Probability of winning the irish lottery

Michael Collins

Not sure I'd say the letter is too far off. How many times would you expect to play the lotto before you win it, on average?

Let me put this in terms of throwing a die, to make the numbers more manageable:

How many times would you expect to have to roll a 6-sided die before you get a 2, say? I'd say 6 times is a fair enough bet. That is to say the statistically expected number of times you have to roll your die before you get a 2 is actually 6.

This means, if you keep trying this experiment, the average number of throws before a 2 shows up will approach 6 as the number of trials becomes large.

I don't think it's too unacceptable to use the term 'statisically' to mean the expected value in this context.

Of course, if you want to be 95% sure of getting a 2 in a specific trial, you will need more throws.

about 3 fity

locum-motion wrote: »

Something that annoys me:
Ads for the National Lottery's 'Millionaire Raffle' are currently stating something like

"You're six times more likely to win a million than be attacked by a bear"

I think that this 'bears' (pun intended) some closer scrutiny.

There were 180,000 tickets for the Easter Millionaire Raffle, and one Millionaire winner drawn out. So, your chance of winning is 1 in 180,000. (or 179,999 to 1 against)

If this is "six times more likely" than being attacked by a bear, then your chances of being attacked by a bear must be 1 in 1,080,000. (or 1,079,999 to 1 against)

But they hold these raffles about 4 times a year, so to truly compare like with like, then we should say that - in any given year - you have a 4 in 1,080,000 chance of being attacked by a bear. That's the same as 1 in 270,000. (or 269,999 to 1 against)

All of these raffles are held in the Republic of Ireland. The population of the Republic of Ireland is 4,588,252.

Therefore, if the claim made in the ad is true, then 17 people have been attacked by bears in Ireland in the last year.

Isn't it amazing how none of those 17 attacks made it into the newspapers?

simply swap the word "bear" with "elephant" and it all makes sense

ArmCandyBaby

Michael Collins wrote: »

Not sure I'd say the letter is too far off. How many times would you expect to play the lotto before you win it, on average?

Let me put this in terms of throwing a die, to make the numbers more manageable:

How many times would you expect to have to roll a 6-sided die before you get a 2, say? I'd say 6 times is a fair enough bet. That is to say the statistically expected number of times you have to roll your die before you get a 2 is actually 6.

This means, if you keep trying this experiment, the average number of throws before a 2 shows up will approach 6 as the number of trials becomes large.

I don't think it's too unacceptable to use the term 'statisically' to mean the expected value in this context.

Of course, if you want to be 95% sure of getting a 2 in a specific trial, you will need more throws.

Then what you are looking for is the point where the probability of success equals 0.5 which I guess for your dice example is 3.5 throws. If you want to give me 6 throws to roll a 2 then I'll happily take the under all day long! If you want to calculate the average length of time someone will go until they have one jackpot you can use the negative binomial distribution with the probability = 0.5. I might have a look at it later but it'll be far less than the figure in the letter.

EDIT: Actually the probability of rolling a 2 in the first 4 throws is 0.518 so the expected number of throws is just less than that.

MathsManiac

ArmCandyBaby, you are choosing to interpret the letter in a particular way that is not justified. The letter does not, to my mind, incorporate any assertion of the kind you claim it does. (i.e., that the probability of winning at least once in the specified time is 0.5, or 1, or any other number.)

I tend to agree with Michael Collins. The reality is that, if you play the lottery with the same six numbers for the specified number of times, the expected value of the number of Jackpot wins is indeed 1. (This is simply an example of the fact that the expected value of a binomial distribution is Np.)

In writing in plain language for a lay audience, it is quite reasonable for the letter writer to descibe this in the way she did (although one might quibble over using "should win", rather than, say, "would expect to win").

ArmCandyBaby

MathsManiac wrote: »

ArmCandyBaby, you are choosing to interpret the letter in a particular way that is not justified. The letter does not, to my mind, incorporate any assertion of the kind you claim it does. (i.e., that the probability of winning at least once in the specified time is 0.5, or 1, or any other number.)

