Lottery number odds.

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There are 39 numbered ping pong balls. The people who control the handling of the balls have a very complex way of making the draws 100% random.
The cash rewards are for 2-3-4 or 5 numbers.

The chances of one of our 5 numbers drawn in the first draw is 5 in 39 or 12.82% = ?
Then the chances of one of our other 4 numbers being drawn in round two would be 4/38 = 10.52% = ?

Their site says the chances for match 2 is 9.2%
3 = 1 in 95
4 = 1 in 3042
5 = 1 in 501.942

If four of our numbers have been drawn there are only 35 numbers left to be drawn. Why isn't the chance on the 5th number 1/35?
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Commented:
It would be helpful if you pointed to the site making the claims or at least copy it here.  I think you're comparing different odds.

Your numbers appear to be correct for the first and second tries (5/39 and 4/38).  If you want to know what the odds are of having two matches on the first two picks, you have to multiply those percentages 12.82% x 10.52%=1.34%.  Divide 1 by the percentage and you'll get the odds: 1/1.34%=74.6.  That tells me you have a 1 in 74.6 chance in having the first two lottery picks match your numbers.

To continue:
Third pick is 3/37, which is about 8.1%.  Odds of getting success on first three picks is about .109% or 1 in 914
Fourth pick is 2/36, which is about 5.5%.  4 success is about .006% or 1 in 16,450
Fifth pick is 1/36, which is about 2.9%.  5 successes is about .0002% or 1 in 575,757

I believe that the site you're referring to is showing you the cumulative odds (the odds of getting the first three successfully, for example), not the individual odds.  What's not clear to me is why my numbers are off from theirs.

Commented:

Commented:
If I'm reading it correctly, they are using 38, not 39 numbers.

I've attached a spreadsheet to let you see the calculations.

What's interesting is that I get the same last answer (1 in 501,942) as the site, but the other odds aren't the same.  I'll leave it to someone else to sort that out!
Lottery.xlsx

Commented:
A Lottery drawing is one of those cases where order does not matter.  You pick five numbers, they pick five balls, and then we look at the results.

Presumably you are wondering: If the odds of picking four balls are 1 in 3042, why aren't the odds of picking five 1 in 34x3043 or 103,428?
The 1 in 3042 odds is for the case where you have four of the five winning numbers and one losing number.

The odds that the first four balls drawn match your five numbers are 5 in C(38,4)  or  1 in 14,763.

And 34 x 14,763 does equal 501,942.

**

Commented:
@d-glitch:

Thank you!  That's exactly what I was missing.  My stats show the odds of having the first x correct choices.  The published odds are for EXACTLY x correct choices after all 5 are done.  That's why my result matched for the last one (5 out of 5) but not for the rest.

Commented:
Are the odds the same if the first 4 numbers they draw match 4 of our 5 numbers?
Once we have the first 4 what are the odds of getting the 5th one. It seems like it would be 1 in 34 or around 3%.

But before any balls are drawn what are the odds of getting all 5 when the number of balls in the draw are 38.

Commented:
If there are 38 balls total and you pick 5, the odds of you getting all 5 right are 1 in 575,757 or about 0.0002%.

Commented:
Thanks.

Commented:
>>  If there are 38 balls total and you pick 5, the odds of you getting all 5 right are 1 in 575,757 or about 0.0002%

That should be  C(38, 5)  =  38! / (33! * 5!)  =  501,942

Commented:
>>  Once we have the first 4 what are the odds of getting the 5th one. It seems like it would be 1 in 34 or around 3%.

Yes.  If the first four balls drawn match four of your five numbers, the odds of matching the final number are exactly 1 in 34.

Commented:
Yes, my statement with 575,757 was completely incorrect.  My spreadsheet had it correct (501,492) as well as the lottery site.  I quoted the wrong number with my answer.  Thank you for the correction.

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