Please check my math - pass/come strategy

Hello,

Please help me with my math, I’m sure there is an error in here, or perhaps I’m not understanding the rules of craps and the payouts correctly.

There is a new roller. I make a $5 Pass bet. On the come out roll, I actually have a positive expected value.

Chance of a 7 or 11 = 8/36 --> win $5
Chance of a 2, 3, or 12 = 4/36 --> lose $5

$5*(8/36) - $5*(4/36) = $.56


Now, assume a point was made and the point is 8. I back my pass bet with odds. Assume I’m playing a 3-4-5 table and back my bet with maximum odds, so, $25. I now have the following situation on the second roll.

Chance of a 7 = 6/36 --> lose $30
Chance of a 6 = 5/36 --> win $35

$35*(5/36) - $30(6/36) = -$.14


I’m expecting to lose $.14 on this roll, but we already knew that. What if, instead, I add a come bet for the second roll. Following the math above, on the next roll I expect to make $.56 on the Come bet and to lose $.14 on the Pass bet with odds. I’m expecting to make $.42 on the second roll. I seem to confirm this with the following math.

Chance of a 7 = 6/36 --> lose $25
Chance of a 2, 3 or 12 = 4/36 --> lose $5
Chance of 11 = 2/36 --> win $5
Chance of a 6 = 5/36 --> win $35

$35*(5/36) + $5*(2/36) - $30*(6/36) - $5*(4/36) = $.42


So, I figure this must break down at some point. Let’s say the second roll turns out to be an eight and I back the eight with max odds also. This gives me the following.

Chance of a 7 = 6/36 --> lose $60
Chance of a 6 = 5/36 --> win $35
Chance of an 8 = 5/36 --> win $35

$35*(10/36) - $60 *(6/36) = -$.28


So, I have a negative expected value again. But what if I add another Come bet? Then I reduce the amount I would lose on a 7 by $5, take a chance of losing on a 2, 3, or 12 but gain the opportunity to win $5 on 11. I have 12 chances to win and 10 ways to lose, but would lose big on a 7.

Chance of a 7 = 6/36 --> lose $55
Chance of a 2, 3 or 12 = 4/36 --> lose $5
Chance of an 11 = 2/36 --> win $5
Chance of 6 or 8 = 10/36 --> win $35

$35*(10/36) + $5*(2/36) - $5*(4/36) - $55*(6/36) = $.28

So, I still have positive expected value on the third roll. This will change if the third roll is a 4, correct? Let’s see. Assume the third roll is a four and I back the 4 with max odds, $15.

Chance of a 7 = 6/36 --> lose $80
Chance of a 6 or 8 = 10/36 --> win $35
Chance of a 4 = 3/36 --> win $35

$35*(11/36) - $80*(6/36) = -$.69


As anticipated, I have negative expected value and the negative number is getting larger as I put more money on the table. Certainly the number is too large to overcome with another Come bet right? Let’s see…

Chance of a 7 = 6/36 --> lose $75
Chance of a 2, 3, or 12 = 4/36 --> lose $5
Chance of a 6, 8 or 4 = 13/36 --> win $35

35*(13/36) - $5*(4/36) - $75*(6/36) = $.14


I still have positive expected value!

At this point my head hurts and I’m sure I have made at least one error so I won’t do the math of continuing the strategy of adding a Come bet with every roll and backing every number with max odds. Another interesting piece of math would be to assume the 4/10 are the first numbers established.

So, where did I go wrong?

well I really didn't read it that thoroughly....But you can leave the free odds bets out of the equation...becasue they dont add or take away from expected loss.....If one man made a $5 P/L and always backed with $100 free odds bet and one man bet $5 P/L bet and never backed with odd....their expected loss would be the exact same.....

Secondly if the points rolled were always 5,6,8,9.... craps would be a player advantage game(and big one at that)....If you just took out the point of 4.....and played craps with the sames rules but if a 4 was rolled on the come out, it was just a "push" and no point was established. and the shooter just rolled another "come out"....craps would also be a player advantage game....

