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iPod classic on shuffle
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wracket
wracket
1427 posts

Re: iPod classic on shuffle

Harold Bissonette wrote:
I think the difference between our arguments is that you are arguing for what should happen mathematically/logically, whereas I am arguing for what tends to happen in practice.

I guess I've just failed to see anything which convinces me that anything happens in practice which is in any way contrary to what should happen in theory. Oh well, no harm done.

Jul 10, 2011, 17:34


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wracket
wracket
1427 posts

Re: iPod classic on shuffle

cyberpainter wrote:
Interesting, thanks wracket. But even though each flip is independent, can't you look at odds over a longer span? Let's say you did 100 flips. The first 4 were even. You've already "beaten" the odds. But I thought maybe the odds would get greater as time went on, that it would indeed get close to 50 odd and 50 even. I'm thinking there might be some magic number that would bring the odds back to 50/50?

:)


Scratch that last bit, magic number that would make the results split evenly, is what I meant

Indeed, the more flips you do after point 4 in this scenario the closer your expected distribution gets to 50/50...but without ever reaching it.

Jul 10, 2011, 17:37


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s_lush_s
s_lush_s
7383 posts

Re: iPod classic on shuffle

pulce7-shoe wrote:
i dont have an ipod classic.
tho i have a nano, dont use it. run out of memory.

i could list my spotify listens here.
very cool stuff.

i saw a good ipod classic down computer exchange for about 80. (80 gigs.)

actually i might add that i'm listening to the soundcarriers.

whom i've seen post on here!


I don't either. I had the AIDS one but I sold it because I couldn't reset the music and I was a fool. Currently I have the screen one and I haven't loaded it. I want to get just a regular ipod classic and load it. I only have about 350 songs but I like to listen for something new preferably classical music as that's my preference. But I haven't heard anything good in a while except some Phillip Glass which is G-r-E-a-T =s great.

Jul 10, 2011, 19:32


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s_lush_s
s_lush_s
7383 posts

Re: iPod classic on shuffle

I listen to the satellite radio on the classical stations.

Jul 10, 2011, 19:34


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c in dc
c in dc
1032 posts

Let's Make a Deal

wracket wrote:
cyberpainter wrote:
Hmm, I'm no mathematician, but wouldn't the odds change with each flip? Each flip has a 50% chance of being odd or even. But the chances of all being even in a row gets statistically lower each flip. Is that right?

Well, even if you've already had 7 straight heads, the 8th flip is clearly still a 50/50 scenario. Of course, the odds are pretty low that you would get 10 straight heads (hhhhhhhhhh), but are the exact same as that of any other single combination (hththththt, hhhhhttttt, hhtththttt, etc.) Not sure if that's what you're asking about.


Wracket, you are a man after my own heart.
Here is a probability puzzle for you to ponder:

Say you are on a Let's Make a Deal type game show. There are 3 doors to choose from; one has the grand prize, the other 2 nothing. You choose one door. The host then reveals one of the 2 remaining doors, and it has nothing behind it. The host then offers to trade you the other unchosen door for your chosen door.
Should you accept his offer?

Jul 10, 2011, 20:50


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wracket
wracket
1427 posts

Re: Let's Make a Deal

c in dc wrote:
Say you are on a Let's Make a Deal type game show. There are 3 doors to choose from; one has the grand prize, the other 2 nothing. You choose one door. The host then reveals one of the 2 remaining doors, and it has nothing behind it. The host then offers to trade you the other unchosen door for your chosen door.
Should you accept his offer?

Ahh, the good old Monty Hall problem. I know it well--so maybe I'll just leave this for someone else to answer?

Jul 10, 2011, 20:59


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cyberpainter
cyberpainter
5933 posts

Re: Let's Make a Deal

c in dc wrote:


Wracket, you are a man after my own heart.
Here is a probability puzzle for you to ponder:

Say you are on a Let's Make a Deal type game show. There are 3 doors to choose from; one has the grand prize, the other 2 nothing. You choose one door. The host then reveals one of the 2 remaining doors, and it has nothing behind it. The host then offers to trade you the other unchosen door for your chosen door.
Should you accept his offer?


Wouldn't you have a 50/50 shot either way at that point?