I tend to agree with Michael Collins. The reality is that, if you play the lottery with the same six numbers for the specified number of times, the expected value of the number of Jackpot wins is indeed 1. (This is simply an example of the fact that the expected value of a binomial distribution is Np.)

In writing in plain language for a lay audience, it is quite reasonable for the letter writer to descibe this in the way she did (although one might quibble over using "should win", rather than, say, "would expect to win").

The expected value of 1 comes from:

0.368 ie. the probability of 1 win times the value of 1 win, plus
0.184 ie. the probability of 2 wins times the value of 2 wins, plus
0.061 ie. the probability of 3 wins times the value of 3 wins, plus etc.

and this calculation has no relevance to what she's trying to say in the letter. I'm happy to interpret her sentance as the expected number of times the average person has to play before they win the jackpot but in that case you are using the negative binomial distribution like I said above.

MathsManiac

ArmCandyBaby wrote: »

The expected value of 1 comes from:

0.368 ie. the probability of 1 win times the value of 1 win, plus
0.184 ie. the probability of 2 wins times the value of 2 wins, plus
0.061 ie. the probability of 3 wins times the value of 3 wins, plus etc.

and this calculation has no relevance to what she's trying to say in the letter. I'm happy to interpret her sentance as the expected number of times the average person has to play before they win the jackpot but in that case you are using the negative binomial distribution like I said above.

But you're still interpreting the letter in a very bizarre way. Why would you think that she is saying anything about the expected number of plays to the first win, when what she has actually said is what you should expect to happen in 8,145,060 plays of the game? Taking the rolling of a die as an equivalent example, and taking "your favourite number" to correspond to your chosen set of lotto numbers, the following are all saying the same thing. The last of them is perhaps a little sloppier than the others, but it's hard to justify asserting that it's wrong:

• "If you roll a die many times, you should expect your favourite number to come up about one-sixth of the time."
• "If you roll a die six times, the expected value for the number of times your favourite number comes up is 1."
• "Statistically, if you roll a die six times, you should get your favourite number once."

People often use "statistically" in this way, to capture the idea that they are referring what is expected to happen in the long-run average.

Brussels Sprout

MathsManiac wrote: »

But you're still interpreting the letter in a very bizarre way. Why would you think that she is saying anything about the expected number of plays to the first win, when what she has actually said is what you should expect to happen in 8,145,060 plays of the game?

+1 to this

There's a famous question that is known to come up in technical job interviews that asks "On average how many times would you have to randomly flip open the Manhattan telephone book (500 pages) to find a specific name?".

Now in this particular case it would be correct to use ArmCandyBaby's method of working out the probability of finding the probability of finding the name in a particular number of flips and then returning as the answer the number that yields a probability of 50% (or whatever confidence level you think is most appropriate).

That letter though would be the same as saying "If you randomly flipped open the Manhattan phone book 500 times, statistically how many times would you flip it open to a particular name?" And the answer to that is 1.

ArmCandyBaby

If you are using the binomial distribution like this then you are already having to make the assumption that they are playing the lotto the following week whether they win this week or not which is a fairly incredulous and unnecessary assumption to make. Fair enough maybe I misinterpreted the letter on first reading it but I do regard using the expected value here as being completely irrelevant to the point she's making and of having very little descriptive value (it doesn't address the point she's making in the first paragraph which is a man expecting to win 1 jackpot). I'm not expecting huge sums in a letters page or using confidence intervals or anything like that but I just believe her point could have been explained 100 times better in the second paragraph.

locum-motion

Brussels Sprout wrote: »

...
That letter though would be the same as saying "If you randomly flipped open the Manhattan phone book 500 times, statistically how many times would you flip it open to a particular name?" And the answer to that is 1.

Surely that would depend on the name. If there were 5 pages of Smiths, for example, then you'd open the book at Smith 5 times, wouldn't you?*

* assuming all the usual stuff like the process of choosing the page is truly random so that page 1 and page 499 have exactly the same chance of being picked as page 250 and yada yada yada

locum-motion

Michael Collins wrote: »

...How many times would you expect to have to roll a 6-sided die before you get a 2, say? I'd say 6 times is a fair enough bet. That is to say the statistically expected number of times you have to roll your die before you get a 2 is actually 6.