Quote: craigrow

Hello,

Please help me with my math, I’m sure there is an error in here, or perhaps I’m not understanding the rules of craps and the payouts correctly.

There is a new roller. I make a $5 Pass bet. On the come out roll, I actually have a positive expected value.

Chance of a 7 or 11 = 8/36 --> win $5
Chance of a 2, 3, or 12 = 4/36 --> lose $5

$5*(8/36) - $5*(4/36) = $.56

...

So, where did I go wrong?

You didn't factor in the chances to roll one of the point numbers on the first roll, nor the accompanying EVs (all of which are negative). On the come out roll, you have an EV of -$0.0707 on a $5 bet. Try to replicate that number with your calculations. Here's a hint: how would you calculate the EV if the rules of craps were (a) if you establish any point number, then on the next roll you automatically win 10x your wager; and (b) if you establish any point number, wait one more roll and then automatically lose? Obviously (a) and (b) have very different house edges. If your calculations would arrive at the same figure, something's fishy.

Quote: MathExtremist

You didn't factor in the chances to roll one of the point numbers on the first roll, nor the accompanying EVs (all of which are negative). On the come out roll, you have an EV of -$0.0707 on a $5 bet. Try to replicate that number with your calculations. Here's a hint: how would you calculate the EV if the rules of craps were (a) if you establish any point number, then on the next roll you automatically win 10x your wager; and (b) if you establish any point number, wait one more roll and then automatically lose? Obviously (a) and (b) have very different house edges. If your calculations would arrive at the same figure, something's fishy.

Thanks MathExtremist. I'm sure there is an error in my logic somewhere, but I don't think it's leaving out the points. I'm calculating the expected value for a given roll of the dice. On a come out roll there are 8 chances to win $5, 4 chances to lose $5 and everything else is a push. Adding $0*24/36 would not change the expected value. How do you get expected value of $0.0707 on a come out roll?

When you say "everything else is a push" that's not entirely true. In blackjack, when you push you can remove your wager. In craps, when you "push" as you call it, your money is now locked in to a bet that has a distinct and measurable disadvantage. You've got to weight that disadvantage post come-out to any advantage you may perceive pre come-out. Take both into consideration and you'll see that they are are losing bets.

Quote: TheNightfly

When you say "everything else is a push" that's not entirely true. In blackjack, when you push you can remove your wager. In craps, when you "push" as you call it, your money is now locked in to a bet that has a distinct and measurable disadvantage. You've got to weight that disadvantage post come-out to any advantage you may perceive pre come-out. Take both into consideration and you'll see that they are are losing bets.

Thanks TheNightfly. I understand the ability to remove a bet changes the math. However, I'm calculating the EV for a single roll. On the second roll I will leave the Pass bet on the table but I will also have additional bets on the table which change the EV of the second roll. It doesn't matter to me that $5 was already on the table for the first roll.

My understanding of the math is not good enough to calculate any sort of interactions between rolls, but since I NEVER remove a bet I don't think I need to.

The theory I'm testing is, if I can place my bets so that I have a positive EV on as many rolls as possible I should have better results. Consider the following.


Option 1: Place a $5 Pass bet, assume the number is 6, back it with max odds and wait for the point or a 7.

Roll #1: EV = $0.56
Roll #2: EV = -$0.14
Roll #3: EV = -$0.14
Roll #4: EV = -$0.14
Roll #5: EV = -$0.14

Total EV after 5 rolls = $0.00.


Option 2: Play the scenario above where I start with a $5 Pass bet then bet the Come on every roll and back everything with max odds.

Roll #1: EV = $0.56
Roll #2: EV = $0.42
Roll #3: EV = $0.28
Roll #4: EV = $0.14
Roll #5: EV = -$0.56

Total EV after 5 rolls = $0.84


Now, of course, if we continue this then the large negative EVs on Option 2 start to take over...


Option 1:

Roll #6: EV = -$0.14 Total EV = -$.14
Roll #7: EV = -$0.14 Total EV = -$0.28
Roll #8: EV = -$0.14 Total EV = -$0.42
Roll #9: EV = -$0.14 Total EV = -$0.56


Option 2: Let's just assume the next number is 9 but then I stop placing bets and the shooter rolls 5 repeatedly.

