tag:blogger.com,1999:blog-22557003.post6802984972277452638..comments2013-04-17T19:36:48.021+01:00Comments on Graham Poulter's Blog: Tales from ObzFestGraham Poulternoreply@blogger.comBlogger3125tag:blogger.com,1999:blog-22557003.post-61928805847250385912008-01-03T08:03:00.000+00:002008-01-03T08:03:00.000+00:00Thanks Graham... i feel so honored. Philip OwiraThanks Graham... i feel so honored. Philip OwiraPhiliphttps://www.blogger.com/profile/05135099036565142871noreply@blogger.comtag:blogger.com,1999:blog-22557003.post-80260830455874695842007-12-15T07:31:00.000+00:002007-12-15T07:31:00.000+00:00Hello. From http://www.salottery.co.za/lotto/faq....Hello. From http://www.salottery.co.za/lotto/faq.aspx<BR/>we see that the odds of a division 1 win is 1 in 14 million probability and division 2 is 1 in 2.3 million <BR/><BR/>Suppose you would like a 5% chance of winning division 2, then you need 117,974 different entries (see end). At R2.50 per entry, that's almost R300,000 for a 5% chance. From http://www.salottery.co.za/lotto/results.aspx, we see the division 2 payout is R60000. Your "expected value" is probability of win times value of win = 0.05 * 60000 = R3000. R3000/R300,000 = 1%.<BR/><BR/>So, after playing the lottery for a long time, your division 2 payout would be roughly 1% of the money you put in.<BR/><BR/>[Math: to find N (number of tickets), to have a q=0.05 chance of winning given p=1/2.3e6 chance per ticket, solve 1-q = (1-p)^N. The answer is N = log(1-q)/log(1-p)]Graha/V\https://www.blogger.com/profile/16072516650932490004noreply@blogger.comtag:blogger.com,1999:blog-22557003.post-44456055195966811742007-12-14T17:41:00.000+00:002007-12-14T17:41:00.000+00:00This is a rather arbitrary place to leave this mes...This is a rather arbitrary place to leave this message, because it does not relate to the topic of the blog entry.<BR/><BR/>Your profile pic is very nice. I think you should get a further iteration of that haircut.<BR/><BR/>What I wanted to ask you was this (since you know stats better than I do): It is my opinion that if you want to win the lottery, you must enter the lottery. Whereas it is a mathematical fact that you increase the probability of winning by submitting additional entries, you do not really have a "better chance of winning" if you submit two entries instead of one, because two in a million is not significantly more than one in a million. Therefore, you are wasting money buying three or five lines when you could just buy one. You could only significantly raise your probability of winning if you had multiple entries in an entirely different order of magnitude, say, a hundred in a million. Now: Is there any mathematical way of expressing what I am trying to say?Tania.co.zanoreply@blogger.com