## Stubborn bettor

For the discussion of math. Duh.

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### Stubborn bettor

Hi everybody,

Here are the bet rules.
We toss a coin. Bettors bet k dollars on head (or tail). If head (or tail) appears all those who bet head (or tail) wins 2k dollars (profit = k dollars). Otherwise they loss k dollars if the opposite of their bets appears (head or tail).
One bettor is stubborn.
His strategy is to always bet 1 dollar (not more or less) on head at each tossing or rebet its current win.
As long as he is loosing money he rebets his current wins. Otherwise he bet one dollar and take his profits until the goal is reached.
His goal is to reach 10 dollars profit or more.
Once he reaches his goal he stop playing.

He does not know how much money he need to reach his goal in 98% of the cases

Can you help him?

Thanks for any clue

measure
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### Re: Stubborn bettor

I faster way (assuming no maximum bet and no borrowing limit) would be to always bet enough (borrowing if needed) that a win will allow him to immediately reach his goal after repaying what he borrowed. ex:

1. Goal = 10,000, Have/Owe = 0/0
2. Borrow 10,000 (10,000/10,000)
3. Bet 10,000, Toss 1 = loss (0/10,000)
4. Borrow 20,000 (20,000/30,000)
5. Bet 20,000, Toss 2 = loss (0/30,000)
6. Borrow 40,000 (40,000/70,000)
7. Bet 40,000, Toss 3 = win (80,000/70,000)
8. Repay debt of 70,000 (10,000/0)
9. done

Using this method, the bettor will always reach his goal after the first successful bet.

thefargo
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### Re: Stubborn bettor

I took a shortcut and just coded it up. I show he needs \$71 for a 98% chance of hitting \$10 before going broke.

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### Re: Stubborn bettor

He can not bet more than one dollar.
So doubling the bet is not allowed.

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### Re: Stubborn bettor

thefargo wrote:I took a shortcut and just coded it up. I show he needs \$71 for a 98% chance of hitting \$10 before going broke.

As an example :
A tossing sequence :

TTTTTHHHH

wiil allow him to reach his goal :
-5 loss
by rebetting 3 times he can NOT reach but rebetting 4 times he will win 15 dollars - 5 = 10 dollars.
Last edited by Goahead52 on Fri May 29, 2015 3:18 pm UTC, edited 1 time in total.

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### Re: Stubborn bettor

thefargo wrote:I took a shortcut and just coded it up. I show he needs \$71 for a 98% chance of hitting \$10 before going broke.

If it is true then I will go to the Montreal casino tonight with 71000 dollars as capital to win 10000 dollars!

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### Re: Stubborn bettor

Simulating the case is easy but finding a close formula to compute the capital needed to reach a goal m (instead of 10 dollars) with 98% chances is hard.

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### Re: Stubborn bettor

thefargo wrote:I took a shortcut and just coded it up. I show he needs \$71 for a 98% chance of hitting \$10 before going broke.

Hi,

Because if it was proven true then many people will invade the casinos with 7100 dollars as capital to win an average of 1000 dollars a day. That`s 30.000 dollars a month!!!!????
Any proof other than simulation?

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### Re: Stubborn bettor

\$71 doesn't sound high enough.

I would think a valid method to calculate this would be to assume that your average winnings is 0. There are two outcomes, either you gain \$10 (98% of the time) or lose k dollars (2% of the time).
Average winnings = 0 = 10*.98 - k*.02
k = 10*.98/.02 = 490

So the stubborn bettor would need \$490.

This doesn't take into account the betting style and I'm not sure if it matters.

measure
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### Re: Stubborn bettor

Goahead52 wrote:He can not bet more than one dollar.
So doubling the bet is not allowed.

My mistake. I took "can you help him" to mean "can you improve his strategy".

So to be clear, after entering the game with capital C and goal G, does his bet differ from 1 dollar:

A.) only if he currently has more than C dollars.

- or -

B.) only if he won last bet.

For example, if he losses the first three bets (net -3), then wins once (net -2), does he then bet 1 dollar or 2?