Roll #6: EV = -$1.39 Total EV = -$0.55
Roll #7: EV = -$1.39 Total EV = -$1.94
Roll #8: EV = -$1.39 Total EV = -$3.33
Roll #9: EV = -$1.39 Total EV = -$4.72


So, clearly my EV is getting very negative. However, I had positive EV through five rolls, which is the average number per shooter, correct? I only get these really big negative EVs when the roll goes long.

Quote: craigrow

Thanks MathExtremist. I'm sure there is an error in my logic somewhere, but I don't think it's leaving out the points. I'm calculating the expected value for a given roll of the dice. On a come out roll there are 8 chances to win $5, 4 chances to lose $5 and everything else is a push.

No it's not. It's an unresolved bet. A "push" is a resolution with a value of 0. There is a strongly negative value to rolling any point number on the comeout, but you didn't factor any of that in.

Quote: MathExtremist

No it's not. It's an unresolved bet. A "push" is a resolution with a value of 0. There is a strongly negative value to rolling any point number on the comeout, but you didn't factor any of that in.

Exactly. You never have a positive EV. The game is setup that way so that you don't. No way around it.

Quote: YoDiceRoll11

Exactly. You never have a positive EV. The game is setup that way so that you don't. No way around it.

OK, I know I've seen this but can't remember where and can't seem to find it again. What is the process of factoring in unresolved bets?

In other words...


Come out roll EV = -$5(4/32) + $5(8/32) + (EV of a push)


What is the formula for EV of a push?

Quote: craigrow

Quote: YoDiceRoll11

Exactly. You never have a positive EV. The game is setup that way so that you don't. No way around it.

OK, I know I've seen this but can't remember where and can't seem to find it again. What is the process of factoring in unresolved bets?

In other words...


Come out roll EV = -$5(4/32) + $5(8/32) + (EV of a push)


What is the formula for EV of a push?

Name a push bet in craps.

Quote: craigrow

Come out roll EV = -$5(4/32) + $5(8/32) + (EV of a push)
What is the formula for EV of a push?

1) Stop calling it a push -- it's not.
2) What's the formula for the EV of rolling a 5 on the comeout roll? Well, how much is a $5 passline bet worth if the point is 5? How often do you roll a 5 on the comeout roll? Repeat for 4, 6, 8, 9, 10.

Quote: MathExtremist

1) Stop calling it a push -- it's not.

Quote: YoDiceRoll11

Name a push bet in craps.

Exactly. No such thing. I'll answer my own question. There are no push bets in craps. You have a chance to lose them all.

Example: Playing Pass and Don't Pass at same time for $10 each. Push right?
Wrong.

7/11: Win The Pass, Lose The Don't Pass
2,3: Lose The Pass, Win The Don't Pass
12: Lose The Pass, NO action on the Don't Pass
Any box Number: You're on the hook to lose.

Quote: YoDiceRoll11

Exactly. No such thing. I'll answer my own question. There are no push bets in craps. You have a chance to lose them all.

Example: Playing Pass and Don't Pass at same time for $10 each. Push right?
Wrong.

7/11: Win The Pass, Lose The Don't Pass
2,3: Lose The Pass, Win The Don't Pass
12: Lose The Pass, NO action on the Don't Pass
Any box Number: You're on the hook to lose.

Since I was asking for assistance with math, I'd actually prefer we focus on the math instead of the specific terms used.

1. What is the correct formula used to determine the EV of a Pass line bet?
2. What is the correct formula used to determine the EV of a Pass line bet plus max odds?
3. What is the correct formula used to determine the EV of a Pass line bet plus max odds plus a Come bet?
4. Etc.

Then, how does the math change if we assume I always leave all bets on the table until they're resolved? Or does it?

Or, to re-state my original question, does the EV of #1 change if you know that on the second roll I'm going to have max odds against the Pass line bet and a Come bet?

