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Old 04-01-2006, 02:31 PM   #101
TredWel
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I agree with $9,375 for part 2, just worked it out myself.
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Old 04-01-2006, 02:32 PM   #102
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Originally Posted by EagleFan
With 28 people you get:
27/365 + 26/365 + 25/365 ..... 1/365 = 378/365 or a 103% chance

with 27 people you get 351/365 or a 96% chance.

But the 28 people answer is plainly wrong - a 103% chance makes no sense in the context of the problem, and it's perfectly possible to have 28 people with different birthdays so it has to be less than 100%.

The factorial way of calculating (365/365*364/365*...) really is the proper way of figuring the odds.
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Old 04-01-2006, 02:32 PM   #103
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So the bridge and the poker problem seem to be set answers. I had hoped that the birthday one was as well, but that doesn't seem as true anymore, though there seem to only be 2 possiblities. That gives us 10 of the 18 numbers. Wouldn't that make brute forcing it somewhat more realistic?
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Old 04-01-2006, 02:32 PM   #104
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We're getting close

http://www.solecismic.com/371937568314517711.php
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Old 04-01-2006, 02:33 PM   #105
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Edit - Tred wins.


http://www.solecismic.com/371937568314517711.php

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I've been watching the discussion, and you're close, but you're making an assumption about one of the problems that is not quite correct. Keep at it, you'll get there eventually.

Last edited by VPI97 : 04-01-2006 at 02:34 PM.
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Old 04-01-2006, 02:34 PM   #106
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The bridge, poker, and betting problems seem set. The birthday one's down to two cases - 365 or 366 - and I think the grades one's down to three. At this point it should be possible just to try all the possible combinations.
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Old 04-01-2006, 02:35 PM   #107
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Katon: Are we set on 378 and not 371 as the right answer to the birthday?
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Old 04-01-2006, 02:35 PM   #108
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Originally Posted by cthomer5000
I'm with SkyDog here. I'd at least like to hear some further explanation from albion before I'm willing to go with his answer.

Never trust a lawyer to do math. But here's what I did, as best as I can explain it.

Going into game 7, you need 0 in your account. That means that going into game 6 (where Boston will be up 3-2)*, you need to bet a certain amount such that if your bet pays off, you win $30,000 and if your bet loses, you need 0 in your account (to set up game 7).

GOING INTO GAME 6:

So, going into game 6 (where X is what you have in your account and Y is what you bet):

X+Y = 30,000
X-Y = 0

Solving for X gives us 15,000. So, going into Game 6, we need 15,000 in our account.

GOING INTO GAME 5:

Going into Game 5, the series will either be 3-1 or 2-2.

If the series is 3-1, then

X+Y = 30,000
X-Y = 15,000

Solving for X gives us 22,500.

If the series is 2-2, then

X+Y =15,000
X-Y = -15,000

Solving for X gives us 0.

So, going into game 5, you will need 22,500 in your account if the series is at 3-1 and 0 in your account if the series is tied.

GOING INTO GAME FOUR:

Going into Game 4, the Series will either be 3-0 or 2-1.

If the series is 3-0, then

X+Y = 30,000 (you win the bet)
X-Y = 22,500 (you go to 3-1)

Solving for X gives us 26,250

If the series is 2-1, then

X+Y = 22,500 (you go to 3-1)
X-Y = 0 (you go to 2-2)

Solving for X gives us 11,250.

GOING INTO GAME THREE:

Going into game 3, the series will either be 1-1 or 2-0.

If the series is 2-0 then

X+Y = 26,250 (you go to 3-0)
X-Y = 11,250 (you go to 2-1)

Solving for X gives us 3,750

If the Series is 1-1 then

X+Y = 11,250
X-Y = -11,250

Solving for X gives us Zero

GOING INTO GAME 2:

Going into Game 2, the series will be 1-0

X+Y = 3,750
X-Y = 0

Solving for X = 1,875.

So you need to bet 1,875 in order to have 1,875 going into Game 2.

Feel free to pick apart while I try to format it a little better.


*You can just make all of the bets negative and have Boston losing. It works out the same.
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Old 04-01-2006, 02:35 PM   #109
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DOLA -- And it's nice to know whatever the message is that Jim does at least want us to find the answer.
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Old 04-01-2006, 02:36 PM   #110
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Old 04-01-2006, 02:37 PM   #111
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Originally Posted by Katon
But the 28 people answer is plainly wrong - a 103% chance makes no sense in the context of the problem, and it's perfectly possible to have 28 people with different birthdays so it has to be less than 100%.

