thefinalsix
Joined: May 18, 2020
• Posts: 7
May 18th, 2020 at 10:19:23 AM permalink
Hi all,

It's safe to say that making a Pass Line bet after the point is already established is a bad idea. Of course the casino will let you do it because it's a sucker move, but why would you? You could just make a Come bet! (I've seen people join a table and simultaneously make a Come bet and a Pass Line bet with an established point before... and I can't say I have any understanding what's going through that person's head...)

Anyway, I was thinking... if you're going to be enough of a sucker to make a mid-point Pass bet, at least put some odds on it. But then I did the math and it seemed that the odds weren't actually reducing the House Edge. I'm thinking that can't be right... but I can't find the flaw in my math.

Say the point is already set at 4. We know that 4 is rolled before 7 one-third of the time. Some guy shows up to bet Pass Line. Pass Line will be paid 1:1 and any odds will be paid 2:1...

\$1 Pass Line bet, no odds

Edge is (1/3)*(1) + (2/3)*(-1) = -1/3 = -33.3% = 33.3% in favor of the house

\$1 Pass Line bet with single odds

Edge is (1/3)*(1+2) + (2/3)*(-1-1) = (1/3)*(3) + (2/3)*(-2) = 1 - 4/3 = -1/3 = -33.3% = 33.3% in favor of the house

\$1 Pass Line bet with double odds

Edge is (1/3)*(1+4) + (2/3)*(-1-2) = (1/3)*(5) + (2/3)*(-3) = 5/3 -6/3 = -1/3 = -33.3% = 33.3% in favor of the house

\$1 Pass Line bet with 10x odds

Edge is (1/3)*(1+20) + (2/3)*(-1-10) = (1/3)*(21) + (2/3)*(-11) = 21/3 -22/3 = -1/3 = -33.3% = 33.3% in favor of the house

My question is... what is going on here? I thought taking the free odds was supposed to reduce the house edge of your total bet! Maybe it is different because the point is established? But that doesn't seem right since weighing more money on the side of the table that pays better odds should always help you.

To be clear... I have no intention of making such a silly bet :). I'm just wondering why the math doesn't match up with my intuition.

Thanks!
unJon
Joined: Jul 1, 2018
• Posts: 1626
May 18th, 2020 at 10:24:18 AM permalink
Quote: thefinalsix

Hi all,

It's safe to say that making a Pass Line bet after the point is already established is a bad idea. Of course the casino will let you do it because it's a sucker move, but why would you? You could just make a Come bet! (I've seen people join a table and simultaneously make a Come bet and a Pass Line bet with an established point before... and I can't say I have any understanding what's going through that person's head...)

Anyway, I was thinking... if you're going to be enough of a sucker to make a mid-point Pass bet, at least put some odds on it. But then I did the math and it seemed that the odds weren't actually reducing the House Edge. I'm thinking that can't be right... but I can't find the flaw in my math.

Say the point is already set at 4. We know that 4 is rolled before 7 one-third of the time. Some guy shows up to bet Pass Line. Pass Line will be paid 1:1 and any odds will be paid 2:1...

\$1 Pass Line bet, no odds

Edge is (1/3)*(1) + (2/3)*(-1) = -1/3 = -33.3% = 33.3% in favor of the house

\$1 Pass Line bet with single odds

Edge is (1/3)*(1+2) + (2/3)*(-1-1) = (1/3)*(3) + (2/3)*(-2) = 1 - 4/3 = -1/3 = -33.3% = 33.3% in favor of the house

\$1 Pass Line bet with double odds

Edge is (1/3)*(1+4) + (2/3)*(-1-2) = (1/3)*(5) + (2/3)*(-3) = 5/3 -6/3 = -1/3 = -33.3% = 33.3% in favor of the house

\$1 Pass Line bet with 10x odds

Edge is (1/3)*(1+20) + (2/3)*(-1-10) = (1/3)*(21) + (2/3)*(-11) = 21/3 -22/3 = -1/3 = -33.3% = 33.3% in favor of the house

My question is... what is going on here? I thought taking the free odds was supposed to reduce the house edge of your total bet! Maybe it is different because the point is established? But that doesn't seem right since weighing more money on the side of the table that pays better odds should always help you.

