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**grifta67****Guest**

Hey everyone.

I need some help to solve a debate between my roommate and myself. We're doing a raffle drawing this coming weekend for a pro wrestling event (please save the comments . We have 10 people who will each get three numbers from a pool of 1-30. Last year we did it by simply putting the numbers 1-30 into a hat and went around the table 3 times. That seemed to work just fine.

This year, my roommate is arguing that it isnt fair to the people at the end of the rotation. He wants to put the 10 names in a hat, three times each. Then just go down the list of one through 30, pulling names to see who gets the next number.

Without going into a full explanation of the event, it is generally believed that the closer you get to 30, the better chance you have to win. Of course we're talking about wrestling, so there really is no true advantage to which number you draw, the winner is chosen ahead of time. We're just watching the event, waiting to see which number it ends up being.

My arguement is that this is a raffle amongst friends at a bar. As long as the first method is within a minor degree of error, I think it works just fine and is more fun. Part of the fun is reaching in and pulling out a number, with no idea what it could be. At that instant you could get number 1 or number 30, you just don't know. I tend to think that the ordered method of going down the list number by number kind of takes some of the excitement and the unknown factor out of it.

Any thoughts? If more explination about the event/raffle is needed, just ask.

Thanks for the help

-Sean

**irspow****Member**- Registered: 2005-11-24
- Posts: 456

I am not great with probabilities, but I have an idea that would be much fairer to everyone involved than either of the two methods above. It might take a little longer however, but then again you guys are going to be in a bar anyway.

Have eleven hats or whatever, each containing 10 numbered whatevers, and have drawings for every number. All you have to do is pick a designated "picker" for the eleventh hat. He'll draw a number, as will everyone else from their own individual hats. The one who has the same number as the "picker" will be awarded the designated number for that drawing. It is a lot harder to explain than to actually pull off.

The point is, everyone there will have the same exact odds of getting any number as anyone else. (1 in 10)

In case of more than one person picking the "correct" number, just have them go head to head again.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I think there's an equal probability of everyone winning, it's just that it wouldn't be as exciting for the people who go last because they'll know what they're going to get.

Maybe if you went around in a circle, then reversed directions and went around the other two circles in opposite directions, everyone would think it's fairer. It's fair already, but they'd *think* it's fairer.

Why did the vector cross the road?

It wanted to be normal.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

mathsyperson wrote:

I think there's an equal probability of everyone winning, it's just that it wouldn't be as exciting for the people who go last because they'll know what they're going to get.

Well Stated!

How about the guys who pick last can also be the last to buy a round of drinks ... ?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**irspow****Member**- Registered: 2005-11-24
- Posts: 456

Guys, I think your missing the point. grifta67 made it pretty clear that these guys really want to get the number 30. If your the last in rotation, your odds are significantly lower (before anyone has picked) than the guy who picks first. Obviously if the first nine guys miss it you will technically have the best statistical chance on your particular pick out of the ten.

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**grifta67****Member**- Registered: 2006-01-23
- Posts: 2

irspow is right. Even though there is no true advantage, mentally everyone wants 30, or as close as they can get.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

irspow wrote:

Guys, I think your missing the point. grifta67 made it pretty clear that these guys really want to get the number 30. If your the last in rotation, your odds are significantly lower (before anyone has picked) than the guy who picks first. Obviously if the first nine guys miss it you will technically have the best statistical chance on your particular pick out of the ten.

No, the chance of anyone getting it is always 1/30.

The first person has 30 to choose from, so his chance is 1/30.

The second can only choose if the first has missed it (29/30) and he has a 1/29 chance of getting it himself. 29/30*1/29 = 1/30. This argument continues all the way down.

Why did the vector cross the road?

It wanted to be normal.

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**irspow****Member**- Registered: 2005-11-24
- Posts: 456

Thanks, I realized that after posting and then punching the numbers into my calculator. I never would have thought that instinctively. Maybe the guys he hangs out with think the same way at a gut level.

I calculated it a little differently however. I just looked at the probability of each person missing and finally the last guy getting 30. Like this;

29/30*28/29*27/28*26/27*25/26*24/25*23/24*22/23*21/22*1/21 = 1/30

*Last edited by irspow (2006-01-23 11:18:11)*

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**grifta67****Member**- Registered: 2006-01-23
- Posts: 2

Well this is the first time I've ever been to this forum, and I must say you've all amazed me. That you all would put a fair amount of thought (actually breaking out calculators!) into my comrades' silly little bar drawing is astounding!

Thank you all for the input. I think I'm going to go with my old way of everyone drawing their own numbers, with the added idea of reversing the rotation every round.

The drawing isn't until Sunday, so if anyone else comes up with ideas or comments, feel free to share them.

Thank you

-Sean

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