One for the maths experts...

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graeme
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Re: One for the maths experts...

Post by graeme » Tue Sep 24, 2013 11:23 pm

Ignoring the fact that "Klaxon" is a registered trademark and one of the QI questions was all about how the QI klaxon is not a klaxon for this reason, consider it unleashed. :)

I really think this question is at the heart of why anyone would ever buy a lottery ticket.

Two hints:

What are the odds of each ticket NOT winning?
What influence does owning one ticket have on the total number of possible combinations for the choice of numbers for the second ticket?
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pete
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Re: One for the maths experts...

Post by pete » Wed Sep 25, 2013 12:10 am

graeme wrote:Ignoring the fact that "Klaxon" is a registered trademark and one of the QI questions was all about how the QI klaxon is not a klaxon for this reason, consider it unleashed. :)

I really think this question is at the heart of why anyone would ever buy a lottery ticket.

Two hints:

What are the odds of each ticket NOT winning?
What influence does owning one ticket have on the total number of possible combinations for the choice of numbers for the second ticket?
Odds of ticket not winning (given one prize) 13,983,816 - 1.
Odds of second ticket are the same. But as you own 2 then the odds of YOU not winning are 13,983,816 - 2.


Oooo. That's good. I'll think about that tomorrow.
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rossybee
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Re: One for the maths experts...

Post by rossybee » Wed Sep 25, 2013 12:42 am

So 6,991,908 - 1
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robin
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Re: One for the maths experts...

Post by robin » Wed Sep 25, 2013 8:03 am

In my (no doubt wrong) view of it, the two events are mutually exclusive - by definition because you've said that the tickets are different. The probability of both winning is zero - thus they are mutually exclusive.

In the case of mutually exclusive events the probability of one or other happening is the probability of the first plus the probability of the second.

Hence for the lottery with two different tickets you get 1/big-number + 1/big-number = 2/big-number.

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pete
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Re: One for the maths experts...

Post by pete » Wed Sep 25, 2013 8:12 am

rossybee wrote:So 6,991,908 - 1

(Klaxon).

Nah I wasn't writing 13,983,816 "to" 2. I was writing 13,983,816 "minus" 2. If you buy 2 tickets it does double your chances of having a winning ticket, but that doesn't halve the odds of winning.
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robin
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Re: One for the maths experts...

Post by robin » Wed Sep 25, 2013 8:34 am

We should stick to quoting probabilities as fractions (0 <= p <= 1) and avoid "odds speak" as it is so ambiguous.
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pete
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Re: One for the maths experts...

Post by pete » Wed Sep 25, 2013 9:27 am

robin wrote:We should stick to quoting probabilities as fractions (0 <= p <= 1) and avoid "odds speak" as it is so ambiguous.
I was just discussing this with Mrs Pete and we reached the same exact conclusion. (ie we were arguing about the answer until we realised we were saying the same thing, but differently).
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graeme
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Re: One for the maths experts...

Post by graeme » Wed Sep 25, 2013 11:17 am

robin wrote:In my (no doubt wrong) view of it, the two events are mutually exclusive - by definition because you've said that the tickets are different. The probability of both winning is zero - thus they are mutually exclusive.
Correct assertion...
robin wrote: Hence for the lottery with two different tickets you get 1/big-number + 1/big-number = 2/big-number.
...wrong conclusion. (IMO, I think, perhaps...)

So, sticking with the more narrative "odds" notation, and just looking at jackpot odds:

You pop down to the corner shop and buy a ticket at odds of 1 in 13,983,816. You're about to leave, but decide to go back and buy a second ticket with different numbers to double your chance of winning. You go home happy, you place the two tickets in front of you on the table and stare at them. Each ticket has an equal chance of winning, and each ticket has odds of 1 in 13,983,816.

Two tickets twice as "good" as one, right? But each ticket has an equal 1 in 13,983,816 chance. But you've got two tickets, so we're no longer interested in the odds of an individual ticket winning. Lets change the question to what we really want to know... What's the chance of YOU winning? Or, what's the chance of EITHER ticket winning? Ahh, now we're asking the right question. (Most of you will have got this far already... if you win, you don't care which ticket it was).

Analogy time...

I've lost my wedding ring (the jackpot) somewhere on a football field (the number space of all possible lottery ticket combinations). I (Mr Ticket) have to search for it (pick the winning numbers).

If I have a friend to help me (another ticket) we'll probably find it in half the time, right? Correct, we will, but how?

By communicating. By saying "You search that half, and I'll search this half." and understanding what that means and sticking to our own half. How long would it take if we couldn't talk or communicate? Well, the only thing either of us would know for sure is that the ring is NOT where the other person is standing, but could be anywhere else. We're both back to searching the full football field, minus the bit that has a person on it. Add another person to help (3 tickets) and how much field do we have to cover? Still almost all of it. Add a dozen people. Have we significantly divided the labour if we can't communicate? Nope, we've barely made a dent on the problem.