To be clear... I have no intention of making such a silly bet :). I'm just wondering why the math doesn't match up with my intuition.

Thanks!

You are dividing each of your examples by 1 as if you had \$1 at risk. But that only works for no odds. At single odds, divide by 2. At double odds, divide by 3. Etc.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
odiousgambit
Joined: Nov 9, 2009
• Posts: 8260
May 18th, 2020 at 10:27:40 AM permalink
delete
Last edited by: odiousgambit on May 18, 2020
the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder
Mission146
Joined: May 15, 2012
• Posts: 12598
May 18th, 2020 at 10:31:30 AM permalink
Quote: thefinalsix

Hi all,

It's safe to say that making a Pass Line bet after the point is already established is a bad idea. Of course the casino will let you do it because it's a sucker move, but why would you? You could just make a Come bet! (I've seen people join a table and simultaneously make a Come bet and a Pass Line bet with an established point before... and I can't say I have any understanding what's going through that person's head...)

I would say that the person either does not care, or does not understand, how the probabilities of the Pass Line function relative to the Expected Value of the bet. In some cases, perhaps they just want to be playing, 'With the table,' right off the bat.

Quote:

Anyway, I was thinking... if you're going to be enough of a sucker to make a mid-point Pass bet, at least put some odds on it. But then I did the math and it seemed that the odds weren't actually reducing the House Edge. I'm thinking that can't be right... but I can't find the flaw in my math.

Say the point is already set at 4. We know that 4 is rolled before 7 one-third of the time. Some guy shows up to bet Pass Line. Pass Line will be paid 1:1 and any odds will be paid 2:1...

\$1 Pass Line bet, no odds

Edge is (1/3)*(1) + (2/3)*(-1) = -1/3 = -33.3% = 33.3% in favor of the house

\$1 Pass Line bet with single odds

Edge is (1/3)*(1+2) + (2/3)*(-1-1) = (1/3)*(3) + (2/3)*(-2) = 1 - 4/3 = -1/3 = -33.3% = 33.3% in favor of the house

\$1 Pass Line bet with double odds

Edge is (1/3)*(1+4) + (2/3)*(-1-2) = (1/3)*(5) + (2/3)*(-3) = 5/3 -6/3 = -1/3 = -33.3% = 33.3% in favor of the house

\$1 Pass Line bet with 10x odds

Edge is (1/3)*(1+20) + (2/3)*(-1-10) = (1/3)*(21) + (2/3)*(-11) = 21/3 -22/3 = -1/3 = -33.3% = 33.3% in favor of the house

My question is... what is going on here? I thought taking the free odds was supposed to reduce the house edge of your total bet! Maybe it is different because the point is established? But that doesn't seem right since weighing more money on the side of the table that pays better odds should always help you.

To be clear... I have no intention of making such a silly bet :). I'm just wondering why the math doesn't match up with my intuition.

Thanks!

I'm just going to do this how I would approach it because it's faster than finding something I might not like about the way you did it, so you can apply what I'm about to do.

First, I want to point two things out:

1.) It doesn't reduce the house edge of the Pass Line bet itself, it reduces the expected loss (percentage) relative to how much is being bet, in total. The total expected loss (\$\$\$) remains the same based on the Pass Line bet, but the percentage relative to total bet amount goes down.

2.) Similarly, Odds bets have an expected value of zero.

Situation: Put bet on point of four v. Put bet plus 2x odds:

Relevant probabilities: 4 (3/9) or 7 (6/9).

I am going to do \$5 Put Bet and \$10 Odds.

