Help with Formulas

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

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?

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

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.

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

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.

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.

Quote: unJon

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).

Quote: Ace2

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

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.

Quote: OnceDear

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: OnceDear

I could hear a thousand mathematicians groan with despair.
Tuttigym has no concept of house edge and this statement of his is absolutely wrong.

Quote: pwcrabb

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

Quote: OnceDear

True enough. The house edge acts on the total of wagers placed. All those little winning bets may not cancel all those little losers. Run your bankroll through with very, few very large bets, has lower expected loss value. But the time it saves you might leave you inclined to just try 'one more hit and run' and then just one more... which might bust you.

Quote: Ace2

I ran a simulation in Excel and the chance of achieving the goal is around 63.7%. That's slightly higher than my Markov chain method of 62%, probably because I lowered the payout from 9.24 to 9, which also lowered the variance (higher variance generally gives a better chance of achieving a goal).

By the way, the 2X odds variance figure of 8.14 means you're laying 4 units on the 4&10, 3 units on 5&9, and 2.4 units on the 6&8, with a win on any point paying 3 total units (2 units on odds plus 1 unit on the flat bet). All my calculations are for that scenario.

If you're talking about laying 2 units for every point, then the variance is 4.99 and will be a significantly different chance of success. Though laying $10 would not be good idea on the 4,10,6 and 8 since there will be breakage. You need to lay in multiples of $3 and $6 to avoid that.

Quote: tuttigym

Mr.Allistah: You got that? Beyond simple, so be sure to memorize it or write it down and bring it to the tables. 600 Iterations huh? How long would that take to complete? What is an iteration, and where does one find it? 62% chance of winning? Sounds incredibly good to me. I think you should meet Mr. Ace2 and have him show you how it all works and comes together.

tuttigym

Quote: Ace2

You just don't understand the calculations, and that's fine, but don't mock the people that do understand. Incidentally, these calculations are not difficult...really just algebra.

Quote: tuttigym

Mr. Ace2: As you are probably aware, I do NOT believe in THAT HE. So here is a simple challenge for you and OnceDear, memorize that convoluted set of plays and wagers as shown above; go to a table near you; and play. If you can not memorize the above, write it down and take it to the table and try and play without everyone including the dealers screaming at you to hurry up and do something. Make sure you publish your results.

Quote: tuttigym

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."

Quote: ThatDonGuy

You mean my "convoluted set of plays"? Here they are:

1. Bet $12 on Don't Pass (it's $12 instead of $10 so you get full payouts on your odds bets)
2. If a point is established, bet $24 Odds.

That's it.

Quote: tuttigym

What you have posted above is great. It is simple, readable, understandable, and doable. The rest of the stuff posted and below is just "noise." I do not intend to critique it because you actually answered the original inquiry with some prodding by me. My only question is: What does producing all the other "junk" (my words) do for the reader or user? It is not necessary, in my opinion. You did well and thank you.

Quote: ThatDonGuy

What "other junk" is that? You mean the part that shows that terms like "66.67% success rate" that you threw around when mentioning your "50 on 5, 6, 8, 9, then take them down after two hits" are pretty much meaningless in terms of what you are expected to win?

Quote: tuttigym

No, the garbage of line after line of numbers that have no meaning to someone trying to play the game. You provided clarity of your original post so that any individual who had a mind to try it could without OVERTHINKING every little nit picky fraction of win/loss possibilities. If your goal is confusion, then you have achieved it big time. If you remember, I did NOT criticize the plan only the mind numbing and unnecessary overall approach. Is your ego so large that somehow producing line after line of numbers to "prove a point" makes you feel superior. The simplicity of the resultant "plan" gives you credibility.

tuttigym

Quote: ThatDonGuy

What "other junk" is that? You mean the part that shows that terms like "66.67% success rate" that you threw around when mentioning your "50 on 5, 6, 8, 9, then take them down after two hits" are pretty much meaningless in terms of what you are expected to win?

Here's a system that lasts no more than two comeouts that has a better than 70% chance of making $100 profit:
1. Bet $100 on the pass line. If you win, take the $100 profit and walk.
2. If you lose, bet $200 on the pass line.
Best case: more than 70% of the time, you win $100.
Worse case: you lose both bets, but that's less than 30% of the time.