This is the problem lottery tickets have. They are totally independent as Robin asserted, and can't communicate. The have no effect on each other. No way of dividing the work. They can't decide to cover half the possibilities each, so each operates independently in the entire space. Except that's not quite true... they did communicate once, through you just once when you decided not buy two tickets with the same numbers. (The same as not looking for the ring in any spot that has a friend standing on it). So, they each know they're not the same as the other, but there's no communication mechanism between tickets that says, "I'll cover this 6,991,908, you cover that 6,991,908". So the odds are NOT 1 in 6,991,908, or 2 in 13,983,816.

So, what are they?

Well, that second ticket isn't the same as the first because when we picked it, it was chosen from a slightly smaller pool of possibilities (13,983,816 minus the first ticket combination, equals 13,983,815 remaining possibilities). We can shuffle the tickets around and it doesn't make any difference which is which or which order we bought them in, so while the odds of a particular ticket winning remain at 1 in 13,983,816, the odds of YOU (either ticket) winning have dropped to the odds of your "second" ticket, or 1 in 13,983,815. Buy another (different) ticket and it will drop to 1 in 13,983,814.

The odds of any ticket (i.e. you) winning the jackpot are simply 1 in (total possibilities - number of tickets bought). There's no division of labour. No dividing up the search space. Every ticket is scrabbling around looking for the jackpot in the entire space of possibilities, except where there's already another of your tickets standing.

So, in the case of two tickets, for that extra £1 (effectively doubling your stake), you increased your chances by a whopping... 0.0000071524%.
Last edited by graeme on Wed Sep 25, 2013 11:23 am, edited 1 time in total.
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whaleys
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Re: One for the maths experts...

Post by whaleys » Wed Sep 25, 2013 11:21 am

All -

Please disregard my incorrect post, I'll go back to keeping quiet...

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Re: One for the maths experts...

Post by j2 lot » Wed Sep 25, 2013 12:06 pm

[quote="graemeI've lost my wedding ring (the jackpot) somewhere on a football field (.[/quote]


Are you offering a reward .................


:mrgreen:


:wink:
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graeme
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Re: One for the maths experts...

Post by graeme » Wed Sep 25, 2013 12:07 pm

I'm offering the chance of a reward, for a small administration fee... :D
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robin
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Re: One for the maths experts...

Post by robin » Wed Sep 25, 2013 2:50 pm

Graeme,

You are wrong.

The probability of two mutually exclusive outcomes is always the probability of one plus the probability of the other. If there is a probability of both happening then the sums are different - for example if you wanted to know the probability of ticket 1 AND ticket 2 winning any sort of prize.

So the probability is 2/13,983,816 or 1/6,991,908.

Try thinking through the problems with a pack of 52 cards with one card being drawn at random from the deck. Before the card is drawn you have to choose one or more cards and hope one of them comes up. If you pick just the king of hearts, you know p=1/52. If you pick the king of hearts and the king of spades, you know p=1/26. If you say it'll be a spade, you know p=1/4. Etc.

Why do you think it is any different with the lottery - it's just a very big deck of 13,983,816 cards ...

Finally, if you needed more proof consider the odds offered on the roulette wheel for the various types of bet.

Cheers,
Robin
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pete
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Re: One for the maths experts...

Post by pete » Wed Sep 25, 2013 2:56 pm

robin wrote:
Finally, if you needed more proof consider the odds offered on the roulette wheel for the various types of bet.

Cheers,
Robin
Dammit. I started a post that involved a roulette wheel.

Imagine the roulette wheel has 10 slots. I bet on <1>. In every 10 spins I will win.

I then also start betting on <2>. There are now eight spins of the wheel that I will lose on, but the probability of me winning is 2 in every 10. Or 1 in every 5. Or p=0.2.

The lottery is the same.
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'03 - '10 Starlight Black S2 111S
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Re: One for the maths experts...

Post by tut » Wed Sep 25, 2013 3:18 pm

Just wait till we get onto Physics.

Now that really is my speciality, though I expect Robin would avoid it. :D

tut

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Re: One for the maths experts...

Post by Corranga » Wed Sep 25, 2013 3:25 pm

robin wrote:Try thinking through the problems with a pack of 52 cards with one card being drawn at random from the deck. Before the card is drawn you have to choose one or more cards and hope one of them comes up. If you pick just the king of hearts, you know p=1/52. If you pick the king of hearts and the king of spades, you know p=1/26. If you say it'll be a spade, you know p=1/4. Etc.
This is right.

The key thing with this pack of cards analogy of course is that all selections are made prior to the card being extracted, so if you choose a 2nd card (or buy another ticket) the probability is still 1/52 of that card coming up

1/52 + 1/52 = 1/26.

Much easier to wrap your head around with suits. There are 4. You select hearts, 1/4. You then select spades as well, another 1/4. 1/4 + 1/4 = 1/2. You have a 50/50 chance of having selected the right suit of the drawn card.

If each lottery ticket could be sold only once (ie, the cards are extracted at each selection) then the probability changes to:
Select card 1 = 1/52
Select card 2 = 1/51
Select card 3 = 1/50
etc..
since selecting a card then effects the available pool.


Right? :blackeye

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