the NY Lotto asks that you pick 6 numbers from the numbers 1 to 59. If your 6 selected numbers are drawn out... you win....woohoooo (a) what is the probability of winning if you play a single game and your 6 numbers are drawn out. (b) if you round the probability of winning to the 6th decimal places, would you have the same probability if you didn't buy a ticket?
@saifoo.khan
@ganeshie8
@Abhisar
well this problem is phrased strange. usually by odds we mean 1:x where 1/(1+x) is the probability of winnin
there you go @inkyvoyd
so given that using the typical definition of "odds" doesn't make sense, I might choose to do probability instead
there you go @inkyvoyd
Also, this problem is cruddy because it doesn't tell you the winning conditions lol
@Zarkon
is this all the information you were given?
yes
Well, that's dreadful. have you learned about both combinations and permutations yet
*in class I mean
one is a C and one is a P
The probability that the first number drawn matches one of your six numbers is:\[\frac{6}{59}\]You then have 5 numbers left and there are 58 remaining number in the loot. The probability that the next number drawn matches one of your remaining 5 numbers is then:\[\frac{5}{58}\]So the probability of both events happening is:\[\frac{6}{59}\times\frac{5}{58}\]Continue this to cover all 6 numbers
cool.... thanks.... what about part (b)
work out the answer to part (a) and that will let you answer part (b)
but to have a chance of winning don't you need to buy a ticket..?
you'll need to work out part (a) as a decimal number - if this number rounds to zero when you take it to just 6 decimal places then it will effectively mean you have the same chance as if you didn't buy a ticket
but isn't that a direct answer... and my question is still... don't I need to buy a ticket to have a chance of winning..?
"if you round it to 6 decimal places" <-- that is the key
but you need a ticket to win
if I told you that the probability of winning something is zero - then does it matter whether or not you buy a ticket?
the "actual" chances of winning are very very slightly above zero
so then if I don't win... I may or may not have a ticket..
what the question is saying is what would you conclude "if" you rounded the probability to 6 decimal places
ok... the if thing... is that like man is a mammal and elephants are mammals therefore man is an elephant
I remember if if you build it they will come
let me try a different approach...
cool
suppose I told you that I am going to toss a coin - if it lands "heads" I win - if it lands "tails" I win - if it lands on its edge you win $1000000 would you take the bet? and what would you say is the probability of you winning?
wow.... thats a great game... I don't lose... I can only win... given there appears to be no cost..?
you have to bet "$10" say :)
what is the financial expectation of your game..? I read about financial expectation on here
so what are the odds... ?
I expect to keep all the money collected from everyone placing the bet :)
my question is - what do YOU think are the odss of you winning?
well you need to advise me of the probabilities before I play... I'm not silly and can we toss the coin on grass
on a flat table only :)
well I need to probabilities to calculate the financial expectation to decide if I'll play or not, is that ok..?
@inkyvoyd can you help..?
the point I am trying to make is that "most" people would not take the bet as they would think the odds of them winning is zero. but the odds are just above zero - you would probably need to do some very complex calculations involving gravity, friction, etc - but you would end up with a very very very small number - still greater than zero
why.. that's not correct... there is a chance... which you won't reveal..
now imagine you had a calculator that only calculated to 6 decimal places. you use that calculator to work out your odds of winning the lotto and it comes up with zero. would you buy a ticket?
well that's not very handy... when was the last time you wrote an answer to 6 decimal places..
I won't buy the calculator
I am obviously not explaining this very well. I am trying to make the point that the question is asking you to conclude what you would do "if" your probability calculations were only accurate to 6 decimal places. Obviously in the "real" world" the odds would be greater than zero for the lotto (however small that number might be) - that is why some people still play the lotto :)
I think the only time I use 6 decimal places is when I find the growth or decay constant is exponential models..
well you have received 3 medals... so you must be doing something right
but I am not happy because you haven't been fully satisfied with the answer. and - I don't pay much attention to medals any way - I am here to help rather than to collect medals :)
I think my smart score will go down after this...
lol! :)
that should help :)
so now.... do I have to be 18 or 21 to play the lottery in New York..?