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### Re: Stubborn bettor

measure wrote:
Goahead52 wrote:He can not bet more than one dollar.
So doubling the bet is not allowed.

My mistake. I took "can you help him" to mean "can you improve his strategy".

So to be clear, after entering the game with capital C and goal G, does his bet differ from 1 dollar:

A.) only if he currently has more than C dollars.

- or -

B.) only if he won last bet.

For example, if he losses the first three bets (net -3), then wins once (net -2), does he then bet 1 dollar or 2?

If he losses 3 dollars (3 losses in row) then he will rebet his 2 dollars after winning them. He will continue to rebet what he wins win one dollar until his goal is achieved.
As long as he is far from his goal either he bets 1 dollar either he rebets what he wins (not what he has cumulated as profits).
He is stubborn.
Stubborn strategy if we want to label it.

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### Re: Stubborn bettor

Is this method of betting worth a deep analysis?
We have to keep in mind that all the casinos rule a limit to bet (from 1000 dollars to 6000 dollars or maybe more).
At some point we need to define precisely the capital needed to bet in real. The goal then will have a limit too.
Finding a close formula seems to be hard anyway.

>-)
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### Re: Stubborn bettor

what do you mean by "rebets his current wins"?

measure
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### Re: Stubborn bettor

Okay, as I understand it, the "stubborn strategy" uses this basic algorithm:

Code: Select all

`if(net_profit >= goal)    Success;if(won_last_time){    Bet(Min(goal - net_profit, net_profit_last_round - net_profit)); // bet twice as much as last time unless close to goal}else // lost last time or first round{    Bet(1);}if(net_profit <= -capital)    Failure;`

This strategy gives a 98% success rate for a goal of \$10 with starting capital of at least \$439, which is noticeably less than the expected \$490. This gives an expected gain of \$1.02 per \$10 goal.

Code: Select all

`Capital: 440Goal: 10Trials: 10000000Success: 9806092 (0.980609)Fail: 193908 (0.0193908)Capital: 439Goal: 10Trials: 10000000Success: 9806095 (0.980609)Fail: 193905 (0.0193905)Capital: 438Goal: 10Trials: 10000000Success: 9798890 (0.979889)Fail: 201110 (0.020111)`

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### Re: Stubborn bettor

>-) wrote:what do you mean by "rebets his current wins"?

It is easy to understand it.
Assume that at some stage the bettor has lost 20 dollars.
He bet then 1 dollar on the next tossing.
Head! he wins 2 dollars (profit +1 dollar).
So his situation is now +1-20=-19
He does not need to take money from his pocket.
So he re-bets the 2 dollars.
now he has 4 dollars to re-bet
So he is now at -15 loss.
He does it.
Waouh! He is has to re-bet the 8 dollars.
Tail!
He has lost his rebet.
So his siuation will -20-1=-21 (loss)
As long as he did not reach his goal (10 dollars global profit) he will continue either to bet 1 dollar in case of loss or re-bet what he has currently won.

In this case as the loss it 20 he has to win his bet 6 times in row to reach his 32-20=12.

Understood now?

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### Re: Stubborn bettor

measure wrote:Okay, as I understand it, the "stubborn strategy" uses this basic algorithm:

Code: Select all

`if(net_profit >= goal)    Success;if(won_last_time){    Bet(Min(goal - net_profit, net_profit_last_round - net_profit)); // bet twice as much as last time unless close to goal}else // lost last time or first round{    Bet(1);}if(net_profit <= -capital)    Failure;`

This strategy gives a 98% success rate for a goal of \$10 with starting capital of at least \$439, which is noticeably less than the expected \$490. This gives an expected gain of \$1.02 per \$10 goal.

Code: Select all

`Capital: 440Goal: 10Trials: 10000000Success: 9806092 (0.980609)Fail: 193908 (0.0193908)Capital: 439Goal: 10Trials: 10000000Success: 9806095 (0.980609)Fail: 193905 (0.0193905)Capital: 438Goal: 10Trials: 10000000Success: 9798890 (0.979889)Fail: 201110 (0.020111)`

440 dollars as capital seems to me more logical.
But you do not need to fix the number of trials at 10000000.
The number of tossing needed to reach the goal is changing too.

