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18-11-2019, 11:58   #1
marieholmfan
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Factorials and Probabilities - Worked Lottery example

I was doing a factorial for the Euromillions:


It is effectively two 'draws'
In the first there are fifty numbers and the player must select the first five chosen.

50!÷((50-5)! x 5!)

I get 2,118,760

In order to win the main jackpot the player must now win a second draw
There are twelve numbers and the player must select the first two chosen.
12!÷((12-10)! x 2!)
66

2,118,760 x 66 = 139,838,160

And this is the value given on the wikipedia page.


But when I attempt to calculate the second prize probability I can't get it to come out.

The second prize is given to the players who select only one of the two numbers drawn from the second pool of twelve.

12!÷(((12-10)! x 2!) + (12-10)! x 2!))) =33
Because you get two goes with the second draw I sum the factorials on the right hand side and 33 has a nice connection to 66 and 'looks good' to me (though I have no intuitive sense for numbers).

2,118,760 x 33 =69,919,080
Similar to 1 in 6,991,908 on the wikipedia page for the Euromillions EXCEPT for the fact that I am out by a factor of 10!

SO where is my mistake?
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18-11-2019, 17:20   #2
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Focusing on the 2 bonus numbers:
There are 66 possible combinations of choosing 2 from 12 (12*11 / 2).

Of those combinations, there is only 1 where both numbers are correct.
To get the first number correct and the second number incorrect, it can be only 1 number, and of the remaining 11 (remember you've already chosen 1 from the 12), the other 10 must be wrong. This means there are 10 combinations of numbers where you match the first chosen number and don't match the 2nd.
Similar logic applies for the second bonus number, there are also 10 combinations of the 66 where the "first" number is wrong and the second number is correct.
Lastly, to have zero numbers correct, from the 12 possible numbers to chose from, you must pick from one from the 10 remaining "wrong" numbers and one more from the (now) remaining 9 "wrong numbers", which is 10*9 or 90 permutations. You must divide by 2 as the order is irrelevant (selecting 2,12 is the same as 12,2).

In summary, of the 66 combinations of the bonus numbers there are:
1 combination with both correct
20 combinations where exactly 1 number is correct
45 combinations where both numbers are wrong.

So (as you note above), there are 2,118,760 * 66 or 139,838,160 combinations of both main numbers and bonus numbers.
Of these, there is only 1 combination of numbers with both the main and bonus numbers correct (so odds of 1 : 139,838,160)
There are 20 combinations of numbers with the main numbers correct and 1 bonus number correct, so odds of 20 : 139,838,160 (or 1: 6,991,908)
There are 45 combinations of numbers with the main numbers correct and zero bonus numbers correct, so odds of 45 : 139,838,160 approx 1 : 3,107,514.67)

I hope that helps.

Last edited by Yakuza; 18-11-2019 at 20:04.
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18-11-2019, 18:14   #3
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Just to follow up, for lower prizes you can use the following logic:

For matching 4 main numbers and both bonus numbers, you split up the main numbers into winning and losing sets (5 and 45 respectively).
To match 4 numbers, you have 5C4 (or 5) possible ways of getting 4 of the five winning numbers right and 45C1 (or 45) ways of picking 1 number from the "losing" 45.
This means there are 5 * 45 * 1 combinations of getting 4 main numbers and both bonus numbers, so 225:139,838,160 or approx. 1:621,502.93

Similarly, to match 3 main numbers and 1 bonus number, you have 5C2 (or 10) ways of getting the main ones right and 45C2 (or 990) ways of getting the "wrong" numbers, leading to a total of 10 * 990 * 20 (from my previous post) or 198,000 combinations, so odds of 198,000:139,838,160 or approx. 1:706.25

(I leave the rest as an exercise!)

Last edited by Yakuza; 18-11-2019 at 23:15.
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18-11-2019, 19:18   #4
marieholmfan
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Quote:
Originally Posted by Yakuza View Post
Focusing on the 2 bonus numbers:
There are 66 possible combinations of choosing 2 from 12 (12*11 / 2).

