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In 6/49 lottery 6 numbers are drawn from 49, and jackpot prize is win if ticket matches all 6 numbers. Probability of winning on one play is 1/13983816.

What are the odds of wining jackpot prize in case of:

  1. Buying X tickets for 1 draw.
  2. Buying 1 ticket (with the same numbers each time) for X draws in a row.
  3. Buying 1 ticket (with random numbers each time) for X draws in a row.

Lets assume $X = 8$

  1. First is trivial: $8/13983816 = 1/1747977$
  2. I'm not sure but think that it will be: $1-(1-1/13983816)^8 = 1/1747977.43835098$

  3. And here I'm confused. Will it be the same as in 2nd case or maybe some other way ?

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    $\begingroup$ Yes, we have 3. = 2. $\endgroup$ – Hagen von Eitzen May 5 '13 at 22:30
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Yes, the answer for $3$ is the same as $2$, which you have correct. The logic is the same-you have $8$ successive shots with probability $1/13983816$. The reason the chance is (marginally) lower in $2$ and $3$ is that in those you have (a tiny) chance for winning twice, so the chance of losing all must go up.

Some of us would take exception to the last equals sign. The number is calculator accurate, but not exact equality. Alpha gives me $1.747977437500046929251891230733416853156459995356317343093051705432886833053224746096036022393413364855116416926038... \times 10^6$ and isn't done yet.

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  • $\begingroup$ Ok, thanks. (I have calculated in Excel :) $\endgroup$ – anth May 5 '13 at 22:44

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