# odds of winning **URL:** \ **Category:** In My Humble Opinion\ **Created:** [July 17, 2004, 1:13am UTC](https://boards.straightdope.com/t/odds-of-winning/255522 "2004-07-17T01:13:25Z")\ **Posts on this page:** 13\ **Page:** 1
**Author:** ![vanilla](https://avatars.discourse-cdn.com/v4/letter/v/f19dbf/32.png) [@vanilla](https://boards.straightdope.com/u/vanilla)\ **Post date:** [July 17, 2004, 1:13am UTC](https://boards.straightdope.com/t/odds-of-winning/255522/1 "2004-07-17T01:13:25Z")
Okay. Our library is having a reading fun thing. It lasts 5 weeks. You, an adult, read books, lists the books on little yellow squares, with name of book, author and short comment on one side and name and address on the other. You put them in a box. Each week they draw winners, but the grand Winner is drawn July 31. I am up to 50 books already. My question is, would I have a better chance of winning if I put them ALL in one weeks box, or divided them up weekly, where there would be less entries, but more chances weekly. ? ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [July 17, 2004, 1:27am UTC](https://boards.straightdope.com/t/odds-of-winning/255522/2 "2004-07-17T01:27:04Z")
When they draw a card for each week’s winners, do they throw it out, or do they put it back in? And where do they get the pool of grand prize winners? If they’re just putting the weekly boxes together, any difference in odds is more or less negligible and really difficult to calculate. Are you thinking of dividing them up evenly, or is that up for grabs. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [July 17, 2004, 1:34am UTC](https://boards.straightdope.com/t/odds-of-winning/255522/3 "2004-07-17T01:34:40Z")
OK, I gave this a little thought, and I think you have better odds of winning the grand prize if you dump all your entries in one week’s basket. Say you have k entries. If you put some entries in each basket, you could win up to 5 times, which means that you could have k - 5 entries in the final basket. But if you put all of them in one basket, you can win at most one week’s prize, leaving you at least k - 1 entries. It’s a negligible difference if there are a lot of entries, but it’s real. ---
**Author:** ![Duke](https://avatars.discourse-cdn.com/v4/letter/d/e8c25b/32.png) [@Duke](https://boards.straightdope.com/u/Duke)\ **Post date:** [July 17, 2004, 1:39am UTC](https://boards.straightdope.com/t/odds-of-winning/255522/4 "2004-07-17T01:39:03Z")
I could be totally off-base here, but I think I’m right. What I seemed to find out when I ran the numbers is that it is actually slightly to your advantage to spread the tickets out. Here’s how it works. Let’s say that, apart from your tickets, there are 1,000 tickets in the draw every week. In Scenario A, you add your 50 tickets to the first week’s pile. Your chances of winning are now 50/1050, or 4.7619%. In Scenario B, you add 10 tickets to the 1,000 other weekly tickets. Your chances of winning every week are 10/1010, or 0.9901%. If you add up your chances of winning each week, your chances become 4.9505% minus the miniscule chance that you will win more than once (your chances of winning twice are a mere 0.009803%). Therefore, your chances in winning in Scenario B are increased by 0.1788%. The more tickets in the drawing every week, the less advantageous Scenario B becomes. If there are 10,000 tickets in the drawing other than yours, your chances of winning under Scenario A are 0.4975%; under Scenario B, 0.4995%–an increase of a mere 0.002%. Why would it even matter whether you split up the tickets? If the library’s drawing stated “All tickets have a 1 in X chance of winning,” it wouldn’t. But because the chance of winning depends on the number of entries, the tickets you add to the pile lengthen the odds of winning. In the above Scenario A, each of your tickets has a 1 in 1050 chance of winning. In Scenario B, each of your tickets has a 1 in 1010 chance of winning. Therefore, although the difference is slight, it will pay off to spread out your tickets! ---
**Author:** ![Duke](https://avatars.discourse-cdn.com/v4/letter/d/e8c25b/32.png) [@Duke](https://boards.straightdope.com/u/Duke)\ **Post date:** [July 17, 2004, 1:43am UTC](https://boards.straightdope.com/t/odds-of-winning/255522/5 "2004-07-17T01:43:00Z")
> [@ultrafilter](#): > > OK, I gave this a little thought, and I think you have better odds of winning the grand prize if you dump all your entries in one week’s basket. Actually, I dunno about that one, either. How is the grand prize winner being determined? If they put together all the losing entries from the previous weeks, then actually Scenario A above gives you a better chance (because you will have less chance at a winning entry from previous weeks!) If it’s just picked from the last week’s entries, then by all means throw all your tickets in the last week’s pot. But if you want a chance at “winning,” as opposed to “winning the grand prize,” spreading out the tickets makes more sense. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [July 17, 2004, 1:46am UTC](https://boards.straightdope.com/t/odds-of-winning/255522/6 "2004-07-17T01:46:08Z")
> [@Duke](#): > > If they put together all the losing entries from the previous weeks, then actually Scenario A above gives you a better chance (because you will have less chance at a winning entry from previous weeks!) Does it? Putting no tickets in a given week’s box makes you less likely to win that week’s prize than putting some tickets in. > [@](#): > > But if you want a chance at “winning,” as opposed to “winning the grand prize,” spreading out the tickets makes more sense. Agreed. ---
