# Roulette strategy probability question

Here's the basic question:  I want to walk into a casino with a given bankroll and keep betting on red or black on a roulette wheel, each time betting, probably, 75% of however much money I have on hand (net of wins or losses).  I plan to stop when I reach my goal or lose it all.  My question is, what is the probability of reaching my goal?

Just to make my strategy clearer:  Say I walk into the casino with \$1,000.  My 1st bet would be 75% of that, or \$750.  Say I win that first bet; the casino pays me, therefore, my bet back plus \$750.  So, now, getting ready for the 2nd bet, I have a bankroll of \$1,750.  My 2nd bet would be 75% of that, or \$1,312.50.  If I should lose that 2nd bet, then I'd be down to a bankroll of \$437.50 (i.e., \$1,750 - \$1,312.50) , and my 3rd bet would be 75% of that, or \$328.13.  Etc., Etc.

I'l elaborate and define terms in the next post.  I'd do so here, but if I make this first one too long, nobody will read it.
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Commented:
What is your goal?  \$2000??   \$10,000??
And what are the red/black/green odds?
And how do you want to deal with minimum bets and rounding?

Roulette is a pretty fair game.  You can come up with a reasonable answer.
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Author Commented:
Now, I'll elaborate a bit.  This post is to define terms, that is to say names of variables.  My next, and final (question-clarifying) post will discuss why this problem is not straight-forward.

First, I'm not sure that roulette is the game I'll end up playing, but if it is, then P(victory on any given spin) = 18/38 = 0.4736842105 (It actually goes on, but that should be precise enough).  Let's call P(victory on a given turn) V; this should be expressed as a fraction or decimal, not a percentage.  As I said, I'm not sure I'm going to play roulette, so it would be better if I could get a more general formula as an answer.  Similarly, I don't know what my initial bankroll is going to be, so let's call that B, or what my goal is, so let's call that G, or what percentage of my bankroll I'd be wagering on each turn, so let's call that W, and W should also be expressed as a decimal, not as a percentage.

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Commented:

if minimum bet is \$50 and you lose down to \$49.  You haven't lost it all, but you can't play anymore either.

looks like you're rounding to the nearest penny based on your example above
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Commented:
Without any real world restrictions, the astonishing answer is: You will reach your goal with probability 1 (i.e. almost surely), simlpy because you can never lose it all.
However with real world restrictions (minimum coins of 0.01\$ or posibly minimum jetons of 2\$ say?) the answer is different - not to mention the slight bias towards the bank introduced by the zero (and slightly decreased again especially for simple chances like red/black by certain special rules of roulette).
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Commented:
The most important piece of information is the ratio of  Goal to Bankroll.

If Goal= \$4000,   then Goal/Bankroll = 4 = 2^2    and the P of Success = (18/38)^2

You could set this up to do a Monte Carlo Simulation in Excel without too much trouble and run a few thousand cases.
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Author Commented:
I just realized, as I've posed the question, I'll never fully bust, so let's define having my bankroll, B, dropping to less than \$10 as busting out, total defeat.

Also, I should say that, let's assume I stick to the betting strategy even if a smaller bet woluld let me reach my goal.  For example, say that my bankroll has reached \$5,000, and I set a goal of getting my bankroll up to \$5,500, and my strategy has been to bet 75% of my existing bankroll (not my starting bankroll, just to be clear).  Assume that I don't merely bet \$500, but instead I stick to the strategy and, therefore, bet \$3,750.  Now, if I win that bet, the game's over, and I won, as I'll have driven my bankroll up to \$8,750 (\$5,000 + \$3,750), which is greater than or equal to my goal of walking away with \$5,500 including my inital bankroll.

Also, assume no maximum bet is imposed by the casino.
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Commented:
I don't believe W (the fraction of your bankroll) matters at all.

I think the general formula you are looking for is:

Probability P of walking away with \$G with an initial bankroll of \$B  is  (18/38)^log2(G/B)
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Commented:
OK, so let's assume there is no minimum and no maximum bet and arbitrary amounts, not necessarily multiples of 0.01 \$, can be placed.

Then we should apply lograrithms and obtain a random walk:

You start at a certain point  x_0 = ln(B_0)  and at each roll you change x by either the positive amount ln(1.75) with probability V or by the negative amount  ln(0.25) with probability 1-V. (The ln(0.25) comes from you possibly losing 75%, the ln(1.75) comes from winning 75%, you may need to adjust the number accordingly if the payout is not 1:1).
(Here I use the fact that according to your post you seem to talk about the American variation of roulette, i..e. with an additional double-0 and without special treatment for simple chance bets)
We let the random walk end, when it passes either the left boundary at ln(10\$) or the right boundary at ln(G).
This should be doanly (i.e. a more or less well-known problem in random walks).
I'll have to look it ab, though
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Author Commented:
This post explains why this is not an easy question, and I also pose a follow-up question at the end.

