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Mugaaz
RealPoor Guru
Joined: 16 Oct 2002
Posts: 3576
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 Posted: 08/23/05 - 04:27 Post subject: Since everyone on Realpoor has an IQ of 160, easy question!
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If you start with 1 trillion dollars and play a gambling game where your average positive expectation is 3 dollars every 100 games, with a standard devation of 20 dollars every 100 games.
Summary:
Bankroll = 1 Trillion
EV = +3/100 games
SD = 20
If you played this game for an infinite amount of time, would you eventually bust out?
If anyone is able to provide a proof of their answer which is not identical to what I find doing random google searches, I will paypal or neteller you 5 dollars.
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Mugaaz
RealPoor Guru
Joined: 16 Oct 2002
Posts: 3576
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Posted: 08/23/05 - 04:54 Post subject:
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I'm going to eventually post some explanation for this problem, but as a general hint I don't think the exact SD really matters as long as it's within certain bounds, I'm about 95% certain of that.
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Soriak
Toomuchtimeonhands
Joined: 11 Oct 2002
Posts: 952
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Posted: 08/23/05 - 04:55 Post subject:
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This solution seems way too easy, but:
If your average expectation is +$3, then the longer you play, the closer you will get to that +$3 figure. The SD is low enough not to bankrupt you on a bad streak, so, you'll never bust out.
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Mugaaz
RealPoor Guru
Joined: 16 Oct 2002
Posts: 3576
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Posted: 08/23/05 - 04:59 Post subject:
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| Soriak wrote: | This solution seems way too easy, but:
If your average expectation is +$3, then the longer you play, the closer you will get to that +$3 figure. The SD is low enough not to bankrupt you on a bad streak, so, you'll never bust out. |
You're free to post any thoughts about whatever you think, but I'm not paying the money without proof. I'm not accepting an axiom as an answer since I believe it is not self evident.
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Silvermouse
RealPoor Jedi
Joined: 12 Oct 2002
Posts: 11015
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Posted: 08/23/05 - 05:51 Post subject:
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lol, nobody wants your 5 dollars. Just post the answer.
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Jakanden
RealPoor Master of Posts
Joined: 11 Nov 2003
Posts: 5334
Location: Fuck if I know - I am always lost
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Posted: 08/23/05 - 05:51 Post subject:
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42 - thank you
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Fattguyy
RealPoor Master of Posts
Joined: 09 Sep 2003
Posts: 9911
Location: Shreveport, Louisiana
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Posted: 08/23/05 - 05:53 Post subject:
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| Silvermouse wrote: | | lol, nobody wants your 5 dollars. Just post the answer. |
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Mugaaz
RealPoor Guru
Joined: 16 Oct 2002
Posts: 3576
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Posted: 08/23/05 - 06:23 Post subject:
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I don't know the answer yet, I have a very strong suspicion but I don't want to have people agree with me for no reason.
The pure theory aspect of this question is if there is a possibility of something happening, is it destined to happen over an infinite amount of time? I happened to disagree with someone about this, I think the actual answer is a little suprising.
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khrath
RealPoor Master of Posts
Joined: 11 Oct 2002
Posts: 8750
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Posted: 08/23/05 - 06:43 Post subject:
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Mug you'll never break $10,000, so forget about all this trillion shit.
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Soriak
Toomuchtimeonhands
Joined: 11 Oct 2002
Posts: 952
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Posted: 08/23/05 - 07:02 Post subject:
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| Mugaaz wrote: | | The pure theory aspect of this question is if there is a possibility of something happening, is it destined to happen over an infinite amount of time? I happened to disagree with someone about this, I think the actual answer is a little suprising. |
If there is a possibility for something to occur, then, given an infinite amount of attempts, it will happen. That also means the lottery could spew up the numbers 1, 2, 3, 4, 5 and 6 - but the odds are so small (for any number) that it'd take ten thousands of years to actually occur. But technically, your odds of winning with that combination are the same as any other.
Not sure what this has to do with your example though. You say that on "average" you make $3 profit after 100 games. The budget and the low margin of error essentially provide you with infinite attempts, removing any blockade from ending up with said average.
