Lottery how?

Can you win the local lottery even if you do not have a.r.c.?
I mean only on a visitor visa.
Do they still pay you?

p.s.: I haven’t won, but I was wondering if it is worth the effort…

You can collect the smaller prizes that you get from the bank (take your passport), but only Taiwanese can collect the biggies (not that it should eb a problem to get someone to help you).

bri

A while ago, they had a lucky draw at Nova where they especially mentioned that there would be x% of tax to be paid while “foreigners would have to pay y% (y>x) tax” if they win one of the main prices - and in case of a foreigner, the tax would be withdrawn directly, so you would never see those y%.
That was an interesting statement (and it was the first time I saw the “foreigner tax” publicly mentioned), because it didn’t make any difference between tourists, foreign residents within their first 183 days and long-time residents. When I asked the girls at the service desk where that differece comes from when the tax law does not distinct between foreigners and locals, they called their supervisor and were told “that is the law”.
So, it seems you can be a lucky winner even as a tourist, but there is still the same “foreigner tax” thinking you encounter on income questions…

You can collect the smaller prizes that you get from the bank (take your passport), but only Taiwanese can collect the biggies (not that it should eb a problem to get someone to help you).

I did win a measily NT$200 on the sales receipt lottery and went to a Taipei Bank branch.
First you need to bring your passport and secondly, as foreigner, they wanted to tax me 20%.
My friends told me however that you can go to 7-11 and claim such prices there or redeem the receipt for goods purchase, i.e. no tax would apply. Or use, as Bri suggested, a friend to claim the price.

Interestingly the receipt I have had no (company) stamp and thus I have not been able to claim the “great price”.

As for the really big prizes: didn’t a Thai couple win last time the main prize, i.e. it seems to be possible for non-citizens to collect those!?

I’ve often collected the 200 or 1000 prizes at the bank (Land Bank or something I think) with no taxes even when I was on a visitor visa. 7-11 will jsut let you use it to buy stuff, not give you cash. That Thai couple won the Lotto not the receipt lottery.

Bri

Ok, all you math wizz kids out there can you help me out.

I have a little disagreement over the odds of striking some numbers in the new 4 star lotto.

What are the odds of getting the last 2 numbers in the string of four, in the correct order?
Let’s say I choose XX12 as my numbers - what would the odds be for getting those numbers and winning a prize?
Another person told me that she thinks it is 1 in 100, but I feel that because you are also betting on the position being the last 2 digits in a series of four numbers then the odds of winning must be considerably higher than that. Am I wrong? I find it difficult to believe that the Lotto people would give odds somewhere like 1 in 100.

If you can help me out I’d appreciate it and if you can show me the math involved it would also be appreciated.

If they are not sorted, and numbers can be reused, eg: 1234, 1111, 4321 are all valid then:

with 4 digits there are 10000 total outcomes.

how many of these outcomes end in XX12 ? starting from (00)12 up to (99)12, there will be 100, so 100 winning outcomes. You have to get one of these 100 tickets to win.

100 out of 10000 = 100

Or look at it another way. The first 2 digits are irrelevant, so then you have 100 tickets, so 1:100 of picking your tickets. Multiply everything by 100, 100 copies of 00, 100 of 01 etc… 100 of 12. You still have 1:100 chance.

They would give those odds, depending on how much they pay you.
eg: if you pay $100 for a ticket, and if you win they pay you $1000

1 times out of a hundred they pay you $1000
99 times they keep your $100

they take in $9900, pay out $1000

However, you are also selecting the position. ie. I want the numbers that I selected to be the last 2 numbers in the combination and NOT the be in the first 2 numbers of the 4 number combination. So the posible outcomes of my 1,2 selection could be xx12 (win) 12XX (loss) or x12x. This is only for the exact combination of 1,2 in that order and not any variations such as xx21.

So, would the fact that you are nominating the placement of the numbers factor in the odds of you winning with your number combination in the position that you select?

BH, if i’m reading this right, and i’m not sure that i am, your friend is correct. It’s one in ninety-nine.
M

Numbers are from 00 to 99 that’s 100.

Any other takers?
My position theory?

