Weird craps bet you might of never heard of question!!

Ok so i heard of this weird bet thing called the "bar" I don't under why they called this that Cuz it made it confusing ... So say you bet $5 against a person and you roll a 6 ... Ok so now there saying they want to bet $5 on the "bar" and this is the definition of that to them ... You both bet an additional $5 if the 6 is hit they win all $20... If a 6 1 is rolled the roll is a push you both take $10 back if a 5 2 or 4 3 is rolled the person betting in place of the dont pass wins...U can make this bet no matter what there point is whether it's 4 5 6 8 9 or 10... Who has the advantage on this bet the roller or the person betting in place of the dont pass?? Is it different depending on what there point is??


Please provide mathematical proof of which it Is in favor of ! Love this site ... It took a lot for me to realize the martingale didn't work , but this site proved it to me!

Quote: roudy12345

Ok so i heard of this weird bet thing called the "bar" I don't under why they called this that Cuz it made it confusing ... So say you bet $5 against a person and you roll a 6 ... Ok so now there saying they want to bet $5 on the "bar" and this is the definition of that to them ... You both bet an additional $5 if the 6 is hit they win all $20... If a 6 1 is rolled the roll is a push you both take $10 back if a 5 2 or 4 3 is rolled the person betting in place of the dont pass wins...U can make this bet no matter what there point is whether it's 4 5 6 8 9 or 10... Who has the advantage on this bet the roller or the person betting in place of the dont pass?? Is it different depending on what there point is??


Please provide mathematical proof of which it Is in favor of ! Love this site ... It took a lot for me to realize the martingale didn't work , but this site proved it to me!

Let me make sure I understand your description.

You roll the dice and set a point, in this case a six: Now, having established a point of six with a $5 bet, you can essentially double your bet and you will win $10 if another six comes.

Now, this six bet BARS losses on Sevens containing a six, so you only lose on 4-3 or 2-5, but not 6-1.

If this is all correct:

POINT 6/8

(10 * 5/36) - (10 * 4/36) = 0.27777777777 EXPECTED WIN, PLAYER EDGE .27777777777/10 = 2.78%

Point 5/9

(10 * 4/36) - (10 * 4/36) = 0

Point 4/10

(10 * 3/36) - ( 10 * 4/36) = -.27777777777777 EXPECTED RETURN, HOUSE EDGE 2.78%

Conclusion

Simply put, you would ALWAYS double your bet on a point of 6/8 because you have the advantage, and you would also double on 5/9 because that has zero expectation, which is better than you're going to get on any amount of odds.

You would also double on Points of 4/10 because you have a lower expected loss, at that point, than with the $5 flat bet. Specifically, you expect to lose 27.8 cents on the $10 bet and you expect to lose 41.7 cents on a five dollar bet under normal rules with no odds.

HOWEVER:

There is a bit of a caveat to that 4/10. Some people consider the Odds towards the House Edge of the Pass Line bet, I personally don't agree, I just look at it as 1.41% and 0%. However, if you wanted to look at the Odds as mitigating the House Edge, then:

(25 * 3/36) - ( 15 * 6/36) = -0.41666666666/15 = -0.02777777777 House Edge

What that represents is the same 2.78% House Edge on your overall action on 4/10 if you are able to take 2x Odds (Hence the $15 bet for a $25 win in parentheses), so if you do look at the House Edge from a combined standpoint, you would be equal to take 2x Odds as to make this bet and it would be superior to take 3x Odds (or better) than make this double bet.

Personally, I just look at the House Edge in terms of the Pass Line bet made, so I would probably just double my bet because, in my view, that has the lower House Edge.

OVERALL HOUSE EDGE

Okay, so let's look at this in terms of a $5 Base Bet. You're going to win 8/36 on the CO and lose 4/36:

(8/36 * 5) - (4/36 * 5) = 0.55555555555

That done, you have a 10/36 chance of rolling a Point of 6 OR 8, which results in:

0.27777777777 * 10/36 = 0.07716049382

With that, you have a 8/36 chance of rolling a Point of 5 OR 9 which results in:

0---The Doubled Bet is zero EV

Finally, you have a 6/36 chance of rolling a Point of 4 OR 10, your only negative result, which is:

6/36 * (-0.27777777777) = -0.04629629629

If we add all that up: 0.55555555555+0.07716049382-0.04629629629 = 0.58641975308

0.58641975308/5 = 0.11728395061 Which represents a PASS LINE BETTOR ADVANTAGE of 11.728395061%

I take it this is some kind of home or party game? Mind PMing me an address and schedule?