Of those combinations, there is only 1 where both numbers are correct.
To get the first number correct and the second number correct, it can be only 1 number, and of the remaining 11 (remember you've already chosen 1 from the 12), the other 10 must be wrong. This means there are 10 combinations of numbers where you match the first chosen number and don't match the 2nd.
Similar logic applies for the second bonus number, there are also 10 combinations of the 66 where the "first" number is wrong and the second number is correct.
Lastly, to have zero numbers correct, from the 12 possible numbers to chose from, you must pick from one from the 10 remaining "wrong" numbers and one more from the (now) remaining 9 "wrong numbers", which is 10*9 or 90 permutations. You must divide by 2 as the order is irrelevant (selecting 2,12 is the same as 12,2).

In summary, of the 66 combinations of the bonus numbers there are:
1 combination with both correct
20 combinations where exactly 1 number is correct
45 combinations where both numbers are wrong.

So (as you note above), there are 2,118,760 * 66 or 139,838,160 combinations of both main numbers and bonus numbers.
Of these, there is only 1 combination of numbers with both the main and bonus numbers correct (so odds of 1 : 139,838,160)
There are 20 combinations of numbers with the main numbers correct and 1 bonus number correct, so odds of 20 : 139,838,160 (or 1: 6,991,908)
There are 45 combinations of numbers with the main numbers correct and zero bonus numbers correct, so odds of 45 : 139,838,160 approx 1 : 3,107,514.67)

I hope that helps.

It does help alot!
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18-11-2019, 23:16   #5
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It does help alot!
I would never have thought of doing the complement where the complement has more events.

Stupid I know but true unfortunately.
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18-11-2019, 23:33   #6
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Couple of random Euromills questions for math lovers.

Q1. How come different odds are shown for '5main' when playing regular EM vs EM-Hotpicks (without stars), as below?
Q2. What are the odds for EM 5main, if using the spread (rule): 'one ball from each decade only' 0x,1x,2x,3x,4x?
Q3. (Optional) What would be the 'guesstimate' (P+) advantage for applying the LOLN with 2/7/14 type yrs worth of (5main) data at hand.



I'll admit am more interesting in picking lower hanging fruit, rather than digging down to study the roots.
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19-11-2019, 10:38   #7
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Quote:
Originally Posted by marieholmfan View Post
I would never have thought of doing the complement where the complement has more events.

Stupid I know but true unfortunately.
Don't be hard on yourself, a lot of probability/odds type questions are can be best answered by working out the chances of something not happening and taking that answer from 1 - just come at it from another angle.
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19-11-2019, 10:52   #8
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Hi Accumulator,




For your first question

The odds for picking five numbers on their own are different from the odds for picking five numbers with no lucky stars because there are effectively two draws.


First we draw five numbers from fifty the chance here is the hotpicks value
2,118,760 to 1 against

Then we draw two numbers from twelve
The chance of getting 2 numbers from 12 is 66 to 1 against

The chance of getting 0 numbers from 12 is 45 to 1 on

So the odds of getting 5 lucky stars and 0 numbers

Is (2,118,760 x 66) ÷ 45 = 3,107,515

Your second question is I suggest irrelevant at least in theory the odds of 1,11,21,31,41 don't differ from 1,2,3,4,5
With access to inside information you could identify the 5 least picked numbers and back them as then you won't split the winnings; but again in theory all numbers are equally likely.

Coming to the third theory you could use this information to identify the least picked numbers and pick them ; again in theory all numbers in each pool are likely.









Quote:
Originally Posted by Accumulator View Post
Couple of random Euromills questions for math lovers.

Q1. How come different odds are shown for '5main' when playing regular EM vs EM-Hotpicks (without stars), as below?
Q2. What are the odds for EM 5main, if using the spread (rule): 'one ball from each decade only' 0x,1x,2x,3x,4x?
Q3. (Optional) What would be the 'guesstimate' (P+) advantage for applying the LOLN with 2/7/14 type yrs worth of (5main) data at hand.



I'll admit am more interesting in picking lower hanging fruit, rather than digging down to study the roots.

Last edited by marieholmfan; 19-11-2019 at 10:56.
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19-11-2019, 10:55   #9
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Quote:
Originally Posted by Accumulator View Post
Couple of random Euromills questions for math lovers.

Q1. How come different odds are shown for '5main' when playing regular EM vs EM-Hotpicks (without stars), as below?
Q2. What are the odds for EM 5main, if using the spread (rule): 'one ball from each decade only' 0x,1x,2x,3x,4x?
Q3. (Optional) What would be the 'guesstimate' (P+) advantage for applying the LOLN with 2/7/14 type yrs worth of (5main) data at hand.



