Another Drive4show 'Anything goes' thread

by **JuliusHMarx** Thu 29 Oct 2020, 8:39 am

First topic message reminder :

How have some individuals got to the point where all they do is whinge and moan and complain about other people without any sense of compassion for those less well off than themselves through no fault of their own?

by Soul Requiem Thu 24 Dec 2020, 3:53 pm

Two in 14 million is 1 in 7 million.

by super_realist Thu 24 Dec 2020, 4:05 pm

Soul Requiem wrote:Two in 14 million is 1 in 7 million.

I see what your saying but buying two lottery tickets doesn't mean you now have a one in 7million chance of winning.

by Soul Requiem Thu 24 Dec 2020, 4:08 pm

super_realist wrote:
Soul Requiem wrote:Two in 14 million is 1 in 7 million.

I see what your saying but buying two lottery tickets doesn't mean you now have a one in 7million chance of winning.

It doubles your odds of winning so from 1 in 45mil to 1 in 22.5mil.

by super_realist Thu 24 Dec 2020, 4:19 pm

You have two tickets out of a possible 14,000,000 combinations (or however many it is)

You haven't eradicated 7,000,000 possible ticket combinations by buying 2 tickets.

Your chances have improved fractionally by having two tickets instead of one, but your chance of winning has still only improved by 1 chance out of 14,000,000.

by Soul Requiem Thu 24 Dec 2020, 4:35 pm

That's not how odds work.

by super_realist Thu 24 Dec 2020, 4:42 pm

Soul Requiem wrote:That's not how odds work.

It's called game theory. If I buy 100 tickets I don't suddenly divide the odds by 100, so in my example of 14 million combinations 100 tickets doesn't reduce odds to 1 in 140,000.

If I have 100 tickets, I know AT BEST that 99 of those tickets are losing tickets. Therefore my odds of winning with 100 tickets are are 1 in 13,999,901

That's how little buying extra tickets does for your chances. Buying 100 tickets can only ever eradicate 99 of the 14,000,000 combinations.

by McLaren Thu 24 Dec 2020, 5:10 pm

Super

This might help

If the lottery has 14 million possible unique draws and you buy 7 million (unique combinations) of them, would you agree your probability of winning the lottery is 0.5?

Or to put it another way 7m/14m. Which can be reduced to 1/2 or 0.5.

Or what about if you bought all 14 million possible tickets. Your probability of winning would be 14m/14m (1).

or what about if you bought 3.5m tickets (25%) of the tickets. Would the probability of you winning not be 3.5m/14m(0.25)?

You can probably by now see where this is going.

What if you wanted a 0.00001428571% chance of winning the lottery? How many tickets would you need to by

number of tickets to be bought = 0.00001428571% of 14000000

And to top it off subtract your answer from 1/7000000.

by Soul Requiem Thu 24 Dec 2020, 5:11 pm

super_realist wrote:
Soul Requiem wrote:That's not how odds work.

It's called game theory. If I buy 100 tickets I don't suddenly divide the odds by 100, so in my example of 14 million combinations 100 tickets doesn't reduce odds to 1 in 140,000.

If I have 100 tickets, I know AT BEST that 99 of those tickets are losing tickets. Therefore my odds of winning with 100 tickets are are 1 in 13,999,901

That's how little buying extra tickets does for your chances. Buying 100 tickets can only ever eradicate 99 of the 14,000,000 combinations.

Not how it works in the slightest.

by super_realist Thu 24 Dec 2020, 5:15 pm

In terms of a lottery that is exactly how it works.

You can't seriously think that if you buy 1 ticket and your chances of winning are 1 in 14,000,000 that buying another means they drop to 1 in 7,000,000.

You have two tickets out of a possible 14,000,000. You know one must be wrong so your chances are 1 in 13,999,999

by Soul Requiem Thu 24 Dec 2020, 5:16 pm

The point is you do not know which one is a losing ticket so you must apply an equal chance to both.

by super_realist Thu 24 Dec 2020, 5:19 pm

McLaren wrote:Super

This might help

If the lottery has 14 million possible unique draws and you buy 7 million (unique combinations) of them, would you agree your probability of winning the lottery is 0.5?

Or to put it another way 7m/14m. Which can be reduced to 1/2 or 0.5.

Or what about if you bought all 14 million possible tickets. Your probability of winning would be 14m/14m (1).

or what about if you bought 3.5m tickets (25%) of the tickets. Would the probability of you winning not be 3.5m/14m(0.25)?

You can probably by now see where this is going.

