An UNLUCKY Lottery Win!? - Can anyone work out the odd's..

[The Optimist]

I know what your point is, but I’m trying to explain that your workings are wrong. You’re saying that there are roughly 700,000 (699191) ways of getting 3 numbers in the first ticket, so the odds of a person buying 1 ticket and getting 3 numbers is 1 in 20, purely because there are 20 ways of having 3 numbers from a choice of 6. That’s obviously wrong. The odds of getting 3 numbers on 1 ticket is 56.66.

You haven't understood the method that DTES is using. DTES's method is:
- Buy two tickets and assume no number appears on both tickets.
- There are 20 different sets of 3 numbers possible on the first ticket.
- There are 20 different sets of 3 numbers possible on the second ticket.
- Therefore there are 20x20=400 different ways of picking 3 numbers from the first ticket and 3 numbers from the second ticket.
- When the draw is made we require one of these 400 ways of picking 3 from each ticket.
- Therefore the odds of this occurring are 400/13,983,816.

To use this method to calculate the odds of matching two numbers from each ticket:
- Buy two tickets and assume no number appears on both tickets.
- There are 15 different sets of 2 numbers possible on the first ticket.
- There are 15 different sets of 2 numbers possible on the second ticket.
- Therefore there are 15x15=225 different ways of picking 2 numbers from the first ticket and 2 numbers from the second ticket.
- This only gives us 4 numbers but 6 numbers will be drawn.
- The other 2 numbers can be any 2 that don't appear on the two tickets.
- There are 37C2 = 666 such sets of 2 numbers that don't appear on the first two tickets.
- Therefore there are 225x666=149850 ways of generating 6 numbers that contain 2 numbers from each of the tickets plus another 2 numbers not on either ticket.
- Therefore the odds of this occurring are 149,850/13,983,816, which is significantly higher than the odds of matching three numbers from each ticket.

To use this method to calculate the odds of matching 3 numbers from just one ticket:
- Buy one ticket.
- There are 20 sets of 3 numbers possible on this ticket.
- This gives us 3 numbers but 6 will be drawn.
- The other 3 numbers need to be numbers not on the ticket.
- There are 43C3 = 12341 different sets of 3 numbers that don't appear on the ticket.
- Therefore there are 20x12341 = 246820 ways of generating 6 numbers with exactly 3 from the ticket.
- Therefore the odds of winning are 246820/13983816 = 1/56.66.

 

[SeagullSongs]

You haven't understood the method that DTES is using. DTES's method is:
- Buy two tickets and assume no number appears on both tickets.
- There are 20 different sets of 3 numbers possible on the first ticket.
- There are 20 different sets of 3 numbers possible on the second ticket.
- Therefore there are 20x20=400 different ways of picking 3 numbers from the first ticket and 3 numbers from the second ticket.
- When the draw is made we require one of these 400 ways of picking 3 from each ticket.
- Therefore the odds of this occurring are 400/13,983,816.

I understand what Triggaaar is trying to say.
For simplicity's sake, let's say the winning numbers were 1, 2, 3, 4, 5 and 6.

If you get 1, 2 and 3 on the first ticket, the only numbers on the second ticket that will satisfy our winning condition are 4, 5 and 6. So for each of the 20 successful outcomes on the first ticket, there is only one corresponding set of numbers on the second ticket, so 20 combinations in total.
Although technically, 10 of these pairs will be duplicates.

Ticket A - Ticket B
123 - 456
124 - 356
125 - 346
126 - 345
134 - 256
135 - 246
136 - 245
145 - 236
146 - 235
156 - 234
234 - 156
235 - 146
236 - 145
245 - 136
246 - 135
256 - 134
345 - 126
346 - 125
356 - 124
456 - 123​

The ones in bold are duplicates.

It's far too complex for me to work out at the moment, I'm out of practice

 

[The Optimist]

I understand what Triggaaar is trying to say.
For simplicity's sake, let's say the winning numbers were 1, 2, 3, 4, 5 and 6.

If you get 1, 2 and 3 on the first ticket, the only numbers on the second ticket that will satisfy our winning condition are 4, 5 and 6. So for each of the 20 successful outcomes on the first ticket, there is only one corresponding set of numbers on the second ticket, so 20 combinations in total.
Although technically, 10 of these pairs will be duplicates.

