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December 20th, 2016, 05:05 AM  #1 
Newbie Joined: Dec 2016 From: Scotland uk Posts: 4 Thanks: 0  Just joined (raffle probability problem)
Ok so here in Scotland we run a raffle to raise funds for our Village Hall. We have a variety of prizes and we sell raffle tickets numbered from 001, 002 etc upwards and which are printed in strips of 5 tickets (i.e 001  005, 006  010 etc). We sell as many tickets as people are willing to buy at £1 per strip of 5 (we don't sell less than a full strip). We can then run the raffle in two ways; first we put every ticket individually into the bucket and draw a ticket one at a time. Whoever has the matching ticket wins a prize. Secondly we put the whole strips of 5 tickets into the bucket and draw one strip at a time. Whoever has the matching strip wins a prize. Drawn tickets or strips are not returned to the bucket and there's no limit to how many prizes people can win. So if for example we have 30 prizes and we sell 1000 tickets (and people usually buy around £5 worth) there will either be 1000 tickets in the bucket with 1000 matching individual tickets in the audience, or 200 strips of tickets in the bucket with 200 matching strips in the audience. Question is  given that with either system you have the same chance of winning one prize (I think!), are you more or less likely to win more than one prize with the ticket system or the strip system. I was always sure that it makes no difference but my maths is not up to proving it to people. Hope someone can help. Last edited by greg1313; December 20th, 2016 at 11:05 AM. 
December 20th, 2016, 07:28 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,388 Thanks: 849 Math Focus: Elementary mathematics and beyond 
Welcome Chaddie! You are correct. Strip scenario, player has 1 strip entered: 1/200. Single ticket scenario, player has one strip entered: 5/1000 = 1/200. Conclusion: No difference. Now it's basic math to show the odds are the same in any scenario, i.e. no matter how many strips the player has entered. Let x be the number of strips entered: x/200 compares to 5x/1000 = x/200. Got it? 
December 21st, 2016, 03:54 AM  #3 
Newbie Joined: Dec 2016 From: Scotland uk Posts: 4 Thanks: 0 
Thanks for that but I'm still confused. Let's assume 1000 tickets/200 strips sold, 30 prizes and I've bought 25 tickets/5 strips. I'm told that the chance of anyones ticket winning the first prize is 30/1000 and if I have 25 tickets then my chance of winning that prize is 25 x 30/1000 = 0.75. Likewise for strips it would be 5 x 30/200 = 0.75. But then the chance of any ticket winning the second prize is 29/999 and I have 24 tickets left so my chance of the second prize is 24 x 29/999 = 0.6966, but with strips it's 4 x 29/199 = 0.5829 so I have less chance of a second prize with strips?? And continuing that logic to the fifth prize my chance with tickets is 26/996 x 21 = 0.5819 and with strips it's 26/196 x 1 = 0.1326 so a very much smaller chance with strips. Or is the logic here wrong? And of course with single tickets then in theory I could win 25 prizes, one with every ticket, but with strips I can only win a maximum of 5 prizes. 
December 21st, 2016, 09:47 PM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,388 Thanks: 849 Math Focus: Elementary mathematics and beyond 
The number of prizes has no effect on the probability of winning a draw  that depends solely on the number of strips/tickets a player has entered. On the inital draw the probability of a player winning a prize is the same whether you are using strips or tickets. In subsequent draws, the single ticket system gives more chances of winning. For example five strips would give a player 5 chances of winning under the strip system and 25 chances under the single ticket system. Also, the probability of winning is greater for the single ticket system. Consider the expression below (where $t$ is the number of tickets, $s$ is the number of strips and $d$ is the number of draws that have previously been held) $$\frac{td}{1000d}\frac{sd}{200d}$$ which is the probability of winning with a ticket less the probability of winning with a strip, from which we may derive the inequality $$(200d)t+800d(1000d)s\ge0$$ I wish for a simpler way of putting it but I haven't found one. If you have any questions please don't hesitate to ask. 
December 22nd, 2016, 01:27 AM  #5 
Newbie Joined: Dec 2016 From: Scotland uk Posts: 4 Thanks: 0 
Hi, so many thanks for that which I now understand better. Am I correct in assuming that the expressions you give assume that you win; i.e (t  d)/(1000d) where your number of tickets reduces each draw implying you must have won and discarded one ticket each draw. If you did not win the first draw then you would still have tickets 't', not 'td' so the probability of a win in the next draw becomes t/(1000d) or s/(200d) in which case the strip option gives a slightly higher chance of success? So if you win the first draw you're better with tickets, if you don't win the first draw you're better with strips? 
December 22nd, 2016, 03:52 AM  #6 
Newbie Joined: Dec 2016 From: Scotland uk Posts: 4 Thanks: 0 
And assuming that I have bought less than half of all the tickets/strips sold then my chances of winning the first draw are less than even, so I'm likely to lose the first draw so overall I would be better with a strip system? Or would I? 
December 22nd, 2016, 08:13 AM  #7 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,388 Thanks: 849 Math Focus: Elementary mathematics and beyond 
It gets complicated and the odds vary from person to person depending on whether they have won. I personally favour the single ticket system  which would you rather have  five chances of winning or one chance of winning for the same price?


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