:: farm
Mr. Cheney the Saint
on Tue Jun 2 18:07:51 2009, Tyrone hummed the following to the tune of 'Ice Ice Baby'
In talking about terrorists and Guantanamo Bay, Cheney said, "You know, if you don't have a place where you can hold these people the only other option is to kill them. And we don't operate that way."

No we don't operate that way Mr Vice President. We could never be THAT kind. We operate in the idea that making people think we are going to kill them over and over again. Now that is the way to be pure evil.
Pabst
on Thu May 28 00:22:52 2009, Tyrone filabustered vigilantly
Yes Pabst. The beer of choice for this unemployed guy.
Yeah!
on Sat May 16 02:53:25 2009, Tyrone gibbered intensely
That's all I gots ta say bout that.
on Fri May 1 01:59:18 2009, Arlen Spector snarfled angrily
Dear Magic Peace Farm,

After having been a Democrat for nearly three days, it is becoming clear that the Democratic party is drifting away from my long-held beliefs.

For this reason, I am announcing my shift to the Magic Peace Farm Party. Now, where's the beer?
rolled
on Sat Apr 11 19:11:21 2009, j raged intensely
first time evar

thx
$36 computer
on Wed Apr 8 23:19:09 2009, L spouted quietly



I'm not sure how much RAM it can take, tho
Re:
on Wed Apr 8 03:48:23 2009, Ty deciphered ancient Swahili texts reading
Wow. I'd need a computer that can hold that much RAM. I wonder if I could find a computer for $36.
wow
on Wed Apr 8 00:38:28 2009, L posited ecstatically
I just bought 4 gigabytes of RAM for $36. Go recession!
on Sat Apr 4 00:06:30 2009, L hired a team of highwaymen
Noooooooooooooooooooooooooooooooooooo
You know it.
on Wed Apr 1 22:11:17 2009, Shaft remarked ecstatically
on Mon Mar 23 18:15:04 2009, L pondered consummately
The win percentage would have to offset the fact that wins are only 90% profit while losses are 100% loss.

So the win percentage would need to be > 10/19 ~= 52.6% in order to make a profit in expectation.

This is independent of the number of consecutive wins you're targeting... so that strategy is a bust; it doesn't matter if you want to wait for 1 or 3 or 8 wins.

on Sat Mar 21 05:59:22 2009, T spouted intensely
Just so we're clear, he is an example of how a string of this betting strategy might go:

bet 1
win .9 win 1
total 1.9
bet 1.9
win 1.71 win 2
total 3.61
bet 3.61
loss 3.61
reset
bet 1
win .9 win 1
total 1.9
bet 1.9
win 1.71 win 2
total 3.61
bet 3.61
win 3.25 win 3
total 6.86
bet 6.86
win 6.17 win 4
total 13.03
bet 13.03
win 11.73 win 5
total 24.76
bet 24.76
win 22.28 win 6
total 47.04
bet 47.04
win 42.34 win 7
total 89.38
bet 89.38
win 80.44 win 8
total 169.82
8 wins reached so now we bet 1
win .9
total 170.72
bet 1.9
win 1.71
total 172.43
bet 3.61
loss 3.61
total 168.82
now see the dollar is subtracted from our total after the 8 win streak
and so on...
more %
on Sat Mar 21 05:43:48 2009, T boomed ragefully
So then is there a win percentage that works? lets say 51% of the bets are winners rather than 50%. or higher?
hmmm
on Sat Mar 21 05:41:42 2009, T pondered ecstatically
Yes, 8 is part of the strategy. So what you are saying is that for every 510 bets, you could expect that 8 of those will be correct in a row?

In that case, this strategy does not seem profitable since a run of 8 wins profits 170 but the next possible 8 win streak could be 510 away thus putting you right back to zero, right? I assume any number below 8 also doesn't work.
(combinatorial clusterfuck)
on Sat Mar 21 05:17:42 2009, L hired a team of highwaymen
Ah, so 8 is a part of your strategy, not a part of the rules.

I ran a quick simulation to test different "cutoff" values from 1 to 10, over one year of betting. 1000 trials for each cutoff value.

