Probability of winning the irish lottery

in laymans terms (being a non mathematician mesel') what this means is that it appears harder to avoid getting a lucky star when one matches five in the main draw than simply matching five in the plus draw.

by the way, despite the astronomical odds, can't resist the temptation to play as jackpot heading for €90,000,000 !

and zeroish by playing

I'm afraid that it's not as simple as that. The four numbers you select must all come out whereas with lotto it's 4 from 6.
Otherwise we'd all be down the bookies gettin' €6,000 for four and there'd be no National Lottery at all!:eek:

Have to ask, consider the odds are massive, and put that to one side for a moment. Using a distribution curve, obviously normality/non-normality would have to be determined first, could the bell curve give and reasonable ideas of probabilities on sequences? For example, the numbers are 1 to 49 (I think), but if say the most recurring combinations with respect to time/number of events add up to say 120 [20 + 32 + 38.....] kind of a median so to speak... could one determine a reasonable sequence? Now it would be crazy to assume 6 from the 49, but perhaps three, and using the bookie [the bookie gives better odds on smaller sequences than the national lotto. Head scratching.....

28064212 wrote: »

Simple answer: no. The probability of 1-2-3 coming out will always be the same as the probability of 47-48-49 coming out. Analysing past draws will never, ever, ever change that

I appreciate that and agree completely. What I am wondering is if patterns could be of use? The bell curve is an extremely powerful tool, used extensively in industry to estimate error, reliability etc.

You could make a good prediction for the sum of the lotto numbers but it wouldn't increase your chances of winning. There are far more combinations which would sum to 120 than there are that'd sum to 30.

sham69 wrote: »

Sorry to resurrect this.

Has there been any online winners of large amounts in the lotto?

Reason I ask is, I always do mine online but havent heard of any winners that werent associated with shops in particular areas?

Maybe i am being paranoid but have you a better chance doing it in a shop or is it exactly the same?
Cheers.

Exactly the same. Far less people do it online, so it's much less likely to have a winner from online. Also, even if someone does win online, they're likely to just refer to them as a "winner from ...". When it's won in a shop, the shop is going to actively look for their name to be associated with the winner

I think that the statistical experience of large lotteries is something like this -- when they sell about as many tickets as the number of combinations, then about two-thirds of the time there will be one winner, and about half of those will yield two winners, three would be unusual.

When ticket sales are larger, let's say four times the number of random combinations, then the chance of at least one winner is way up to close to 90% and there is a much greater chance of two winners or more. At some point, the chance of 2,3 or 4 winners would probably exceed the chance of only one winner. So given that the number of sales often rises with a larger jackpot in these lotteries, an outcome like six winners is not that anomalous. The percentage chance of it, exactly, would always be small. But there might be some number of sales, probably in the order of ten times random combinations, where six might be near the top of the bell curve.

You can calculate these things if you have a lot of time and a powerful computer, the formula for probability of having a winner or excluding cases of no winner is basically this:

1 - [(x-1)/x]^n

where x is the number of combinations in the winning lottery ticket and n is the number sold.

Note that when the number sold is equal to combinations, then x equals n.

For increasing values of n (combinations), this case of sales equalling combinations starts at 0.75 for 2 (the chances of throwing a head or a tail in the first two tries) and slowly decreases towards a presumed limit around 0.6, the values being 0.703 for three, 0.684 for four, .670 for five, .665 for six, and .651 for ten. By one hundred combinations, it only drops a bit further to around .634.

In other words, when you get up into the size of a national lottery of any kind, with millions of combinations, and sales equal the number of combinations, then there will be better than a 50-50 chance of a winner. Now if you double that amount of sales, there will be a squared value of the non-winner fraction (the minus part of the equation) so if that was 0.4, it ends up at 0.16 giving us an expectation that 84% of the time, a double-sale of combinations will produce at least one winner.

So what's the math for a second winner? That is even more complicated and it's not as simple as saying 60% of the 60% or 80% of the 80%, that would only be the case if the excluded first winner was always the first case selected, and there's a random chance of it sitting anywhere from first to last. My impression from a rather casual inspection of actual results of our own similar lottery here, is that when sales reach about three or four times the number of combinations, you are past the 50-50 point for a second winning ticket.

