The Geek Forum
Main Forums => Homework Help => Topic started by: allfactorsnerd on October 03, 2009, 04:18:47 AM
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how many people do you need in a room to have a 50/50 chance of 3 of them having the same birthday????
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More than the number of computers you'd need to have in a room such that you'd have a 50/50 chance that one of them would have an application with a spell check component.
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This sounds like an extra credit problem to me.
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This sounds like an extra credit problem to me.
damn, you got me
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I must be really smart.
Tell you what. YOU figure it out and tell us and I'll give you extra credit too! Real life EC plus fake internet EC...who could ask for more?
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how many people do you need in a room to have a 50/50 chance of 3 of them having the same birthday????
There's not enough information to answer, unless you're a terracentric.
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There's not enough information to answer, unless you're a terracentric.
well, that's not really an answer, is it?
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Sure it is.
You can postulate a planet X that has one day a year and then you'd need 3 people for a 100% chance of them having the same birthday. Assuming of course that birthday refers to only the same month and day and not year, because they would all have the birthday 1/1/x (or choose your own syntax, after all you wouldn't want to be terracentric)...
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Been there, done that. At least, for 2 people.
Picture a bowl of 365 balls. Person 1 picks a ball, writes his name on it and puts it back.
Then Person 2 picks a ball.
The chance person 2 picks the ball person 1 picks is 1/365.
So far for 2 people.
For 3 people, the chance person 2 picks the ball person 1 has is 1/365 (0,2740%)
The chance that its not the case is 1-1/365=(364/365) =
In that case person 3 picks a ball. For person 3 there are 2 balls forbidden, so his chance to pick something bad is 2/365. So the chance that 3 persons have the same ball is: 1/365 + (364/365))(2/365) = 0,2740% + 0,5464% = 0,8204% note, 0,5464 != (does not equal) 2*0,2740)
for 4 people the chance of 2 people having the same ball is 1/365
in case not, the chance of 3 people having the same ball 1/365 + (364/365))(2/365)
in case not, the chance of 4 people picking the same ball is 1/365 + (364/365))(2/365) + (1-(1/365 + (364/365))(2/365)))(3/365) = 0,2740% + 0,5464% + 0,8152% = 1,656%
For 5: (1-1,1656%)(4/365) = 1,0831%
Total chance: 0,2740% + 0,5464% + 0,8152% + 1,0831% = 2,739%
For 6: (1-2,739%)(5/365) = 1,332%
Total chance: 4,071%
For 7: (1-4,071%)(6/365) = 1,577%
Total chance: 5,648%
For 8: (1-5,648%)(7/365) = 1,809%
Total chacne: 7,457%
For 9: (1-7,457%)(8/365) = 2,028%
Total: 9,485%
For 10: (1-9,485%)(9/365) = 2,231%
Total: 11,716%
For 11: (1-11,716%)(10/365) = 2,419%
Total: 14,135%
For 12: 2,588%
16,938%
13: 2,731%
19,67%
14: 2,861%
22,53%
15: 2,97%
25,50%
16: 3,06%
28,56%
17: 3,13%
31,69%
18: 0,318%
34,78%
19: 3,21%
37,99%
20: 3,23%
41,22%
21: 3,22%
44,44%
22: 3,20%
47,64%
23: 3,16%
51,0%
Yay, 23 people.
I might be off 1 or 2 because of roundings
Some stuff need to be added to make this work for 3 people which. I estimate 62 people. I'd have to calculate it numerically because for 3 people things get a lot longer, but the idea is the same as above.
Would be interesting to see what function it is for N amount of people to have 50/50 chance to have the same birthday.