The result suggests that your expected profit at the end is independent of the cutoff value, and grows slowly with the number of bets made. That's kind of surprising, I'm not sure why. It's not a proof, just a simulation, so I could be wrong.

I will point out that to have a >50% chance of winning N consecutive bets, you need to play at least 2^(N+1)-2 bets total. So for N=8, you would need 510 bets before you could expect to get even one run of 8 consecutive.
maths
on Sat Mar 21 03:15:35 2009, Tyrone raged gallantly
I guess the reason I say the dollar is replenished is that you can't run a negative balance. The minimum balance is 1. Like what you said Lonnie, you can't start losing dollar bets until you win eight in a row.
math
on Sat Mar 21 03:11:47 2009, Tyrone spouted incomprehensibly
it is kind of like this, someone continuously replenishes your dollar every time your balance is zero. But only zero. I guess my point is is that if you win three in a row, you can't pocket the money and get another dollar. You must start with the dollar and as long as you loose all, you will get a restart, but to earn any continual profit, you must then make it from your own winnings.

As for eight in a row, I mean eight continuous plays, meaning 2.66 straight days of winning.

And I also really don't know if 8 in a row is the optimal number. It might be more or less. Or like I said, is this system profitable.

I know this is confusing. Don't ask me why I really care, just an interesting idea I guess. I'll keep answering questions as they come.
on Fri Mar 20 23:47:52 2009, L sarcastically proffered the alibi
and also, by eight in a row, do you mean:

A. You win at least one (of three) bets per day, for eight days

B. You win 8 bets in a row (2.66 days of solid win)
on Fri Mar 20 23:45:52 2009, L probed wrathfully
I'm confused.

You don't lose $1 bets unless you have previously won eight in a row?


And: If I win three times in a row, then I have $6.85. If I stop the streak there, do I go back to betting $1 and keep the $6.85?
a little math problem
on Fri Mar 20 22:27:49 2009, Shaft snarfled angrily
Ok, a hypothetical situation that I would love some help on. Here is the premise.

I would like to know the chance that one could get 8 correct guesses in a row based on a coin flip, 50% scenario. To take this farther, lets say each day a person is able to make three guesses each one having a 50% probability of being right. How often can we assume that an 8 correctly picked streak will happen, once a month? once every two months?

Now lets take it farther.

A person bets one dollar per 50% coinflip. If a loss happens, the result goes back to $1, the person can never go below $1, if the person wins, they win 90 cents. Thereby giving them $1.90 This $1.90 is then bet again at 50% and if won, gets a 90% return, if lost, goes back to $1. A four game win streak would look like this, 1, 1.9, 3.61, 6.86, 13.03. If the person ever wins eight guesses in a row, they stop and start at $1 again, however, that dollar is not replenished if they lose again, so if they lose, now their dollar is taken away from any total winnings. They continue this pattern on and on hoping to have a profitable situation.

My question is if it is possible to have a profitable situation with this scenario, or, how many wins to stop a streak is profitable? 4, 5, 10, anything?

Let me know if this is confusing. Not that I do plan on implementing any of said strategy because one would need a 50% situation which is nearly impossible to find, but I am just curious and would appreciate the math help.
ashamed.
on Sun Mar 15 04:19:18 2009, JF led an rabid mob of pro-child barbary apes, chanting
I just stopped checking. Glad you're back, MPF (although I know I am late to say so)
on Sun Mar 1 03:25:45 2009, L. Brezhnev probed cheerfully
That's a good point. Perhaps they are not nationalizing the Bank of America because the government believes it is in SOVIET RUSSIA! (Obamunists?)

On an unrelated note, here is the most exciting thing to EVER HIT THE INTERNETS:

The On-Line Encyclopedia of Integer Sequences

Communism!
on Fri Feb 27 22:12:41 2009, Shaft ventured ecstatically
I the government was going to go ahead an nationalize banks, shouldn't they have started with Bank of AMERICA? not citi group.
curses!
on Sat Feb 21 05:59:04 2009, Tyrone spouted glibly
5at5.com is already registered!
We're in it together
on Wed Feb 18 20:39:46 2009, Shaft boomed glibly
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