In fact a close formula will give not only the capital needed but the maximal number of tossing to reach the goal.

Thank you very much for your feedback.

PeteP
What the peck?
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### Re: Stubborn bettor

The number of trials isn't the number of tosses. Measure determines the result by playing 10000000 times until success or failure is reached.

Btw mind posting your whole code, measure? I got 980004/19996 w/l ratio for 490 and only 977911 for 439.

measure
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### Re: Stubborn bettor

PeteP wrote:The number of trials isn't the number of tosses. Measure determines the result by playing 10000000 times until success or failure is reached.

Btw mind posting your whole code, measure? I got 980004/19996 w/l ratio for 490 and only 977911 for 439.

Sure thing:

Spoiler:

Code: Select all

`#include "stdafx.h"#include <iostream>using namespace std;const bool DEBUG = false;const unsigned SEED = 447328;const int START = 440;const int STEP = -1;const int GOAL = 10;const int TRIALS = 1000000;const int DEFAULT = 1;bool Game(int, int);int _tmain(int argc, _TCHAR* argv[]){   srand(SEED);   int capital = START;   while(true)   {      int success = 0;      int fail = 0;      for(int i = 0; i < TRIALS; i++)      {         if(Game(capital, GOAL))            success++;         else            fail++;      }      cout << "Capital: " << capital << endl;      cout << "Goal: " << GOAL << endl;      cout << "Trials: " << TRIALS << endl;      cout << "Success: " << success << " (" << success/(float)TRIALS << ")" << endl;      cout << "Fail: " << fail << " (" << fail/(float)TRIALS << ")"  << endl;      cin.get();      capital += STEP;   }   return 0;}bool Game(int capital, int goal){   int current = capital;   int winnings = 0;   if(DEBUG) cout << "Current\tGoal\n" << current + winnings << "\t" << capital + goal << endl;   while(current + winnings > 0)   {      if(DEBUG) cin.get();      int difference = (capital + goal) - (current + winnings);      if(difference <= 0)      {         return true;      }      else      {         if(winnings > 0)         {            if(winnings > difference)            {               current += winnings - difference;               winnings = difference;            }            if(rand()%2 < 1) // bet winnings               winnings *= 2;            else               winnings = 0;         }         else         {            current -= DEFAULT;            if(rand()%2 < 1) // bet 1               winnings += 2*DEFAULT;         }         if(DEBUG) cout << current + winnings << "\t" << capital + goal << endl;      }   }   return false;}`

By the way, for larger goal and capital amounts, the default bet of \$1 should be increased proportionally as well. Otherwise, while the expected return and chance of success increase, the expected number of bets also increases dramatically.

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### Re: Stubborn bettor

Thank you.
What will happen to the capital needed if we adopt a new strategy " half-stubborn-half-reasonable"?
In this new strategy a bettor re-bet his current wins ONLY if he has a global loss, otherwise he bets 1 dollar an take immediately his profit.

There is another stubborn strategy used by 2 friends on the same table in the same casino. One play head and the other play tail but the loss taking into account is equal to the sum of their losses. They will play until reaching their goal (let us say 10 dollars). Is the capital needed going to change?

Anyway I will try this stubborn strategy starting with 500 dollars as capital to see if it is interesting.

measure
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### Re: Stubborn bettor

Goahead52 wrote:Anyway I will try this stubborn strategy starting with 500 dollars as capital to see if it is interesting.

Be prepared to be very stubborn. The average number of rounds is about 90, but about 5% of games reach 500 rounds or more.
Spoiler:

PeteP
What the peck?
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### Re: Stubborn bettor

measure wrote:
PeteP wrote:The number of trials isn't the number of tosses. Measure determines the result by playing 10000000 times until success or failure is reached.

Btw mind posting your whole code, measure? I got 980004/19996 w/l ratio for 490 and only 977911 for 439.

