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Seagull58
In the Algarve
In the case of a 6 number lottery only 1 number needs to be different on the two tickets for the odds against winning to halve. Only if all 6 numbers are the same do the odds remain unchanged.
Is that for real? ha.
Seagull58
In the Algarve
This is rubbish. Each ticket has the same odds but you now have 2 tickets so your odds have halved. Stop digging!
EDS
Banned
- Nov 11, 2012
- 2,040
Absolute shite. If I buy two tickets for the lottery the odds of me winning do not suddenly become 7 million/1? so how do I halve them odds again?
Seagull58
In the Algarve
What's that for? the calculation is correct when you're talking about odds of 120 to 1 on a single ticket.
EDS
Banned
- Nov 11, 2012
- 2,040
Are you sure of that, genuine question
True. I'm not sure of your point though.
My point is that odds can be used in the context which Eaglesdestroyseagulls used it, as well as by bookmakers. Unless I missed something El Pres alluded to the fact they could not and that Eagles was getting confused.
Seagull58
In the Algarve
Yes, they do. To halve the odds again to 3.5 million to 1 you have to buy another 2 tickets. To halve the odds again to 1.75 million to 1 you have to buy another 4 tickets, so 8 tickets in total. To halve the odds again to 0.875 million to 1 you have to buy another 8 tickets, so 16 in total. Do you get it yet? At this point you have 16 chances in 14 million or 1 chance in 0.875 million. Keep up at the back!
Seagull58
In the Algarve
Yes! 7/120 = 0.0583333333333 = 5.83333333%
6/120 = 0.05 = 5% or 1 in 20 if you prefer.
12/120 = 0.100000 = 10%
Are you convinced yet?
jgmcdee
New member
- Mar 25, 2012
- 931
Out of curiosity, and in a vague attempt to bring this thread to some sort of conclusion, what do you think the odds become if you have two tickets in this situation?
EDS
Banned
- Nov 11, 2012
- 2,040
Yes I get it, I also know that 7 into 120 is 17.14
Yes, he's right. 5.8% (Not 1 in 5.8)
EDS
Banned
- Nov 11, 2012
- 2,040
I believe I have two chances of winning the lottery at 14m/1 I do not believe I now have a 1/7m chance of winning it
Seagull58
In the Algarve
Yes but you have to calculate 120 into 7, not 7 into 120!
Seagull58
In the Algarve
You have 2 chances each at 14 million to 1. That is the same as 1 chance at 7 million to 1. If you don't believe me ask your maths teacher when you go back to school tomorrow.
Seagull58
In the Algarve
Sorry for hijacking the thread for a maths lesson. I'll retire and let it get back on topic.
jgmcdee
New member
- Mar 25, 2012
- 931
Okay, so some basic probability.
Your probability of a successful outcome is (number of successful outcomes)/(number of possible outcomes).
If you have one ticket then your probability of a successful outcome is 1/14,000,000.
If you have two tickets then your probability of a successful outcome is 2/14,000,000. Which is twice as much as if you have one ticket.
Happy now?
(As has been mentioned previously, this assumes that you don't have the same numbers on both tickets).
EDS
Banned
- Nov 11, 2012
- 2,040
Okay let me get this straight then! If I have one ticket then my odds are 14,000000/1. If I have two tickets my odds are 7,000000/1. Seagull58 said that when I double the amount of tickets again the odds half so four tickets 3,500000/1, eight tickets 1,750000/1...and so on and so forth, is that correct?
jgmcdee
New member
- Mar 25, 2012
- 931
Yes, although you might be better off leaving your number of possible outcomes the same when writing the odds (so 14,000,000:4 rather than 3,500,000:1); same resultant number but easier to read to yourself as "this many outcomes, that many chances of winning".