This means, if you keep trying this experiment, the average number of throws before a 2 shows up will approach 6 as the number of trials becomes large..

I'm no mathematician (I only did pass @ Leaving Cert, and that was >20 yrs ago) but it seems to me that that isn't right. Forgive me if I don't use the correct mathematical expressions; I'm going to try to explain what I mean in english!

Approx 1/6 of the time, you'll get a 2 on the first throw. This can also be expressed as 60/360. In 300 out of 360 trials, you'll progress onto a second throw.

Approx 1/6 of these 300 trials, you'll get a 2 on the 2nd throw. This is 50/360. In 250 out of 360 trials, you'll progress onto a third throw.

In 1/6 of these, you'll get a 2. That's 42/360. In 208 trials, you'll proceed to a fourth throw.

In 1/6 of these, you'll throw a 2. That's 35/360. 173 times, you'll progress yada yada yada.

Isn't the average of these going to tend toward 3 or 3.5 or something?





In fact, now that I've asked the question, I'm going to try to work it out. I'll probably prove myself wrong in a minute. So far, I'm typing as I think, so bear with me!

So, we're now progressing to a fifth throw, 173 times. 1/6 of these will be a 2. That's 29. 144 left.

We'll get a 2 on 24 of those 144, 120 left.
We'll get a 2 on 20 of those 120, 100 left.
We'll get a 2 on 17 of those 100, 83 left.
We'll get a 2 on 14 of those 83, 69 left.
We'll get a 2 on 12 of those 69, 57 left.
We'll get a 2 on 10 of those 57, 47 left.
We'll get a 2 on 8 of those 47, 39 left.
We'll get a 2 on 7 of those 39, 32 left.

yada
yada
yada


Taking a few iterations of these, I'm going to try to calc the average.

(1x60)+(2x50) = 1.4545
60+50

(1x60)+(2x50)+(3x42)+(4x35) = 426/187 = 2.278
60+50+42+35

(1x60)+(2x50)+(3x42)+(4x35)+(5x29)+(6x24) = 715/240 = 2.979
60+50+42+35+29+24

(1x60)+(2x50)+(3x42)+(4x35)+(5x29)+(6x24)+(7X20)+(8x17)
60+50+42+35+29+24+20+17

= 991/277 = 3.5776

(1x60)+(2x50)+(3x42)+(4x35)+(5x29)+(6x24)+(7X20)+(8x17)+(9x14)+(10x12)
60+50+42+35+29+24+20+17+14+12

= 1237/303 = 4.08

OK, so I've proven myself wrong already, because I though it was going to tend to about 3.5 and already it's gone past that.

So, if I were to continue, would it tend to 6? Or have I completely misunderstood it?

Brussels Sprout

locum-motion wrote: »

Surely that would depend on the name. If there were 5 pages of Smiths, for example, then you'd open the book at Smith 5 times, wouldn't you?*

* assuming all the usual stuff like the process of choosing the page is truly random so that page 1 and page 499 have exactly the same chance of being picked as page 250 and yada yada yada

The "particular name" bit implies that you're looking for an individual
e.g. James P. Smith

locum-motion

Brussels Sprout wrote: »

The "particular name" bit implies that you're looking for an individual
e.g. James P. Smith

So, the question should say "particular person" or "particular entry" then, shouldn't it?

Stripey Cat

locum-motion wrote: »

Therefore, if the claim made in the ad is true, then 17 people have been attacked by bears in Ireland in the last year.

Isn't it amazing how none of those 17 attacks made it into the newspapers?

I don't think are any bears in Ireland, are they? Certainly no wild ones, but as far as I know, there aren't even any in captivity.

So your chance of being attacked by a bear , in this country, is incredibly slight.

You could, I suppose, import a bear just to be attacked by it...

