.
Derek, suppose using the old number of balls, that there are 14m combinations. (There are now 45 million). Anyway, suppose that you buy every combination except 1. Using spreadsheet I get, using your method, a winning chance of 1 in 1.0000000714. But since only 1 combination can win, the reality is that there are 2 combinations left and you can only have one of them. I think that actually leaves you with a 1 in 2 chance. This works all the way back to show that buying 2 tickets only reduces the odds by 1 in 14 million.
The real message is that it’s a lousy bet to buy lottery tickets and its a even more of a con if people think that their is any big change in winning chances by buying multiple tickets.
Willie,
The fact that someone could buy the same ticket combination as you is irrelevant. It would simply reduce the prize money.
With regards to the buying-all-but-1 combination scenario (13,999,999 tickets in the old system) you would have a 1 in 1.000000007143 chance in winning.
Look, I don’t mean to be rude, but it’s simple maths.
If there were only 10 possible combinations and you bought 9 then there is a 9 in 10 chance you are going to win. It’s not a 50/50 chance. It would be a 50/50 chance if you bought half the combinations (i.e. 5)
Similarly if there were 14,000,000 combinations and you bought 13,999,999 then there is a 13,999,999 in 14,000,000 you are going to win. Or 1 in 1.000000007143. It’s almost certain you will win. There is a 1 in 14 million chance you won’t win as there is only 1 possible winning combination that you haven’t bought.
Derek
I’m afraid that I don’t think that your maths are correct. Assuming different numbers on each, if you bought 9 out of ten possible tickets you still can only have one that is the winner and eight that definitely can’t. The person who owns the tenth has a ticket that could be a winner.
All you have done is remove 8 tickets that can’t win.
If you bought half of the 14 million there are 7 million and 1 tickets that could win so your chances are 1 in 7,000,001.
Best wishes
P’s identical tickets don’t reduce the prize money, it just means it is shared if that’s the lucky combination. It also means that anyone buying all of the possible combinations can’t guarantee they get all of the prize money (and it is setup so that the prize money varies with entries anyway) I only mentioned it cos it was a previous query btw
Willie,
Imagine there are 10 combinations (I’m not drawing 14 million):
[ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] [ 6 ] [ 7 ] [ 8 ] [ 9 ] [ 10 ]
In the draw, 1 of these combinations will be drawn.
Imagine, you buy 1 of these combinations (e.g. the 6th combination, marked with x):
[ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] [x6x] [ 7 ] [ 8 ] [ 9 ] [ 10 ]
When the lottery is drawn you have a 1 in 10 chance of winning (The fact that you have bought the 6th combination does not affect the fact that there are still 10 potential combinations).
Possible combinations:
[ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] [ 6 ] [ 7 ] [ 8 ] [ 9 ] [ 10 ]
Combinations you own (marked with x):
[ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] [x6x] [ 7 ] [ 8 ] [ 9 ] [ 10 ]
It’s obvious that there is a greater chance that a combination you don’t own will be drawn out. There is a 9/10 chance you will lose.
Now…
Imagine, you buy 9 of these combinations (e.g. all but the 8th combination, marked with x):
[x1x] [x2x] [x3x] [x4x] [x5x] [x6x] [x7x] [ 8 ] [x9x] [x10x]
When the lottery is drawn you have a 9 in 10 chance of winning (The fact that you have bought all but the 8th combination does not affect the fact that there are still 10 potential combinations).
Possible combinations:
[ 1 ] [ 2 ] [ 3 ] [ 4 ] [ 5 ] [ 6 ] [ 7 ] [ 8 ] [ 9 ] [ 10 ]
Combinations you own (marked with x):
[x1x] [x2x] [x3x] [x4x] [x5x] [x6x] [x7x] [ 8 ] [x9x] [x10x]
You seem to think that the 8th combination now has a 50/50 chance of winning. Why?
It doesn’t. There is a 1 in 10 chance that the 8th combination will be drawn. There is a 9 in 10 chance that any of other combinations will be drawn. You own the other combinations. You have a 9 in 10 chance of winning.
Do you understand?
This is great, I’d still be confused if it weren’t for you guys.
I think the confusion is identifying the difference between number of combinations left and the odds before the initial drawing and after a drawing.
Please can one submit this to a professor of Maths at Oxford to quash the issue?
Derek please just take note from those who oppose your view and accept that maths can be used to explain the most challenging problems and theories that man has and ever will have to endeavour to explain and deduce. With that in mind you can not just to rely on basic schoolboy arithmetic and what we perceive as logic.
You Jan 9th post stated that:
1 ticket. = 1 in 14,000,000 chance
2 tickets = 1 in 7,000,000 chance
By buying two tickets you have immediately halved your odds, apparently wiping out 7 million chances of losing in to thin air. The logic here is that this should then continue in the same basic format. So surely buying 3 tickets should be even more advantages. However by your calculations buying 3 tickets =
1 in 4,666,667
Hang on! Why are my chances diminishing the more tickets I buy?
4 tickets = 1 in 3,500,000 chance. I’ve just
Lost little over a million more chances to lose.
How come I saw off 7 million by buying 1 ticket?
There are not 14 million tickets out there that is how many variations there are. Therefore buy two tickets each ticket still has 1 chance in whatever remains.
The playing card analogy is a good one. If there are 14 million cards including just one ace the chance of finding that ace is 1 in 14 million. You now have the chance to buy any number of the face down cards in the hope you will find the ace. Buy 1 card = 1 in 14 million chance of turning the correct card over. Buy two cards there are still 13,999,999 cards. Another simple example is the lotto has 1-49 numbers and only one number is drawn, that is the winning number. Like the normal lotto you can have 1 go, 2 goes or as many as you like. Buy one go and you have a 1 in 49 chance. What happens if you now buy another ticket with a different number do you have a 1 in 24.5 chance NO!
“With that in mind you can not just to rely on basic schoolboy arithmetic and what we perceive as logic.”
But it is schoolboy arithmetic.
I cannot be bothered to explain it to you anymore.
http://jlotterysbimonthlyarticles.blogspot.co.uk/2008/04/how-lottery-odds-change-when-buying.html
Read the above article. It will explain it to you much better than I can.
omg you started this argument back in 2011, up to 2016.
That’s 5 years : Does anyone know the answer after 5 years, because I need it for my college work.
I sent an email to Camelot
“Hi
I am trying to make and informed choice on how to spend my monthly budget for playing games . If I buy on lotto ticket the odds of winning are 1 in 45 million if I buy two tickets are the odds 1 in 22.5 million and if I by three are the odds 1 in 11.25 million and so on.
Thanks for your help Brian”
After pestering them for a reply this is what I got back
Dear Sir/Madam
Thank you for your email dated 29 March 2016 regarding Lotto odds
Please accept my apologies for the delay in my response.
I can advise that if you buy 1 Lotto ticket you would have a 1 in 45,057,474 chance of winning the jackpot. However, if you bought 2 Lotto tickets you would have a 2 in 45,057,474 chance of winning the jackpot and a 3 in 45,057,474 chance of winning if you bought 3 Lotto tickets.
I hope the above has been helpful but if you have any further queries, please do contact us again.
Yours faithfully
Melanie Garwood
National Lottery Customer Care Team
Yes! Camelot are correct. By the way a 3 in 45,057,474 chance is the same as 1 in 15,019,158 not 1 in 11.25 million.
Some times its a pain in the ass to read what blog owners wrote but this website is very user pleasant!