What if you wanted a 0.00001428571% chance of winning the lottery? How many tickets would you need to by

number of tickets to be bought = 0.00001428571% of 14000000

And to top it off subtract your answer from 1/7000000.

Mac, if you buy half the tickets, you have a 50% chance of winning as you rightly point out. If you buy 2 tickets that does not mean the odds drop to one in 7,000,000 as Soul claims.

Souls chances of winning have increased because he has increased his tickets by 100%, but this does not mean the odds of him having the winning numbers have halved. Surely you see that.

by super_realist Thu 24 Dec 2020, 5:21 pm

Soul Requiem wrote:The point is you do not know which one is a losing ticket so you must apply an equal chance to both.

No, it doesn't matter which one is the losing/winning number.
You know that one of them is definitely a loser so that means you can exclude ONE of the 14,000,000 tickets from your possible winning combinations.

That means you can only have a 13, 999,999 chance of either of the two being correct.

by Soul Requiem Thu 24 Dec 2020, 5:25 pm

super_realist wrote:
Soul Requiem wrote:The point is you do not know which one is a losing ticket so you must apply an equal chance to both.

No, it doesn't matter which one is the losing/winning number.
You know that one of them is definitely a loser so that means you can exclude ONE of the 14,000,000 tickets from your possible winning combinations.

That means you can only have a 13, 999,999 chance of either of the two being correct.

So which one are you discounting?

by super_realist Thu 24 Dec 2020, 5:30 pm

It doesn't matter, you don't need to know as it makes no difference.
It's enough to know that one of them HAS to be wrong.

If you buy two tickets you can only be sure that one of the numbers is wrong, it doesn't matter which one.

They still have the same odds of winning at 1 in 13,999,999

by McLaren Thu 24 Dec 2020, 5:32 pm

Super

If you by a 100% of the tickets you probability of winning is 1
buy 50% its 0.5
buy 25% its 0.25
buy 10% its 0.1

So why would buying 0.00001428571% not give you a 0.0000001428571 chance of winning

(To be clear 0.0000001428571=2/14000000=1/7000000)

You are essentially saying that at some point buying X fraction of unique tickets does not give you an equivalent probability of winning. You accept this at 0.5 and 0.25 but not at 0.0000001428571.

by Soul Requiem Thu 24 Dec 2020, 5:36 pm

super_realist wrote:It doesn't matter, you don't need to know as it makes no difference.
It's enough to know that one of them HAS to be wrong.

If you buy two tickets you can only be sure that one of the numbers is wrong, it doesn't matter which one.

They still have the same odds of winning at 1 in 13,999,999

If I'm thinking of a number between 1 and 10, you can choose two of them, what odds of you getting it right?

by Shotrock Thu 24 Dec 2020, 5:37 pm

Always fun, for me, to throw a few dollars at a lottery on occasion.

But I still believe in that old maxim: "A lottery is a tax on people who are bad at math!"

by super_realist Thu 24 Dec 2020, 5:38 pm

McLaren wrote:Super

If you by a 100% of the tickets you probability of winning is 1
buy 50% its 0.5
buy 25% its 0.25
buy 10% its 0.1

So why would buying 0.00001428571% not give you a 0.0000001428571 chance of winning

(To be clear 0.0000001428571=2/14000000=1/7000000)

You are essentially saying that at some point buying X fraction of unique tickets does not give you an equivalent probability of winning. You accept this at 0.5 and 0.25 but not at 0.0000001428571.

Mac,
My point is that if there are 14,000,000 combinations then 1 ticket means you have 1 of the available 14,000,000 combiantions, if you buy two tickets that does not mean that the odds drop to 1 in 7,000,000.

by super_realist Thu 24 Dec 2020, 5:39 pm

Soul Requiem wrote:
super_realist wrote:It doesn't matter, you don't need to know as it makes no difference.
It's enough to know that one of them HAS to be wrong.

If you buy two tickets you can only be sure that one of the numbers is wrong, it doesn't matter which one.

They still have the same odds of winning at 1 in 13,999,999

If I'm thinking of a number between 1 and 10, you can choose two of them, what odds of you getting it right?

1 in 9

Please don't tell me you think it's 50%?

by McLaren Thu 24 Dec 2020, 5:40 pm

If I have 7000000 tickets, I know AT BEST that 6999999 of those tickets are losing tickets. Therefore my odds of winning with 7000000 tickets are are 1 in 7000001.

