Formula for nobody winning is ((N-1)/N)^T where N is ...

Reply #3

Gamma Ray — Fri, 19 Apr 2002 17:04:01 GMT

Your formula is creative, but I don't think it is correct!Assume there is a lottery with 1/10 odd of win (9/10 Odd of lose), and suppose 10 different tickets sold (all combinations). With your formula, chance of nobody win= ((N - 1)/N)^T= ((10 -1)/10^10= 0.347In simple logic, the winning ticket is sold. It is because all possible pattern sold. Therefore, the formula is false

Reply #2

Guest — Fri, 19 Apr 2002 12:49:36 GMT

Well, if one person buys one ticket the chance that that person will lose is 76,275,359/76,275,360. Almost 100%. I f two tickets are sold the chance that they will both lose is (76,275,359/76,275,360)*(76,275,359/76,275,360) or (76,275,359/76,275,360)^2. If there are three tickets it will be (76,275,359/76,275,360)^3. So it is 76,275,359/76,275,360 raised to the power of the number of tickets sold.

Reply #1

Guest — Thu, 18 Apr 2002 20:19:46 GMT

How did you arrive at the equation

Formula for nobody winning is ((N-1)/N)^T where N is ...

Guest — Wed, 17 Apr 2002 21:02:05 GMT

A few messages ago I was wondering how to calculate the odds that nobody would win. Looks like it's ((N - 1)/N)^T where N is the total pool of numbers and T is the number of tickets sold. Let's apply this to the drawing just held: 2,744,447 won $1 with the odds of 1 in 62. So there were approximately 170,155,714 tickets sold. (76,275,359/76,275,360)^170,155,714 gives 10.75% chance that nobody wins.