Jul 10, 2011, 23:00


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Mu Mu
Mu Mu
2779 posts

Re: Let's Make a Deal

cyberpainter wrote:
c in dc wrote:


Wracket, you are a man after my own heart.
Here is a probability puzzle for you to ponder:

Say you are on a Let's Make a Deal type game show. There are 3 doors to choose from; one has the grand prize, the other 2 nothing. You choose one door. The host then reveals one of the 2 remaining doors, and it has nothing behind it. The host then offers to trade you the other unchosen door for your chosen door.
Should you accept his offer?


Wouldn't you have a 50/50 shot either way at that point?



Hmmm, each door has a 1/3 chance of containing the prize. When one door is shown to have no prize, the two remaining doors have a 1/2 chance of containing the prize. I agree with CP, your chances remain 1/2 no matter which door you choose from the remaining 2 doors. Correct?

Jul 11, 2011, 19:33


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c in dc
c in dc
1032 posts

Re: Let's Make a Deal

Mu Mu wrote:
cyberpainter wrote:
c in dc wrote:


Wracket, you are a man after my own heart.
Here is a probability puzzle for you to ponder:

Say you are on a Let's Make a Deal type game show. There are 3 doors to choose from; one has the grand prize, the other 2 nothing. You choose one door. The host then reveals one of the 2 remaining doors, and it has nothing behind it. The host then offers to trade you the other unchosen door for your chosen door.
Should you accept his offer?


Wouldn't you have a 50/50 shot either way at that point?



Hmmm, each door has a 1/3 chance of containing the prize. When one door is shown to have no prize, the two remaining doors have a 1/2 chance of containing the prize. I agree with CP, your chances remain 1/2 no matter which door you choose from the remaining 2 doors. Correct?


Nope. You wanna splain this one, Wracket?

Jul 11, 2011, 19:44


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wracket
wracket
1427 posts

Re: Let's Make a Deal

c in dc wrote:
Mu Mu wrote:
cyberpainter wrote:
c in dc wrote:


Wracket, you are a man after my own heart.
Here is a probability puzzle for you to ponder:

Say you are on a Let's Make a Deal type game show. There are 3 doors to choose from; one has the grand prize, the other 2 nothing. You choose one door. The host then reveals one of the 2 remaining doors, and it has nothing behind it. The host then offers to trade you the other unchosen door for your chosen door.
Should you accept his offer?


Wouldn't you have a 50/50 shot either way at that point?



Hmmm, each door has a 1/3 chance of containing the prize. When one door is shown to have no prize, the two remaining doors have a 1/2 chance of containing the prize. I agree with CP, your chances remain 1/2 no matter which door you choose from the remaining 2 doors. Correct?


Nope. You wanna splain this one, Wracket?

Sure. You should take the host up on changing, and here's the "short" answer why. Mu is right that there is a 1/3 chance of each door containing the prize, so that means that a third of the time your initial choice will be the one with the prize and 2/3 of the time it won't be. Therefore, when the host shows one of the two remaining doors, since he has to open a door without a prize, he has effectively doubled your chances of winning by changing...which can be demonstrated with a basic three-part binomial diagram (which I can't be bothered to try to recreate here). So here's a quick-and-dirty written version:

Assume for simplicity's sake that you've chosen door 1. There is still an equal 1/3 chance of the prize being behind any door, so bring out your diagram to show the three possible locations of the prize. Now, from each of those three possibilities, let's look at what can happen next. If you've chosen door 1 and the prize is indeed behind door 1, the host can either open door 2 or door 3. As there is an equal probability of either of these outcomes, each individual scenario has the probability of 1/3 x 1/2 = 1/6. Either way you lose by switching here, so that means you have a 1/6 + 1/6 = 1/3 contribution to your overall chance of losing by choosing to switch from door 1 if the prize is behind door 1.

If, however, the prize is behind door 2, the host has no choice as to which door he will open--he can only open door 3. Therefore, you don't need to divide this probability by two...it is simply a 1/3 contribution to your overall chance of winning by choosing to switch from door 1 if the prize is behind door 2.

Likewise, if the prize is behind door 3, the host can only open door 2...once again, 1/3 contribution to your overall chance of winning by choosing to switch from door 1 if the prize is behind door 3.

So now that we have covered all the possible scenarios, we see that we have a total of 1/3 chance of losing by switching and a 1/3 + 1/3 = 2/3 chance of winning.

I know it sounds weird at first, but do the binomial tree and see!

Maybe C can do a better explanation if mine was too convoluted...

Jul 11, 2011, 20:15

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