A straight-forward, similar problem might be, what is the probability that I reach my goal with only consecutive wins.  In this example, let's assume my strategy is to bet 60% of my bankroll on each bet, my initial bankroll, B, or Bsub1 if you prefer, is \$2,000, and my goal is to walk away with \$5,000 including my inital \$2,000 bankroll.  Now in this example, the probability of winning the first bet and pushing my bankroll, B, or Bsub2 if you prefer, up to \$3,200 is whatever the value of V is, assuming roulette is the game, then it's 18/38.  The probability of winning the first and second bets, pushing my bankroll up to \$5,120 (and ending the game in victory for me) is (18/38)^2.

The game I proposed is more complicated, because I might win the first round, lose the next 5 rounds, win another round, lose another two rounds, and then stay on a winning streak that lets me reach my goal, thus ending the game.  There are an infinite(?) number of scenarios.

My follow-up question is:  What is the value of W (again, expressed as a decimal) that minimizes my chances of losing completely, which, again, is defined as letting the banroll drop below \$10?

That's the end of my question.  I'll now review the comments made and answer and relevant questions (in my next post).

Thanks.
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Commented:
>>> (18/38)^log2(G/B)

don't you need a win factor in there?    Is 2 correct?

log base-"factor" of the ratio?

(odds of winning once) ^ log( (win factor), (ratio of goal/bankroll))

so,  if you have a 25% chance of winning
and you bet 75%  your win factor is 1.75  (assuming 1:1 payout)
and you have a goal of 3000 starting with 1000,

0.25 ^ log(1.75, 3) = .06578

by my trials that seems close but a hair optimistic,  I've been hitting about 0.064

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Commented:
Reformulation of the problem:
Given constants a,b,c,d,V, we are looking for a function F with the following properties:
F(x) = 0 for x <= a
F(x) = 1 for x >= b
F(x) = V * F(x + c) + (1-V) * F(x-d)  for  a < x < b

Here a = ln(10\$), b = ln(G), c = ln(1+W), d = -ln(1-W).
We want to evaluate F(ln(B)), which is the desired probability of winning in the end.

F is obviously monotonous (or is it?), but not contoinuous (it is at least not continuous at a and b, and this spreads to dicontinuities at a+d, b-c, etc.)
As F is not continuous, d-glitch's fomrula cannot be correct (at least not exactly).
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Author Commented:
d-glitch:

You asked what my goal is.  I don't know yet.  That's why I'm trying to get at a general formula.  I can impose some limits, if somebody wants to write a computer program that will answer this question.  My starting bankroll will definitely be at least \$300, and my goal will definitely be less than \$10,000.

You asked how do I wan't to deal with minimum bets and rounding.  My answer is that the reason I defined total defeat as letting the bankroll drop below \$10 was to deal with the minimum bet issue; I'm assuming the minimum bet is \$10.  There's a related issue and this gets at your rounding question: I don't believe that they let you make any bet you want so long as it's above \$10.  For example, you can't bet \$14.56.  I think the problem is going to become way too complicated if I start imposing such rules.  It's complicated enough, so let's assume a \$14.56 bet is allowed.  If you have a solution that comes closer to real-world rules, please, by all means, lay it on me.  Did I answer the rounding question fully?  (No need to reply if I did.)  If anybody wants to assume a \$10 incremental-bet rule (i.e., you can only make bets in multiples of \$10), feel free to do so.

Thanks.
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Commented:
The simple fact is in the long run you will loose all your money. (you may have to play several times).
The wheel operators are not in the business of giving you money. The rules are such that they will always make money.
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Author Commented:
thehagman:

I don't know what a "jetson" is.

I can lose it all, as I've defined losing it all:  My bankroll falls below \$10.  So P(total victory) is absolutely not 1.

I already addressed the house advantage by pointing out that P(winning any given round) = 18/38, which is less than 50%.

I have no idea what you're referring to when you say, "...and slightly decreased again especially for simple chances like red/black by certain special rules of roulette."  If you want to elaborate, please feel free.

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Author Commented:
d-glitch:

You said: "You could set this up to do a Monte Carlo Simulation in Excel without too much trouble and run a few thousand cases."  I'm pretty good at Excel, but I can't see how to structure, what is essentially an enormous truth tree.  It gets unwieldy, really fast.

If you can see how to do this, I'd be most interested.

Thanks.
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Author Commented:
By the way, I'm just responding to these responses in the order I read them, so it's possible that somebody may say something, and somebody else then corrects it.  I'll still end up responding to the original point.  You're all coming at me so fast.  I gave the big point value not because I need the answer in the next 30 minutes, but becasue I think it's a very difficult problem, at least at my level of mathematical knowledge.

d-glitch:

You wrote:

"I don't believe W (the fraction of your bankroll) matters at all.

"I think the general formula you are looking for is:

"Probability P of walking away with \$G with an initial bankroll of \$B  is  (18/38)^log2(G/B)"

A couple of things.

I highly doubt you're correct that W makes no difference.  Every time I place a bet, I have a negative expected return, so it's best to minimize the number of bets placed.  On the other hand, if I make big bets, and I loose, I'll end up busting out faster.  So maybe you're right.  Maybe these two effects exactly offset each other, but I doubt it.  Anybody else have an opinion on this issue?  Anybody want to make a statement on this issue, making my question conform more closely to the real world?