no idea - I am in the UK :)
well it has a big bearing... I maybe to young to play... so the question is a waste of time.
but if you can give me the odds on the coin toss game... I think I'll play that instead
that was just a random game I invented to try and explain this - obviously I failed and it looks like you need to talk to a gambling counsellor to remove your addiction from gambling! :)
no... I'm thinking of a career in gambling...
he he - I better leave now before I totally corrupt you my friend! :)
@campbell_st I think one of the ways to put it is that your chances of winning are so close to zero that we can say they are approximately zero. We usually approximate things that we think will or won't happen. For instance, on a monday night, we might assume that the teacher will be there the next day at school to collect homework.
but sometimes the teacher isn't there, and we have a sub instead, and if that's the case, we mnight get lucky and not have to do the homework and can turn it in a day late. But the chances of that happening are so low that we just assume that the teacher will be there 100% of the time.
but don't I need to buy a ticket... how can they be the same
They aren't the same, but they are approximately the same
so that's like saying a person who is 151 cm tall is 2 metres tall if you use round to the nearest metre
is that what you mean..?
no, it's like saying a person who is 199.99cm tall is 2 metres tall
the difference is so small it's not significant
but isn't the person who is 151 cm tall 2 metres tall if you round to the nearest metre..?
yes, but this is where rounding errors come in. usualy when we talk about height we think that 1cm is a significant difference, but .1 cm is not a significant difference. Thus we round to the nearest centimeter. However, we don't round to the nearest meter because that is too much to round to, because it exceeds the 1 cm resolution by a factor of 100
why not.... you can round anything
in the same sense, we usually calculate probabilities to 2, 3, or in soem cases, 4 decimal places. In other words, a difference in 1/100, 1/1000, or even 1/10000 chances are significant to us. But the chances of winning the lottery are so low that if we round even to the 6th decimal palce we still get zero. That is our way of saying the chances of winning the lottery are approximately zero, or zero as far as we care.
what is rounding error....
I've not heard of that
@campbell_st rounding is not arbitrary. usually in the context of a problem we pick our rounding such that the maximum error obtained from the rounding (rounding error) will not significantly change our answer
think of it this way. If you're asked to measure something with a ruler in cm, the smallest measurement it can give you is a difference of 1mm accurately
so then the 151 cm person can be 2 metres tall
maybe I should say they are 16 decimetres tall
but who uses decimetres?
@campbell_st , he can be 2 metres tall, but that depends on how much you are willing to allow your rounding error to be. Most people think that 50cm (the rounding error if you round to the nearest meter) is significant because it is a high proportion of the original number (151cm)
so what is a non significant amount...?
the idea is that in the real world our measurements aren't infinitely accurate. Your calculator is limited to 9 decimal points, your ruler is limited to the nearest milimeter, and your stopwatch is limited to the nearest milisecond. Any number after this obtained in a calculation is not significant because you have reached the limitations of the accuracy of your measurement device
so a person who is 165 cm is 2 metres... is less signifcant
so does that mean a light year is accurate to the nearest metre..?
no... it means that when you use a light year for calculations, depending on what you are doing (designing a space ship, doign gravitational calculations, esimating the distance) you care about a different number of decimal pionts in your light year
so if it takes light about 8.5 minutes to travel from the sun to earth, it could be quicker or slower..?
depending on the decimal places
then why do astronomers ignore pi... I read that once
I think we are talking about observed versus expected... when we apply mathematics to science, we make mathematical models that predict the outcome or the condition of a system or events. we can calculate our mathematical model usually to a high degree of accuracy, sometimes with arbitrary preceision (we can do it to 100,000 decimal places), but in the real world, our measurements are limited by the acuracy and precision of our instrument. when we say that it takes light about 8.5 minutes to travel from the sun to the earth, it could indeed be quicker or slower, depending on the distance the sun is from the earth (the earth's orbit is not perfectly spherical). But how much quicker or how much slower is not something we are concerned about because we are just using the rough figure of 8.5 minutes in whatever context.
thank's for the information, I'll tell my teacher that gambling is wrong... I have to go and do the dishes for mum. thanks
hahaha, farewell my friend :)
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