Greetings everyone, I'm new on this forum with 1 question for now

[Marten]

I could use some help with a problem I have but cannot find a solution, maybe it's even impossible. We bought an old wheel of fortune and it has 64 fields/segments. Our class has 35 kids in it. Sometimes we receive gifts for the kids and we want to give every kid an equal change to obtain a gift playing with the wheel of fortune. If it's possible, can we make a calculation or use a formula using the wheel of fortune with the 64 fields and give all the 35 kids an equal change of winning? Maybe with more than one turn. (I'm not a math teacher btw haha).

 

[Cubist]

we want to give every kid an equal change to obtain a gift
..
(I'm not a math teacher btw haha).

I assume that you're not an English teacher either because you said "change" instead of "chance" .
I'm completely joking! No insult intended. I make lots of mistakes too

If it's possible

Yes it's possible. Just have (64-35) = 29 segments that say "spin again". This might also extend the fun

 

[topsquark]

I could use some help with a problem I have but cannot find a solution, maybe it's even impossible. We bought an old wheel of fortune and it has 64 fields/segments. Our class has 35 kids in it. Sometimes we receive gifts for the kids and we want to give every kid an equal change to obtain a gift playing with the wheel of fortune. If it's possible, can we make a calculation or use a formula using the wheel of fortune with the 64 fields and give all the 35 kids an equal change of winning? Maybe with more than one turn. (I'm not a math teacher btw haha).

This is also not a differential equations problem... (thread moved to appropriate forum section)

-Dan

 

[BigBeachBanana]

I assume that you're not an English teacher either because you said "change" instead of "chance" .
I'm completely joking! No insult intended. I make lots of mistakes too


Yes it's possible. Just have (64-35) = 29 segments that say "spin again". This might also extend the fun

I like Cubist's idea but having 29 "spin again" segments will give each spin a probability of 29/64= 45.3% to land on them. Since there are 35 kids, it might take a while to get through all of them.

 

[Cubist]

I like Cubist's idea but having 29 "spin again" segments will give each spin a probability of 29/64= 45.3% to land on them. Since there are 35 kids, it might take a while to get through all of them.

Very good point. If there are multiple prizes then let us know - there might be a better way that will also ensure that no-one gets two prizes.

#spins  Probability of a winner at, or before, this spin (approx)
=================================================================
   1      0.546875
   2      0.794677
   3      0.906963
   4      0.957842
   5      0.980897
   6      0.991344
   7      0.996077
   8      0.998222
   9      0.999194
  10      0.999635

The "spin again option" could be made more interesting by changing it to "everyone does a star jump and then spin again". There's no downside! (Unless the classroom floor isn't very sturdy )

 

[Steven G]

I like cubists method--it's obvious and simple.
If you want to have less spin again situations you can give each student 1 1/2 spaces instead of just one. You get to decide which half the marker is on. If you can't determine, then let that be a spin again situation. This way you have (64-52.5) = 11.5 spin again situations

 

[Marten]

I believe my reply earlier wasn't posted. I try again. Thank you very much for all the advise. And indeed I'm not an English teacher heh, but don't worry, I'm far away from being offended. I can take a joke. I'm a comedy teacher Haha, just kidding.
The floors are sturdy enough to resist bouncing sweet little devils, all good. Yes, there're different gifts to hand out. Thanks again everyone