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  #16   Spotlight this post!  
Unread 24-02-2004, 18:39
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Re: Odd Escape

Quote:
Originally Posted by Jones571
yea and after playing seventy millinon diffrent combos in one week you will need to win the lotto after spending 70 million dollars on tickets.
Yea really! You gotta win big.

Which brings up another interesting point:

Option B guarantees the win in this scenario, but you only win on one set of #s.

Option A does not gaurantee the win, however you do have the possibility of winning multiple times.

So, in the end: if you are concerned with just the probability of winning once and only once, then Option B is better. If you are concerned total payoff probability, both options WOULD work out to be the same. An interesting twist to an already interesting riddle.
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Unread 24-02-2004, 18:50
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Re: Odd Escape

But other tickets may not be "big" winners but they still have a smaller pay out. You would have to find out how much you would win form the jack pot plus all the other smaller pay outs for having a portion of the numbers correct. then you could find out if it would be worth it.
You also have the possibility were some one else ends up playing the same number so you have to split it so really your screwed no matter what

yea but that is alot of work and u need 70million to start with so really we have learned just save up your money and buy a segway lotto is bad

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Unread 08-04-2004, 22:51
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Re: Odd Escape

This problem might cause heated debate, but it's hard to argue against mathematical fact. In fact, to prove disbelievers wrong, I wrote a program to do exactly this:

Three doors, randomly picked one to be correct.
Randomly pick a door.
Randomly choose a door that is incorrect and not picked to be opened.
Switch to the other door and see if it is correct.

The program won 66.7% of the time, after enough trials.

But I have better problems to solve :
http://www.chiefdelphi.com/forums/sh...ad.php?t=27647

Have fun!
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Unread 09-04-2004, 18:23
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Re: Odd Escape

Quote:
Originally Posted by Grommit
This problem might cause heated debate, but it's hard to argue against mathematical fact. In fact, to prove disbelievers wrong, I wrote a program to do exactly this:

Three doors, randomly picked one to be correct.
Randomly pick a door.
Randomly choose a door that is incorrect and not picked to be opened.
Switch to the other door and see if it is correct.

The program won 66.7% of the time, after enough trials.

But I have better problems to solve :
http://www.chiefdelphi.com/forums/sh...ad.php?t=27647

Have fun!
That's funny I did the same thing (and got the same results of course.) Score one cookie for Grommit.

Here's a new one:

Is it possible to place four lamp-posts of different colors on the inside of a circle so that a person walking all the way around the outside of the circle would see all 24 (4!) combinations of colored posts from left to right at some point along the circle? Obviously, a proof is required either way.
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Unread 09-04-2004, 18:44
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Re: Odd Escape

No, it isn't possible; here's my geometric proof. Take any four points representing the four lights. Draw lines connecting every pair of points, extending off to infinity in both directions; since the observed order of the lights only changes when one light appears to the observer to pass behind another, crossing a line represents a change in observed order. Since only six lines can be drawn (4 choose 2), and the observer can only cross each one twice in one pass around the circle, the observed order can only change 12 times (one of these being a change back to the originally observed order), so there is a maximum of 12 orders that can be observed (not the full 24).
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Unread 09-04-2004, 20:05
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Re: Odd Escape

Quote:
Originally Posted by Ian Mackenzie
No, it isn't possible; here's my geometric proof. Take any four points representing the four lights. Draw lines connecting every pair of points, extending off to infinity in both directions; since the observed order of the lights only changes when one light appears to the observer to pass behind another, crossing a line represents a change in observed order. Since only six lines can be drawn (4 choose 2), and the observer can only cross each one twice in one pass around the circle, the observed order can only change 12 times (one of these being a change back to the originally observed order), so there is a maximum of 12 orders that can be observed (not the full 24).
Absolutely perfect! Score one cookie for Ian Mackenzie!
How long did that take you? For me, I drew some pictures and started rotating them around in front of me, keeping track of all the combinations for about a half hour. Then I realized that you could just draw lines like you said. Congrats, that's not an easy one to figure out.
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  #22   Spotlight this post!  
Unread 09-04-2004, 20:37
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Re: Odd Escape

Sweet, I love cookies. (Just ask anyone on my team...)

Actually, I got it pretty quickly, through a combination of laziness and luck. I thought about doing some actual rotations as you described, but I really didn't want to, and figured there should be a more elegant method.

Here's another really cool problem for you: You have two ropes, and the only thing you know about them is they take exactly 1 hour each to burn (i.e. if you lit either end, the flame would take exactly 1 hour to reach the other end). The ropes are not of constant cross-section, so half a rope will not necessarily take half an hour to burn, and the ropes are not the same (i.e. not only do they burn unevenly, they don't even burn equally). Given these two ropes and a lighter, how do you measure exactly 45 minutes?
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Unread 09-04-2004, 21:31
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Re: Odd Escape

Well I think if you burn the first rope at both ends, their flames will meet and fizzle at 30 minutes, right? Because their uneven burning will still make them meet somewhere along that length. So if you also burned the second rope at the start (i.e. the same time as you lit the first rope), when the first rope's flames die out, it'll be 30 minutes, at which point you could light the other end of the second rope. When both ends of the second rope meet, it'll be another 15 minutes. This adds up to 45 minutes, I think. Unless I'm horribly missing something.
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Unread 09-04-2004, 21:35
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Re: Odd Escape

Quote:
Originally Posted by jonathan lall
Well I think if you burn the first rope at both ends, their flames will meet and fizzle at 30 minutes, right? Because their uneven burning will still make them meet somewhere along that length. So if you also burned the second rope at the start (i.e. the same time as you lit the first rope), when the first rope's flames die out, it'll be 30 minutes, at which point you could light the other end of the second rope. When both ends of the second rope meet, it'll be another 15 minutes. This adds up to 45 minutes, I think. Unless I'm horribly missing something.
That sounds right. I was on the right track of burning both ends of the first rope for 30 minutes but couldn't get the 15 second part...I ventured down the blind alley of trying to light the second rope in four places...but it only works when you light the ends I guess.
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Unread 09-04-2004, 21:50
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Re: Odd Escape

Quote:
Originally Posted by jonathan lall
Well I think if you burn the first rope at both ends, their flames will meet and fizzle at 30 minutes, right? Because their uneven burning will still make them meet somewhere along that length. So if you also burned the second rope at the start (i.e. the same time as you lit the first rope), when the first rope's flames die out, it'll be 30 minutes, at which point you could light the other end of the second rope. When both ends of the second rope meet, it'll be another 15 minutes. This adds up to 45 minutes, I think. Unless I'm horribly missing something.
Exactly right. (Your solution, not the statement that you're horribly missing something ).
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Unread 09-04-2004, 22:14
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Re: Odd Escape

Well...I think I've got the paper solution to the first one...

Given that there are 3 doors and 1 of them is "escape", you can:

A) Choose door 1
B) Choose door 2
C) Choose door 3

Now given that 3 is the "correct" door, if you pick 1, 2 will be taken away and vice versa and if you pick 3, either 1 or 2 will be taken away.

A1) Stick with door 1
A2) Go with door 3
B1) Stick with door 2
B2) Go with door 3
C1) Stick with door 3
C2) Change to door 1 or 2 (whichever is not eliminated)

Now, as you can see, if you stick with your first choice (A1, B1, C1), you have a 1 in 3 chance (C1) of getting it correct while if you change it (A2, B2, C2) you have a 2 in 3 chance of getting it right (A2, B2).
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