Thanks.
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Author Commented:
thehagman:

You said:

"Then we should apply lograrithms and obtain a random walk:

You start at a certain point  x_0 = ln(B_0)  and at each roll you change x by either the positive amount ln(1.75) with probability V or by the negative amount  ln(0.25) with probability 1-V. (The ln(0.25) comes from you possibly losing 75%, the ln(1.75) comes from winning 75%, you may need to adjust the number accordingly if the payout is not 1:1).

"(Here I use the fact that according to your post you seem to talk about the American variation of roulette, i..e. with an additional double-0 and without special treatment for simple chance bets)
We let the random walk end, when it passes either the left boundary at ln(10\$) or the right boundary at ln(G).

"This should be doanly (i.e. a more or less well-known problem in random walks).

"I'll have to look it ab, though[.]"

I've got a few things to say:

(1) You're math is beyond me.  Did you actually even answer the question?  Your last sentence seems to suggest that you didn't.  I don't know what you mean by "ab" in the last sentence.

(2) Yes, I am assuming the double-0 possibility.  However, see my next point.

(3) I'm trying to stear this answer away from specific numbers (e.g., bet .75 of net bankroll each round) and toward a general answer expressed in variables (same e.g., bet W of net bankroll each round).  The real-world complication I may be facing is that roulette often has low maximum bets.  I'm trying to learn Baccarat; if the rules there are simple enough that you can make bets just like betting on red or black in roulette (where there is no further decision to be made once your bet is placed), then that's probably the game for me, because the per round limit is so high.

(4) I don't know what "special treatment for simple chance bets means."

Thanks.
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Author Commented:
sdstuber:

You lost me.

However, you seem to be on to something here.

Can you restate what you're saying, and can you or somebody, explain how logs enter the picture here?

Thanks.
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Commented:
Below are simulation results of running a Monte Carlo simulation in Excel.
The actual spreadsheet went to 100 trials with up to 35 spins per trial.\

I ran the sheet ten times for 1000 trials.  You win 10.3% of the time.
4 of the 1000 trials were still running after 35 spins.

The basic formula for every cell except the first (which is the bankroll) is

=IF(B4>=\$B\$1,B4, IF(B4<10, B4,IF(RAND()>(18/38),0.25*B4,1.75*B4)))

where B4 is the Bankroll    and   \$B\$1 is the Goal
``````Goal =>	5500

1	2	3	4	5	6	7	8	9	10
1	1000	250.00	437.50	765.63	1339.84	2344.73	586.18	1025.82	256.45	64.11
2	1000	250.00	437.50	109.38	191.41	334.96	586.18	1025.82	1795.18	448.80
3	1000	250.00	62.50	109.38	27.34	6.84	6.84	6.84	6.84	6.84
4	1000	250.00	437.50	109.38	27.34	47.85	83.74	20.94	36.64	9.16
5	1000	250.00	62.50	15.63	3.91	3.91	3.91	3.91	3.91	3.91
6	1000	250.00	62.50	15.63	27.34	47.85	83.74	146.55	36.64	9.16
7	1000	1750.00	3062.50	5359.38	1339.84	334.96	586.18	1025.82	1795.18	448.80
8	1000	1750.00	3062.50	765.63	191.41	334.96	586.18	1025.82	1795.18	3141.57
9	1000	250.00	437.50	109.38	191.41	47.85	83.74	20.94	36.64	64.11
10	1000	1750.00	437.50	765.63	1339.84	2344.73	586.18	1025.82	256.45	64.11
``````
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Author Commented:
thehagman:

The math of the post: 10/07/11 01:54 PM, ID: 36933794 is beyond me.

I did follow it enough to see that it was not an answer, but a proposed strategy to attack the problem.

Anybody want to follow-up on it?

Thanks.
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Author Commented:
aburr:

This is demonstrably false.  In fact I've demonstrated that above.  In fact, I'll give you an even simpler example.  I could just keep winning bet after bet.  The probability of such a streak (in the roulette example) is (18/38)^n, where n is the number of rounds played.  No sane mathematician would disagree with me here.

You're the kind of person who, I'm betting, sees the world as black and white.  I'm asking about a shade of grey.

The fact might be that the probability of winning is low in my particular case, but it's very clear that this probability depends, at the very least on how big a bankroll I start with--the bigger the better--and how much money I want to walk away with before calling it quits--the less the better.  I'm inclined to think that the fraction of my bankroll I bet on each round matters too, but I'll admit that's subject to dispute.

If you're going to be arrogant, at least be correct.
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Author Commented:
d-glitch:

Thanks for the Excel suggestion.  I'll have to think about that one.

The values in row 3 represent the spin number, right?

Each row, starting with row 4, represents a new series of test games, right?

This looks, on the face of it at least, very interesting.

I'd like to evaluate your reasoning right now, but my back is killing me.  I've to get out of this chair.

But very interesting.  You get at least some points, even if your answer is false, for serious and understandable effort.

