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# Help me beat the odds.

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# Help me beat the odds.

August 12, 2011 3:22 PM Subscribe

How do you improve your odds of winning at a drawing with multiple prizes?

There's a prize drawing at work. There are six prizes, all employees have multiple tickets, depending on OT hours worked during July. So I have 13 tickets. I like three of the prizes (iPad 2, NYC Broadway trip, or 1000$), and I would like to know the numbers/odds of increasing my chances of winning. I don't

Keep in mind that everyone has completely different amounts of tickets, and 13 could be a pretty low number in comparison.

I don't actually know if I can win more than one prize, lets go with the assumption if I'm drawn more than once I have to choose between prizes.

Should I put all 13 in my favorite? Split them up? Try to guess which has the least people gunning for it?

There's a prize drawing at work. There are six prizes, all employees have multiple tickets, depending on OT hours worked during July. So I have 13 tickets. I like three of the prizes (iPad 2, NYC Broadway trip, or 1000$), and I would like to know the numbers/odds of increasing my chances of winning. I don't

*care*that much, but I was curious.

Keep in mind that everyone has completely different amounts of tickets, and 13 could be a pretty low number in comparison.

I don't actually know if I can win more than one prize, lets go with the assumption if I'm drawn more than once I have to choose between prizes.

Should I put all 13 in my favorite? Split them up? Try to guess which has the least people gunning for it?

Sorry, I knew I would forget something. There are six separate prize boxes, one for each. And I know I want to put them only in the ones I want. But there are three I'd be okay with. I'm trying to see A)Putting 13 in one prize box is better or B)putting 4 or 5 in each.

posted by trogdole at 3:29 PM on August 12, 2011

posted by trogdole at 3:29 PM on August 12, 2011

You will need to account for 2 parameters:

1. Your preference: do you prefer all three prizes equally? or do you prefer one over another. One could use the prize value as weights, so: w1, w2 and w3.

2. The number of tickets for each of the prizes: if you can see how many tickets in each boxes, then you can calculate your odds. If you can't see inside the boxes, then you just have to guess how many tickets each box contains; let them be n1, n2 and n3.

Let the number of tickets you put in each box be t1, t2 and t3:

What you want to do is maximize your return, which is calculated as follow:

return = w1*t1/n1 + w2*t2/n2 + w3*t3/n3 and

t1 + t2 + t3 = 13.

This function is linear, so ideally, you want to put all your ticket into the one box with the highest w/n ratio; or: put all your ticket into the item you estimate to like the most which you think will get the least number of tickets.

posted by curiousZ at 3:43 PM on August 12, 2011 [1 favorite]

1. Your preference: do you prefer all three prizes equally? or do you prefer one over another. One could use the prize value as weights, so: w1, w2 and w3.

2. The number of tickets for each of the prizes: if you can see how many tickets in each boxes, then you can calculate your odds. If you can't see inside the boxes, then you just have to guess how many tickets each box contains; let them be n1, n2 and n3.

Let the number of tickets you put in each box be t1, t2 and t3:

What you want to do is maximize your return, which is calculated as follow:

return = w1*t1/n1 + w2*t2/n2 + w3*t3/n3 and

t1 + t2 + t3 = 13.

This function is linear, so ideally, you want to put all your ticket into the one box with the highest w/n ratio; or: put all your ticket into the item you estimate to like the most which you think will get the least number of tickets.

posted by curiousZ at 3:43 PM on August 12, 2011 [1 favorite]

Ok, napkin math time.

So, assuming there are 200 tickets already in each box (100 employees with 12 tickets each, divided evenly among the 6 prizes), you would have about a 6% chance of winning if you put all your tickets in one box (13/213=6.1%). If you put 4 in each, you would have a 1.96% chance of winning each (4/204=1.96%). Odds of losing all boxes is 94% (1-(.98^3))! So you're about even either way it looks like. Keep in mind though, popularity of various prizes will change up those odds. Also, perception of popularity of those prizes will change the odds too.

Having said that though, you have the chance to win all three prizes and go home with a huge haul if you split your tickets. Also, you have the suspence of three drawings where you're listening intently for your name. Not exactly math based, but that's what I'd do.

posted by Garm at 4:08 PM on August 12, 2011

So, assuming there are 200 tickets already in each box (100 employees with 12 tickets each, divided evenly among the 6 prizes), you would have about a 6% chance of winning if you put all your tickets in one box (13/213=6.1%). If you put 4 in each, you would have a 1.96% chance of winning each (4/204=1.96%). Odds of losing all boxes is 94% (1-(.98^3))! So you're about even either way it looks like. Keep in mind though, popularity of various prizes will change up those odds. Also, perception of popularity of those prizes will change the odds too.

Having said that though, you have the chance to win all three prizes and go home with a huge haul if you split your tickets. Also, you have the suspence of three drawings where you're listening intently for your name. Not exactly math based, but that's what I'd do.

posted by Garm at 4:08 PM on August 12, 2011

Heh. My dad always swore that if he folded over or crinkled up his raffle tickets, he'd win. He claims that whenever the person picking the winning ticket would reach in, their fingers would "catch" on the folded over or crinkled ticket and they'd tend to pick it up.

He won prizes alot. (I once tried this at a Rotary Club raffle and won $100. YMMV.)

posted by HeyAllie at 4:36 PM on August 12, 2011

He won prizes alot. (I once tried this at a Rotary Club raffle and won $100. YMMV.)

posted by HeyAllie at 4:36 PM on August 12, 2011

Thanks guys, I knew the math would be a bit over my head. I will be splitting my tickets, and folding them over.

Bonus question: Should my 13th ticket go to the iPad, Broadway, or 1000$ visa card?

posted by trogdole at 6:29 PM on August 12, 2011

Bonus question: Should my 13th ticket go to the iPad, Broadway, or 1000$ visa card?

posted by trogdole at 6:29 PM on August 12, 2011

Well, I'd bet that the Visa card -- which is, after all, worth more than the other two -- will have the most entries of those three. So I'd put more entries on the other two than the Visa. Probably something like 5/5/3.

posted by jeather at 7:18 PM on August 12, 2011

posted by jeather at 7:18 PM on August 12, 2011

Seconding HeyAllie. I always crumple up my ticket if it's going into a drawing. It seems to win more often than it should. I was told curious fingers are drawn towards novelty.

posted by troublewithwolves at 8:19 PM on August 12, 2011

posted by troublewithwolves at 8:19 PM on August 12, 2011

I seem to be involved in a disproportionate number of raffles, and a couple of times I've heard the ticket-folders half-jokingly referred to as "cheaters," and the person drawing the tickets admonished not to pick to a folded one. So it can backfire, too.

posted by HotToddy at 7:03 AM on August 13, 2011

posted by HotToddy at 7:03 AM on August 13, 2011

This thread is closed to new comments.

For instance, if they are going to draw 13 winning tickets from a single bowl, then assign those winners one of the 13 prizes each (or let them pick) then you would increase your odds by buying more tickets, period.

If there are bowls of tickets for each prize, you'd want to buy more tickets only for the things you want to win, not the others.

posted by odinsdream at 3:25 PM on August 12, 2011