That's how little buying extra tickets does for your chances. Buying 7000000 tickets can only ever eradicate 6999999 of the 14,000,000 combinations.

by super_realist Thu 24 Dec 2020, 5:44 pm

McLaren wrote:If I have 7000000 tickets, I know AT BEST that 6999999 of those tickets are losing tickets. Therefore my odds of winning with 7000000 tickets are are 1 in 7000001.

That's how little buying extra tickets does for your chances. Buying 7000000 tickets can only ever eradicate 6999999 of the 14,000,000 combinations.

Mac you're trying to argue something different from me.
Soul is trying to argue that buying two tickets eradicates 7,000,000 numbers, you have argued that you need to buy 7,000,000 numbers to do so, which is correct.

by McLaren Thu 24 Dec 2020, 5:48 pm

No, he is saying that the probability of winning if you buy 2 unique tickets is 1/7000000. Which it is.

by Soul Requiem Thu 24 Dec 2020, 5:48 pm

super_realist wrote:
Soul Requiem wrote:
super_realist wrote:It doesn't matter, you don't need to know as it makes no difference.
It's enough to know that one of them HAS to be wrong.

If you buy two tickets you can only be sure that one of the numbers is wrong, it doesn't matter which one.

They still have the same odds of winning at 1 in 13,999,999

If I'm thinking of a number between 1 and 10, you can choose two of them, what odds of you getting it right?

1 in 9

Please don't tell me you think it's 50%?

It's 1 in 5.

by beninho Thu 24 Dec 2020, 5:48 pm

Your chance of winning will improve by almost 100% your odds of winning improve slightly. Easy to get them mixed up.

https://math.stackexchange.com/questions/1369659/if-i-buy-2-lottery-tickets-do-i-double-my-chance-of-winning

by super_realist Thu 24 Dec 2020, 5:52 pm

beninho wrote:Your chance of winning will improve by almost 100% your odds of winning improve slightly. Easy to get them mixed up.

https://math.stackexchange.com/questions/1369659/if-i-buy-2-lottery-tickets-do-i-double-my-chance-of-winning

Yes, your chance of winning increase by virtue of the number of tickets you hold, but your odds do not halve every time you buy double the tickets.

by super_realist Thu 24 Dec 2020, 5:53 pm

Soul Requiem wrote:
super_realist wrote:
Soul Requiem wrote:
super_realist wrote:It doesn't matter, you don't need to know as it makes no difference.
It's enough to know that one of them HAS to be wrong.

If you buy two tickets you can only be sure that one of the numbers is wrong, it doesn't matter which one.

They still have the same odds of winning at 1 in 13,999,999

If I'm thinking of a number between 1 and 10, you can choose two of them, what odds of you getting it right?

1 in 9

Please don't tell me you think it's 50%?

It's 1 in 5.

Please explain.
If I have two chances then I know at least one is wrong. That leaves 9 possibilities left, not 5.

by Soul Requiem Thu 24 Dec 2020, 5:55 pm

That logic only works if you know which one is incorrect.

by super_realist Thu 24 Dec 2020, 6:00 pm

Soul Requiem wrote:That logic only works if you know which one is incorrect.

No, it doesn't matter at all.
You know for a fact there is only One winning answer. If you ask me to pick two numbers I know before I've even picked them that one WILL be wrong as there is only one winning number.
You have told me one will be wrong

Therefore if one is definitely wrong before I've even begun, then that leaves only 9 possibles.

You are falling into the trap of adding intrinsic value to a number. I don't need to know anything about it to know one must be wrong because that is automatic in your rules. One is wrong by definition.

by McLaren Thu 24 Dec 2020, 6:04 pm

By the way you can get even more bonkers results with your method.

Which appears to be

If I have x tickets, I know AT BEST that (x-1) of those tickets are losing tickets. Therefore my probability of winning with x tickets is 1 in 14000000-(x-1).

By your calculation the probability of winning the lottery if you buy 13999999 tickets is 0.5.

x-1=13999999-1 = 13999998 losing tickets

so odd of winning are 1 in 14000000-13999998

which means probability of winning is 1 in 2 (or 0.5)

by super_realist Thu 24 Dec 2020, 6:05 pm

McLaren wrote:By the way you can get even more bonkers results with your method.

Which appears to be

If I have x tickets, I know AT BEST that (x-1) of those tickets are losing tickets. Therefore my probability of winning with x tickets is 1 in 14000000-(x-1).