That is, of course, assuming a low bet per game. If the bet is high (ie the whole trillion) then the average profit doesn't really do you a whole lot of good, considering the odds for that bet would be only slightly better than tossing a coin.
If I understand your premesis right, then your "game" would be comparable to roulette with only one 0. Even if everyone only played black (or red), never changing their bets, the casino would, of course, make a profit, if everyone played for an infinite number of time.
And you're looking at your game from the side of the casino - slight odds in your favor and effectively unlimited amount of play.
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Mugaaz
RealPoor Guru
Joined: 16 Oct 2002
Posts: 3576
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Posted: 08/23/05 - 08:02 Post subject:
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| Soriak wrote: | | Mugaaz wrote: | | The pure theory aspect of this question is if there is a possibility of something happening, is it destined to happen over an infinite amount of time? I happened to disagree with someone about this, I think the actual answer is a little suprising. |
If there is a possibility for something to occur, then, given an infinite amount of attempts, it will happen. That also means the lottery could spew up the numbers 1, 2, 3, 4, 5 and 6 - but the odds are so small (for any number) that it'd take ten thousands of years to actually occur. But technically, your odds of winning with that combination are the same as any other.
Not sure what this has to do with your example though. You say that on "average" you make $3 profit after 100 games. The budget and the low margin of error essentially provide you with infinite attempts, removing any blockade from ending up with said average.
That is, of course, assuming a low bet per game. If the bet is high (ie the whole trillion) then the average profit doesn't really do you a whole lot of good, considering the odds for that bet would be only slightly better than tossing a coin.
If I understand your premesis right, then your "game" would be comparable to roulette with only one 0. Even if everyone only played black (or red), never changing their bets, the casino would, of course, make a profit, if everyone played for an infinite number of time.
And you're looking at your game from the side of the casino - slight odds in your favor and effectively unlimited amount of play. |
Are you familiar with the term non sequitur? Hint: Paragraph 1 into 2.
The game example I gave was poker, and the SD is more like 17, but both of these are totally irelevant, it's just an example I'm familiar with.
The last hint I'm going to give before I spill what I think about this is that an idiot will never get the answer (wihout proof) to this question, unless purely by chance, and that chance is much less than 50%, and more like less than 1%. Smart people should see where I'm heading with this, Occulis will probably know what my suspected answer is right away.
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motherface
RealPoor Guru
Joined: 12 Mar 2003
Posts: 3407
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sinrakin
RealPoor Master of Posts
Joined: 11 Oct 2002
Posts: 7044
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Posted: 08/23/05 - 08:34 Post subject:
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Heh don't answer it yet, I haven't had a chance to think about it.
First though, why did you quote expectation value and standard deviation over 100 games rather than one game? That throws what could be an irrelevent complication into the problem, of whether you could have pathological reward structures where there were huge disparities over single game rewards, i.e. almost identical probabilities of losing or winning a trillion dollars per game. The SD over 100 games might still be very low, but the SD per game could be very high. I'd have to think about that for awhile before I could move on to the main problem.
If you changed it to an expectation value of $.03 with SD of $.20 every game, would you still consider it to be the same problem? Or do you consider the 100 game stipulation an essential component?
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Mugaaz
RealPoor Guru
Joined: 16 Oct 2002
Posts: 3576
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Posted: 08/23/05 - 09:13 Post subject:
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| sinrakin wrote: | Heh don't answer it yet, I haven't had a chance to think about it.
First though, why did you quote expectation value and standard deviation over 100 games rather than one game? That throws what could be an irrelevent complication into the problem, of whether you could have pathological reward structures where there were huge disparities over single game rewards, i.e. almost identical probabilities of losing or winning a trillion dollars per game. The SD over 100 games might still be very low, but the SD per game could be very high. I'd have to think about that for awhile before I could move on to the main problem.
If you changed it to an expectation value of $.03 with SD of $.20 every game, would you still consider it to be the same problem? Or do you consider the 100 game stipulation an essential component? |
You can change it to those values if you like, I have no problem with that. All of the numbers were chosen at random.