Thanks.
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Author Commented:
Whoops, I meant to say, "I've got to get out of this chair.

Forgive me?
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Commented:
There are an infinite(?) number of scenarios.

It approaches infinity because of the bet amounts, but you can clearly see that the number of win/lose combinations is finite.

I might be wrong, but whether you win or lose X times in a row can be calculated by a simple permutation of 2 raised to the Nth, with N being the number of times you bet in a row. In other words, if you bet 4 times in a row you have 16 different win/lose combinations. To determine the chances of whether you reach your goal would be a matter of plugging in your bet amount to a set of permutations and then determine how many of those permutations would reach your goal and how many would not reach your goal.

Let's keep it simple: two bets = 4 outcomes where 0 is a loss, 1 is a win, you have 00, 01, 10, and 11 as a binary repesentation of all the possible outcomes. Goal is 6000.00 with 4000.00 initially, 75% bet of bankroll each time.

00 is a loser, you end with 375 dollars
01 is a loser, you end with 2125 dollars
10 is a loser, you end with 2625 dollars
11 is a winner, cha ching 18375 dollars

In this scenario, you have a 25% chance of reaching the goal.

Is this correct?

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Commented:
Let me correct my previous post, I used 6000.00 dollars as the initial bankroll (duh) instead of 4000.00 dollars.

00 is a loser, you end with with 1000 if you bet once or 250 if you bet twice
01 is a winner, you end with 7000 if you stop after the first bet (bet twice you lose though, won't reach goal)
10 is loser, you end with 1750 because you won the 2nd bet
11 is a big win, you end with 12,250 because you won both spins

In this scenario, you have a .50 chance of reaching the goal, with a .25 chance of winning big. A .50 chance of losing more than 50% of the bankroll.
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Commented:
therearestupidquestions:

My apologies, my posts above did not consider you keep going until you lose it all, I just wanted to give you a better picture of the outcomes of the win/lose combinations for a given number of consecutive bets.
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Author Commented:
PCableGuy:

1st, let me applaud your overall clarity.

Finally, there's what you said in your earlier post:

"I might be wrong, but whether you win or lose X times in a row can be calculated by a simple permutation of 2 raised to the Nth, with N being the number of times you bet in a row. In other words, if you bet 4 times in a row you have 16 different win/lose combinations. To determine the chances of whether you reach your goal would be a matter of plugging in your bet amount to a set of permutations and then determine how many of those permutations would reach your goal and how many would not reach your goal."

Well, some of this is correct, and this kind of reasoning--I'd have to give it more thought to be sure--probably applies to a question similar to mine, one where the dollar amount bet--that might prove to be a useful variable, so, everybody, let's call that D--in each round is constant (e.g., always bet \$100 each round until the game is over).  The problem with your reasoning in the above paragraph, as I see it--and I'd love to be shown I'm wrong--is that in my question the bet in each round is a function of the permutations of wins and losses in previous rounds.  The more I think about it, I don't see anything in the above paragraph that is actually wrong; it's just too vague.  Maybe if you could be more specific about what you mean by "plugging in your bet amount to a set of permutations," there's an answer there.  I suspect not, though.  I think you're just giving me a very basic introduction to probability and expected return.  Truth trees--I think that's the right term--and such.  You're also giving a lesson in the concept of diversification and why it's valuable; it reduces the risk of being wiped out, even if, in the case of being a player in a casino (i.e., in playing a game where I'm facing a negative expected return) it lowers (makes more negative) my overall expected return.  Perhaps I'm missing something.  Again, I'd love to be shown that I am.

So, if you think you're on to something, please try again.  Maybe somebody else's thinking has been stimulated by your response, as well.

In any event, thank you for the effort.
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Author Commented:
Ha!

PCableGuy:

We both corrected you in the same way at the same time!

Great minds think alike, or at least reasonably intelligent minds do.

Anyway, do you now feel that my correction totally addressed your posts, or is there something I'm missing, where they lead us closer to the answer?
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Author Commented:
I'm gong to keep a close eye on this thread, so anybody, please keep up the responses.

The only possible winner, so far, is d-glitch, with his spreadsheet.

If those of you who used more advanced math can clarify your reasoning for me, then maybe you're winners as well.

Thank you all for playing.  The game's not over.  (Sorry I don't have any showgirls to go along with my Las Vegas question here.  :-)  )
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Commented:
Yes you are right, my last post was not a definite answer, but rathe a reformulation of the problem to (from a mathematical standpoint) better descriptions. It did give no definite answer to the problem at all, but at least made clear (to me) that a continuous function is *not* an answer and that there are some wild dependencies among the givens involved.

However, your ultimate question "What is the optimal choice of W?" can be answered much more easily, I guess (though not with a water-tight proof here).
I just try to make something of what I remember of the famous book "How to gamble if you must" (you may want to dig that book in a librry).

