Allistah
Allistah
Joined: Apr 23, 2011
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February 14th, 2021 at 6:59:49 PM permalink
Hey there,

I can't seem to locate the math to figure out how much of a bankroll I should have, and with said bets, what my chances of winning and losing are?

I found someone that posted some information using Don't Pass with 2x and 345x odds.

Don't pass w/2x odds has a variance of 8.13971304. It has a standard deviation of 2.85301823. (if these aren't true, let me know, found those numbers someone posted here in another post from some table)

For a $5 table with $5 on the DP, and $10 in odds, how would I go about calculating my chances of winning X amount or losing my Y amount in bankroll? Once I have that out, I could be able to make a spreadsheet that would give me a variety of bankroll numbers and win goals that I could choose from so that I can give myself the best chance to walk away a winner.

For example, If I walked up to the table with $200 and played until I made $100 (50% profit) or went broke, what are my chances (in %) of hitting that $100 win goal, and what are my chances (in %) of busting the bank?

Thanks,

-Alli
OnceDear
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OnceDear
Joined: Jun 1, 2014
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February 15th, 2021 at 2:21:03 AM permalink
Quote: Allistah

Hey there,

For example, If I walked up to the table with $200 and played until I made $100 (50% profit) or went broke, what are my chances (in %) of hitting that $100 win goal, and what are my chances (in %) of busting the bank?

Thanks,

-Alli

https://wizardofvegas.com/member/oncedear/blog/5/#post1370
Beware. The earth is NOT flat. Hit and run is not a winning strategy: Pressing into trends IS not a winning strategy: Progressives are not a winning strategy: Don't Buy It! .Don't even take it for free.
tuttigym
tuttigym
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July 21st, 2021 at 9:56:59 AM permalink
Quote: Allistah

For example, If I walked up to the table with $200 and played until I made $100 (50% profit) or went broke, what are my chances (in %) of hitting that $100 win goal, and what are my chances (in %) of busting the bank?



I do not know what "formula" OnceDear" linked to you or how in earth your "variance" and "standard deviation" numbers you posted have anything to do with reality, so try this simple mathematically correct approach.

1. Do not play the PL at Come Out.
2. Wait for a point to be established, any point.
3. Put $50 each on the 5, 6, 8, & 9. That gives you 18 ways to win and only 6 ways to lose or a 3 to 1 advantage OVER the House on any given roll of the dice.
4. Best case scenario: two hits and win $100+
5. Worst case scenario: one roll 7 out lose it all.

Longer one plays, especially at a $5 table, betting at one or two units at a time, the more likely one will become frustrated. The PL with the associated FO plus place bets are bankroll killers since point conversions happen only 27% of the time according to Mr. W's estimates or very vague "probabilities."

tuttigym
ThatDonGuy
ThatDonGuy
Joined: Jun 22, 2011
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July 21st, 2021 at 3:05:45 PM permalink
Where's MustangSally when we need her?

Anyway, since we can calculate (a) the probability of a win, (b) the mean value of a win, and (c) the mean value of a loss, you would think it could be reduced to a Gambler's Ruin problem, but I can't think of a straightforward formula that could be plugged into Excel at the moment. I would probably create a set of 301 equations (one for each possible bankroll condition from -200 to +100) in 301 unknowns and calculate the solution for bankroll condition zero.
We know that if the current bankroll is B, and the bets are $10 pass and $20 odds, the possible results are:
2/9: B+10 (7, 11 on the comeout)
1/9: B-10 (2, 3, 12 on the comeout)
2 x 1/12 x 1/3: B+50 (make a point of 4 or 10)
2 x 1/9 x 2/5: B+40 (make a point of 5 or 9)
2 x 5/36 x 5/11: B+34 (make a point of 6 or 8)
2 x 1/12 x 2/3 + 2 x 1/9 x 3/5 + 2 x 5/36 x 6/11: B-30 (miss any point)
So, if P(B) is the probability of reaching your goal with current bankroll B:
P(B) = 2/9 P(B+10) + 1/9 P(B-10) + 1/18 P(B+50) + 4/45 P(B+40) + 25/198 P(B+34) + 588/1485 P(B-30)
where P(B) = 0 if B <= -100 and P(B) = 1 if B >= 200
tuttigym
tuttigym
Joined: Feb 12, 2010
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July 21st, 2021 at 3:49:31 PM permalink
Quote: ThatDonGuy

Where's MustangSally when we need her?

Anyway, since we can calculate (a) the probability of a win, (b) the mean value of a win, and (c) the mean value of a loss, you would think it could be reduced to a Gambler's Ruin problem, but I can't think of a straightforward formula that could be plugged into Excel at the moment. I would probably create a set of 301 equations (one for each possible bankroll condition from -200 to +100) in 301 unknowns and calculate the solution for bankroll condition zero.
We know that if the current bankroll is B, and the bets are $10 pass and $20 odds, the possible results are:
2/9: B+10 (7, 11 on the comeout)
1/9: B-10 (2, 3, 12 on the comeout)
2 x 1/12 x 1/3: B+50 (make a point of 4 or 10)
2 x 1/9 x 2/5: B+40 (make a point of 5 or 9)
2 x 5/36 x 5/11: B+34 (make a point of 6 or 8)
2 x 1/12 x 2/3 + 2 x 1/9 x 3/5 + 2 x 5/36 x 6/11: B-30 (miss any point)
So, if P(B) is the probability of reaching your goal with current bankroll B:
P(B) = 2/9 P(B+10) + 1/9 P(B-10) + 1/18 P(B+50) + 4/45 P(B+40) + 25/198 P(B+34) + 588/1485 P(B-30)
where P(B) = 0 if B <= -100 and P(B) = 1 if B >= 200



Hey Mr. Allistah: Make sure you memorize every nit picking move above and be sure not to get confused or transpose anything. Looks like a sure winner or not.

