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Kedas

Lottery real example chance?

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Hi,

 

In the paper today they mentioned 800 people that played together for a total of about 40.500 Euro, max to win if they are the only one is 7.000.000 euro.

 

I want to know the chance that they win.

you have 42numbers (order doesn't matter) 6 numbers are drawn.

if you only choose 6numbers of the 42 than your change is simple.

6/42*5/41*4/40*3/39*2/38*1/37= 1/5.245.786

 

But they paid a lot more so they could choose 14 numbers of the 42.

and they did that 27times (27*1500euro)

 

Now how to calculate the chance that they win (or lose)?

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The probability of winning once with 14 numbers will be (14 choose 6)/(42 choose 6) = 33/57646. The probability of losing every time after 27 (equally random) trials will be (probability of losing once)^27. Thus, the probability of winning at least once will be 1-(1-P(winning once))^27 = 1-(1-33/57646)^27 = 0.01534 = 1.534%. So actually, it seems like their "investment" may not have been as terrible an idea as it normally is.

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(Just moved this to applied math, since it doesn't really fit in well here :))

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The probability of winning once with 14 numbers will be (14 choose 6)/(42 choose 6) = 33/57646

 

how is (42 choose 6) = 57646 ?

 

I got 42!/(6!*36!) = 5245786

 

3003/5245786=1/1746.85 (=33/57646) => 27/1746.85 = 0.01546

 

edit:

Any ideas why 0.01546 <> 0.01534

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(42 choose 6) isn't 57646, but (14 choose 6)/(42 choose 6) = 33/57646, as you've shown. I simply reduced the fraction before writing it out (that is, my calculator reduced it before it printed the answer on its screen).

 

I don't know what you mean by "<>", but I assume that the approximate equality is simply coincidence. You can't find the probability of X after n trials with n*P(X), because otherwise X would have a probability of 1 (or higher!) after a certain number of trials, and that's (somewhat) intuitively not the case. The formula that I used is P(at least one X after n trials) = 1-(1-P(X))^n, which merely approaches 1 as n increases, which makes sense.

 

If you want to see it visually, pick a value of P(X) -- say, 0.4, and graph y = 0.4*n and y = 1-(1-0.4)^n on a graphing calculator, if you have one (or work it out by hand if you're REALLY interested). The two functions obviously don't behave the same way.

 

As for why the formula I used is the correct one, think of it this way: the probability that X will happen at least once is "one minus the probability that X will not happen at all." The probability that X will not happen at all is equal to "the probability that it will not happen the first time" times "the probability that it will not happen the second time," etc. Finally, the probability that it will not happen at any given single time is "one minus the probability that it will happen."

 

Even though the above paragraph is pretty tedious, if you slog through it and write it all down algebraically, you ought to end up with the same thing I did.

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I don't know what you mean by "<>"' date=' but I assume that the approximate equality is simply coincidence. You can't find the probability of X after n trials with n*P(X), because otherwise X would have a probability of 1 (or higher!) after a certain number of trials, and that's (somewhat) intuitively not the case. The formula that I used is P(at least one X after n trials) = 1-(1-P(X))^n, which merely approaches 1 as n increases, which makes sense.

[/quote']

 

"<>" means different (programming)

You'r right, but for very small chances n*P(X) gives a very good aproximation.

 

BTW the easiest way to explain that 1-(1-P(X))^n is correct.

is using a change of 1/2 that you do two times.

1-(1-0.5)^2 = 0.75 (0 1;0 0;1 0;1 1) 3 out of 4 have a 1 in it => 0.75

2*0.5 = 1 (wrong)

 

edit:

here a graph: (the second one is to show how it keeps going)

The yellow full line is the correct one. 1-(1-P(X))^n

The one above it is the n*P(X) (linear)

The purple one is the difference (error)

The red one the % error almost 6% for 200 (less than 1% for 27)

lottograph.GIF

lottograph2.GIF

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