Choose W such that you barely reach G with as few wins as possible.
That is:
First determine how many successive wins with full risk would reach the goal, i.e. find the smalles n such that B * 2^n >= G.
Then determine W such that B*(1+W)^n = G, i.e. let W = (G/B)^(1/n) -1

Why should this be a good idea?
If you start with a winning streak of n wins, you are done, which gives you a winning chance of AT LEAST  2^(-n). No *smaller* choice of W can guarantee that so easily.
Of course you can also win along different routes, not just from an immediate streak.
Say, there is one possibility where you win a times and lose b times in some order and reach the goal with my proposed value of W.
That means that
(1+W)^a * (1-W)^b >= G/B
As said, with W chosen as above, this inequality holds whereas with W=0 this would become 1 >= G/B, i.e. false.
It is thus at least slightly plausible that decreasing W decreses the left hand side.
That is: If a slightly maller value of W manages to reach the goal, then my propsed W will also. On the other hand, a slightly smaller W will not reach the goal in the situation of an immediate winning streak of n games. Therefore, a slightly smaller W would be a worse choice.

It would win, too, if you start with a streak, but it would win too much, i.e. it would "waste some of your luck", which you need to recover from an initial lose.

You may also note that in the above inequality, a must be bigger than b, in fact bigger by at least n.
One may thus rewrite it as
(1+W)^(a-b) * (1-W^2)^b >= G/B
So we see: The bigger b gets, the bigger a-b must be, i.e. a few loses canceling your lucky streak mean that you nead even more wins.
As a consequence, long-term playing will hardly contribute to your overall winning probability.The most likely route along which you might win *is* the immediate streak.
That's why my choice of W optimizes that.
It may be advisable to check against the next best paths of winning: To exactly reach G with a winning streak that is interrupted by a single loss.
With that ansatz, one would try to find W such that B*(1-W) * (1+W)^(n+1) = G holds with n as small as possible.
However, neithr is it possible to determine W from this easily (except numerically).
Nor would I dare to chose etween these candidate values for W without a simulation.

Hope that helps
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Commented:
You are correct about the rows being separate trials and the columns being spin numbers within a trial.

In Excel test, the chance of reaching (or exceeding) the oal was ~10%.
My earlier calculation suggested it should be ~16%.

This discrepancy arises because in some cases when you win you exceed the goa,
and in some cases when you lose you have money left.

If you modify your strategy slightly my formula will work.
When you are close to goal, bet just what it takes to reach
When you are close to minimum bet, put it all on the line.

le final winnings exceed the goal.

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Commented:
When you win at roulette, you double your money.
If you keep winning, your money grows exponentially.

log2(Goal/Bankroll) is the number of wins you need in a row to reach your goal if
you bet W=100% every time.

If you bet W=75% (or any other number), you will need more wins, but you can also survive a bad spin or two.  But the time/work/risk require to meat the goal
is something like (# of bets)*(wager per bet) and it is constant.

I remain quite sure that the value of W does not matte:   0 < W <= 100%

I will try to modify my spread sheet and run a few cases. It may take a while.
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Commented:
Interesting,  I ran the spreadsheet above and I get around 8.9% (with W=.75) I ran 1500 trials at a time letting it run for 75 spins.

This matches the results I got from the program I mentioned earlier too (after changing the parameters to match roulette).
I ran my program for 500,000 iterations with a 100 spin limit and came up with the same ratio.
The longest number of turns required to either win or run out was 70.

Also interesting, 10.3% is the expected success rate from the formula I gave above, but as in my previous tests
The expected result seems optimistic.

My only explanation is my math is wrong somewhere but I'm not sure where.
I ran my program with fictitious money that didn't require rounding (subject to system limits) and I let it go as low as  0.000000000000001 before stopping and let it run 1000 spins and it never went more than 204 spins.
It still hovered around 8.9% win rate.
So, I don't think the missing 1.4% is from rounding or minimum bet limits

earlier the question was asked why are LOG's involved.

Others have gone through the logic as well, but I'll put in terms of LOG
We're trying to find the number of times you have to win to get to the goal.
so... when does this pattern 1000 * 1.75 * 1.75 * 1.75 * ...... * 1.75 reach 5500?

or written another way

1000 * 1.75^N = 5500

solve for N

N = log base-1.75 ( 5500/1000)   = ln(5500/1000) / ln(1.75) = 3.046

So, what are the odds of winning 3.046 times?  Answer: (18/38)^3.046 = .103  or 10.3%

Of course,  you can't win 3.046 times,  because there aren't fractional spins.
You'll really need to win 4 times.

I tested the spreadsheet above and my own program with W of various values and it definitely does impact the results

I've attached my version of the spreadsheet for anybody to check my work and/or save yourselves some keystrokes.

