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How High Would You Go


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#21 Zach6668

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Posted 25 September 2006 - 03:28 PM

View PostActuary, on Monday, September 25th, 2006, 7:27 PM, said:

lolyou know first thing I did was verify that it's on the sequence
Obv.
QUOTE (serge @ Tuesday, May 12th, 2009, 7:20 PM) <{POST_SNAPBACK}>
LETS GO PITTSBURGH
QUOTE (Acid_Knight @ Monday, March 10th, 2008, 4:15 PM) <{POST_SNAPBACK}>
Zach is right about pretty much everything.

#22 RodReynolds

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Posted 25 September 2006 - 03:33 PM

View PostActuary, on Monday, September 25th, 2006, 7:27 PM, said:

Bascomeb, I wan't heading to a bankroll question; but I see the analogy.************Danny,that seems like a gambling mindset there.even if you made 60k per yr, it's not as if an additional 60k above that is meaningless.************I can't prove we will go broke eventually, but surely we do.any math wizards?LLY ?Copernicus?Rod?others?I get (5/3) ^ x as our Amount, unbounded as x goes to infinity. X being the number of rolls we takeOk, so the EV is unbounded but the likelihood of going broke is still 100% as x goes to infinity.***lolyou know first thing I did was verify that it's on the sequence
Yeah, I don't know how to make sense of this intuitively. I think infinity is screwing with our minds again. I think what you said is right though.Our EV approaches infinity while our chances of growing broke approach 1. It is easier to understand if we just let x = 1000 or something.

#23 dapokerbum

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Posted 25 September 2006 - 03:33 PM

$1024 the first time. If there was a second and third i would try for 16,384 and then of course 1,048,576.Interesting question.Seems I don't have as much gamble as the rest of ya'll
There was madness in any direction, at any hour…You could strike sparks anywhere. There was a fantastic universal sense that whatever we were doing was right, that we were winning…. And that, I think, was the handle-that sense of inevitable victory over the forces of Old and Evil. Not in any mean or military sense; we didn’t need that. Our energy would simply prevail. There was no point in fighting-on our side or theirs. We had all the momentum; we were riding the crest of a high and beautiful wave….So now, less than five years later, you can go up on a steep hill in Las Vegas and look West, and with the right kind of eyes you can almost see the high-water mark-that place where the wave finally broke and rolled back.

#24 PrtyPSux

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Posted 25 September 2006 - 03:33 PM

Id say 65k, maybe 130k....not much more ...

#25 Actuary

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Posted 25 September 2006 - 03:42 PM

View PostRodReynolds, on Monday, September 25th, 2006, 3:33 PM, said:

Our EV approaches infinity while our chances of growing broke approach 1. It is easier to understand if we just let x = 1000 or something.
even at x=478 (477 dbls)EV = 6,635,193,109,306,990,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 lol.that's nonsenseI"m missing sometthing

#26 CrazyJoe

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Posted 25 September 2006 - 03:43 PM

i know it would suck super hard nuts to roll a 6 on the first roll.

#27 RodReynolds

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Posted 25 September 2006 - 03:46 PM

View PostActuary, on Monday, September 25th, 2006, 7:42 PM, said:

even at x=478 (477 dbls)EV = 6,635,193,109,306,990,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 lol.that's nonsenseI"m missing sometthing
I don't think you are. Perform the experiment n = (5/6)^478 times. You expect one success. Divide your earnings of 2^478 by n. You get your number. (I think, I didn't actually do the math... I'm just reassuring your troubled mind.)Discussions with you really pad my post count. Thanks for that.Edit: n=1/(5/6)^478 = (6/5)^478

#28 Actuary

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Posted 25 September 2006 - 03:50 PM

chime in dorks:13 Members: kkcountry, joshrogo, Vyszion, RodReynolds, FatBurger, mase_gotsem, davezz5, madams01, TheCinciKid, Viking145, bursell, pragtyro

#29 davezz5

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Posted 25 September 2006 - 03:56 PM

View PostRodReynolds, on Monday, September 25th, 2006, 3:46 PM, said:

I don't think you are. Perform the experiment n = (5/6)^478 times. You expect one success. Divide your earnings of 2^478 by n. You get your number. (I think, I didn't actually do the math... I'm just reassuring your troubled mind.)Discussions with you really pad my post count. Thanks for that.Edit: n=1/(5/6)^478 = (6/5)^478
I would roll untili i hit a reasonable amount i.e 5k +. Then on the second roll i would take a chance to see if i could reach a life changing amount of money, knowing i can still take 5k as a consolation if I hit that six.
Freedoms just another word for nothing left to lose.

#30 76169

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Posted 25 September 2006 - 04:12 PM

View PostActuary, on Monday, September 25th, 2006, 4:27 PM, said:

I can't prove we will go broke eventually, but surely we do.any math wizards?
Chance of going broke each time we play is:(1/6) + (5/6)(1/6) + (5/6)(5/6)(1/6) + (5/6)(5/6)(5/6)(1/6) + ...= (1/6)(1/(1-(5/6))) = (1/6)(1/(1/6)) = 1 = 100%And EV of the game is positive infinity

#31 Actuary

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Posted 25 September 2006 - 08:23 PM

View Post76169, on Monday, September 25th, 2006, 4:12 PM, said:

Chance of going broke each time we play is:(1/6) + (5/6)(1/6) + (5/6)(5/6)(1/6) + (5/6)(5/6)(5/6)(1/6) + ...= (1/6)(1/(1-(5/6))) = (1/6)(1/(1/6)) = 1 = 100%And EV of the game is positive infinity
I think chance probability of going broke is Limit 1-(5/6)^n as n goes to infinity. Which equal 1-0 = 1.And the EV = Integral 2^n * (5/6)^n = (5/3)^n as n goes from 0 to Infinity. Which, although I don't remember my integrals for powers, has to be Infinity.