I'll admit am more interesting in picking lower hanging fruit, rather than digging down to study the roots.
Q1) Those odds are for if you only pick 1 or 2 or 3 or 4 or 5 numbers only (ignoring the bonus).
I.e. there are 5C4 (or 5) ways to pick 4 from the winning 5, and 50C4 (or 230300) ways to pick 4 numbers from 50, so the hot pick odds are 5:230300 or 1:46060. Same for the others.

Q2) (Need to have a think about this)

Q3) Not sure what you're trying to say here - but if you mean trying to guess numbers based on past frequency, this will give you zero help as draws are memoryless.

Last edited by Yakuza; 19-11-2019 at 11:03.
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19-11-2019, 11:05   #10
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By the way Yakuza am I overcomplicating things with the factorials?
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19-11-2019, 12:46   #11
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Hi Accumulator,
The odds for picking five numbers on their own are different from the odds for picking five numbers with no lucky stars because there are effectively two draws.
See what you and Yakuza both mean on the Q1 for 2 draws. Guess my early assumption on the odds presentation, was that it didn't matter about the stars, as one would have gained 5main anyway (which was only ever the only intended target). So yes, the 2m/1 or so, would stand (for the objective in this case).

Quote:
Originally Posted by marieholmfan View Post
Your second question is I suggest irrelevant at least in theory the odds of 1,11,21,31,41 don't differ from 1,2,3,4,5
This Q2 is a bit more tricky, as i) using 'only' (5)decade spreads, you void any selections such as 1-5, even 1,11,21,31 and 39 would also never be selected, as it also doesn't meet the strict 'spread selection criteria'. Thus, the total variants would be much less.

Likewise you could apply additional rule(s) such as ii) must be 3e/2o or vice-versa. Typically (on average), there is either 3evens or 3odds per draw. Again reduces total available line selections.

Additionally...Saw a bell curve of the 14yrs worth (LOLN) of EM data, showing 125 stands as the 'total' average count (again main balls onl). This could be a third rule iii) t=125 (+/-5) to apply.

- All in order to grind down the (5m) 2,118,760/1 odds way, way, down into the sub-750,000/1 region, for the (guaranteed 1m) per 1.50 stake.

Quote:
Originally Posted by marieholmfan View Post
again in theory all numbers in each pool are likely.
A coin flip may well display highly random or skewed results for the 1st 20 or so throws, perhaps even 10:90 h/t. Now do this 1,000 flips, and you will almost certainly get at very least 40:60, or more likely 45:55, or ideally 49:51 due simply, to the law of large numbers.

If at the 250/500/750 point of flipping, you notice any large standard curve deviance, you could assume it will readdress itself naturally, a bit like the Vegas card-counters playing 'Jack.
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19-11-2019, 13:02   #12
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This means there are 10 combinations of numbers where you match the first chosen number and don't match the 2nd.




Similar logic applies for the second bonus number, there are also 10 combinations of the 66 where the "first" number is wrong and the second number is correct.

So because I don't include the pair of winners I am no longer choosing from 12 but from 10 and I can do something like :



10!÷(10-2! x 2)) =45


Then I can use ( 66 - (1+45) )=20 for the chance of getting any one 'lucky star')


(50!÷((50-5)! x 5!)) x ((12!÷((12-10)! x 2!)-(1+10!÷(10-2! x 2!)) )

Last edited by marieholmfan; 19-11-2019 at 13:15.
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19-11-2019, 13:20   #13
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Quote:
Originally Posted by marieholmfan View Post
So because I don't include the pair of winners I am no longer choosing from 12 but from 10 and I can do something like :



10!÷(10-2! x 2)) =45


Then I can use ( 66 - (1+45) )=20 for the chance of getting any one 'lucky star')


(50!÷((50-5)! x 5!)) x ((12!÷((12-10)! x 2!)-(1+10!÷(10-2! x 2!)) )
Yep, that works too.
The state space (the full set of possible results / states) is {'0 correct', '1 correct', '2 correct').
It's relatively easy to work out the odds of '0 correct' and '2 correct', so whatever's left over must be '1 correct'
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