Roulette.xls
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Commented:
OK, I've made a spread sheet of my own in the mean time (see attached)
You can feed it with V (winning probability per round), W (relative wager), L (money amount considered total loss), B (beginning amount), and G (goal amount) in B1 - B5.
In cell H1 you are shown the resulti, i.e. the probability of reaching the goal.
Some precautions:
1) The "array size" in E6 must not exceed 50 for the calculation to work. This happens if you have relatively small values of W (in fact as soon as it takes 50 consecutive wins to reach G when starting from L). You may of course extend the helper tables to the right in the obvious manner in order to raise the limit
2) The result is given with a possible absoulute error. This is the case because I calculate a lower and an upper bound. Again, unless W is relatively small, the error is negligible. In order to decrease the error introduced for small W, one would have to extend both helper tables vertically (and start with a bigger number in "losses"). However, this would require adjusting some of the references (though in an almost obvious manner)

What we can observe is that the result definitely depends on W, contrary to d-glitch's gut feeling.
A few sample calculations with V = 18/36, L = 10\$, B = 1000\$, G = 2000\$ and varying W show the following:
For W = 0.99999, we get P = 0.224377 (inserting W=1 gives an error, sorry)
For W = 0.9, P = 0.2534
For W = 0.8, P = 0.2925
For W = 0.7, P = 0.3129
For W = 0.6, P = 0.3282
For W = 0.5, P = 0.3600
For W = 0.4, P = 0.3139
For W = 0.3, P = 0.3281
For W = 0.2, P ~ 0.30 ... 0.39 (here the problem with small W giving inexact results raises its ugly head)
So there seems to be a maximum betwen 0.4 and 0.5
It turns out that 0.414 maps to 0.3111 and 0.415 maps to 0.3808 (yes, there is one of these discontinuities I predicted).
Apparently, this looks like sqrt(2)-1, i.e. the best choice of W (unless there are better choices in the range of small W, which I cannot examine without increasing the helper tables) is such that an initial streak of two wins exactly reaches our goal. This is not exactly what I said above but somewhat close to it. ;)
Actually, my prediction from the previous post would have been that W = 1 should be optimal (because it allows to possibly reach G in a single round and is the minimal W with this property). Was I wrong? No, the problem is that the spread sheet gives an error when W=1. But slightly wiggling with the values shows that W = 1 should produce a value significantly bigger than the 0.3808 obtained with W = sqrt(2)-1. For example, if I set G = 1999.99 and W = 0.999999, the result is 0.473684!

Have fun playing with the spread sheet.
Explaining its calculations would take yet another post, I'm afraid.
I hope you can work with it - I created it with openoffice, but assuming you rather use Excel exported it as xls file

roulette-calculation.xls
0
Author Commented:
thehagman:

You addressed my follow-up question, where I asked what is the ideal value for W.

You said, "Choose W such that you barely reach G with as few wins as possible."

You also said:

"Why should this be a good idea?

"If you start with a winning streak of n wins, you are done, which gives you a winning chance of AT LEAST  2^(-n). No *smaller* choice of W can guarantee that so easily."

I need to evaluate the math you used above, which I'm not in the greatest state of mind to do, at this moment.  However, I really like your reasoning.  The proof that definitively answers my follow-up question smells to me like one that is Calculus-based, and I've undoubtedly forgotten enough Calculus to evaluate such a proof.  I really like your cutting-the-Gordian-Knot approach to attacking the problem.  It seems correct.  The player, ideally, wants to place as few bets as possible, since the expected return, owing to the house's advantage, is negative on each bet.

I'm very impressed.  You definitely get some of the points for introducing this idea.

Well done.
0
Author Commented:
Everybody:

I haven't yet read the responses between my last post, made a couple of minutes ago, and my prior post.  I just skimmed far enough to see hagman's point that prompted me to post to him a couple of moments ago.

My back is killing me.  I've got to get out of this chair.  I will read what's already been said, just not now.  I invite you all to continue replying.

Thank you.
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Commented:
It does 1000 trials of 100 spins each.
If the trial finishes, you walk away with your goal (exactly) or zero.

You can change the Goal, Wager, or Bankroll.
The probability of Success does seem higher as Wager approaches 1,
but the scatter is significant even with 1000 trials.

Once you have entered the data, you can rerun the simulation by hitting F9.

Go wild.
ExEx-Roulette--3.xls
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Commented:
Here's another way to look at it.   Rather than simulating it, calculating it directly (or at least close to it)

In any 6 spins,  if 5 of them are losses, you will drop below \$10 minimum.  Even if you started just shy of your goal (5499.99),  5 losses in the next 6 spins will end the game (assuming the first spin isn't a win).
If you have a smaller bank then the number of losses required will go down.

Odds of 5 losses from 6 spins is 88.5%  So 11.5% "maximum" chance of success.  Empirically we already know this to be true from tests reported above.
We can solve this asymptotically by extending the loss per total spin counts.  I'll work on expanding those later.
0
Commented:
oops, I wrote the percentages backwards above

it should have been...