Lol it's just a once a year thing! I'm sorry for the confusion if the point is set a 6 1 results in a push 5 2 or 4 3 results in the player covering the bet winning

Quote: roudy12345

Lol it's just a once a year thing! I'm sorry for the confusion if the point is set a 6 1 results in a push 5 2 or 4 3 results in the player covering the bet winning

Oh, it goes until resolved like a standard PL bet. Let me re-do that.

POINT 6/8

(10 * 5/9) - (10 * 4/9) =1.11111111111 EXPECTED WIN, PLAYER EDGE 1.11111111111/10 = 11.11

Point 5/9

(10 * 4/8) - (10 * 4/8) = 0

Point 4/10

(10 * 3/7) - ( 10 * 4/7) = -1.42857142857 EXPECTED RETURN -1.42857142857/10 = 14.2857% House Edge

Conclusion

Simply put, you would ALWAYS double your bet on a point of 6/8 because you have the advantage, and you would also double on 5/9 because that has zero expectation, which is better than you're going to get on any amount of odds.

You would also double your bet on 4/10 because the Expected Loss of $1.43 is lower than: (5 * 3/9) - (5 * 6/9) = -1.66666666667 $1.67 if you do not double.

OVERALL HOUSE EDGE

Okay, so let's look at this in terms of a $5 Base Bet. You're going to win 8/36 on the CO and lose 4/36:

(8/36 * 5) - (4/36 * 5) = 0.55555555555

That done, you have a 10/36 chance of rolling a Point of 6 OR 8, which results in:

1.11111111111 * 10/36 = 0.3086419753

With that, you have a 8/36 chance of rolling a Point of 5 OR 9 which results in:

0---The Doubled Bet is zero EV

Finally, you have a 6/36 chance of rolling a Point of 4 OR 10, your only negative result, which is:

6/36 * -1.42857142857 = -0.23809523809

If we add all that up: 0.55555555555+0.3086419753-0.23809523809 = 0.62610229276

0.62610229276/5 = 0.12522045855 Which represents a PASS LINE BETTOR ADVANTAGE of 12.522045855%

That Six & Eight became even more powerful, so your edge is a little better.

Good to know for next year ! I never roll so I'm only gonna let them "bar" when there point is 4 or 10 ... If Im up any ill let them "bar" on a 5 or 9 since there is 0 advantage either way

Quote: roudy12345

Good to know for next year ! I never roll so I'm only gonna let them "bar" when there point is 4 or 10 ... If Im up any ill let them "bar" on a 5 or 9 since there is 0 advantage either way

(8/36 * 5) - (4/36 * 5) = 0.55555555555

((5 * 5/11) - (5 * 6/11)) * 10/36 = -0.12626262626

((5 * 4/10) - (5 * 6/10)) * 8/36 = -0.22222222222

6/36 * -1.42857142857 = -0.23809523809

Conclusion

0.55555555555-0.12626262626-0.22222222222-0.23809523809 = -0.03102453102 Expected Loss

In this case, you bet $5 30/36 times and $10 6/36 times, so your average bet is:

(5 * 30/36) + (10 * 6/36) = 5.83333333333

Which means the House Edge is:

0.03102453102/5.83333333333 = 0.00531849103 or 0.531849103%

NOTE:

The House Edge will turn into a PLAYER edge if your BAR the 6/1 on the five and nine. While this is technically an even overall bet, it still eliminates the House Edge you would normally face on a Point of 5 or 9, thereby resulting in an advantage on the Pass Line bet if you do the BAR thing. To wit, an Expected loss of zero on a $10 bet is better than an expected loss of 22.2 cents on a $5 bet.