Sure thing:

Spoiler:

Code: Select all

`#include "stdafx.h"#include <iostream>using namespace std;const bool DEBUG = false;const unsigned SEED = 447328;const int START = 440;const int STEP = -1;const int GOAL = 10;const int TRIALS = 1000000;const int DEFAULT = 1;bool Game(int, int);int _tmain(int argc, _TCHAR* argv[]){   srand(SEED);   int capital = START;   while(true)   {      int success = 0;      int fail = 0;      for(int i = 0; i < TRIALS; i++)      {         if(Game(capital, GOAL))            success++;         else            fail++;      }      cout << "Capital: " << capital << endl;      cout << "Goal: " << GOAL << endl;      cout << "Trials: " << TRIALS << endl;      cout << "Success: " << success << " (" << success/(float)TRIALS << ")" << endl;      cout << "Fail: " << fail << " (" << fail/(float)TRIALS << ")"  << endl;      cin.get();      capital += STEP;   }   return 0;}bool Game(int capital, int goal){   int current = capital;   int winnings = 0;   if(DEBUG) cout << "Current\tGoal\n" << current + winnings << "\t" << capital + goal << endl;   while(current + winnings > 0)   {      if(DEBUG) cin.get();      int difference = (capital + goal) - (current + winnings);      if(difference <= 0)      {         return true;      }      else      {         if(winnings > 0)         {            if(winnings > difference)            {               current += winnings - difference;               winnings = difference;            }            if(rand()%2 < 1) // bet winnings               winnings *= 2;            else               winnings = 0;         }         else         {            current -= DEFAULT;            if(rand()%2 < 1) // bet 1               winnings += 2*DEFAULT;         }         if(DEBUG) cout << current + winnings << "\t" << capital + goal << endl;      }   }   return false;}`

By the way, for larger goal and capital amounts, the default bet of \$1 should be increased proportionally as well. Otherwise, while the expected return and chance of success increase, the expected number of bets also increases dramatically.

I wondered why we got different results and checked the script and found nothing. Then I remembered vaguely that the rand()% method introduced a light skew. So first I added variables to compare the amount of negative and positive results. Predictably that didn't really help. But then later I replaced that with this http://stackoverflow.com/questions/2999075/generate-a-random-number-within-range/2999130#2999130 which gives a range without skew and used rand_lim(1) instead of rand%2. And now it give the expected result that you need 490 for 98%

Lesson learned even a very tiny amount of skew in your random numbers can fuck with your simulations.

>-)
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### Re: Stubborn bettor

shouldn't he overshoot \$10 at least some of the time, making k > 490?

PeteP
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### Re: Stubborn bettor

>-) wrote:shouldn't he overshoot \$10 at least some of the time, making k > 490?

The script measure vote checks whether a win would bring you over 10 and reduces the bet so that winning will land you exactly on 10.

measure
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### Re: Stubborn bettor

PeteP wrote:Lesson learned even a very tiny amount of skew in your random numbers can fuck with your simulations.

Good to know!

PeteP
What the peck?
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### Re: Stubborn bettor

Btw if anyone else was curious, about the difference between the rand_lim method from StackOverflow and rand%2. Some messy c code i tested some things with:
Spoiler:

Code: Select all

`#include <stdlib.h>#include <time.h>#include <math.h>#include <iostream>using namespace std;const unsigned SEED = 447328;const int TRIALS = 10000000;int rand_lim(int limit) {/* return a random number between 0 and limit inclusive. */    int divisor = RAND_MAX/(limit+1);    int retval;    do {         retval = rand() / divisor;    } while (retval > limit);    return retval;}int main(){   srand(SEED);   int capital = START;   int seq_length = 0;   int seq_length_sum = 0;   int seq_length_quad_sum = 0;   int sequences = 0;   int res_0 = 0;   int res_1 = 0;   int seq_l1=0;   int seq_l2=0;   int seq_l3=0;   int seq_l4=0;   int seq_l5=0;   int seq_l6=0;   int seq_l7=0;   int seq_l8=0;   int seq_l9=0;   int seq_l10=0;   int last = -1;   int random;   while(true)   {      seq_length = 0;     seq_length_sum = 0;     seq_length_quad_sum = 0;      sequences = 0;     last = -1;     seq_l1=0;     seq_l2=0;     seq_l3=0;     seq_l4=0;     seq_l5=0;     seq_l6=0;     seq_l7=0;     seq_l8=0;     seq_l9=0;     seq_l10=0;     res_0 = 0;     res_1 = 0;      for(int i = 0; i < TRIALS; i++)      {         random=rand_lim(1);       if(random==0)res_0++;       else res_1++;                if(random==last){         seq_length++;       }       else{          if(seq_length==1)seq_l1++;          if(seq_length==2)seq_l2++;          if(seq_length==3)seq_l3++;          if(seq_length==4)seq_l4++;          if(seq_length==5)seq_l5++;          if(seq_length==6)seq_l6++;          if(seq_length==7)seq_l7++;          if(seq_length==8)seq_l8++;          if(seq_length==9)seq_l9++;          if(seq_length==10)seq_l10++;          seq_length_sum += seq_length;          seq_length_quad_sum += pow(seq_length,2);          seq_length = 1;           sequences++;                 }       last=random;      }      seq_length_sum += seq_length;      cout << "Average sequence length: " << seq_length_sum/(float)sequences << endl;      cout << "Average (sequence length)^2: " << seq_length_quad_sum/(float)sequences << endl;      cout << "Zeroes " << res_0 << " Ones " << res_1 << endl;   /*    cout << seq_l1 << endl;      cout << seq_l2 << endl;      cout << seq_l3 << endl;      cout << seq_l4 << endl;      cout << seq_l5 << endl;      cout << seq_l6 << endl;      cout << seq_l7 << endl;      cout << seq_l8 << endl;      cout << seq_l9 << endl;      cout << seq_l10 << endl; */            seq_length = 0;     seq_length_sum = 0;     seq_length_quad_sum = 0;      sequences = 0;     last = -1;          seq_l1=0;     seq_l2=0;     seq_l3=0;     seq_l4=0;     seq_l5=0;     seq_l6=0;     seq_l7=0;     seq_l8=0;     seq_l9=0;     seq_l10=0;     res_0 = 0;     res_1 = 0;      for(int i = 0; i < TRIALS; i++)      {         random=rand()%2;       if(random==0)res_0++;       else res_1++;      if(random==last){         seq_length++;       }       else{          if(seq_length==1)seq_l1++;          if(seq_length==2)seq_l2++;          if(seq_length==3)seq_l3++;          if(seq_length==4)seq_l4++;          if(seq_length==5)seq_l5++;          if(seq_length==6)seq_l6++;          if(seq_length==7)seq_l7++;          if(seq_length==8)seq_l8++;          if(seq_length==9)seq_l9++;          if(seq_length==10)seq_l10++;          seq_length_sum += seq_length;          seq_length_quad_sum += pow(seq_length,2);          seq_length = 1;           sequences++;       }       last=random;      }      seq_length_sum += seq_length;      cout << "Average sequence length: " << seq_length_sum/(float)sequences << endl;      cout << "Average (sequence length)^2: " << seq_length_quad_sum/(float)sequences << endl;      cout << "Zeroes " << res_0 << " Ones " << res_1 << endl;/*       cout << seq_l1 << endl;      cout << seq_l2 << endl;      cout << seq_l3 << endl;      cout << seq_l4 << endl;      cout << seq_l5 << endl;      cout << seq_l6 << endl;      cout << seq_l7 << endl;      cout << seq_l8 << endl;      cout << seq_l9 << endl;      cout << seq_l10 << endl; */      cin.get();         }   return 0;}`

For some reason rand_lim seems to produce slightly longer sequences of consecutive zeroes or ones.
Spoiler:
Look at the average of the square of the sequence lenghts. For the other things there doesn't seem to be a consistent difference but it's consistently higher for that. (10000000 trials each)
Well not that I know enough about probability to say how significant that is.