Quote: craigrow

Since I was asking for assistance with math, I'd actually prefer we focus on the math instead of the specific terms used.

1. What is the correct formula used to determine the EV of a Pass line bet?
2. What is the correct formula used to determine the EV of a Pass line bet plus max odds?
3. What is the correct formula used to determine the EV of a Pass line bet plus max odds plus a Come bet?
4. Etc.

Then, how does the math change if we assume I always leave all bets on the table until they're resolved? Or does it?

Or, to re-state my original question, does the EV of #1 change if you know that on the second roll I'm going to have max odds against the Pass line bet and a Come bet?

The EV of 1 and 2 are the same. that's the nature of the odds bet.

Quote: rdw4potus

The EV of 1 and 2 are the same. that's the nature of the odds bet.

In money, not as a percentage. $5 line bet costs 7.07c in expectation, regardless of odds multiple, but that expected cost is a much smaller percentage of a flat+20x odds wager than just flat alone.

Formula for EV of pass bet (and most other bets, really):
p(2) * EV(2) +
p(3) * EV(3) +
p(4) * EV(4) +
...
p(12) * EV(12)

The OP has the EV of 2, 3, 7, 11, and 12 figured already (-1, -1, 1, 1, and -1, respectively). The probabilities are also already known. What's missing from the OP's calculation so far is the EV of rolling a point number. Here's one example to get the ball rolling:
If I come out on a 4, what's my EV? I have 1/3 chance to win 1, 2/3 chance to lose 1, my net is -1/3. So where it says p(4) * EV (4) above, you'd use 3/36 * -1/3.

Rinse and repeat for the rest of the points, then add everything up.

Quote: craigrow

Since I was asking for assistance with math, I'd actually prefer we focus on the math instead of the specific terms used.

1. What is the correct formula used to determine the EV of a Pass line bet?
2. What is the correct formula used to determine the EV of a Pass line bet plus max odds?
3. What is the correct formula used to determine the EV of a Pass line bet plus max odds plus a Come bet?
4. Etc.

Then, how does the math change if we assume I always leave all bets on the table until they're resolved? Or does it?

Or, to re-state my original question, does the EV of #1 change if you know that on the second roll I'm going to have max odds against the Pass line bet and a Come bet?

Quote: Wikipedia

"In probability theory the expected value of a random variable is the sum of the probability of each possible outcome of the experiment multiplied by the outcome value (or payoff).

Pass line probabilities
(6 ways for a seven, 2 ways for an eleven = 8 ways to win)(1 way for a two, 1 way for a 12, 2 ways for a 3 = 4 ways to lose) 8/4 or 2/1 odds of winning an even money bet on the come out roll.

An "EV" can also be expressed as a house edge in a percentage.

1. Pass line for $100. House edge: 1.41% (EV: -1.41). You will lose $1.41 for every $100 wagered. Here is how we get there.

Probability of winning on come out roll with 8 ways to win:
P(Win on first roll) = 8/36 = 0.222222

Probability of winning on a point
What if the first roll is not a 2,3,7,11, or 12? For example, if the first roll is an 8 (called your point), then you win if an 8 comes up before a 7 on any of the future rolls. To compute the probability of winning given that the shooter rolled an 8 on the first roll, notice that there are 11 outcomes that make you win or lose on the next roll. Of these outcomes, 5 make you win and 6 make you lose. Hence this probability is 5/11. Since the probability of getting an 8 on the first roll is 5/36, we have

P(Win via a point of 6) = (5/36)(5/11) = 0.063131

Similarly, we can find the probabilities of winning with other points: 4,5,8,9,10:

P(Win via a point of 4) = (3/36)(3/9) = 0.027778

P(Win via a point of 5) = (4/36)(4/10) = 0.044444

P(Win via a point of 8) = (5/36)(5/11) = 0.063131

P(Win via a point of 9) = (4/36)(4/10) = 0.044444

P(Win via a point of 10) = (3/36)(3/9) = 0.027778

To find the probability of winning, just add up all these probabilities:

P(Win) = 0.222222 + 0.027778 + 0.044444 + 0 .063131 + 0.063131 + 0.044444 + 0.027778

= 0.493

We can calculate the probability of losing by subtracting from 1:

P(Lose) = 1 - 0.493 = 0.507

And finally, with $100 on the Pass Line, then the expected value is

(100)(0.493) + (-100)(0.507) = -1.40

Because you are rounding, the actual expected value is -1.41.

