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I thought the question was what is the probability of 2 people winning the lotto on the same draw. If my understanding is correct, the chances for 2 people winning at the same time becomes slimmer than for one person winning it.

 

one person has a one in 42!/36! chance of winning the draw.

 

two people winning at the same time is one chance in 2*42!/36! to occur.

 

The above is from the perspective of the lotto organizer, checking what are the chance of having multiple winners.

 

Let's now take it from the perspective of the chances of ONE person winning.

 

For him or her to win, he or she has to choose the correct 6 numbers out of 42. It doesn't matter how many people enter, becuase what matters is for the entrant to get all 6 numbers correct.

 

If we examine the probability of one person winning, the equation is still 42!/36!, regardless of the number of people entering the lotto. I would assume that a person's chance of winning is the same. He would receive a smaller sum, but his chances are still the same.

 

What do you think, Sir Boy?

 

Off topic, your picture looks very familiar... Are you from Jakarta? :D

oh yes, you are right I think we share the same view that the chance of any person to win a lotto draw is the same assuming everybody has the same number of bets. It is also true that the chance of a single winner in a draw is higher than the chance of multiple winners. But your eqn says otherwise, the more winner the bigger the chance. I think it should be the reciprocal.

 

1

------------- or 6! where n = the number of winners

n* 42! ----------

------- n*42!

6!

 

but on the other hand, according to some experts the one's chance to win a 6/42 lotto draw is 1 is to 5.3 million, because there are approximately 5.3 million possible combinations for the 6/42 lotto. this means you need at least 53million pesos to bet for you to be a sure winner. Now, I can not derived these figures from your eqns. Any more thoughts?

 

(off the record, oh yes i'm from jakarta :rolleyes: :rolleyes: )

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oh yes, you are right I think we share the same view that the chance of any person to win a lotto draw is the same assuming everybody has the same number of bets. It is also true that the chance of a single winner in a draw is higher than the chance of multiple winners. But your eqn says otherwise, the more winner the bigger the chance. I think it should be the reciprocal.

 

1

------------- or 6! where n = the number of winners

n* 42! ----------

------- n*42!

6!

 

but on the other hand, according to some experts the one's chance to win a 6/42 lotto draw is 1 is to 5.3 million, because there are approximately 5.3 million possible combinations for the 6/42 lotto. this means you need at least 53million pesos to bet for you to be a sure winner. Now, I can not derived these figures from your eqns. Any more thoughts?

 

(off the record, oh yes i'm from jakarta :rolleyes: :rolleyes: )

Sorry for the mess! What you see is not what you get. This is the edited post.

 

 

oh yes, you are right I think we share the same view that the chance of any person to win a lotto draw is the same assuming everybody has the same number of bets. It is also true that the chance of a single winner in a draw is higher than the chance of multiple winners. But your eqn says otherwise, the more winner the bigger the chance. I think it should be the reciprocal.

 

1

-------------

n* 42!

-------

6!

 

or

 

6!

---------

n*42!

 

 

where n = the number of winners

 

 

 

but on the other hand, according to some experts the one's chance to win a 6/42 lotto draw is 1 is to 5.3 million, because there are approximately 5.3 million possible combinations for the 6/42 lotto. this means you need at least 53million pesos to bet for you to be a sure winner. Now, I can not derived these figures from your eqns. Any more thoughts?

 

(off the record, oh yes i'm from jakarta :rolleyes: :rolleyes: )

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It is also true that the chance of a single winner in a draw is higher than the chance of multiple winners. But your eqn says otherwise, the more winner the bigger the chance.

I think the numbers are getting confusing. I think it's just a matter of interpretation like specific gravity and density - it means the same relationship but read in different ways.

 

The chances for one winner is higher than for multiple winners - this we agree.

 

But it can be seen from the equations I wrote that this is so:

 

For one winner he has a 1 in 5.3 million chance.

 

For two winners, there is a 1 in 10.6 million chance. So the there really is a slimmer chance of having multiple winners.

 

The numbers are growing but it is in the interpretation that we differed.

 

This is the same as your equation which is the reciprocal. :blush:

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I think the numbers are getting confusing. I think it's just a matter of interpretation like specific gravity and density - it means the same relationship but read in different ways.