Code: Select all

`Average sequence length: 2.00002Average (sequence length)^2: 5.99707Zeroes 5001023 Ones 4998977Average sequence length: 2.00019Average (sequence length)^2: 5.98716Zeroes 4999994 Ones 5000006Average sequence length: 1.99886Average (sequence length)^2: 5.99256Zeroes 5000607 Ones 4999393Average sequence length: 2.00016Average (sequence length)^2: 5.98688Zeroes 4999947 Ones 5000053Average sequence length: 2.0002Average (sequence length)^2: 6.00324Zeroes 4999087 Ones 5000913Average sequence length: 2.00018Average (sequence length)^2: 5.98704Zeroes 5000160 Ones 4999840Average sequence length: 1.99993Average (sequence length)^2: 6.00026Zeroes 4999031 Ones 5000969Average sequence length: 2.00021Average (sequence length)^2: 5.98726Zeroes 4999820 Ones 5000180Average sequence length: 2.00027Average (sequence length)^2: 6.00194Zeroes 5002390 Ones 4997610Average sequence length: 2.00019Average (sequence length)^2: 5.98718Zeroes 5000136 Ones 4999864`

Kinda pointless to be honest but whatever.

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### Re: Stubborn bettor

measure wrote:
Goahead52 wrote:Anyway I will try this stubborn strategy starting with 500 dollars as capital to see if it is interesting.

Be prepared to be very stubborn. The average number of rounds is about 90, but about 5% of games reach 500 rounds or more.
Spoiler:

Do you mean tosses when you talk about rounds?
500 tosses is too much.

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### Re: Stubborn bettor

PeteP wrote:
>-) wrote:shouldn't he overshoot \$10 at least some of the time, making k > 490?

The script measure vote checks whether a win would bring you over 10 and reduces the bet so that winning will land you exactly on 10.

I did not understand "reduces the bet" because it is assumed that the bettor either bet 1 dollar either 2^k dollars depending on his current win (if there is a win).

PeteP
What the peck?
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### Re: Stubborn bettor

Read the pseudocode example measure posted to see what they are doing. If you remove that property and leave the chance to over shot the goal in, 490 only gives you about 94.4% and you need about 2090 to reach 98%. (Ouch. That means on average you win about 41.8 per not lost game for an expected value of zero.)

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### Re: Stubborn bettor

There is misunderstanding here.
You do not need to reach exactly the goal G.
The only thing you need is to have G profit or more. Then it will be a success.
What is the trick of this method "stubborn strategy"?
Your maximal losses depend only on the number of tosses.
If you play 10000 tosses your losses will not exceed 10000*(1/4)=250.
Instead of using a martingale 1,2,4,8,...,2^k by picking from your own money you are just using the wined money.
For sure you will always reach your goal no matter how your level goal is.
You have 2 problems to solve :
- the number of tosses needed to reach the goal.
- the bet limit ruled by the casinos. If the limit is 1000 then you can not bet more than around 2^10 . Even if you succeed of having head 10 times in a row you need to remove and take your profits over 1000 dollars. It will surely change the capital level needed.

PeteP
What the peck?
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### Re: Stubborn bettor

As I said with your method without the modification to hit the goal exactly you need an start capital of about 2090 to reach or exceed 10 in 98% of the cases.

The method sounds a bit boring to be honest. You end 98% of the games with a small win sure, but it's small compared to the money you brought with you and there still is a 2% chance to lose everything if you follow the method to the end. Lower chances but higher rewards sounds more exiting and the expected value remains the same either way.

But well to each their own I guess it gives a relatively reliable series of small successes with the occasional crushing failure.

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### Re: Stubborn bettor

Thank you very much for your comments.
I think that there is still a room for improving the method.
I designed in fact the method for another kind on gambling games NOT the roulette.
I have made simulations for 2 players one playing head and the other tail. I did modify some constraints to adapt the method to the roulette.
Is is a win win method.

Good sunday!
Bon dimanche!

It is raining here.

measure
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### Re: Stubborn bettor

measure wrote:
Goahead52 wrote:Anyway I will try this stubborn strategy starting with 500 dollars as capital to see if it is interesting.