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sinrakin
RealPoor Master of Posts
Joined: 11 Oct 2002
Posts: 7044
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Posted: 08/23/05 - 09:41 Post subject:
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Not an answer yet, but just to try to put it in numerical rather than intuitive concepts:
What we have is an infinite series of probabilities. That series could have two results: it could converge to 1, which would mean you definitely bust at some point, or it could converge to a nonzero number, like .2 or something, which would mean if you could play infinitely long, you'd have a 2/10 chance of going bust. The probability is clearly not zero, because at any point, if you take the amount of money you have at the moment, there would be a tiny probability that you could lose every bet from then on until you ran out of money. For instance, when you start the game, there's a miniscule probability that you could lose every single bet. The probability of going bust can never be less than that, so it can't be zero.
Naively, you could say that if at any point there's a fractional probability of going bust, and there are an infinite number of times you could make that check, then it must be guaranteed that you'll eventually lose. However, since your balance is increasing over time because of the positive expectation value, it's possible that that infinite series will in fact converge to a value less than one because the individual probabilites will get smaller fast enough (the longer you play, the more losses in a row you need to bust you).
Maybe that was obvious, just trying to point out that you can't simply say "if the probability is non-zero and you do it forever you're guaranteed to lose". I still suspect the answer is you'll lose because your expected bankroll appears to be going up "slowly" (i.e. linearly), but you have to compute it to be sure.
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Occulis
RealPoor Jedi
Joined: 11 Oct 2002
Posts: 13293
Location: Moral Relativity Central
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Posted: 08/23/05 - 10:10 Post subject:
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You always lose in gambling, but it's nice to fantasize that somehow you won't! There you go sir, an IQ-unrelated answer!
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Mugaaz
RealPoor Guru
Joined: 16 Oct 2002
Posts: 3576
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Posted: 08/23/05 - 10:55 Post subject:
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| sinrakin wrote: | Not an answer yet, but just to try to put it in numerical rather than intuitive concepts:
What we have is an infinite series of probabilities. That series could have two results: it could converge to 1, which would mean you definitely bust at some point, or it could converge to a nonzero number, like .2 or something, which would mean if you could play infinitely long, you'd have a 2/10 chance of going bust. The probability is clearly not zero, because at any point, if you take the amount of money you have at the moment, there would be a tiny probability that you could lose every bet from then on until you ran out of money. For instance, when you start the game, there's a miniscule probability that you could lose every single bet. The probability of going bust can never be less than that, so it can't be zero.
Naively, you could say that if at any point there's a fractional probability of going bust, and there are an infinite number of times you could make that check, then it must be guaranteed that you'll eventually lose. However, since your balance is increasing over time because of the positive expectation value, it's possible that that infinite series will in fact converge to a value less than one because the individual probabilites will get smaller fast enough (the longer you play, the more losses in a row you need to bust you).
Maybe that was obvious, just trying to point out that you can't simply say "if the probability is non-zero and you do it forever you're guaranteed to lose". I still suspect the answer is you'll lose because your expected bankroll appears to be going up "slowly" (i.e. linearly), but you have to compute it to be sure. |
Pretty close to what I'm thinking. I think that even over an infinite amoutn of time there is only a finite chance of going bust.
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Frax
RealPoor Master of Posts
Joined: 11 Oct 2002
Posts: 8489
Location: Fuck yoiu fucking fuckers
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Posted: 08/23/05 - 11:23 Post subject:
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You'd die before you went broke.
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Plat4PoP
RealPoor Jedi
Joined: 12 Oct 2002
Posts: 14376
Location: USA
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Posted: 08/23/05 - 11:31 Post subject:
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the rake will eat your bankroll up eventually
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motherface
RealPoor Guru
Joined: 12 Mar 2003
Posts: 3407
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Posted: 08/23/05 - 12:13 Post subject:
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I like (1+(1/n))^n !
