kingofpain

I will offer you 1000 shares of KOOC for your stash of 500 JKEL. If you would like cash, name your price...

@Dee: It looks like your purchases are not feasible given that we start with $10000...

stubble, anthropogenic, iris, lye, plectrum

Renaming the Picometer Precision Ping-Pong Paddle Polymer Producers' Panel by measuring the gauges in Angstrom apparently almost alleviated alliterative allergies and elicited a request for God's blessings amongst the members of the conservative confederate criticism consortium.

I was very tempted to be sure (which explains the KOOC purchase a little later compared to the ASRE purchase). But I think I will wait for day 2 or 3 and prefer to buy from other players . Also, a big factor was that buying ASRE derivatives might be questionable (in terms of conflict of interest) due to lack of plausible deniability for these stocks if I trade in them and hold derivatives.

The probability that the cut us exactly in the middle is precisely 0. However, if you impose a tolerance, then you will receive a finite answer. e.g. Rope is 10m long. The equality is defined as length of left side is equal to right side within 1 mm. This means that cutting anywhere from 4999.5mm to 5000.5mm will be ok. Therefore, the probability will be 1 in 10,000 if the probability of the cut at any point on the rope is equal. Cheers!

1. Aaryan 2. Panther 3. dee_tot 4. Kikacat123 5. Araver 6. Vommack 7. Kingofpain 8. 9. 10. 11.

BUY 1700x KOOC

BUY 10000x ASRE

1. Molly Mae 2. Panther 3. Flamebirde 4. Kingofpain 5. 6. 7. 8. 9. 10.

Lets go again! I feel theres potential beyond just buying and selling normally...in the early days, I thought of floating derivatives and such... If the moderator offers to enforces agreements between parties (to allow shorting and such), this could turn deviously entertaining!

Sell 1000x HORID

Can you send a screenshot of what you are trying? this might help matters! In any case, I was just guessing what the puzzle wanted and actually found after calculations that indeed the 41 approach works... It could be wrong, but I would say a variant is likely to be right...

Yeah, buying the cheap stocks early on was a gamble that paid off. I actually thought I had it sown up on day 6 when I realized that I could afford to mimic everyone's portfolio and basically have a few dollars left over which meant I would win if I had a standing order to buy/sell anything anyone bought or sold.(we were not even close to buying 5000 stocks then). But I figured it would be more fun to experiment and find more about the algorithm, Big risk, big reward is the message I guess. So much for a distributed portfolio, eh?

BUY 9730x PUNTY BUY 10000x HOTY

perfect shuffle is when the cards are dropped perfectly alternately. riffle shuffle is where the probability of dropping the next card depends on the size of the remaining undropped deck on that size.

7 + 3 +15+ 12 +13 +21 +3 +5 = 79 19+ 2 +14 +9 +4+ 7+ 2 + 12= 69 10+6 + 8+ 14 +16 +11 +8 +18= 91 17+ 3+ 8+ 20 + 9 +18 +8+ 6 = 89 Total sum = 328 I would say your target for the sum in the middle should be 41. - This means that under the arrow you should have either only 1 odd number out of 4 or only one even number out of 4. - Starting clockwise from the green arrow in your picture, first circle is oooeoooo second circle is oeeoeoee third circle is eeeeeoee fourth circle is .ooeeoeee first and the third circle will line up to sum to an odd number in either 6 positions or 8 positions. so we need to line up the 2nd and 4th circle so that the sum is even for most part (in all 8 positions is not possible due to 3 consecutive evens in the fourth) Let us check how many evens we get when we shift the 4th ring clockwise oeeoeoee 0: ooeeoeee 4 1: eooeeoee 4 2: eeooeeoe 4 3: eeeooeeo 4 4: oeeeooee 6 5: eoeeeooe 4 6: eeoeeeoo 2 7: oeeoeeeo 6 We need options 4 or 7. Let us try 7 turn of 4th ring clockwise and addit to the second ring 19 2 14 9 4 7 2 12 3 8 20 9 18 8 6 17 SUM: 22 10 34 18 22 15 8 29 Focus on the odd numbers here. Those will have to correspond with the even numbered sums from rings 1 and 3. Let is pick the 11 from ring 3. It obviously cannot match the 29 since there is no '1' on the first ring and we want the sum to be 41. 12 also cannot match.from the 1st ring since there is no 0 on ring 3. So we try the option of 4 turns clockwise of ring 4. 19 2 14 9 4 7 2 12 9 18 8 6 17 3 8 20 SUM: 28 20 22 15 21 10 10 32 Again let us try to match the 11 on the third ring with the 21. this means from the first ring we need a 9 which does not exist. So let us match the 11 with the 15. This means we need the 15 on the first ring which means the rings look as below first to 4th: 5 7 3 15 12 13 21 3 19 2 14 9 4 7 2 12 9 18 8 6 17 3 8 20 8 14 16 11 8 18 10 6 SUM: 41 41 41 41 41 41 41 41 <<<<!!!!! YAY! Just spin the dials so that the above numbers are lined up as shown and you should be fine. If it does not work, turn ALL the rings together so that a different set of 4 numbers is under the green arrow!

Depends on if a lie means the person evaluates the statement and then inverts his answer (For an out of town liar: You: Are you from out of town? OOTL: *I am indeed from out of town. So I will lie and say NO* You: If you raised your hand last time, then raise it again. OOTL: *I did not raise my hand last time, so this time I should do so* ) However, a perfect liar would give you the answer that will prevent you from knowing the truth by withholding information. In this case, that would mean telling the truth when it is likely to lead you to misinterpret the truth or at least not know it.

*whistles, taps foot impatiently* Waiting for another day to dawn... *runs away realizing he might be annoyingly impatient*

Cheers!