So, you should expect to lose on average, $1.41 per bet.


Now, on to 2:
The odds reduce the house edge by not being subject to the negative expectations of losing on the come out roll (on the pass line: 2,3,12).

1x odds reduces overall house edge to 0.85%
2x: 0.61%
3x: 0.47%
3x-4x-5x: 0.37%
10x: 0.18%
100x: 0.02%

3: This is the easiest question. The come bet has the exact same odds as the pass line bet, it just can be made at any time.

Each bet has it's own individual house edge. If you want to know what they are combined, take a look at each one. Let's say you are just betting the pass line no odds and the field every time. You have a house edge of 1.41% for the pass line and 5.56% for the field (assuming 2 and 12 pay 2x).

If you just bet $5 on the pass, and $5 on the field, 10 times, you bet $100 (regardless of winnings). What is your EV or house edge? Well, the House edge is constant for each individual bet. You would expect to lose $0.71 (rounded up) for the $50 passline and $2.78 for the $50 on the field. Add those two house edges and you get a combined house edge of 3.49. But wait YoDr, can't we just take 1.41 and add 5.56 and divide by 2.....yep. That comes out to 3.485% (we rounded up to 3.49%).

So the short answer to question 3 depends on the odds that you take. Than you just add the 1.41% for the come bet, and average it out.

The house edge remains constant for whatever bets are "in action" at any time.


The EV of a pass line bet DOES NOT change just because later you are going to put odds on it. Why? Because you can lose that bet and never have a chance to take odds on it. They are two separate events.

Let me know if you have any further questions.

YoDR11

Thanks YoDr11.

So...


1. Come out roll, $5 bet, house edge = 1.41%
2. Assume roll #1 is an 8 and I place 5x odds, house edge on roll #2 = 0.37%
3. Assume roll #1 is an 8 and I place 5x odds plus a Come bet. The Come bet has a house edge of 1.41%. So, I have $30 with a house edge of 0.37% and $5 with a house edge of 1.41%. How exactly do I get the combined house edge? Is it (5/35)1.41% + (30/35)0.37% ?
4. Assume roll #2 is a 6 and I back that with max odds. Now I'm back to 0.37%, correct?

Quote: craigrow

Thanks YoDr11.

So...


1. Come out roll, $5 bet, house edge = 1.41%
2. Assume roll #1 is an 8 and I place 5x odds, house edge on roll #2 = 0.37%
3. Assume roll #1 is an 8 and I place 5x odds plus a Come bet. The Come bet has a house edge of 1.41%. So, I have $30 with a house edge of 0.37% and $5 with a house edge of 1.41%. How exactly do I get the combined house edge? Is it (5/35)1.41% + (30/35)0.37% ?
4. Assume roll #2 is a 6 and I back that with max odds. Now I'm back to 0.37%, correct?

No problem.

1: Correct.
2: Correct.
3: You just average the house edge by number of bets you have. (Really if you want to get technical, you would have to weight the average based on the amount of money per bet for it's individual house edge, but averaging gives you a good enough approximation). So it would be 1.41+.37/2= 0.89%
4. Correct, but now that percentage applies to more money, since you have $5 pass, $25 odds on winner 8, $30 at 0.37%, and now you have $30 ($5 come +$25 odds) on 6, for $30 at 0.37. So for every $100 you wager, regardless of which bet, you will lose $0.37. So the conclusion is you will come up to this faster by placing $60 in action each time. Two losses in a row, and you are down $120.

House edge is important to understand so that you can try and extend your playing time according to your bankroll. If you are plopping down $20 on the field (5.56% house edge), and you only have a bankroll of $400. You are screwed.