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Silvermouse
RealPoor Jedi
Joined: 12 Oct 2002
Posts: 11015
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Posted: 08/23/05 - 15:38 Post subject:
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| Occulis wrote: | | You always lose in gambling, but it's nice to fantasize that somehow you won't! There you go sir, an IQ-unrelated answer! |
Err...not if he divides by panda, you idiot lol.
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Ishmael
RealPoor Guru
Joined: 03 Jun 2005
Posts: 4446
Location: The US of A
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Posted: 08/23/05 - 15:40 Post subject:
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| Silvermouse wrote: | | Occulis wrote: | | You always lose in gambling, but it's nice to fantasize that somehow you won't! There you go sir, an IQ-unrelated answer! |
Err...not if he divides by panda, you idiot lol. |
lolirl.
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Aeain
RealPoor Sensei
Joined: 22 Oct 2002
Posts: 1973
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Posted: 08/23/05 - 16:54 Post subject:
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Standard Deviation. Shit, I can't remember that and I took in last year....Oh yeah, if I can recall it's what we were doin when I suffered my 2 concussions.
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sinrakin
RealPoor Master of Posts
Joined: 11 Oct 2002
Posts: 7044
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Posted: 08/23/05 - 18:15 Post subject:
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Still not an answer, but some more qualitative stuff.
You can look at it as a one-dimensional random walk with a net drift (equal to the expectation value). There are lots of payoffs that could meet the EV=3, SD=20 criteria. For instance, a 50/50 chance of winning 31 or losing 25 would do it. Average winning is 3, SD is about 20. This is the same as a random walk with steps of plus or minus 28, with the cutoff line starting at minus your initial bankroll and moving to the left by 3 every bet.
The mean distance of a random walk is proportional to the square root of the number of steps, and the distance you move with each step. There's a 50/50 chance that distance could be in the positive (winning) or negative (losing) direction. So your random walk is moving away from the origin at a rate proportional to the square root of the number of bets (actually 28*sqrt(2*N/pi) I think, 28 because that's your average win/loss not counting the drift of 3), but the amount you need to lose to break you is moving away from the origin linearly with the number of bets (3*N).
Intuitively, as time goes on that sqrt function is going to get farther and farther from the linear function, which tends to imply you won't go bankrupt. Still not a proof though.
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IceIsFun
Toomuchtimeonhands
Joined: 11 Oct 2002
Posts: 781
Location: Orlando, FL
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Posted: 08/23/05 - 22:08 Post subject: Re: Since everyone on Realpoor has an IQ of 160, easy questi
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| Mugaaz wrote: | If you start with 1 trillion dollars and play a gambling game where your average positive expectation is 3 dollars every 100 games, with a standard devation of 20 dollars every 100 games.
Summary:
Bankroll = 1 Trillion
EV = +3/100 games
SD = 20
If you played this game for an infinite amount of time, would you eventually bust out?
If anyone is able to provide a proof of their answer which is not identical to what I find doing random google searches, I will paypal or neteller you 5 dollars. |
It's a trick question. Your EV/SD isn't known with 100% confidence.
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sinrakin
RealPoor Master of Posts
Joined: 11 Oct 2002
Posts: 7044
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Posted: 08/24/05 - 11:43 Post subject:
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Actually the math gets complicated surprisngly fast. Qualitatively, I'll wave my hands and stick by my assertion that your expectation value is a gaussian centered on initial_bankroll+3N, with width proportional to sqrt(N), N being the number of bets, which means that with increasing time less and less of the curve appears to the left of the origin, so you don't go bankrupt. The hand-waving part is because some of the left hand part of that Gaussian should have been bankrupt even if it's slightly to the right of the origin, depending on whether you got the losses or the wins first, but we can assume (with no justification) that that's irrelevant 
I got tired of thinking about it so I decided to just test it. Here's my test, don't b***h because I only spent 45 minutes on it so it's not pretty, although it's fairly rigorous.
| Code: |
#include <stdlib.h>
#include <sys/time.h>
/* ========================================================================
WIN_AMOUNT = amount a winning bet pays
LOSE_AMOUNT = amount a losing bet costs
BANKRUPT_COUNT = run simulation until bankruptcy occurs this many times.