**Author:** ![Duke](https://avatars.discourse-cdn.com/v4/letter/d/e8c25b/32.png) [@Duke](https://boards.straightdope.com/u/Duke)\ **Post date:** [July 17, 2004, 1:59am UTC](https://boards.straightdope.com/t/odds-of-winning/255522/7 "2004-07-17T01:59:42Z")
> [@ultrafilter](#): > > Does it? Putting no tickets in a given week’s box makes you less likely to win that week’s prize than putting some tickets in. That’s what I mean. If you don’t spread the tickets around, you have less chance of winning a weekly prize. If you win a weekly prize, and only losing tickets from previous weeks make the grand prize drawing, that ticket will not go into the Grand Prize drawing; hence by maximizing your chances at winning a weekly prize, you lessen your chances of winning the Grand Prize. ---
**Author:** ![wolfman](https://avatars.discourse-cdn.com/v4/letter/w/a8b319/32.png) [@wolfman](https://boards.straightdope.com/u/wolfman)\ **Post date:** [July 17, 2004, 3:01am UTC](https://boards.straightdope.com/t/odds-of-winning/255522/8 "2004-07-17T03:01:36Z")
> [@](#): > > Your chances of winning are now 50/1050, or 4.7619%. In Scenario B, you add 10 tickets to the 1,000 other weekly tickets. Your chances of winning every week are 10/1010, or 0.9901%. If you add up your chances of winning each week, your chances become 4.9505% minus the miniscule chance that you will win more than once (your chances of winning twice are a mere 0.009803%). Therefore, your chances in winning in Scenario B are increased by 0.1788%. Not quite true. You can’t add up odds like that, or else in 110 ten weeks your odds would be over 100%. You have to figure out the odds that you never win. And subtract from 1. Each week odds of not winning = 100.0000-.9901= 99.099% =.99099 Odds of not winning for X weeks in .99099^X, so .099099^5=.9558 Winning at least once in 5 weeks= .04424 or 4.424% which is less than 4.762 ---
**Author:** ![Duke](https://avatars.discourse-cdn.com/v4/letter/d/e8c25b/32.png) [@Duke](https://boards.straightdope.com/u/Duke)\ **Post date:** [July 17, 2004, 8:05pm UTC](https://boards.straightdope.com/t/odds-of-winning/255522/9 "2004-07-17T20:05:14Z")
**wolfman** : Your numbers are right. I did in fact try to subtract the chances of winning more than once during the five-week period, but used the wrong numbers: the percentage I used to designate winning twice (0.009803%) are actually the chances of winning _twice in a row_, not twice in five weeks :smack: My apologies for leading anyone down the wrong path here. Props to **wolfman** , who corrected me. ---
**Author:** ![vanilla](https://avatars.discourse-cdn.com/v4/letter/v/f19dbf/32.png) [@vanilla](https://boards.straightdope.com/u/vanilla)\ **Post date:** [July 17, 2004, 8:26pm UTC](https://boards.straightdope.com/t/odds-of-winning/255522/10 "2004-07-17T20:26:16Z")
I went in and asked them today. They said they use all the cards for the Grand prize, so apparently it would have been better odds had I waited and put them all in at once. ---
**Author:** ![1010011010](https://avatars.discourse-cdn.com/v4/letter/1/ea666f/32.png) [@1010011010](https://boards.straightdope.com/u/1010011010)\ **Post date:** [July 18, 2004, 5:09pm UTC](https://boards.straightdope.com/t/odds-of-winning/255522/11 "2004-07-18T17:09:12Z")
You’ll have even better odds if you _read more books_, which I’m pretty sure is the intent of the whole exercise from the library’s point of view. To determine the best scenario you’d also have to consider the relative values of each week’s prizes… and thus work out the trade off between improving your chances of winning the grand prize by reducing (or removing) the chance of winning a previous week’s prize. Similarly, you’d have to compare the gain from winning a previous week’s prize against the reduction having a ticket removed from the grand prize drawing. If the Grand prize is worth X, and the chance of winning the grand prize with a particular ticket is Y, the value of the ticket is Y_X. If the value of the previous weeks’ prizes is significantly greater than X_Y, then losing a ticket (by winning a previous week’s prize), isn’t a loss… because the prize is worth more than the theoretical value of the ticket. That’s how lotteries make money, BTW- charge more for the ticket than they’re theoretically worth. ---
**Author:** ![Monocracy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/monocracy/32/4493_2.png) [@Monocracy](https://boards.straightdope.com/u/Monocracy)\ **Post date:** [July 18, 2004, 6:26pm UTC](https://boards.straightdope.com/t/odds-of-winning/255522/12 "2004-07-18T18:26:41Z")
> [@wolfman](#): > > Not quite true. You can’t add up odds like that, or else in 110 ten weeks your odds would be over 100%. You have to figure out the odds that you never win. And subtract from 1. > > Each week odds of not winning = 100.0000-.9901= 99.099% =.99099 > > Odds of not winning for X weeks in .99099^X, so .099099^5=.9558 > > Winning at least once in 5 weeks= .04424 or 4.424% which is less than 4.762 Your numbers are a little off. Each week’s odds of not winning, if it’s 10 in 1010, comes out to .990099, not .99099. It brings the final number, winning at least once in 5 weeks, to 4.853%, which is slightly better than 50 in 1050, which is 4.762%. So there are better odds if you divide up the entries, plus you have a small chance of winning multiple times. ---
**Author:** ![wolfman](https://avatars.discourse-cdn.com/v4/letter/w/a8b319/32.png) [@wolfman](https://boards.straightdope.com/u/wolfman)\ **Post date:** [July 18, 2004, 8:26pm UTC](https://boards.straightdope.com/t/odds-of-winning/255522/13 "2004-07-18T20:26:43Z")
> [@](#): > > Your numbers are a little off. Each week’s odds of not winning, if it’s 10 in 1010, comes out to .990099, not .99099. Ahh crap, sorry. :smack: Is there already a mathematical correlary to the web truism that everytime you notice a error in a post, you make really stupid mistake yourself correcting it . 😉 Thanks Mono.