Odds of 5 losses from 6 spins is 11.5%  So 88.5% "maximum"

0
Commented:
I don't know if the method of numerical calculation of P, given W is still of interest, but I found an improvement over the estimates made with my Excel files in post #36936136.
That Excal table uses 0 and 1 as trivial lower and upper bounds and as starting points.
As described above, this gives poor result for small values of W.
The lower and upper bounds 0 and 1 can be replaced by much better bounds (here without proof):
Find the unique positive number k such that f(k) :=  V*exp(k*ln(1+W))+(1-V)*exp(k*ln(1-W)) = 1.
(The fact that exactly one such k exists follows from f(0) = 0, f'(0) < 0 (using V <= 1/2), f''(t) >0 and f(t) -> +oo for t -> +oo.)
Then  a n upper bound u(x) is given by
u(x) = 0  if  x < ln(L), u(x) = 1 if x > ln(G), u(x) = (exp(k*x) - exp(k*(ln(L)+ln(1-W))))/(exp(k*ln(G)) - exp(k*(ln(L)+ln(1-W))))  otherwise
and a lower bound l(x) is given by
l(x) = 0  if  x < ln(L), l(x) = 1 if x > ln(G), l(x) = (exp(k*x) - exp(k*(ln(L)))/(exp(k*(ln(G)+ln(1+W))) - exp(k*ln(L)))  otherwise
In other words, apart from the boundary conditions, these bounds are of the form  alpha*exp(k*x)+beta with alpha, beta chosen such that the exponential passes through (ln(L*(1-W)), 0) and (ln(G), 1) for the upper and through (ln(L), 0) and (ln(G*(1+W), 1) for the lower limtt.
Also note that one can use
l(ln(B)) < P(B) < u(ln(B))
as an immerdiat estimate.
0
progCommented:
I think it can be modelled with a random walk.

starting with capital of X
after n wins and m losses

balance = X * 1.75^n * .25^m

if B is bankrupt and W is won, then

net losses to bankrupt  a
B > X * .25 ^ a

a > log(B/X) / log(.25)

and to win, net wins  b

W < X * 1.75 ^ b
b > log(W/X) / log (1.75)

having worked out the wins and losses required, it becomes gamblers ruin

http://www.mathpages.com/home/kmath084/kmath084.htm

lets say you have 1000 and your target is 3000, but 10 is bust

a = 4  (bust)
b = 2 (won)

for P(win) = 18/38

r = (1 - 18/38) / (18/38) = 10 / 9

in the notation given at the site, this translates to a walk where a gambler has 4 and is aiming for 6, but is bust at zero

P(total win) =  (1 - r ^4) / (1 - r ^6) = .595
0
Author Commented:

I'm wish I could raise the point value to reward experts for their patience, but it's already maxed out.

I'll review answers and award points ASAP.  I may have to post another post along these lines, before that time, to keep the question active.  However, I do appreciate people's responses, and nobody will be cheated out of points they deserve.  You just may have to wait a bit.

0
Author Commented:
Again, I'm just posting a comment here to keep the thread active.
0
Author Commented:
One more time, keeping this active.  Sorry for the delay.
0
Author Commented:
I will award points in the next 2 days.  I'm currently printing all the threads I have going, even as I write this.  I'll award 500 points total.

0
Commented:
I did the simulation a while ago.
I'll post it when I get back to my desktop.
0
Commented:
I know you didn't want any more comments but, after spending a fair amount of time writing a program to make a determination, I thought I should still respond. Assuming an initial bankroll of \$1,000, a goal of \$5,500 and a percentage bet of 75%, let's take a look at the first 4 spins where the outcome of losses(L) and wins(W) would be LLLL,LLLW,LLWL,LLWW,LWLL,LWLW,LWWL,LWWW,WLLL,WLLW,WLWL,WLWW,WWLL,WWLW,WWWL,& WWWW. If we substitute a 0 for an L and a 1 for a W, we would end up with the binary representation from 0 to 15 then if we loop through those outcomes substracting 75% of the existing bankroll for a 0 and adding 75% of the existing bankroll for a W rounded to two places, we end up with the following end results after 4 spins:
0000     3.91 ending bankroll bust
0001    27.34 ending bankroll get more spins
0010    27.34 ending bankroll get more spins
0011   191.41 ending bankroll get more spins
0100    27.34 ending bankroll get more spins
0101   191.41 ending bankroll get more spins
0110   191.41 ending bankroll get more spins
0111  1339.84 ending bankroll get more spins
1000    27.34 ending bankroll get more spins
1001   191.41 ending bankroll get more spins
1010   191.41 ending bankroll get more spins
1011  1339.84 ending bankroll get more spins
1100   191.41 ending bankroll get more spins
1101  1339.84 ending bankroll get more spins
1110  1339.84 ending bankroll get more spins
1111  9378.91 ending bankroll goal
Now if we apply those same four possible outcomes, effectively creating binary representations from 16 to 255 (i.e. 0001000 to 11111111) to the ending balances as the existing bankroll for those that get more spins, we get the following:

For 27.34   - 12 busts and 4 get more spins
For 27.34   - 12 busts and 4 get more spins
For 191.41  - 5 busts and 11 get more spins
For 27.34   - 12 busts and 4 get more spins
For 191.41  - 5 busts and 11 get more spins
For 191.41  - 5 busts and 11 get more spins
For 1339.84 - 1 bust and 2 goals and 13 get more spins
For 27.34   - 12 busts and 4 get more spins
For 191.41  - 5 busts and 11 get more spins
For 191.41  - 5 busts and 11 get more spins
For 1339.84 - 1 bust and 2 goals and 13 get more spins
For 191.41  - 5 busts and 11 get more spins
For 1339.84 - 1 bust and 2 goals and 13 get more spins
For 1339.84 - 1 bust and 2 goals and 13 get more spins