I would say the result of six winning tickets not being extraordinary is about equivalent to selling ten to fifteen times the combinations, just rough numbers, don't have the computer capacity to test it out really. When you get up into large numbers of tickets sold in relation to combinations, the distribution of winners (1, 2, 3, 4 ... etc) begins to look like a very flat bell curve with some chance of a number greater than 1 or even 2 having the highest single probability. But for most cases, the most likely number of winners remains one, although the cumulative probability of 2-6 might overwhelm that single probability.

Since it may be known that this poster has interests in the weather field, I thought it might be interesting to examine whether this approach works as a guide to long-range forecasting (fast forward, not too well).

If monthly weather conditions were entirely independent (i.e., not dependent on any cause and effect whether understood or assumed) then it might be possible to construct statement such as this:

There is an equal chance that one of the three summer months (take this either as June to August or June 21 to July 20 etc) would be warmer than average. To make it more relevant to how we perceive weather, let's say there's an equal one in three chance of each month being notably warm (about 1 deg above normal). So would that not follow the logic of random distribution outlined above, so that we could say,

chance of no warm month is (2/3) cubed or 8/27, therefore chance is 19/27 that any given summer will produce at least one notably warm month.

The problem with this approach is that it assumes no background dependency. There is in fact something like a 0.4 correlation between June and July temperatures and a higher one for July-August. This would suggest that your actual experience would equate to fewer than the randomly predicted 19/27 summers having at least one notably warm month, but a few having 2 or 3. This has indeed been the case in actual statistics. The frequency of summers having at least one notably warm month is closer to 50-50 than the almost three in four chance predicted by 19/27. But when they do occur, two out of three is often the standard achieved.

Lotteries might have some non-random elements not immediately obvious. I gather yours is a 45 choose 6 type lottery, as I mentioned, we have a 6-49 lottery here in Canada (and that's what it is called too). In either case, one form of non-randomness of outcome would be the range of numbers drawn, because many people play birth dates and anniversaries, all of which would have to fall in the range 1 to 31. Even more limiting, half the numbers would fall in the range 1 to 12. This means that the ticket pool is likely to be disproportionately shifted towards those lower numbers, and if the actual winning number draw goes high, the predicted number of winners would very likely be lower than random over time.

You could even make the case that your ticket is marginally more valuable in expected value over time, if you avoid the lower numbers, because you will reduce the chances of duplicating another winner's tickets (this might actually be testable at the more frequently paid out partial sets). However, you are still well into the negative investment range if your technique involves anything other than clairvoyance or a tardis. I prefer a blend of these.

someones off to the bookies

Donn Cuailnge wrote: »

On the 3rd of February lotto draw there were 23,055 match 3 winners and 1,242 match 3 + bonus winners. That means 24,297 lines matched 3 numbers from the six main balls drawn.
My question is this how can the above info be used to calculate how many lines of the lotto were purchased last Wednesday night?

Probability of matching 3 balls is 1 in 50.36385

Estimated number of lines purchased: 24,297 x 50.36385 =1,223,690

Actual number of lines purchased could be up to 300 higher or 300 lower than that estimate.

Donn Cuailnge wrote: »

how did you calculate the 50.36385?
If a line matched 4 balls, that same line matched 3 balls 4 times. Would you need to multiply the winning combinations in the 4, 5 and match 6 also?

Good Question! You are correct that a match-4 includes 4 match-3 instances.

However the calculation I used considers the probability of picking 3 'good' balls out of 6 'good' balls, and 3 'bad' balls out of 41 'bad' balls, so we avoid match-4 instances.

See http://en.wikipedia.org/wiki/Lottery_mathematics for more details!

If you don't want to know all the maths, but just need the answer use http://www.lottogenie.com/html/odds.html

I suspect the only way of applying statistics to the lottery is to study them and then realise the draw is random and the odds are so high it's not worth playing.

The balls have no memory of what numbers were drawn in the previous draw. Perhaps if certain balls weighed slightly differently or were distorted perhaps it might be relevant, but I suspect the balls are designed to be very similar. You would also need to work out which machine and balls were used for each draw if different machines are used each week (as in the UK).

As others have indicated "game theory" might influence how many others play the same numbers, potentially reducing your winnings. e.g. the lottery is probably won more often with all numbers under 30 as many people use birth dates.