 

The chances for one winner is higher than for multiple winners - this we agree.

 

But it can be seen from the equations I wrote that this is so:

 

For one winner he has a 1 in 5.3 million chance.

 

For two winners, there is a 1 in 10.6 million chance. So the there really is a slimmer chance of having multiple winners.

 

The numbers are growing but it is in the interpretation that we differed.

 

This is the same as your equation which is the reciprocal. :blush:

okey. but then again, the eqn 42!/36! = 3,776,965,920. i believe you missed out something.

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okey. but then again, the eqn 42!/36! = 3,776,965,920. i believe you missed out something.

Oops, I did miss something.

 

42!/36! gives all the unique combinations of the 6 numbers out of 42, but the numbers have to be picked at the exact order.

 

Since lotto allows us the combination of the 6 numbers in any order, we have to divide it by the number of possible combinations of the 6 numbers.

 

So it should be:

 

42!/36!

-------- = 5,245,786

6!

 

or 1 chance in 5,245,786

 

since we are after the inverse,

 

6!

-------- = 0.000,000,190,629

42!/36!

 

or 0.0000190629 % probability

or 0.190629ppm

 

Sheesh - in a production system, that's close to impossible!

 

So it must be true - winning the lotto is an act of GOD ...

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Oops, I did miss something.

 

42!/36! gives all the unique combinations of the 6 numbers out of 42, but the numbers have to be picked at the exact order.

 

Since lotto allows us the combination of the 6 numbers in any order, we have to divide it by the number of possible combinations of the 6 numbers.

 

So it should be:

 

42!/36!

-------- = 5,245,786

6!

 

or 1 chance in 5,245,786

 

since we are after the inverse,

 

6!

-------- = 0.000,000,190,629

42!/36!

 

or 0.0000190629 % probability

or 0.190629ppm

 

Sheesh - in a production system, that's close to impossible!

 

So it must be true - winning the lotto is an act of GOD ...

Ok doki. so, For philippine lotto draws. the probability of winnning is

 

 

(L-6)! 6! / W*L! where W is no. of winners, L = lotto game i.e. 42, 45 & 49.

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  • 2 weeks later...

on dividing by zero:

 

to divide by a number x is to multiply by its multiplicative inverse, 1/x

 

the reciprocal/inverse of a number is that which when multiplied by the number is 1

 

so we need the value of 1/x when x=0. however, any number multiplied by zero is zero (zero property of multiplication).

 

and so zero has no inverse, and we can't divide anything by it. ganun na yun, parang langis at tubig na di maghahalo kahit ano pa ang bate mo.

 

 

---------------------------

 

regarding the lotto, let us simplify the game first. hulaan ng labas sa six-sided die, tapos tatlo lang tao sa mundo na tag-isa ang mga hula. kumbaga 1-6 ang laro.

 

chances of winning? 1/6, for each player

 

chances of no person winning: (5/6)(5/6)(5/6) = 125/216

chances of exactly 1 person winning: 3(1/6)(5/6)(5/6) = 75/216

chances of exactly 2 persons winning: 3(1/6)(1/6)(5/6) = 15/216

chances of all persons winning: (1/6)(1/6)(1/6) = 1/216

total probability: 1 (125+75+15+1=216)

 

the point of it all: least likely scenario is that everyone wins, least likely is no one wins, with other possiblities falling inbetween. pero di masasagot yang lotto question kung walang assumption kung ilan ang tumaya, kung ilang numero ang tinaya ng mga tao, kung nandadaya ba yung mga taga-PCSO, etc.

 

-------------------------

 

i move that this thread be closed!

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You are in bar... chillin out with your 7 friends... you table is 10 meters away from the bar. if you can carry 3 bottles of beer with one hand, how many trips do you have to make from your table to the bar if you and your friends want to drink up 14 bottles of beer?

Edited by Killer5
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How about 2, 3 or 4 persons winning the same draw?

Kung ang tanong eh probability na dalawang tao manalo sa isang draw, kailangan natin malaman kung ilan ang tumataya at kung ilang number ang tinataya nung bawat isa.

