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.

Solved

Posted on 2011-10-07

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.

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.

54 Comments

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.

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)

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

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).

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.

Also, I should say that, let's

Also,

I think the general formula you are looking for is:

Probability

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

My

Thanks.

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

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).

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

Thanks.

The wheel operators are not in the business of giving you money. The rules are such that they will always make money.

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.

Thank you for your input.

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.

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.

What's your reasoning?

I highly doubt you're correct that

Thanks.

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

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

Thanks.

You lost me.

However, you seem to be on to something here.

Can you restate what you're saying, and

Thanks.

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*

where

```
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
```

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.

This is demonstrably false. In fact I've demonstrated that above. In fact, I'll give you an even simpler example. I

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

If you're going to be arrogant, at least be correct.

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.

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?

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.

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.

1st, let me applaud your overall clarity.

Next, in your second post (ID: 36934546) what you said is correct, but you threw in the assumption that the game is over, no matter what after 2 bets, tops. This as an answer to

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

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.

PCableGuy:

We both corrected you in the same way

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?

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

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).

You start with amount B and want to reach G.

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.

What about bigger W?

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

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.

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.

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

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

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

You said, "Choose

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

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

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

Well done.

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

Thank you.

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

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.

it should have been...

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

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

(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))))/(e

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(

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.

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/h

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

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.

Sorry about this.

Thank you all for your patience and advice.

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.

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.

```
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

I recommend an even split among the following answers:

d-glitch https:l#a36938793

sdstuber https:#a36936101

deighton https:#a36985859

thehagman https:#a36971217

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