Search results for:

The odds of winning the lottery if you do buy a ticket (and don't loose it)
for a £1 investment is 1 in 13,983,816

Those odds are so high, things like diet and age probably play a large role in your likelihood of hitting the jackpot.

God willing, I'll get there soon.

Once you are fully noncognitivist things get a lot clearer, you will receive your membership card and an approved list of category errors - only then will you be able to go out into the world and deal with shit.

The expected value of the UK lottery is around £0.54 (on the basis of the most common payout profiles).

So if you buy two you are guaranteed to win!

I should write a book.

The odds of winning the lottery if you do buy a ticket (and don't loose it)
for a £1 investment is 1 in 13,983,816

Now of course without a ticket my odds of winning are far lower
but my investment is nill so its actually a far better return

and as I can't visualise the difference in odds once they are over even 1 in 1000 it doesn't make any difference to my dreaming.

Ha.

I like that line: "without a ticket my odds of winning are far lower".

: )

Speaking in tongues, I see.

No, I am genuinely ignostic, it took a Herculean effort to get there.

I think it was Orwell who pointed out that buying lottery tickets is about buying the hope that you will win. The hope makes you feel better and can be worth paying for even if you never do win.

Yep, it's also why religion doesn't need to be true.

Are you a creationist?

Theological noncognitivist.

http://www.youtube.com/watch?v=bx6lupC6WyE

Muslamic rape gangs.

Brilliant, Chris Morris would be proud.

cringe

Reminds me of Marc Wootton (My New Best Friend).

Few people realise how closely linked probability and game theory are to the theory of evolution.

Agreed, all lies.

If the two players are aware . . . .

Cheers for the response.

Makes things clearer

It's witchcraft.

Only if the host is *randomly *picking cards. Remember that the host knows that the cards he is picking are not the X card.

Agreed, we know that the host is not randomly picking cards, he is selecting the non-X cards to leave one card (that we don't know the value of) and your chosen card.

But I can still see a situation where the order of the cards can be identical but depending on your starting state you would be best advised to do two opposing moves ?

. . . or imagine two players rather than one, they both choose a card (one the far left card and the other the card on the far right) the host removes the rest of the cards, do they both improve their chances of finding the card marked X by swapping ?

This is also known as the Monty Hall problem. Easiest way to figure it out is to imagine there are 1000 cards. Imagine you picked one (1/1000 chance of getting the right one) and then 998 of the others were turned over. The person won't turn over the X card as he knows were it is, so there is now a 1/1000 chance your card is correct, and a 999/1000 chance that the other remaining card is; hence you should swap.

I can't really get my head around that one, never have been able to, I remember it from being a kid (I think the version I remember had a gameshow host and three doors, with a prize behind one).

Surely as the 998 cards are turned over, the card you have your finger on will have it's odds changed in proportion to the amount of remaing cards rather than stay at 1/1000 ?

If you imagine the 1000 cards laid out in a row and the person making the choice picks the card on the far left, the host then turns over the middle 998 cards to leave the card on the far right untouched, now the suggestion is that the card on the far right is enormously more likely to be the card with the X so we should always swap if given the choice.

But what if we rewind right to the start of the experiment with all the cards in exactly the same state and when asked to choose we pick the card on the far right, the host then removes the middle 998 cards leaving one untouched card on the far left, at this stage should we, if given the chance, swap our choice ?

Help!'s position seems to be that the only hypothesis allowed is that the coin is fair.

Well I'm not sure about the word 'allowed', but essentially - yes.

The idea that the cliche about tossing a coin for 100 heads is dependent on someone else betting on an alternative equally probable specific sequence is absurd.

Except for this, must reply to this.

[temporary de-flounce]

The probability of flipping a coin and producing a head 100 times in a row is not dependent on someone else betting on an alternative equally probable specific sequence, I am not sure anyone has made the claim that it is.

I think you are misunderstanding what I am saying, that - counterintuitively - the order of this exercise is important in working out the probability.

Is the order that lottery balls are drawn in a factor in calculating the probability of winning ?

No it is not a factor.

In a sequence of coin flips is the order of the 100 results a factor in calculating the probability of 'winning' (getting your 100H sequence) ?

Yes it is a factor.

The reason is that the lottery doesn't judge the order of the numbers (6,5,4,3,2,1 is identical to 1,2,3,4,5,6), but with coin flips the order matters because the probability of a specific sequence can only be measured against the absolute number of possible sequences.

So, whereas with the lottery 6,5,4,3,2,1 can be read as identical to 1,2,3,4,5,6 - with coin flips
H T H H H H H H (etc) is not identical to T H H H H H H H (etc) - if it were you would have much much less than the 1.27n outcomes.

Hopefully that has made what I mean a little clearer.

[/reflounce]

Oh come on this is getting ridiculous... I'd post a troll head here if I had the same forum posting skills as others on the forum. I specifically have said this is NOT what I'm saying... and I reposted precisely to try and stop you cutting my line to fit that interpretation.

I quoted from your own post (verbatim) - this was done before you edited your post and regardless of the edit I can't see how what I quoted from you was NOT what you were saying.

I am going to have to go with teome's 'up my own arse' and with your 'pedantic troll' and call it a day, we have obviously not convinced each other of our positions and as much as I am interested in this - and both your's and teome's posts - once the name calling kicks in these conversations start to lose their attraction for me.

[/flounce]

You're clearly too far up your own arse . . .

You shouldn't let maths anger you, name calling doesn't clarify your position.

Fixed that to make it even clearer what I was saying. : p

Ok, parsing, back in a second.