By your calculation the probability of winning the lottery if you buy 13999999 tickets is 0.5.

x-1=13999999-1 = 13999998 losing tickets

so odd of winning are 1 in 14000000-13999998

which means probability of winning is 1 in 2 (or 0.5)

No Mac, this is the very thing I'm arguing against.
Soul is aiming that if you buy two tickets your odds reduce to 1 in 7,000,000. That's clearly false.

by Soul Requiem Thu 24 Dec 2020, 6:06 pm

That simply is not how it works and I'm baffled you're still arguing the point despite being so wrong it's embarrassing. You still have ten possibles as any of the ten can be drawn and you have two chances to win therefore have odds of 5/1. The only real difference between odds and chance is the ordering of the numbers but the percentile is the same.

by super_realist Thu 24 Dec 2020, 6:08 pm

Ha ha we have eradicated one.

You are confusing the number of tickets you have with the number of available possibilities.

Order is also irrelevant as order is something we imbue to numbers.

Please explain how buying two tickets means your odds are halved?

by super_realist Thu 24 Dec 2020, 6:14 pm

Soul Requiem wrote:
super_realist wrote:
Soul Requiem wrote:Two in 14 million is 1 in 7 million.

I see what your saying but buying two lottery tickets doesn't mean you now have a one in 7million chance of winning.

It doubles your odds of winning so from 1 in 45mil to 1 in 22.5mil.

This is what you're claiming.
How does two tickets mean you can disregard
22.5m possible combinations?
Your chance of winning has increased by having more tickets and is relative ONLY to your ticket holding, your odds have increased by an almost imperceptible magnitude.

You have only two possible combinations of 44.5m possibles. You don't simply divide 45 m by two to work out odds.

by McLaren Thu 24 Dec 2020, 6:16 pm

Odds are the ratio that you would expect an event to occur against not occurring.

So for Supers 100 ticket example

100/14000000/(1-(100-/4000000)

by McLaren Thu 24 Dec 2020, 6:22 pm

odds for 2 tickets would be

2/14000000/1-(2/14000000)

which is the same as

1/7000000/1-(1/7000000)

by Soul Requiem Thu 24 Dec 2020, 6:23 pm

super_realist wrote:Ha ha we have eradicated one.

You are confusing the number of tickets you have with the number of available possibilities.

Order is also irrelevant as order is something we imbue to numbers.

Please explain how buying two tickets means your odds are halved?

I'm not doing any such thing, odds are expressed in the simplest form but that does not mean you are disregarding possibilities.

If I have two chances to win in a ten number game you have a 2 in 10 chance to win therefore your odds are 5 to 1.

by super_realist Thu 24 Dec 2020, 6:32 pm

Please explain why you think buying a second ticket halves your odds of winning the lottery.

There are still 13,999,998 other riders in the race

by Soul Requiem Thu 24 Dec 2020, 6:36 pm

It can't really be explained any simpler.

by beninho Thu 24 Dec 2020, 6:40 pm

If you buy a lottery ticket at odds of 1/14,000,000 and you buy another lottery ticket, therefore doubling your chance of winning your odds take that into account.

"Yes, you do double your chance of winning. There are (496) different possible draws. The probability of winning with one ticket is the chance that the 6 drawn balls are one specific combination, i.e. 1(496). With two tickets, there are two winning combinations, so the probability is 2(496)
Basically, the reason it is doubled is because winning one ticket and winning the other ticket are mutually exclusive, i.e. you can't win both tickets (note the problem statement says "different tickets"). If there was a chance of winning both, the chance would be less than doubled."

Copied from the link I attached - I've not done the maths.

by super_realist Thu 24 Dec 2020, 6:42 pm

Soul Requiem wrote:It can't really be explained any simpler.

Don't go to Vegas then, you'll lose your shirt.

by super_realist Thu 24 Dec 2020, 6:44 pm

beninho wrote:If you buy a lottery ticket at odds of 1/14,000,000 and you buy another lottery ticket, therefore doubling your chance of winning your odds take that into account.

"Yes, you do double your chance of winning. There are (496) different possible draws. The probability of winning with one ticket is the chance that the 6 drawn balls are one specific combination, i.e. 1(496). With two tickets, there are two winning combinations, so the probability is 2(496)
Basically, the reason it is doubled is because winning one ticket and winning the other ticket are mutually exclusive, i.e. you can't win both tickets (note the problem statement says "different tickets"). If there was a chance of winning both, the chance would be less than doubled."]

Copied from the link I attached - I've not done the maths.

There's all sorts of conflicting stuff on that link and what relevance does 496 have to do with anything? Is it missing some notation?

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https://www.606v2.com/viewtopic.php?t=69913

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