#32 Dratj

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Posted 25 September 2006 - 08:43 PM

???

#33 GWCGWC

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Posted 25 September 2006 - 09:42 PM

I'd roll until I went broke3 times

#34 SabaAba

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Posted 25 September 2006 - 10:12 PM

LOL, you guys would NEVER reach 16,000 let alone 1,000,000.Think about it, you roll a 6 once every 6 times, so if you are lucky, you will roll say 12 times before rolling a 6 which is only 2,048.The question is all about greed. When you say you want to win 64,000, you are really saying I'm willing to lose 32,000 16.67% of the time for a chance to win another 32,000. Try this at home and see if you can even roll 11 times to get to 1,024 without rolling a 6. Not very likely.

#35 troyomac

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Posted 25 September 2006 - 10:18 PM

View PostSabaAba, on Monday, September 25th, 2006, 10:12 PM, said:

LOL, you guys would NEVER reach 16,000 let alone 1,000,000.Think about it, you roll a 6 once every 6 times, so if you are lucky, you will roll say 12 times before rolling a 6 which is only 2,048.The question is all about greed. When you say you want to win 64,000, you are really saying I'm willing to lose 32,000 16.67% of the time for a chance to win another 32,000. Try this at home and see if you can even roll 11 times to get to 1,024 without rolling a 6. Not very likely.
Yes but if you're lucky enough to go 12 times without rolling a 6. Then your next roll is still a 5/6 chance to double up. I'd risk it!

#36 Actuary

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Posted 25 September 2006 - 10:21 PM

View PostSabaAba, on Monday, September 25th, 2006, 10:12 PM, said:

LOL, you guys would NEVER reach 16,000 let alone 1,000,000.
true.But would you stop at $8it's not what you expect to win, it's where you would stop.can you figure any reason why the EV is not Infinity? It just seems wrong, but mathematically looks correct

#37 onlyme386

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Posted 25 September 2006 - 10:42 PM

View PostActuary, on Monday, September 25th, 2006, 11:21 PM, said:

true.But would you stop at $8it's not what you expect to win, it's where you would stop.can you figure any reason why the EV is not Infinity? It just seems wrong, but mathematically looks correct
Damn you all! You've set my brain into motion and it's firing 100% now. I shall return in 15-20 minutes with the ultimate answer. Maybe, if you're real lucky, I'll show my work on paper, scan it, and post it here. That way you can all see how it works.
s

#38 onlyme386

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Posted 25 September 2006 - 11:15 PM

OK, so here's what I've come up with.Roll # = n Start Amount = SEnd Amount = EEV = 5E/6 - S (for any single roll)n...S...E...EV1...1....2... .66672... 2 ... 4 ... 1.3333 ... 4 ... 8 ... 2.6674 ... 8 ... 16 ... 5.3335 ... 16 ... 32 ... 10.6676 ... 32 ... 64 ... 21.3337 ... 64 ... 128 ... 42.6678 ... 128 ... 256 ... 85.3339 ... 256 ... 512 ... 170.66710 ... 512 ... 1024 ... 341.33311 ... 1024 ... 2048 ... 682.66712 ... 2048 ... 4096 ... 1365.33313 ... 4096 ... 8192 ... 2730.66714 ... 8192 ... 16384 ... 5461.33315 ... 16384 ... 32768 ... 10922.66716 ... 32768 ... 65536 ... 21845.33317 ... 65536 ... 131072 ... 24690.66718 ... 131072 ... 262144 ... 87381.33319 ... 262144 ... 524288 ... 174762.66720 ... 524288 ... 1048576 ... 349525.33321 ... 1048576 ... 2097152 ... 699050.66722 ... 2097152 ... 4194304 ... 1398101.33323 ... 4194304 ... 8388608 ... 2796202.66724 ... 8388608 ... 16777216 ... 5592405.33325 ... 16777216 ... 33554432 ... 11184810.667So, the question becomes, how much are you willing to risk on one dice throw? For me, I'd have to stop at $1,024 the first game. In fact, it's most likely that you'd never even reach that point. Only 13% of the time would you reach it. That next roll has an expected value gain of $1365 but as a college student, I'd say that $1k on a 1 minute gamble would be enough to make my day. I might even stop below that, maybe at 512 or even 256. The second game and third, if I was able to get there I would stop at $4,096. That next roll has way too much of a risk for me.Anyways, where do you stop given this?
s

#39 Actuary

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Posted 25 September 2006 - 11:26 PM

View Postonlyme386, on Monday, September 25th, 2006, 11:15 PM, said:

Anyways, where do you stop given this?
um.what new info did you provide tha would afect answers?I don't follow your EV = 5E/6 - S (for any single roll) equationoh, wait, E is amount you'd have after the roll, if not a 6.If I"m interpreting that correctly, the EV = 5/6 * ENo need to subtract S

#40 onlyme386

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Posted 25 September 2006 - 11:29 PM

View PostActuary, on Tuesday, September 26th, 2006, 12:26 AM, said:

um.what new info did you provide tha would afect answers?I don't follow your EV = 5E/6 - S (for any single roll) equationoh, wait, E is amount you'd have after the roll, if not a 6.If I"m interpreting that correctly, the EV = 5/6 * ENo need to subtract S
Expected Value is the probability of winning, multiplied by the difference if you win MINUS the probability of losing, multiplied by the amount lost.EV = (5/6)(E-S) - (1/6)SEV = (5/6)E - (5/6)S - (1/6)SEV = (5/6)E - SThat clarifies things, I hope.
s




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