BTW my "formula" is straight forward and quite simple. Two hits bring everything down and walk away with a minimum of $116.00.

tuttigym
Ace2
Ace2
Joined: Oct 2, 2017
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July 21st, 2021 at 6:04:31 PM permalink
Quote: tuttigym



1. Do not play the PL at Come Out.
2. Wait for a point to be established, any point.
3. Put $50 each on the 5, 6, 8, & 9. That gives you 18 ways to win and only 6 ways to lose or a 3 to 1 advantage OVER the House on any given roll of the dice.
4. Best case scenario: two hits and win $100+
5. Worst case scenario: one roll 7 out lose it all.

The house edge of this “strategy” is about 2.76%, nearly double the edge of playing the PL at comeout (1.41%). If you believe in house edge
It’s all about making that GTA
Ace2
Ace2
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Thanks for this post from:
pwcrabbodiousgambit
July 21st, 2021 at 7:02:56 PM permalink
Quote: Allistah

Hey there,

I can't seem to locate the math to figure out how much of a bankroll I should have, and with said bets, what my chances of winning and losing are?

I found someone that posted some information using Don't Pass with 2x and 345x odds.

Don't pass w/2x odds has a variance of 8.13971304. It has a standard deviation of 2.85301823. (if these aren't true, let me know, found those numbers someone posted here in another post from some table)

For a $5 table with $5 on the DP, and $10 in odds, how would I go about calculating my chances of winning X amount or losing my Y amount in bankroll? Once I have that out, I could be able to make a spreadsheet that would give me a variety of bankroll numbers and win goals that I could choose from so that I can give myself the best chance to walk away a winner.

For example, If I walked up to the table with $200 and played until I made $100 (50% profit) or went broke, what are my chances (in %) of hitting that $100 win goal, and what are my chances (in %) of busting the bank?

Thanks,

-Alli

First you put the bet in binary format. We know the variance is 8.140 and the return of the DP bet is 0.9864. Using the formula: payout = variance/return + return, we get a payoff of 9.24 for 1 and a probability of winning of 0.1068.

Since it will be much easier to work with an even payoff of 9, let's use that. Keeping the return at 0.9864, this changes the probability of winning to 0.1096 and the variance to 7.905, but that shouldn't affect the answer too much.

Then set up a markov chain, starting at 40 units, with a 0.1096 chance of advancing 8 units and a .8904 chance of retreating 1 unit. Absorbing states at 0 and 60.

After about 600 iterations, the answer converges to about a 38% chance of busting and 62% of achieving your goal.
Last edited by: Ace2 on Jul 21, 2021
It’s all about making that GTA
OnceDear
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OnceDear
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Thanks for this post from:
odiousgambit
July 22nd, 2021 at 12:38:26 AM permalink
Quote: Ace2

First you put the bet in binary format. We know the variance is 8.140 and the return of the DP bet is 0.9864. Using the formula: payout = variance/return + return, we get a payoff of 9.24 for 1 and a probability of winning of 0.1068.

Since it will be much easier to work with an even payoff of 9, let's use that. Keeping the return at 0.9864, this changes the probability of winning to 0.1096 and the variance to 7.905, but that shouldn't affect the answer too much.

Then set up a markov chain, starting at 40 units, with a 0.1096 chance of advancing 8 units and a .8904 chance of retreating 1 unit. Absorbing states at 0 and 60.

After about 600 iterations, the answer converges to about a 38% chance of busting and 62% of achieving your goal.

Excellent answer and with maths ability way beyond mine.
But my 'OnceDear's Rule of thumb' answer was pretty close.
Probability of success<=100/150
Probability of success<=66.7%
Remember my rule of thumb ignored the house edge, just as Tuttygym does.
with the house edge, we have...
Probability of success < 66.7%
Probability of going bust > 33.3%


Quote: And when Tuttigym said

That gives you 18 ways to win and only 6 ways to lose or a 3 to 1 advantage OVER the House on any given roll of the dice.

I could hear a thousand mathematicians groan with despair.
Tuttigym has no concept of house edge and this statement of his is absolutely wrong.
Beware. The earth is NOT flat. Hit and run is not a winning strategy: Pressing into trends IS not a winning strategy: Progressives are not a winning strategy: Don't Buy It! .Don't even take it for free.
pwcrabb
pwcrabb
Joined: May 15, 2010
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July 22nd, 2021 at 12:51:51 AM permalink
Fine work by Ace2 above. His model with a bankroll of 40 Flat bet units, Dark Side, Lay 2x, can easily be modified to accommodate alternative scenarios.

Thank You Ace2 for your template.
"I suppose I was mad. Every great genius is mad upon the subject in which he is greatest. The unsuccessful madman is disgraced and called a lunatic." Fitz-James O'Brien, The Diamond Lens (1858)
unJon
unJon
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July 22nd, 2021 at 4:24:56 AM permalink
Ace2’s great analysis highlights why if your sole goal is a win/bust goal, you are better off betting bigger and making fewer bets, even if it costs you house edge on individual bets (but ion the aggregate it saves you house edge because you bet fewer times).
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.

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