Since the expectation is that probability of bankruptcy is low, Poisson
statistics apply, and expected error will be 1/sqrt(BANKRUPT_COUNT).
For BANKRUPT_COUNT=10000, computed bankruptcy probability will be plus
or minus one percent of the computed value.
======================================================================== */
#define WIN_AMOUNT 31
#define LOSE_AMOUNT 25
#define BANKRUPT_COUNT 10000
/* ========================================================================
keep some stats to sanity check the random numbers
init_stats: initialise win/loss statistics
stats: update win/loss statistics
print_stats: display win/loss statistics
======================================================================== */
long int stats_wins;
long int stats_losses;
void init_stats()
{
stats_wins=0;
stats_losses=0;
}
void stats(int win)
{
if (win) {
if (++stats_wins <= 0) {
printf("wins overflowed\v");
exit(1);
}
}
else {
if (++stats_losses <= 0) {
printf ("losses overflowed\n");
exit(1);
}
}
}
void print_stats()
{
printf("\t(w/l %ld %ld)\n", stats_wins, stats_losses);
}
/* ========================================================================
reseed - reseed random number generator
======================================================================== */
void reseed()
{
struct timeval tv;
if (gettimeofday(&tv, 0))
perror("gettimeofday");
srand(tv.tv_usec);
}
/* ========================================================================
check_win - return 0 or 1 with 50/50 odds
======================================================================== */
int check_win()
{
if (random() > RAND_MAX/2) {
stats(1);
return 1;
}
else {
stats(0);
return 0;
}
}
/* ========================================================================
one_bet - make one win/lose bet and return new bankroll
======================================================================== */
int one_bet(int bankroll)
{
if (check_win()) {
bankroll += WIN_AMOUNT;
if (bankroll < 0) {
printf("overflow error\n");
exit(1);
}
}
else
bankroll -= LOSE_AMOUNT;
return bankroll;
}
/* ========================================================================
see_if_bankrupts - bet until you either go bankrupt or hit an upper limit
======================================================================== */
int see_if_bankrupts(int bankroll, int termination_bankroll)
{
reseed();
while (bankroll > 0) {
bankroll = one_bet(bankroll);
if (bankroll > termination_bankroll)
return 0;
}
return 1;
}
/* ========================================================================
get_N_bankrupts - start with an initial and termination bankroll,
and bet until you bankrupt or terminate, until you bankrupt N times.
Return fraction of times that we bankrupt as opposed to hitting the upper
termination limit.
======================================================================== */
float get_N_bankrupts(int bankroll, int termination_bankroll, int n)
{
int bankrupts = 0;
int trials = 0;
while (bankrupts < n) {
trials++;
if (see_if_bankrupts(bankroll, termination_bankroll)) {
bankrupts++;
}
}
return (float)bankrupts/trials;
}
/* ------------------------------------------------------------------------
Starting with an initial bankroll value, and placing bets that have
50/50 odds of wining WIN_AMOUNT or losing LOSE_AMOUNT.
Do successive trials to determine the fraction times bankruptcy occurs,
as opposed to hitting an upper limit which is a multiple of the initial
bankroll. For successive trials increase the multiple required for
for termination to observe convergence as time goes to infinity.
Reseed random number generator from system clock for each trial.
------------------------------------------------------------------------ */
int main (int argc, char** argv)
{
int results = 0;
int multiple;
int initial_bankroll;
float result;
float result_total, mean;
if (argc != 2) {
printf("Usage %s <initial_bankroll>\n", argv[0]);
exit(1);
}
initial_bankroll = atoi(argv[1]);
printf ("initial_bankroll %d, even odds, win %d, lose %d\n",
initial_bankroll, WIN_AMOUNT, LOSE_AMOUNT);
result_total = 0;
for (multiple=2; multiple<=(1024); multiple*=2) {
init_stats();
result = get_N_bankrupts(initial_bankroll, initial_bankroll*multiple, BANKRUPT_COUNT);
result_total += result;
mean = result_total/++results;
printf("bankrupties %f, %f from mean, %d*bankroll", result, result-mean, multiple);
print_stats();
}
}
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It assumes the specific case of each bet having a 50/50 chance of paying 31 or losing 25 (average win is 3, S.D. is about 20). It places bets and sees whether you go bankrupt or hit an arbitrary upper limit which is a multiple of your initial bankroll. It does successive runs with the upper limit increasing toward infinity to test if the values converge. You supply your initial bankroll when you run the program. It does each run until it gets 10,000 bankruptcies, which should give the answer to plus or minus one percent of itself.
