Help me clear up some odds and math PLEASE

I am trying to clarify the dark side math.

PROB.WIN
47.93%

PROB.PUSH
2.78%

PROB.LOSS
49.29%

So I will win 47.93% of the time once a point is established if I am on the Don't Pass line?

If my goal was to win on the don't 2 times consecutively before I lost (point achieved) say 5 times, how would that math look?

Thanks

Quote: vegasrvp

I am trying to clarify the dark side math.

PROB.WIN
47.93%

PROB.PUSH
2.78%

PROB.LOSS
49.29%

So I will win 47.93% of the time once a point is established if I am on the Don't Pass line?

If my goal was to win on the don't 2 times consecutively before I lost (point achieved) say 5 times, how would that math look?

Thanks

I'm probably the wrong person to chime in since I don't know craps very well, but your numbers don't make sense to me. How can there be a push on either Pass or DP? Do or do not: there is no try. If it's meant per roll, then that's even more off, as a "push" (= no action on a DP bet?) would happen more than 50% of the time (neither the point nor a 7 was rolled).

More importantly, once your bet survives a come-out 7, you have the advantage regardless of the point. So if these numbers are accurate, (and I'm assuming they're aggregate numbers because each individual DP bet has its own chances for a win, none of them close to what you're showing), they're still -EV, when they should be +EV. So your numbers must include the chance of losing on the come-out 7, or they're simply not correct.

"Dark Side" Math... DP = 1.36% HE.... DC = 1.36% HE...

Quote: beachbumbabs

I'm probably the wrong person to chime in since I don't know craps very well, but your numbers don't make sense to me. How can there be a push on either Pass or DP? Do or do not: there is no try. If it's meant per roll, then that's even more off, as a "push" (= no action on a DP bet?) would happen more than 50% of the time (neither the point nor a 7 was rolled).

More importantly, once your bet survives a come-out 7, you have the advantage regardless of the point. So if these numbers are accurate, (and I'm assuming they're aggregate numbers because each individual DP bet has its own chances for a win, none of them close to what you're showing), they're still -EV, when they should be +EV. So your numbers must include the chance of losing on the come-out 7, or they're simply not correct.

This is why I am here. I have no clue how to figure this out.

I took these numbers from the chart on the Wizard of odds site. It was under craps and called "Summary of Multi-Roll Bets".

That being said I am trying to find out the odds of winning my don't pass bet 2 consecutive times? I can actually figure that out if I know the odds of hitting it once.

Thanks for this info though. I'm learning.

Quote: vegasrvp

I am trying to clarify the dark side math.

PROB.WIN
47.93%

PROB.PUSH
2.78%

PROB.LOSS
49.29%

So I will win 47.93% of the time once a point is established if I am on the Don't Pass line?

No. Once a point is established, Don't Pass is far more likely to win. It depends on the point:
If it is 4 or 10, DP wins 2/3 of the time and loses 1/3
If is is 5 or 9, DP wins 3/5 of the time and loses 2/5
If it is 6 or 8, DP wins 6/11 of the time and loses 5/11

The win/loss/push numbers you have include the 1/3 of the time you don't establish a point.

Quote: vegasrvp

If my goal was to win on the don't 2 times consecutively before I lost (point achieved) say 5 times, how would that math look?

You have to include "7 or 11 on the comeout" in your definition of loss, and remember that a 12 (2 in some places, like Reno) on the comeout is a push.

Ignoring pushes, a DP bet wins 949/1925 of the time, and loses 976/1925 of the time.

To calculate the probability of having 2 consecutive wins on DP before 5 total losses, calculate the probabilities of each way you can do this, and then add them up.
If W is a win and L is a loss:
No losses:
WW
1 loss:
LWW
WLWW
2 losses:
LLWW
LWLWW
WLLWW
WLWLWW
3 losses:
LLLWW
LLWLWW
LWLLWW
LWLWLWW
WLLLWW
WLLWLWW
WLWLLWW
WLWLWLWW
4 losses:
LLLLWW
LLLWLWW
LLWLLWW
LLWLWLWW
LWLLLWW
LWLLWLWW
LWLWLLWW
LWLWLWLWW
WLLLLWW
WLLLWLWW
WLLWLLWW
WLLWLWLWW
WLWLLLWW
WLWLLWLWW
WLWLWLLWW
WLWLWLWLWW

I get about 75.147% "success." However, note that a number of these "successes" lose money - for example, LLLLWW puts you 2 bets behind - while every "failure" except one (WLWLWLWLWL) has you losing money. In terms of EV, each "run" can expect to lose 6.475% of what each bet is (i.e if each DP bet is $10, you can expect to lose 64.75 cents over the length of your run).

Quote: ThatDonGuy

No. Once a point is established, Don't Pass is far more likely to win. It depends on the point:
If it is 4 or 10, DP wins 2/3 of the time and loses 1/3
If is is 5 or 9, DP wins 3/5 of the time and loses 2/5
If it is 6 or 8, DP wins 6/11 of the time and loses 5/11

The win/loss/push numbers you have include the 1/3 of the time you don't establish a point.