The factorial way of calculating (365/365*364/365*...) really is the proper way of figuring the odds.


That's a Statistics 101 problem. The very first question our professor asked in the class.
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Old 04-01-2006, 02:37 PM   #112
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Very good twothree. About what I expected.
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Old 04-01-2006, 02:38 PM   #113
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Originally Posted by albionmoonlight
Going into game 3, the series will either be 1-1 or 2-0.

If the series is 2-0 then

X+Y = 26,250 (you go to 3-0)
X-Y = 11,250 (you go to 2-1)

Solving for X gives us 3,750

Er, no it doesn't. It gives you 18,750. Since Y>0, having X < X-Y is silly.

Apart from the calculation error, this is exactly what I did, only explained more clearly.

Last edited by Katon : 04-01-2006 at 04:04 PM.
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Old 04-01-2006, 02:39 PM   #114
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This is like watching a car wreck. But one where, every so often, someone walks over and punches the driver in the face as he struggles to free himself from the wreckage.
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Old 04-01-2006, 02:39 PM   #115
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Katon: Are we set on 378 and not 371 as the right answer to the birthday?
I'm not sure I'm right, the formula I used was an approximation, I'm checking again...
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Old 04-01-2006, 02:39 PM   #116
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Old 04-01-2006, 02:40 PM   #117
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Old 04-01-2006, 02:40 PM   #118
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Originally Posted by EagleFan
That's a Statistics 101 problem. The very first question our professor asked in the class.

Then how does 103% make any sense? The chance always has to be less than 100% for any number of people less than 365, because it's always possible for everyone to have different birthdays.
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Old 04-01-2006, 02:41 PM   #119
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So, it was 67 and not 68 on the bridge problem.
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Old 04-01-2006, 02:41 PM   #120
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For those who care:

Birthday problem $371

Boston betting: $9375

Bridge 67%

Poker: 3145

Skipping grades: 17711
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Old 04-01-2006, 02:41 PM   #121
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Dola: See my mistake now. I just mis-solved for X at one point. I now agree with $9375
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Old 04-01-2006, 02:42 PM   #122
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Old 04-01-2006, 02:43 PM   #123
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Originally Posted by Katon
Also, has anyone checked what happens to #1 if you use 366 days to account for leap years?
That makes is pretty complicated. You can't just make it 366, because that leap year day is 1/4 as likely to be a birthday as all the other days. It would be 1/1461 compared to 4/1461 for all the other days.
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Old 04-01-2006, 02:44 PM   #124
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EEEVIL!
Not really. I half-wondered if it would be house info, considering the way it was worded, but there were some nice mathematics involved there that made it worth it. Good stuff.
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Old 04-01-2006, 02:45 PM   #125
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That makes is pretty complicated. You can't just make it 366, because that leap year day is 1/4 as likely to be a birthday as all the other days. It would be 1/1461 compared to 4/1461 for all the other days.
More complicated than that, I think. 1900, 2100, 2200, 2300, 2500, 2600, 2700 aren't leap years, no?
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Old 04-01-2006, 02:45 PM   #126
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Maybe all those questions are found on a web site.

Anybody try google the answers?
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Old 04-01-2006, 02:46 PM   #127
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I believe it's real, but since we only have one day to complete all 5 puzzles, Jim knows the probability of all 5 right answers are slim.

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Old 04-01-2006, 02:47 PM   #128
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So, it was 67 and not 68 on the bridge problem.

Yes, the question asked people to round down, and the probability is .6783. So that's why I added the other link, because 68 is essentially correct.

For those wondering about the math, because intuitively the answer is 62.5%, you can't look at the distribution of the hearts as a separate problem. That probability is affected by the other 21 cards remaining.

You have to set it up looking at all [13/26] hands. I expected this one to hang people up more than any, didn't think we'd have a real bridge player here.
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Old 04-01-2006, 02:48 PM   #129
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Then how does 103% make any sense? The chance always has to be less than 100% for any number of people less than 365, because it's always possible for everyone to have different birthdays.


It's statistical probability. Take sets of 28 random people and bet on each one having at least one set of duplicate birthdays and you'll have a 103% return.
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Old 04-01-2006, 02:52 PM   #130
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That makes is pretty complicated. You can't just make it 366, because that leap year day is 1/4 as likely to be a birthday as all the other days. It would be 1/1461 compared to 4/1461 for all the other days.