You select the position. 12xx or xx12 but it cannot be x12x. 3 outcomes for the exact number combination. The chance of getting that number selected is 1 in 100, but the chance of winning with those numbers in the positions that you select are much higher. Not even 50 - 50 as there is the x12x wildcard. Correct? :?:

Yep, you’re right, one in one hundred. You originally asked for the odds on the last two digits of a sequence of four, surely this can only be one in one hundred, regardless of the preceding figures.

Look at the 2 digit numbers 1 to 100. There is only one combition of 12 in that order, ie, twelve;

If they can be in any order, there are 2 possibilites: 12 and 21, so then your odds are halfed. 1:50 chance

if you want to allow 12xx, x12x and xx12, your odds are thirded, you have about 1:33 chance of winning

if you didn’t want to select the position, just the chance that it contains a ‘1’, odds are getting very close to even

You have to take the total number of possible outcomes, (number of tickets in the barrel), divided by the number of winning tickets in the barrel

in the scenario you described, it is 1:100, if the digits have to be in order, you are reducing the number of valid tickets, and your chances go up

10% of all numbers end in a ‘1’, 10% of all numbers end in a ‘2’
10% of all numbers start in a ‘1’
1% of all numbers end in ‘00’
1% of all numbers end in ‘12’
1% of all numbers start in ‘12’
1% of all numbers (over 3 digits of course) have ‘12’ as their 2nd and 3rd digit

Ok, So I have a 1 in 100 chance of 1 and 2 appearing in that order and a 1 in 33 chance of that number appearing in the position the I nominate - as per the rules of 4 Star Lotto.

So what I am asking is - If I combine the chance of hitting the numbers 1 and 2 in that order and in the last pair position in the draw - (XX12) - what are the exact odds of the combination of number AND Position?I would hazzard a guess that the odds of position (1:33) and the odds for the numbers (1 and 2) in that order - could be combined to come up with a completely different set of odds for the possibility of actually winning the lottery. However, I am not a math genious - I flunked math a long time ago, I don’t know what those odds might be.

For 4 star LOTTO you can select 1. all four numbers in the exact order to win. 2. The final 3 numbers in exact order to win. 3. The front pair in exact order to win or 4. The final pair of numbers in exact order to win.

Now, if these are the choices that we have and we select from numbers 0-9 and we just want one of the pairs then we must take a gamble on the position and the number order - both correct and we win - one wrong and we don’t.
There is also another factor - it also appears that there are only a certain number of people allowed to buy any one number combination and position selection of that number combination - or so I have been told. However, on second thoughts this has no bearing on the odds that I am asking for.

The other way around, it’s easier (more chances) if they can appear anywhere

[quote]For 4 star LOTTO you can select 1. all four numbers in the exact order to win. 2. The final 3 numbers in exact order to win. 3. The front pair in exact order to win or 4. The final pair of numbers in exact order to win.
[/quote]

  1. 1:10000
  2. 1:1000
  3. 1:100
  4. 1:100

it’s always 1:100 no matter how you work it out, that’s how maths works. If it works out differently, you got it wrong.

Want to combine them ? 1:33.333 in ‘12’ appearing in any position, but we only want 1 out of 3 positions (the last one), so 1:33.333 divided by 3 = 1:100

We worked out the chances for the UK lottery at work when it started, that’s more difficult because each number can only appear once. So the first has a chance of 1:49, the second 1:48 etc…I seem to remember we had to use factorials to work it out

Not entirely sure what you are asking, but here are the odds for the options you mentioned:

  1. all four numbers in the exact order to win. 1:10,000
  2. The final 3 numbers in exact order to win. 1:1,000
  3. The front pair in exact order to win. 1:100
  4. The final pair of numbers in exact order to win. 1:100

EDIT: Damn, matthew, you type too fast!

Ok, so it looks like a case of eating humble pie then.
Both odds combined still come out to 1:100. Damn it :smiling_imp: I stopped taking Math at 16 to get better grades to get into Uni because my math sucked too bad.

I never expected the odds to work out the same.

This is going to be news to one of the local newspapers though. I remember (? - remember and remember correctly could be different) reading them state that the odds of winning with all 4 numbers are something like 1:250 000 . Must have been a typo or perhaps they had someone like me doing the math. :blush:

Anyone played at all?

Answer: Too lousy to waste your money on.

23 million to 1.