 

For example kung isa lang ang tumataya, kahit ano gawin mo hindi dalawa ang mananalo. Kung dalawa ang tumataya, silang dalawa ang dapat manalo. Kung tatlo ang tumataya, you have to add the possibilities na kung sinong dalawa sa tatlo ang mananalo. Kasama pa sa model mo yung kung ilang number ang tinataya nung bawat isa... there are two levels of complexity here, yung personal odds of winning nung bawat individual and the number of winners... there are two random variables at work here.

 

Anyway, mahabang usapan talaga ang probability atsaka yung mga 'in how many ways'. The most crucial part is kung nagkakaintindihan kung ano ba ang tanong.

 

This is not the right place for questions like this.. patulan na lang nating yung beer bottle problem, ok?

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Kung ang tanong eh probability na dalawang tao manalo sa isang draw, kailangan natin malaman kung ilan ang tumataya at kung ilang number ang tinataya nung bawat isa.

 

For example kung isa lang ang tumataya, kahit ano gawin mo hindi dalawa ang mananalo. Kung dalawa ang tumataya, silang dalawa ang dapat manalo. Kung tatlo ang tumataya, you have to add the possibilities na kung sinong dalawa sa tatlo ang mananalo. Kasama pa sa model mo yung kung ilang number ang tinataya nung bawat isa... there are two levels of complexity here, yung personal odds of winning nung bawat individual and the number of winners... there are two random variables at work here.

 

Anyway, mahabang usapan talaga ang probability atsaka yung mga 'in how many ways'. The most crucial part is kung nagkakaintindihan kung ano ba ang tanong.

 

This is not the right place for questions like this.. patulan na lang nating yung beer bottle problem, ok?

salamat sa advice mo. nasagot na yung tanong ko. ibig sabihin naintindihan nung sumagot. mag aral ka muna nng statistics bago ka kwento ng kwento. sa UP ka mag aral ha.

 

yung beer problems para yun sa puzzles doon sa JOKES SECTION.

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quintix: bago ka kasi sabat ng sabat basahin mo muna ang history ng tanong. basically, ang buod ng tanong ko eh. paano ma-derive ang 1:5.3million chance na lumabas ang isang 6- number combination from a 42 balls lotto draw. yung ibang tanong any nag sanga sanga nalang. nasagot naman yung general formula. ewan ko sayo bakit napunta sa number of bettors and pinagsasabi mo. ang punto ko d2 kung ikaw ang tumaya sa lotto (assuming 1 bet lang) ang chance mo manalo at 1:5.3million. mag isa kaman tumaya o isang million kayong tumaya. totally independent yun.

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I cant believe it ... What's the probability na magkita at magtalo ang dalawang gunggong na sina " BOY HAPON" at "KILLER5" dito sa thread na to ...?

 

Take it to the Fight club and leave this thread alone if you have nothing to contribute or solve.

Edited by chunkyhunk
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You are in bar... chillin out with your 7 friends... you table is 10 meters away from the bar. if you can carry 3 bottles of beer with one hand, how many trips do you have to make from your table to the bar if you and your friends want to drink up 14 bottles of beer?

Two probable answers:

 

1. If onlly you are to carry the beer, you'll need a total of three trips. 6 on the first, 6 on the second, and 2 on the last.

 

2. If this is a trick question, why would I carry all of the beer ten meters away and make three trips when I can ask a couple of my friends to help me carry the rest of the bottles?

 

On a serious note: Please take it easy boys. Take your Fight to the Club...

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Two probable answers:

 

1. If onlly you are to carry the beer, you'll need a total of three trips. 6 on the first, 6 on the second, and 2 on the last.

 

2. If this is a trick question, why would I carry all of the beer ten meters away and make three trips when I can ask a couple of my friends to help me carry the rest of the bottles?

 

On a serious note: Please take it easy boys. Take your Fight to the Club...

more so, you can put the 14bottles of beer in a bucket and ask the waiter to carry it for you. hehehe sabi ko na nga ba pang jokes and more tong tanong eh.

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I lost my cool when a reference was dropped regarding my miseducation, and for this I apologize.