I'll give it one more go... while 100 tails or heads is of equal probability to any ONE other specifically noted sequence

Agreed, 1,27 nonillion sequences available from a 100 bit system (a hectobit) all have equal probability of occurring in a genuinely random system.

this is not what people are betting about or generally referring to.

Oh yes it is.

: )

When you bet someone that 100 head/tails is near impossible or that they couldn't possibily toss a coin 100 times and get 100 heads, what we are betting upon is: that ANY ONE among the astronomic number of different random equally possible other sequences is more likely to come up when you toss the coins, than any one specific sequence that they may choose.

This is where we disagree, it's simply not that case that (and I will quote what you say to avoid colouring your idea):

ANY ONE [I]among the astronomic number of different random equally possible other sequences is more likely to come up when you toss the coins, than any one specific sequence that they may choose.[/I]

Let's try an example, here is our 100 heads:

HHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH
HHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

Can you type out (or cut and modify the above) a sequence from the other sequences more likely to occur ?

I guess what you are trying to say is that the chances of it (100Hs) not occurring is more likely than it occurring, this is true, but this equally true for all sequences.

@eamonnog yes I agree. It's a different interest in the outcomes. All outcomes are equally likely but this tells us nothing useful and is not really worth analysing further unless you are picking lottery tickets, flipping coins

100 coins in a row = order important (counterintuitively)

Lottery (6 numbers from 49) order not important.

^^^^Yes but you're only considering a very specific example of equally probable events and the order matters.

Yep, agreed, a coin flip resulting in 100 heads is the target.

Comparing that to a series of other sequences ([99 heads and 1 tail] in one of 100 combinations) doesn't really work as a guide for probability as the original problem is searching for a specific sequence (1 sequence from 1.27n sequences) as 100 heads in a row has no variations even if we ignore the order of events.

The question as stated what is the probability of flipping 100 heads in a row? is looking for a single event not a group of events, so in that respect the order always matters, otherwise all we are saying is "hey look, these 100 sequences are 100 times more likely to occur that this 1 sequence, which is axiomatic.

Hope that makes sense !!!

If the runs were repeated infinitely many times all outcomes would occur. It's infinite! that's the basis of much of the theory, that for infinite runs all will occur.

I can't see that.

Let's keep the machinery light here - I have a perfect random number generator, switch it on and away it clicks, it produces either a 0 or a 1 every second, ignoring the cold death of the universe, a second term for the coalition or any other physical limitations (let's keep it all in an unchanging, hypothetical sempiternal universe) it keeps clicking away for an enormous amount of time, 0 after 0 after 0 after 0 after 0 after 0 after 0 after 0 after 0, nothing is broken, the machine is remotely checked without interference playing a part and all is well, it just happens to have only produces 0s so far. . .

By what mechanism would it be compelled to produce a 1 ?

S = - sum [p_i * ln(p_i)] for all i

'pips', I read 'pips', also 'sum', but I am going to stick with 'pips', how many points for 'pips' ?

I find it hard to see what you are getting at. Maybe you can clarify.

By saying this:

"The problem (as usually stated) is what are the chances of flipping a head 100 times in a row. Rather than in an arrangement of 100 coins what percentage of possible arrangements feature XXXXX (XXXX being whatever you are measuring)"

I am saying that the problem as stated is asking for the probability of a set sequence (in this case a flip resulting in heads 100 times in a row) rather than the probability of the appearance of one of many sequences that share the ratios (H v T) of the original but not the order.

As Teome suggested, if you think of it as sequences, in a set of 100 restricted coin tosses the chances of ending up with 1 tail at some random point along the line and 99 heads is 100 times more likely (given that it has 100 different sequential possibilities than the single sequence possibility of all 100 heads).

It is a 100 times more likely because we are comparing 1 sequence to 100 sequences.

If this is correct (and unless I've misread, I think we're all agreed it is) then I dont see the relevance of your continuing argument about all sequences being *individually *equal in possibility.

Because they are. And because a sequence of 100 heads only has one order, there are no variations of 100 heads in a row.

Contrary to what you are claiming, I would say that Teome has phrased the argument in the manner in which we face it in everyday life: When we say that 100 heads or 100 tails is near impossible, we don't mean it is less likely as an individual sequence than 99 heads with specifically the second toss being a tail, or 99 heads with the 55th toss being a tail, etc. What we normally mean is that the chance of picking out that one random sequence is astronomically smaller than simply betting on ANY of the massive massive number of other undisclosed random sequences.

You have this part wrong.

The chances of that sequence (100 Hs in a row) occurring from a random selection is identical to ANY of the massive massive number of other sequences.

All of them are 1 in 1,267,650,600,228,229,000,000,000,000,000.

A hundred heads = 1 in 1,267,650,600,228,229,000,000,000,000,000
95 heads with 5 tails in the last 5 places = 1 in 1,267,650,600,228,229,000,000,000,000,000
46 heads and 54 tails bunched at either end = 1 in 1,267,650,600,228,229,000,000,000,000,000
HTHTHTHTHTHTHTHT . . . = 1 in 1,267,650,600,228,229,000,000,000,000,000
TTTHHHTTTHHHTTTHHH . . . = 1 in 1,267,650,600,228,229,000,000,000,000,000
One head at the start, one at the end and 98 tails in between = 1 in 1,267,650,600,228,229,000,000,000,000,000

The likelihood of ANY of the massive number of sequences (1.267 nonillion) occurring is identical.

Currently filling mine in. Wondering how little information I can get away with giving them. To the question "How would you describe your national identity?" I've answered Other, none.

I did exactly the same - ticked 'Other, write in' and wrote NONE.