Brussels Sprout

Michael Collins wrote: »

How many times would you expect to have to roll a 6-sided die before you get a 2, say? I'd say 6 times is a fair enough bet. That is to say the statistically expected number of times you have to roll your die before you get a 2 is actually 6.

If you were to say: "If we roll a die 6 times, how many 2's will we roll on average?" then 1 will be the answer. (this can be proven readily enough)

However this does not mean that the answer to "How many times would you expect to have to roll a 6-sided die before you get a 2" is 6.

To do the calc we need to find out the probability of it not happening and subtract this from one. So:

Prob of rolling a 2 in 1 roll = 1 - Prob of not rolling a 2 in 1 roll = 1 - 5/6 = 0.166

Prob of rolling a 2 in 2 rolls = 1 - Prob of not rolling a 2 in 2 rolls = 1 - (5/6)^2 = 0.306

Prob of rolling a 2 in 3 rolls = 1 - Prob of not rolling a 2 in 3 rolls = 1 - (5/6)^3 = 0.421

Prob of rolling a 2 in 4 rolls = 1 - Prob of not rolling a 2 in 4 rolls = 1 - (5/6)^4 = 0.518

Prob of rolling a 2 in 5 rolls = 1 - Prob of not rolling a 2 in 5 rolls = 1 - (5/6)^5 = 0.598

Prob of rolling a 2 in 6 rolls = 1 - Prob of not rolling a 2 in 6 rolls = 1 - (5/6)^6 = 0.665


What this all means is that after 6 rolls we'll have a 66.5% chance of rolling at least one two.

If we say that 50% chance represents an "on average" value then working backwards that lies somewhere between 3 rolls and 4 rolls. Using a linear interpolation we could say that it'll take us 3.8 rolls on average to roll our first 2.

MathsManiac

Michael Collins wrote: »

How many times would you expect to have to roll a 6-sided die before you get a 2, say? I'd say 6 times is a fair enough bet. That is to say the statistically expected number of times you have to roll your die before you get a 2 is actually 6.

Not sure if this was intuition or a statement of fact, but you are correct. If the random variable X is the number of trials required in a Bernoulli process to achieve the first success (supported on {1, 2, 3, ...}), then it has a geometric distribution, which can be considered as a special case of the negative binomial.

Anyway, E(X) for a geometric distribution is 1/p. So you are correct to state that the expected value of the number of throws required to achieve your first 2 is 6.

Bringing this back to the letter that prompted the discussion, this means that even if you did interpret the letter as referring to the expected number of lotto plays to the first jackpot (as opposed to the expected number of jackpot wins in a given number of plays), the assertion is still correct. That is, the expected number of plays to the first jackpot win is indeed 8,145,060.

(Of course, in all of this, we're ignoring the fact that the minimum allowable play involves two panels of 6 numbers! I guess for the sake of this exercise, you could put the same six numbers on both panels - but then you'd have to share the prize with yourself!)

Brussels Sprout

That result I calculated seemed so unintuitive that I actually got a die out and started rolling. From my crude trials it does actually appear to that average number of throws needed before getting a particular number for the first time is 6. Not sure what error I made.

On a side not I rolled 43 times at first without hitting the number I was going for. Would I be right in saying that the chances of this happening are over 2,500/1 against?

Colonel Sanders

Brussels Sprout wrote: »

That result I calculated seemed so unintuitive that I actually got a die out and started rolling. From my crude trials it does actually appear to that average number of throws needed before getting a particular number for the first time is 6. Not sure what error I made.

On a side not I rolled 43 times at first without hitting the number I was going for. Would I be right in saying that the chances of this happening are over 2,500/1 against?

Assume a fair die. Probability of not getting your choosen number is 5/6

Assuming each throw is independent (which I think is reasonable) the probability of not getting your number 43 times in a row would be (5/6)^43 which I make to be 2,539/1 to the nearest whole number

kingsenny

Sorry to bump an old topic but it was quite an interesting read.

Anybody know the odds for the Euromillions w/ Plus?

locum-motion

kingsenny wrote: »

Sorry to bump an old topic but it was quite an interesting read.

Anybody know the odds for the Euromillions w/ Plus?