 

I shouldn't have to defend myself, knowing my own credentials, accepting my own fallibility, and realizing the limitations of this medium for relevant mathematical discourse.

 

To the cooler heads who have intervened, thanks for pointing out this was getting out of hand.

 

--------------------------------------------------

 

Here's my contribution, observe that:

2 + 2 = 2 x 2

1 + 2 + 3 = 1 x 2 x 3

 

Is there a similar equation with the sum of four numbers being equal to their product?

 

(I don't have any answer to this, but if there is a correct answer, everyone will know without any further discussion necessary)

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-------------------------

 

i move that this thread be closed!

at first I thought your question was meant to put an end to this thread. but wait, to answer your question, yes there are.

 

obvious one, 0 + 0 + 0 + 0 = 0 * 0 * 0 * 0

 

another : 1 + 1 + 2 + 4 = 1 * 1 * 2 * 4 = 8

 

 

and this continues....

 

 

for 5 numbers:

 

1+1+1+2+5=1*1*1*2*5 = 10

 

 

for 6 numbers,

 

1+1+1+1+2+6 = 1*1*1*1*2*6 = 12

 

so on and so forth:

 

the general eqn would look like this:

 

(n-2)+2+n = 1^(n-2)*n

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on dividing by zero:

 

to divide by a number x is to multiply by its multiplicative inverse, 1/x

 

the reciprocal/inverse of a number is that which when multiplied by the number is 1

 

so we need the value of 1/x when x=0. however, any number multiplied by zero is zero (zero property of multiplication).

 

and so zero has no inverse, and we can't divide anything by it. ganun na yun, parang langis at tubig na di maghahalo kahit ano pa ang bate mo.

Here's another way of looking at the problem:

 

Formula: A/x = ? when X = 0

 

There are three solutions:

if the number A is already zero, then the answer is zero.

if the number A is NOT equal to zero, then:

 

The formula A * 1/x = ?

As x approaches zero, the answer increases:

 

x=1, 1/x = 1/1 = 1

x=0.5, 1/x = 1/0.5 = 2

x=0.1 1/x = 1/0.1 = 10

x=0.01 1/x = 1/0.01 = 100

x=0.0001 1/x = 1/0.0001 = 10,000

x=0.00001 1/x = 1/0.00001 = 100,000

and so on.

 

As X approaches 0 (i.e. x=0.00000000001), the answer approaches infinity, Positive or negative, depending on the numerator.

-----------------------

 

I think this is a thread that allows us to exercise our thinking skills, not our debating nor fighting skills. Generating solutions is the goal, not determining who has a bigger ego.

 

------------------------

 

Regarding "langis at tubig na binate", add an emulsifier and it will mix! Example: the surfactants in detergent permit the mixing of oil with water.

 

Life is not always black and white. The enjoyment is in the search for DIFFERENT solutions.

 

Peace People!

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I think this is a thread that allows us to exercise our thinking skills, not our debating nor fighting skills. Generating solutions is the goal, not determining who has a bigger ego.

 

------------------------

 

Regarding "langis at tubig na binate", add an emulsifier and it will mix! Example: the surfactants in detergent permit the mixing of oil with water.

 

Life is not always black and white. The enjoyment is in the search for DIFFERENT solutions.

 

Peace People!

I agree with floppydrive.

 

And before somebody will MOVE to close this thread again, i would like to correct and elaborate my last solution.

 

whereby:

 

the general eqn would look like this:

 

(n-2)+2+n = 1^(n-2)*n

 

it should read like this:

 

 

(n-2) + 2 + n = 1^(n-2) * 2 * n

 

 

where (n-2) should be expanded as: 1 + 1 + ... + 1

 

wherefore: 1^(n-2) = 1^(1+1+1+...+1) = 1*1*1*...*1 where the # of term is equal to (n-2).

 

to illustrate: (example problem) : find the the solution where the sum = product of a say, 12 numbers.

 

solution: let n=12

 

(n-2) + 2 + n = 1^(n-2) * 2 * n

 

(10) + 2 + 12 = 1^10 * 2 * 12

 

1+1+1+1+1+1+1+1+1+1+2+10 = 1*1*1*1*1*1*1*1*1*1*2*10 = 24

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