Probability of Winning the Lottery

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KoalaMaths

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49 to the power of 6 is 13 billion. So why does it say that the chances are 1 in 13 million? 6 balls, each go from 1 to 49, so surely it is (1/49)^6 no?
 
Is each ball returned to the bucket after it is drawn? Said another way, can you get 35, 35, 35, 35, 35, 35?
Does it matter in which order the numbers appear? Said another way, is 1,2,3,4,5,6 the same as 6,5,4,3,2,1?

I get 13.98 million

Truthfully, BTW, if you don't play, your chances of winning are 0. It has nothing to do with the arithmetic at this point.
 
tkhunny said:
Is each ball returned to the bucket after it is drawn? Said another way, can you get 35, 35, 35, 35, 35, 35?
Does it matter in which order the numbers appear? Said another way, is 1,2,3,4,5,6 the same as 6,5,4,3,2,1?

I get 13.98 million

Truthfully, BTW, if you don't play, your chances of winning are 0. It has nothing to do with the arithmetic at this point.
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Order matters and you can get 35 6 consecutive times. When I put (1/49)^6 in my calculator it gives me 13.8 billion???
 
KoalaMaths said:
Order matters and you can get 35 6 consecutive times. When I put (1/49)^6 in my calculator it gives me 13.8 billion???
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You're just repeating yourself. So, who's right, you or the lottery sponsor? The sponsor has a conflict to encourage participation.
 
tkhunny said:
You're just repeating yourself. So, who's right, you or the lottery sponsor? The sponsor has a conflict to encourage participation.
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You said you got 13.8 million previously. How did you reach this conclusion?
 
[MATH]\left ( \dfrac{1}{49} \right )^6 \approx \dfrac{1}{38.8 \text { billion}} \approx 0.[/MATH]
If you are neutral between winning a dollar and losing a dollar, you would have a better chance of amusing yourself by withdrawing your life savings in currency, creating a bonfire in the backyard, and dancing around it naked in the company of someone with a liberated spirit unless the payoff exceeds 38.8 billion.
 
JeffM said:
[MATH]\left ( \dfrac{1}{49} \right )^6 \approx \dfrac{1}{38.8 \text { billion}} \approx 0.[/MATH]
If you are neutral between winning a dollar and losing a dollar, you would have a better chance of amusing yourself by withdrawing your life savings in currency, creating a bonfire in the backyard, and dancing around it naked in the company of someone with a liberated spirit unless the payoff exceeds 38.8 billion.
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But this isn't answering my question as to why Google says the chances are one in thirteen million
 
KoalaMaths said:
49 to the power of 6 is 13 billion. So why does it say that the chances are 1 in 13 million? 6 balls, each go from 1 to 49, so surely it is (1/49)^6 no?
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I am missing something!

Where is the problem statement for which we are calculating these chances?
 
Let's think about two balls numbered 1 through 3.

If order matters, we have 3^2 = 9 possibilities with equal probabilities of 1/9

11
12
13
21
22
23
31
32
33

But if order does not matter, we have six possibilities with unequal probabilities

11: 1/9
12, 21: 2/9
13, 31: 2/9
22: 1/9
23, 32: 2/9
33: 1/9

So is 49, 48, 49, 48 the same as 48, 48, 49, 49 or not?
 
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