Results:
| Code: | initial_bankroll 200, even odds, win 31, lose 25
bankrupties 0.167847, 0.000000 from mean, 2*bankroll (w/l 1432769 1436579)
bankrupties 0.195088, 0.013620 from mean, 4*bankroll (w/l 3868262 3867386)
bankrupties 0.197824, 0.010904 from mean, 8*bankroll (w/l 9203847 9203439)
bankrupties 0.197789, 0.008152 from mean, 16*bankroll (w/l 20015606 20013699)
bankrupties 0.198228, 0.006873 from mean, 32*bankroll (w/l 41512069 41505425)
bankrupties 0.197044, 0.004741 from mean, 64*bankroll (w/l 85288833 85281437)
bankrupties 0.200048, 0.006638 from mean, 128*bankroll (w/l 169015792 169013607)
bankrupties 0.198499, 0.004453 from mean, 256*bankroll (w/l 343053699 343076848)
bankrupties 0.195149, 0.000980 from mean, 512*bankroll (w/l 702033339 701981005)
bankrupties 0.197060, 0.002602 from mean, 1024*bankroll (w/l 1388724259 1388613875)
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You can pretty much see that it's converging, not diverging toward 1. It's also interesting that you don't need anywhere near a trillion dollars; this started with 200 dollars and only banrupts 19% of the time.
Behaviour with different starting bankrolls:
| Code: |
initial_bankroll 100, even odds, win 31, lose 25
bankrupties 0.306636, 0.000000 from mean, 2*bankroll (w/l 251247 250800)
bankrupties 0.406521, 0.049942 from mean, 4*bankroll (w/l 581970 582054)
bankrupties 0.431723, 0.050097 from mean, 8*bankroll (w/l 1382295 1381762)
bankrupties 0.429960, 0.036250 from mean, 16*bankroll (w/l 3152046 3149304)
bankrupties 0.434934, 0.032979 from mean, 32*bankroll (w/l 6548053 6544967)
bankrupties 0.423280, 0.017772 from mean, 64*bankroll (w/l 14114020 14103890)
initial_bankroll 200, even odds, win 31, lose 25
bankrupties 0.167847, 0.000000 from mean, 2*bankroll (w/l 1432769 1436579)
bankrupties 0.195088, 0.013620 from mean, 4*bankroll (w/l 3868262 3867386)
bankrupties 0.197824, 0.010904 from mean, 8*bankroll (w/l 9203847 9203439)
bankrupties 0.197789, 0.008152 from mean, 16*bankroll (w/l 20015606 20013699)
bankrupties 0.198228, 0.006873 from mean, 32*bankroll (w/l 41512069 41505425)
bankrupties 0.197044, 0.004741 from mean, 64*bankroll (w/l 85288833 85281437)
initial_bankroll 300, even odds, win 31, lose 25
bankrupties 0.083014, 0.000000 from mean, 2*bankroll (w/l 5268458 5270004)
bankrupties 0.092298, 0.004642 from mean, 4*bankroll (w/l 14458609 14457088)
bankrupties 0.090396, 0.001827 from mean, 8*bankroll (w/l 34919116 34914789)
bankrupties 0.091216, 0.001985 from mean, 16*bankroll (w/l 74309169 74278113)
bankrupties 0.091971, 0.002192 from mean, 32*bankroll (w/l 152855891 152882923)
bankrupties 0.090390, 0.000509 from mean, 64*bankroll (w/l 316701731 316700003)
initial_bankroll 400, even odds, win 31, lose 25
bankrupties 0.039625, 0.000000 from mean, 2*bankroll (w/l 16033036 16030974)
bankrupties 0.041201, 0.000788 from mean, 4*bankroll (w/l 46388078 46384792)
bankrupties 0.040788, 0.000250 from mean, 8*bankroll (w/l 109626799 109630179)
bankrupties 0.041795, 0.000943 from mean, 16*bankroll (w/l 229141359 229147488)
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Brash
RealPoor Guru
Joined: 20 Oct 2002
Posts: 3958
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Posted: 08/24/05 - 12:26 Post subject:
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| Soriak wrote: | is
If there is a possibility for something to occur, then, given an infinite amount of attempts, it will happen |
this is all you need to know and in itself proof enough ... despite the odds being in your favor gambling is random and on an infinite time line you will eventually loose enough hands to break you.