Adding to the 1 bust and 1 goal from the first 4 spins, we end up with 83 busts and 9 goals with 134 outcomes yet undecided (needing more spins) after 8 spins. This represents a winning percentage of about 10% and a losing percentage of about 90% of the decided outcomes and more than 50% of the spins still undecided. As more and more possible outcomes are added these percentages improve to the point that when the number of possible outcomes represented by 2^22 (4194304) requiring 22 spins, the winning percentage is 18.58% and the losing percentage is 81.42% with only 4.3% undecided. Beyond that (i.e. 23+ spins) the winning and losing percentages remain constant while the percentage of undecideds declines ever so slightly. Those figures are pretty much based on an even number of wins and since the win ratio is 18 out of 38 and the lose ratio is 20 out of 38, the percentages then become approximately 18.32% for attaining the goal and 81.68% for going bust, so the odds of attaining your goal would be approximately 5.46 to 1.
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Commented:
Minor correction (i.e. 0001000 ... should have been (i.e. 00010000
0
Commented:
5.46 to 1 is an answer.
0
Commented:
5.46 to 1 is a really optimistic answer and doesn't take into account all of the rules of the question.

You can't keep losing 75% forever and keep betting.  Even if the house accepts one-penny bets, you still need to have that 1 penny.   So even if you haven't lost it all, you still can't win.  And the author did clarify that there is a \$10 minimum.

Based on the inputs of \$1000 initial bank roll, \$5500 goal, 75% wager, and \$10 minimum.  The probability of winning is approximately 0.089.

The spreadsheets in  ID: 36936101 and ID: 36936136 confirm that

both of those were based on the outline in ID: 36934058

I've also independently confirmed that with another program that ran out a million iterations.
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Commented:
Here is a simple pl/sql implementation of the program that I used to double check the spread sheets and run higher iteration counts

``````CREATE OR REPLACE PACKAGE roulette
IS
--https://www.experts-exchange.com/questions/27386172/Roulette-strategy-probability-question.html

FUNCTION play_it_out(
p_bankroll      IN NUMBER,
p_goal          IN NUMBER,
p_wager_ratio   IN NUMBER,
p_min_bet       IN NUMBER
)
RETURN VARCHAR2;

FUNCTION run_scenario(
p_bankroll      IN NUMBER DEFAULT 1000,
p_goal          IN NUMBER DEFAULT 5500,
p_wager_ratio   IN NUMBER DEFAULT 0.75,
p_min_bet       IN NUMBER DEFAULT 10,
p_iterations    IN INTEGER DEFAULT 100
)
RETURN NUMBER;
END;

CREATE OR REPLACE PACKAGE BODY roulette
IS
--https://www.experts-exchange.com/questions/27386172/Roulette-strategy-probability-question.html

FUNCTION play_it_out(
p_bankroll      IN NUMBER,
p_goal          IN NUMBER,
p_wager_ratio   IN NUMBER,
p_min_bet       IN NUMBER
)
RETURN VARCHAR2
IS
v_money   NUMBER := p_bankroll;
BEGIN
WHILE v_money >= p_min_bet AND v_money < p_goal
LOOP
v_money :=
v_money
+ CASE
WHEN DBMS_RANDOM.VALUE <= 18 / 38 THEN v_money * p_wager_ratio
ELSE v_money * -p_wager_ratio
END;
END LOOP;

RETURN CASE WHEN v_money >= p_goal THEN 'Win!' ELSE 'Lose!' END;
END;

FUNCTION run_scenario(
p_bankroll      IN NUMBER DEFAULT 1000,
p_goal          IN NUMBER DEFAULT 5500,
p_wager_ratio   IN NUMBER DEFAULT 0.75,
p_min_bet       IN NUMBER DEFAULT 10,
p_iterations    IN INTEGER DEFAULT 100
)
RETURN NUMBER
IS
v_win_count   INTEGER := 0;
BEGIN
FOR i IN 1 .. p_iterations
LOOP
IF play_it_out(
p_bankroll,
p_goal,
p_wager_ratio,
p_min_bet
) = 'Win!'
THEN
v_win_count := v_win_count + 1;
END IF;
END LOOP;

RETURN v_win_count / p_iterations;
END;
END;
``````

and then run a simple test scenario of 100 iterations with

select roulette.run_scenario from dual

or to run a bunch of iterations

select roulette.run_scenario(1000,5500,0.75,10,1000000) from dual
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Commented:
I recommend a split on these three

ID: 36934058
ID: 36936101
ID: 36936136
0
Commented:
There is extensive discussion of the problem, including several valuable computer models and simulations, and several valid answers which depend on interpreting the problem details.

I recommend an even split among the following answers:
d-glitch        https:l#a36938793
sdstuber      https:#a36936101
deighton      https:#a36985859
thehagman  https:#a36971217
0
Commented:
I'm ok with that split too
0
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