Any Math Wizards Around?

L Gilbert

Winterized
Nov 30, 2006
23,738
107
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50 acres in Kootenays BC
the-brights.net
Chances of losing!!! That's what I had screwed up, I was thinking (2999-10)/2999 = 99.666556% then (2998-10)/2998 = 99.666444% ........ (2991-10)/2991 = 99.66566% .......... (2981-10)/2981= 99.664542% etc. but it seemed screwy and I was in the middle of figuring out what was screwy and I got a brain fart and then sidetracked by helping someone get unstuck.
 

gc

Electoral Member
May 9, 2006
931
20
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More specifically, the probability of winning with one ticket is a third of a percent, 0.003333 or 1/300, and the probability of winning with four tickets is very slightly less than four times that. I don't see why that should be so, frankly, so I reiterate, if anyone can show something's wrong with my argument... ? Has my creaky old brain blown a gasket somewhere?

Basically speaking, because there is a very small probability that you can win more than once. So, the fact that you have a relatively smaller chance of winning with four tickets is made up for by the fact that you can win two, three or four times. So, multiply the probability of winning twice by two, and the probability of winning three times by three and it "averages out" to be the same.

Probabbility of winning exactly once = 0.013213613
Probability of winning twice with 4 tickets = 0.0000597
winning three times = 0.000000107
four times = 6.23 x 10^-11

Probability of winning AT LEAST once = 0.01327342

0.013213613 + 2(0.0000597) + 3(0.000000107) + 4(6.23 x 10^-11) = 0.0133333333
 
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