Ticket A - Ticket B
123 - 456
124 - 356
125 - 346
126 - 345
134 - 256
135 - 246
136 - 245
145 - 236
146 - 235
156 - 234
234 - 156
235 - 146
236 - 145
245 - 136
246 - 135
256 - 134
345 - 126
346 - 125
356 - 124
456 - 123​

The ones in bold are duplicates.

It's far too complex for me to work out at the moment, I'm out of practice

The confusion lies around the fact that there are two ways to approach the problem. One approach, Triggaaar's, is to think of the winning numbers as fixed and calculate the probability of picking 3 of these fixed numbers on one ticket and three of them on a second ticket. The other approach, DTES's, is to buy the tickets and think of these numbers as fixed. Then DTES is calculating the probability of three numbers from each ticket being drawn. Both approaches, if calculated correctly, will lead to the same answer.

 

[DTES]

I think the most important thing to remember is that we're all sat in the middle of our offices, in front of other people, maths-debating.

Disgusting, really.

 

[Hans Kraay]

A few weeks ago a mate bought two lucky dips and was presented with the same set of 6 numbers twice.

That's the same odds as winning the lottery but unfortunately he gets zilch for it!

Didn't think this was possible

 

[Triggaaar]

I will hit problems because you CANNOT pick 12 different numbers from a list of 10 numbers! How is that difficult to understand?!

Ah yes, fair point. I was trying to use your method for a smaller simpler example so I could make sense of it, but ran out of balls

To use this method to calculate the odds of matching 3 numbers from just one ticket:
- Buy one ticket.
- There are 20 sets of 3 numbers possible on this ticket.
- This gives us 3 numbers but 6 will be drawn.
- The other 3 numbers need to be numbers not on the ticket.
- There are 43C3 = 12341 different sets of 3 numbers that don't appear on the ticket.
- Therefore there are 20x12341 = 246820 ways of generating 6 numbers with exactly 3 from the ticket.
- Therefore the odds of winning are 246820/13983816 = 1/56.66.

That's the same method I used, as you're using the number of balls that are not drawn, ie, the 43 losing balls. So I understand/agree with all that.

You haven't understood the method that DTES is using.

It would appear that way

 

[Triggaaar]

I understand what Triggaaar is trying to say.
For simplicity's sake, let's say the winning numbers were 1, 2, 3, 4, 5 and 6.

If you get 1, 2 and 3 on the first ticket, the only numbers on the second ticket that will satisfy our winning condition are 4, 5 and 6. So for each of the 20 successful outcomes on the first ticket, there is only one corresponding set of numbers on the second ticket, so 20 combinations in total.

That's been half my problem.

 

[The Optimist]

That's been half my problem.

This where the two different approaches comes in. You are thinking of the winning numbers being known and then calculating the probability of picking them. This is not DTES's approach.

Consider buying two tickets on a Wednesday, i.e. we don't know what the winning numbers are yet. The first one has numbers 1 - 6 and the second one has numbers 7 - 12. We want to get 3 numbers from each. There are 20 different sets of 3 numbers on each ticket so a total of 20x20=400 different sets of 6 numbers possible with 3 numbers from each ticket. If any one of these 400 sets of 6 numbers occurs in the draw then we have the required situation. Therefore the odds are 400 out of 13983816.

 

[The Wookiee]

I have a headache now after reading this thread !

 

[Triggaaar]

This where the two different approaches comes in. You are thinking of the winning numbers being known and then calculating the probability of picking them. This is not DTES's approach.

Consider buying two tickets on a Wednesday, i.e. we don't know what the winning numbers are yet. The first one has numbers 1 - 6 and the second one has numbers 7 - 12.

Do you realise what this means? This means I'm a f***ing idiot and owe DTES an apology. Shit way to start the weekend.

 

[The Optimist]

Do you realise what this means? This means I'm a f***ing idiot and owe DTES an apology. Shit way to start the weekend.

Don't feel too bad, it took me a while to get my head round DTES's approach (to start with I thought he was wrong) and I used to be an A-Level Maths teacher! Also, once we've thrashed Birmingham you'll have forgotten all about it.