I believe the technical term for it is "astrofúckingnomical"!:D

locum-motion

kingsenny wrote: »

Sorry to bump an old topic but it was quite an interesting read.

Anybody know the odds for the Euromillions w/ Plus?

The 'with Plus' part of your Euromillions ticket doesn't actually affect your chances of scooping the EM jackpot, I think. It's just that you've spent an extra €1 to enter that same set of numbers into an additional, separate draw.
Correct me if I'm wrong.

Yakuza

locum-motion wrote: »

The 'with Plus' part of your Euromillions ticket doesn't actually affect your chances of scooping the EM jackpot, I think. It's just that you've spent an extra €1 to enter that same set of numbers into an additional, separate draw.
Correct me if I'm wrong.

That's correct.
The odds of winning are [latex]{{50}\choose{5}} \times {{11}\choose{2}} [/latex] or about 116 million to one. Or as locum-motion says, astrofarkingnomical.

kingsenny

Furrrck...

Another question. Is it better to keep the same numbers every time or do a quick pick? I know most people would say both options have an equal chance...

BUT... I read that picking the same numbers every time has a higher chance of success. It was likened to being lost in a supermarket and whether you should stand still or find the person. Apparently, one should stand still and the other should move around. I don't think I've explained this analogy correctly but yeah...

locum-motion

kingsenny wrote: »

... Is it better to keep the same numbers every time or do a quick pick? I know most people would say both options have an equal chance...

Absolutely the same.

Simplify the lottery game as an example. Let's just say you only have to pick one number out of six possibilities. And lets just say that instead of the machine with the balls, you have an ordinary 6 sided die which 'chooses' the winning number. And just for fun, let's pretend that it's red.

What is quickpick? Well, it's just another random number generator that chooses the same number of numbers out of the same number of possibilities. Or, in this case, another die. But let's pretend this one's white.

So, to play quickpick, you roll your white die and write down the number. Then you roll the red die, and if it comes up with the same number, you win. There are 36 possible combinations:
White|Red|Win or Lose
1|1|W
1|2|L
1|3|L
1|4|L
1|5|L
1|6|L
2|1|L
2|2|W
2|3|L
2|4|L
2|5|L
2|6|L
3|1|L
3|2|L
3|3|W
3|4|L
3|5|L
3|6|L
4|1|L
4|2|L
4|3|L
4|4|W
4|5|L
4|6|L
5|1|L
5|2|L
5|3|L
5|4|L
5|5|W
5|6|L
6|1|L
6|2|L
6|3|L
6|4|L
6|5|L
6|6|W

By using a random number generator to choose you number, you win on 6 times out of every 36 on average, or 1 in 6.

If you choose one number, say 4, and play that same number every time, then it will on average come up once out of every six throws.

Therfore, you have exactly the same chance of winning whether you use a quickpick each time or the same number each time.

Yakuza

The above is correct, but you need to take game theory into account; certain sets of numbers will be more likely to have been picked by others (for example 1,2,3,4,5,6 or the numbers from the TV show Lost), so your chances of having to share the prize are higher with them. I generally tend to play number sets with at least one number > 31 (avoiding those who use dates in their sets). Quick Picks are guaranteed unique for a draw (at least when picked as a quick pick ).

I don't play that often, but when I do, my strategy is to try to minimise the chances of having to share. When I win, I want to win big

locum-motion

Yakuza wrote: »

The above is correct, but you need to take game theory into account; certain sets of numbers will be more likely to have been picked by others (for example 1,2,3,4,5,6 or the numbers from the TV show Lost), so your chances of having to share the prize are higher with them. I generally tend to play number sets with at least one number > 31 (avoiding those who use dates in their sets). Quick Picks are guaranteed unique for a draw (at least when picked as a quick pick ).

I don't play that often, but when I do, my strategy is to try to minimise the chances of having to share. When I win, I want to win big

I was gonna make that point, but it was getting very late!

filbert the fox

kingsenny wrote: »

Sorry to bump an old topic but it was quite an interesting read.

Anybody know the odds for the Euromillions w/ Plus?