You have to include "7 or 11 on the comeout" in your definition of loss, and remember that a 12 (2 in some places, like Reno) on the comeout is a push.

Ignoring pushes, a DP bet wins 949/1925 of the time, and loses 976/1925 of the time.

To calculate the probability of having 2 consecutive wins on DP before 5 total losses, calculate the probabilities of each way you can do this, and then add them up.
If W is a win and L is a loss:
No losses:
WW
1 loss:
LWW
WLWW
2 losses:
LLWW
LWLWW
WLLWW
WLWLWW
3 losses:
LLLWW
LLWLWW
LWLLWW
LWLWLWW
WLLLWW
WLLWLWW
WLWLLWW
WLWLWLWW
4 losses:
LLLLWW
LLLWLWW
LLWLLWW
LLWLWLWW
LWLLLWW
LWLLWLWW
LWLWLLWW
LWLWLWLWW
WLLLLWW
WLLLWLWW
WLLWLLWW
WLLWLWLWW
WLWLLLWW
WLWLLWLWW
WLWLWLLWW
WLWLWLWLWW

I get about 75.147% "success." However, note that a number of these "successes" lose money - for example, LLLLWW puts you 2 bets behind - while every "failure" except one (WLWLWLWLWL) has you losing money. In terms of EV, each "run" can expect to lose 6.475% of what each bet is (i.e if each DP bet is $10, you can expect to lose 64.75 cents over the length of your run).

OK, so I see how you get the ways to lose for each of the losses 1-5. Two questions though...
1. How does that equate to 75.14%?
2. is it just a coincidence that it doubles with every number? 0=1, 1=2, 2=4, 3=8 & 4=16. That means 10= 1024?

Quote: vegasrvp

OK, so I see how you get the ways to lose for each of the losses 1-5. Two questions though...
1. How does that equate to 75.14%?

Each letter has a 949/1925 chance of being a W and a 976/1925 chance of being an L.
WW has a probability of 949/1925 x 949/1925 = about 0.243036
LWW has a probability of 976/1925 x 949/1925 x 949/1925 = about 0.123223
And so on
If you do that for all of the winning runs and add them up, you get a total of 0.75147.

Quote: vegasrvp

2. is it just a coincidence that it doubles with every number? 0=1, 1=2, 2=4, 3=8 & 4=16. That means 10= 1024?

No, it is not a coincidence.
There is only one way to have 2 wins in a row with no losses: WW (i.e. you win your first two bets).
The runs of 2 wins in a row with N losses can be divided into two groups:
(a) Start with a loss, then have a run of 2 wins in a row with N-1 losses;
(b) Start with a win, then a loss, then have a run of 2 wins in a row with N-1 losses
You can't start with 2 wins as that would be a run of 2 wins with no losses, and you already counted those.
Thus, the total number of runs of 2 consecutive wins with N losses = 2 x the number of runs of 2 consecutive wins with (N-1) losses (since each (N-1) loss run is in both group (a) and group (b)).

Remember, however, that the probability of winning and losing is not the only thing to consider; how much each win and loss is worth has to be taken into account, especially if some of the "wins" actually lose money.

Quote: ThatDonGuy

Each letter has a 949/1925 chance of being a W and a 976/1925 chance of being an L.
WW has a probability of 949/1925 x 949/1925 = about 0.243036
LWW has a probability of 976/1925 x 949/1925 x 949/1925 = about 0.123223
And so on
If you do that for all of the winning runs and add them up, you get a total of 0.75147.


No, it is not a coincidence.
There is only one way to have 2 wins in a row with no losses: WW (i.e. you win your first two bets).
The runs of 2 wins in a row with N losses can be divided into two groups:
(a) Start with a loss, then have a run of 2 wins in a row with N-1 losses;
(b) Start with a win, then a loss, then have a run of 2 wins in a row with N-1 losses
You can't start with 2 wins as that would be a run of 2 wins with no losses, and you already counted those.
Thus, the total number of runs of 2 consecutive wins with N losses = 2 x the number of runs of 2 consecutive wins with (N-1) losses (since each (N-1) loss run is in both group (a) and group (b)).

Remember, however, that the probability of winning and losing is not the only thing to consider; how much each win and loss is worth has to be taken into account, especially if some of the "wins" actually lose money.

I am not worried about the losses.

I believe, and am looking for mathematical confirmation, that if I played a game. In that game I played on the side that would win 48% of the time. If I were to bet on that side the odds of me losing 15 times before winning 2 CONSECUTIVE times is less than 1%. Do you agree and if so can you show the math.

Quote: vegasrvp

I believe, and am looking for mathematical confirmation, that if I played a game. In that game I played on the side that would win 48% of the time. If I were to bet on that side the odds of me losing 15 times before winning 2 CONSECUTIVE times is less than 1%. Do you agree and if so can you show the math.

This is easy to calculate. I get slightly less than 2%.
Click on the button to see how.

The OP needs to confess he wants to martingale LOL