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

https://wizardofvegas.com/member/oncedear/blog/5/#post1370Quote:AllistahHey 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

Quote:AllistahFor 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

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

Quote:ThatDonGuyWhere'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

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 edgeQuote: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.

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.Quote:AllistahHey 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

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.Quote:Ace2First 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.

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%

I could hear a thousand mathematicians groan with despair.Quote:And when Tuttigym saidThat 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.

Tuttigym has no concept of house edge and this statement of his is absolutely wrong.

Thank You Ace2 for your template.