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

What are the odds of winning jackpot prize in case of buying:

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

Let's assume $X = 8$.

  1. First is trivial: $8/13983816 = 1/1747977$

  2. I'm not sure. I think that this will be: $1-(1-1/13983816)^8 = 1/1747977.43835098$

  3. Here I'm confused. Will this be the same fraction as in case 2 above?

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    $\begingroup$ Yes, we have 3. = 2. $\endgroup$ Commented May 5, 2013 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
    Commented May 5, 2013 at 22:44

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