Just picked up the how to play the Lotto booklet this morning. Forgive me if this has come up before but on p.14 they give the chances of winning each of the games. Euromillions has two columns one for the main draw and one for the plus. Here's the mystery:

Chances of matching 5 numbers in main draw: 1 in 3,236,994
Chances of matching 5 numbers in plus draw: 1 in 2,118,760

Now I could be missing something but, intuitively, they should both be the same. Two drums with 50 balls each from which you've attempted to match 5. How can the chances be different when there's a separate drum for Lucky stars?

28064212

filbert the fox wrote: »

Just picked up the how to play the Lotto booklet this morning. Forgive me if this has come up before but on p.14 they give the chances of winning each of the games. Euromillions has two columns one for the main draw and one for the plus. Here's the mystery:

Chances of matching 5 numbers in main draw: 1 in 3,236,994
Chances of matching 5 numbers in plus draw: 1 in 2,118,760

Now I could be missing something but, intuitively, they should both be the same. Two drums with 50 balls each from which you've attempted to match 5. How can the chances be different when there's a separate drum for Lucky stars?

Your mistake is here:
Chances of matching 5 numbers in main draw: 1 in 3,236,994
That's not what the odds are saying. It's the odds of matching 5 numbers and not matching either of the lucky numbers

filbert the fox

trying to follow the logic of this one.

Does this mean that you're less likely to select 5 correct in the main draw than in the plus draw just because there's a possibility of interference by one of those pesky lucky stars. ?

28064212

filbert the fox wrote: »

trying to follow the logic of this one.

Does this mean that you're less likely to select 5 correct in the main draw than in the plus draw just because there's a possibility of interference by one of those pesky lucky stars. ?

No, you have the exact same probability of having 5 correct in either draw. But the listed probability for standard euromillions is the probability of having 5 correct and no lucky numbers correct.

Using 3 correct numbers as an example (smaller numbers):

• Euromillions Plus: 3 correct numbers: 1/214 or 0.00467289719
• Euromillions:
  • 3 correct, and no lucky numbers: 1/327 or 0.00305810397
  • 3 correct, and 1 lucky numbers: 1/654 or 0.00152905198
  • 3 correct, and no lucky numbers: 1/11,771 or 0.00008495454

To get the probability of 3 correct in the Euromillions draw, regardless of the lucky numbers, you have to sum the probabilities of the three instances where 3 numbers are correct: 1/327 + 1/654 + 1/11771 = 0.00305810397 + 0.00152905198 + 0.00008495454 = 0.00467211049, which is (more or less, allowing for rounding errors) the same probability as listed for 3 correct numbers in the Plus draw

Yakuza

filbert the fox wrote: »

trying to follow the logic of this one.

Does this mean that you're less likely to select 5 correct in the main draw than in the plus draw just because there's a possibility of interference by one of those pesky lucky stars. ?

The leaflet states the odds of matching 5 numbers exactly. In the plus draw, there are only 5 numbers (and my post above is incorrect - I thought (like in the Irish Lotto) that a Plus draw meant two extra full draws (including bonus numbers), but it's only 5 numbers and no bonus numbers)
So, the chances of getting 5 numbers exactly in both draws are different, as in the normal Euromillions draw, you choose 5 regular numbers and 2 bonus numbers, whereas in the bonus draw you're only choosing 5 numbers.

Matching 5 numbers exactly in both draws are two different events.
One is matching the 5 numbers in the main draw AND not matching any bonus numbers, whereas the other is only matching 5 numbers.

As above, the odds of matching 5 numbers from 50 is one in [latex]{{50}\choose{5}} } [/latex], or [latex]\frac{1}{2118760} [/latex] and the odds of not matching 2 from 11 is [latex]\frac{{{11}\choose{2}}-2 \times 9 -1}{{{11}\choose{2}}} [/latex] or [latex]\frac{36}{55} [/latex], so the odds matching exactly 5 numbers (and no lucky star numbers) in the main draw is [latex]\frac{1}{2118760} \times \frac{36}{55}[/latex], or approx 3236994.4 : 1