you owe soriak 5 bucks
even though flipping a coin is 50/50, on a infinite time line sooner or later you will flip heads 8 billion times in a row
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sinrakin
RealPoor Master of Posts
Joined: 11 Oct 2002
Posts: 7044
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Posted: 08/24/05 - 12:46 Post subject:
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| Brash wrote: | | Soriak wrote: | is
If there is a possibility for something to occur, then, given an infinite amount of attempts, it will happen |
this is all you need to know and in itself proof enough ... despite the odds being in your favor gambling is random and on an infinite time line you will eventually loose enough hands to break you.
you owe soriak 5 bucks
even though flipping a coin is 50/50, on a infinite time line sooner or later you will flip heads 8 billion times in a row |
Your premises are right but your conclusion is wrong. It is true that in an infinite amount of time, every random event will occur, e.g. flipping heads 8 billion times in a row.
The reason the conclusion is wrong is there's no finite event that you can state which would be sufficient to bankrupt you, since as time goes to infinity, the number of hands you would have to lose also goes to infinity because of your positive expectation value. So while at some point 8 million losses in a row might break you, by the time you finally get that 8 million loss string, there's a fairly high probability that you would need 8 trillion losses to break you. And by the time you get those 8 trillion losses, it might take 8 million trillion losses to break you.
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Brash
RealPoor Guru
Joined: 20 Oct 2002
Posts: 3958
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Posted: 08/24/05 - 13:34 Post subject:
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finite event = when you run out of money.
on a long enough time line (infinite) you will flip heads 8 trillion times or 800,000,000,000,000,000,000 times, whatever it takes to get you to the finite event of running out of money so you can't play anymore .
*edit*
| Quote: | | It is true that in an infinite amount of time, every random event will occur |
one of the random events will be you running out of money. when you run out of money you are done and have sustained loses ...
f**k i want the $5 now
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Devook
RealPoor Guru
Joined: 31 Mar 2004
Posts: 2374
Location: Ypsilanti or Troy, MI
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Posted: 08/24/05 - 14:34 Post subject:
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| Brash wrote: | finite event = when you run out of money.
on a long enough time line (infinite) you will flip heads 8 trillion times or 800,000,000,000,000,000,000 times, whatever it takes to get you to the finite event of running out of money so you can't play anymore .
*edit*
| Quote: | | It is true that in an infinite amount of time, every random event will occur |
one of the random events will be you running out of money. when you run out of money you are done and have sustained loses ...
f**k i want the $5 now |
This is faulty logic; running out of money is not an event, it's an outcome.
Over an infinite amount of time, there is only one outcome: whichever has the highest probability. I don't remember any formulas since I haven't done this kind of work in about 4 years, but it would seem to me the odds of coming out ahead after 100 hands are 23:17, or 57.5% probability. The odds of bankrupting overall are dependent on the amount of money you hold. Over an infinite timeline, assuming that you didn't hit the infinitesimally small chance that you bankrupt your trillion dollar roll through a series of losses ranging from rougly 1-20 dollars, the probability of never going bankrupt rises to 100% and the probability of losing it all goes to 0.
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