Put Bet Expected Loss: (5 * 3/9) - (5 * 6/9) = -1.66666666667

Expected Loss of total bet: -1.66666666667

Expected Loss Relative to Total Bet: -1.66666666667/5 = -0.33333333333 (Same as House Edge of PUT 4-33.33%)

--------

With Odds:

Odds Bet Expected Loss: (10 * 3/9) - (5 * 6/9) = 0

Put Bet Expected Loss (From Above): -1.66666666667

Expected Loss of All Bets: -1.66666666667

Expected Loss Relative to Total Bets: -1.66666666667/15 = -0.11111111111 (11.11% of all monies bet)

As we would expect, the expected loss on the total action (expressed as a percentage) is 1/3rd as without the odds because 2/3rds of the total exposure is on the zero edge Odds bet.
Vultures can't be choosers.
thefinalsix
Joined: May 18, 2020
• Posts: 7
May 18th, 2020 at 10:34:52 AM permalink
Thanks for the response, unJon! I'm still not sure exactly where the error is. Can you show me the correct Edge calculation for 10x, for example? If you win in that case, you get \$1 for pass bet plus \$20 for your odds. Total win \$21. If you lose, you lose all your \$11 you had at stake.

(1/3)*21 + (2/3)*(-11) = -1/3.

Where is the flaw?
Mission146
Joined: May 15, 2012
• Posts: 12598
Thanks for this post from:
May 18th, 2020 at 10:36:56 AM permalink
Quote: thefinalsix

Thanks for the response, unJon! I'm still not sure exactly where the error is. Can you show me the correct Edge calculation for 10x, for example? If you win in that case, you get \$1 for pass bet plus \$20 for your odds. Total win \$21. If you lose, you lose all your \$11 you had at stake.

(1/3)*21 + (2/3)*(-11) = -1/3.

Where is the flaw?

I'm going to stick with my \$5 base bet, so we'll do \$50 odds.

With Odds:

Odds Bet Expected Loss: (100 * 3/9) - (50 * 6/9) = 0

Put Bet Expected Loss (From Above): -1.66666666667

Expected Loss of All Bets: -1.66666666667

Expected Loss Relative to Total Bets: -1.66666666667/55 = -0.0303030303 (Expected Loss of 3.03% relative to all monies bet)

Oh, and your flaw is you are treating all of this as the same bet when it is not. If you really want to express it in one line, do this:

(((3/9 * 1) - (6/9 * 1)) + ((3/9 * 20) - (6/9 *10)))/11 = -0.0303030303

Otherwise, you just keep recalculating the expected loss (in dollars) rather than calculating the expected loss (percentage) relative to your total bet. That's why you keep getting the same result.
Vultures can't be choosers.
thefinalsix
Joined: May 18, 2020
• Posts: 7
May 18th, 2020 at 10:43:26 AM permalink
Is a Put bet different than a Pass Line bet after point is established? Wikipedia makes it sound like they are the same but I had assumed they were different.

Sorry, not an expert in Craps!
Mission146
Joined: May 15, 2012
• Posts: 12598
May 18th, 2020 at 10:47:48 AM permalink
Quote: thefinalsix

Is a Put bet different than a Pass Line bet after point is established? Wikipedia makes it sound like they are the same but I had assumed they were different.

Sorry, not an expert in Craps!

I added some stuff in my post above and found your flaw, so please make sure to look at that.

A Put Bet is literally the same thing as a Pass Line bet, with the only difference being you make a Put Bet after the point has been established.

The reason for the tremendous disadvantage of a put bet is because you lose the following results on the Come Out roll:

WIN (7/11): 8/36

LOSE (2, 3, 12): 4/36

As you can see, making a normal Pass Line Bet, you are twice as likely to win immediately as you are to lose immediately. If you make a Put Bet after this, it is effectively the same as if you made a Place Bet that pays even money, but Place Bets do not pay Even Money.

While it is still worse than a Pass Line bet (generally) from an Expected Value standpoint, if someone REALLY insists on playing right away and wants to be playing with the table, they should put a Place or Buy bet on the point, assuming the casino permits.
Vultures can't be choosers.
AlanMendelson
Joined: Oct 5, 2011
• Posts: 3222
Thanks for this post from:
May 18th, 2020 at 10:48:44 AM permalink
Quote: thefinalsix

Is a Put bet different than a Pass Line bet after point is established? Wikipedia makes it sound like they are the same but I had assumed they were different.

Sorry, not an expert in Craps!

A put bet is making a pass line bet after the point is established. It's still a put bet whether you bet odds or not.
thefinalsix
Joined: May 18, 2020