Be prepared to be very stubborn. The average number of rounds is about 90, but about 5% of games reach 500 rounds or more.
Spoiler:

Do you mean tosses when you talk about rounds?
500 tosses is too much.

Yes, I am referring to tosses, and an alternating series of win-loss-win-loss could result in a game with as many as 2*C rounds. My implementation also reduces the bet after a win if the net profit is close to the goal, and this can cause even more rounds.

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### Re: Stubborn bettor

Hi Measure,

So what you implemented is not a stubborn strategy.
The idea behind the method is no matter how big is the goal in terms of money you will always reach it.
The questions are : how much money you need in the worst cases and how long it will take.
You give yourself 2% chances to loose which is very low.
More than even in the worst cases you do not loose all of your money.

PeteP
What the peck?
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### Re: Stubborn bettor

No in the worst case you do lose all your money, that is how you reach 98% by risking a high amount of money for a small goal. If you stop before you have used up all your money you don't get the 98% chance.

And you can change measures script to follow your method by removing one if block.

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### Re: Stubborn bettor

PeteP wrote:No in the worst case you do lose all your money, that is how you reach 98% by risking a high amount of money for a small goal. If you stop before you have used up all your money you don't get the 98% chance.

And you can change measures script to follow your method by removing one if block.

You have 98% of chances to win 5 dollars for 1000 invested (10 dollars for capital 2090 dollars). Even in the stock market you will not have that opportunity.
Have you ever seen such method before?
No.
If yes then give me the url or where I can find it.

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### Re: Stubborn bettor

I`m not a programmer.
I did my simulations using Excel 2010.
What I`m trying to do is to explore ways....
My method can be improved.
Lot of scenarios are waiting.

PeteP
What the peck?
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### Re: Stubborn bettor

Do you know the term expected value?
Basically when you play a huge amount of times on average how much do you win per game? For throwing a coin and either getting losing X or gaining 2X the expected value is 0. For your system the expected values is 0.

No matter how you change your system your system you won't get a better expected value.

If casinos offered you to bet 2190 dollar and gain only 42 dollar if you win but with a 98% victory chance that would be about the same as following your method.

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### Re: Stubborn bettor

PeteP wrote:Do you know the term expected value?
Basically when you play a huge amount of times on average how much do you win per game? For throwing a coin and either getting losing X or gaining 2X the expected value is 0. For your system the expected values is 0.

No matter how you change your system your system you won't get a better expected value.

If casinos offered you to bet 2190 dollar and gain only 42 dollar if you win but with a 98% victory chance that would be about the same as following your method.

Even if your expected value is > 0 you COULD loose ALL YOUR MONEY no matter how big is your capital.
The expected value is just an AVERAGE.
The bettor and the mathematician are in fact 2 different persons.
One is playing with the risks the other is not taking the risks.
In fact the real life is not an infinite one. That is the bettor life.
For the mathematician the real life does not exist ....

Bon dimanche donc!

PeteP
What the peck?
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### Re: Stubborn bettor

My point is your method isn't anything special it's just a way to simulate a low risk-low reward bet by making many 50/50 bets. Which is fine of course if that is what you want I just want to make sure that you are aware that you haven't found some sort of superior bet system because some people into gambling like to believe they have clever strategies that improve their chances. If you aren't among that group everything is good.

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### Re: Stubborn bettor

Once you start gambling you know for sure that in the long run the odds are against you.
The casinos are not dumb.
I know how many rounds I`m going to play (to bet).
Let us say 1000.
A capital of 250 dollars will be sufficient if I use my method.
So if I loose that means that unfortunately I fell on the 2% looser.
My hope is not to be as unlucky as the 2% loosers.
Winning does not even matter.
If I reach only one dollar profit after playing 1000 rounds that`s good.
I had a fun without loosing money.
Playing 1000 rounds with 250 dollars is 25 cents each round. Loosing even all the 250 dollars is better than spending 250 dollars in alcohol or drugs etc....

Ps : Nothing can change the expected value but we all and always expect that the expected value will be what we wish deeply (having fun). No one will die from gambling expect if he use all what he owns.

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