Yes, that is how you solve it. But it ends up not changing the final result. The difference is only about a tenth of a percentage point.
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Old 04-01-2006, 03:03 PM   #131
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Old 04-01-2006, 03:05 PM   #132
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Originally Posted by sabotai

Part 5

If you want to receive a PhD in math, all together you're going to go through about 21 grades' worth of school material.
Well, math geeks tend be quite intelligent people overall, especially in ways that impress teachers.
Let's say we have a truly superior math geek student who can skip grades at will. But he can only skip one grade per year. Or not, if he chooses. So he can go one grade at a time and take 21 years to obtain his PhD. Or he can skip as often as possible, and need only 11 years. His choice.
Those are only two of literally thousands of ways he could choose to go. So, how many different combinations of skipping and not skipping are possible?

Another way to view this problem is...

How many distinct ways can you empty a basket of 21 apples, if you can only take one apple or two apples from the basket at a time.
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Old 04-01-2006, 03:40 PM   #133
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It's statistical probability. Take sets of 28 random people and bet on each one having at least one set of duplicate birthdays and you'll have a 103% return.

But that's not the question. The question is what is the possibility that at least two of them have the same birthday. Your system gives extra weight to the cases where four or five or all 28 of them have the same birthday, because that makes sense for the problems it's designed to solve, but in the context of this question all 28 of them sharing a birthday is exactly the same as just two sharing a birthday.
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Old 04-01-2006, 03:47 PM   #134
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Wow, this is crazy.
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Old 04-01-2006, 04:06 PM   #135
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Wow, this is crazy.
Yes, indeed. One crazy signature you've got there.


Nice to see the progress of the project, Jim.
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Old 04-01-2006, 04:34 PM   #136
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Still some fun puzzles. And Jim did hint that he wanted to announce this way. But even if it is all a big joke, what's the harm in letting Jim have his fun and us try to solve some puzzles?

Knowing what the final answer was and what we got out of it, I still stand by this. 'twas fun Jim.
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Old 04-01-2006, 08:27 PM   #137
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Yeah, I'm sure of it now. You're going straight to hell for that.


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Old 04-01-2006, 09:03 PM   #138
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The answer will be a redirect to tubgirl.
This brings to mind a question that is very perplexing: Would it be possible for Jim to be put in the box?
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Old 04-01-2006, 09:10 PM   #139
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Old 04-01-2006, 09:17 PM   #140
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Hmmm "The house that Jim built" .... "The house that Ruth built" It's obvious, he's doing a baseball sim.

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Old 04-01-2006, 11:07 PM   #141
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Old 04-02-2006, 03:16 AM   #142
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So I'm Confused... Jim is making a Virtual reality version of the sims?
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Old 04-02-2006, 10:45 AM   #143
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Glad I missed this thread in its popularity.

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Old 04-02-2006, 11:29 AM   #144
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i rest assured knowing that i'm not nerdy enough to have even attempted to answer those questions. too much work for a saturday.
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Old 04-02-2006, 02:59 PM   #145
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i rest assured knowing that i'm not nerdy enough to have even attempted to answer those questions. too much work for a saturday.

The rest of us rest assured as well. We'd be disappointed if you weren't too cool to hang with us.

Now, where are Lindsay Weir and the rest of the Mathletes? We have a competition coming up and need to study differential equations.
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Old 04-02-2006, 03:07 PM   #146
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Now, where are Lindsay Weir and the rest of the Mathletes? We have a competition coming up and need to study differential equations.

I heard she went to a Grateful Dead concert.
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Old 04-02-2006, 03:20 PM   #147
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I heard she went to a Grateful Dead concert.

She had better not be hanging out with that stoner who thinks he going to make a living as a drummer. He's only going to break her heart once he meets that girl from band camp.
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Old 04-02-2006, 10:42 PM   #148
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"Mathletes".

HAHAHAHAHAHA. llol (literally laughed out loud). sounds like a mega-nerd, or some type of bionic geek.
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Old 04-02-2006, 11:02 PM   #149
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She had better not be hanging out with that stoner who thinks he going to make a living as a drummer. He's only going to break her heart once he meets that girl from band camp.
She's my Lady L.
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Old 04-01-2007, 02:15 AM   #150
M GO BLUE!!!
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Gridiron Essentials!

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