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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%)

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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.

Anyway, that’s the correct way to solve the problem, or one of many correct ways. One of the great things about probability math is the fact that there are usually several different ways you can do something.

TheFinalsix remained curious as to where his mistake was made, so I helped him out:

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?

Brandon James: 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.

The one problem with probability math is that, with several ways to solve something correctly; there are also several ways to make a mistake! My advice is if you have a result that doesn’t pass the, “Smell test,” just make sure that you’re asking the question correctly.

Elementary math is about figuring out how to solve stuff, but knowing what question to ask and how to ask it (though not extremely advanced) is one of the things you’re going to need to know how to do to be successful at Statistics and Probability.

Also, use the different ways to solve a problem to your advantage. If you’re not getting the answer you expect, then stretch those brain muscles and try to figure out a different approach!

Answer: 1st Paragraph: There are some houses that at least claim to factor odds into your rating in some way. I assume this varies from casino to casino and would not assume that it is universally not the case.

The Rest: As you have already stipulated, the Odds bet has no house edge. In the case of what you're discussing, you are either betting the $10 on the Pass Line and Taking $100 in odds, or you are betting $10 on the Pass Line and the bet is being resolved on the first roll. In any case, the only amount that comes with any sort of expected loss is the $10 bet on the Pass Line, so let's figure out our average total wager:

(10 * 12/36) + (110 * 24/36) = 76.6666666667

The average total wager is $76.6666666667, so now the next thing that we want to do is divide our expected loss of 14.1 pennies by this amount:

0.141/76.666666667 = 0.00183913043

Thus, we get a combined House Edge of 0.183913043%, as you have already determined.

However, if we isolate only those situations in which you are betting $110 in total, same expected loss:

.141/110 = 0.00128181818

Thus, you could say that your combined House Edge when you do in fact Take Odds is 0.128181818%

So, that's what it comes down to, you don't always bet the $110, so you would not divide the expected loss by 110.

ADDED: It’s important to remember that math is not only about getting the right answer, you must first start by asking yourself the question the right way, often the hardest part! As you see, MrVegas’ math was perfectly fine, but the error was in the (unintentional) mathematical assumption that a Craps player can Take Odds on every single round. He cannot, and in the long run, 1 out of 3 Come Out rolls, he won’t.

Mission Says: The player has the advantage at that point, you said so yourself. The only time I've ever seen someone attempt it, that person was allowed to take it down after a point was established. He didn't like the point because it was a six...lol

Lesson: The lesson is simple: You never want to take down or reduce a Don’t Pass bet after a point has been established. Pretending that the point is Six, the math behind this is simple. There are eleven total results that matter, five ways to make a six and the six ways to make a seven. We’ll pretend the bet is ten bucks:

(6/11 * 10) - (5/11 * 10) = 0.9090909090909091

Therefore, the player has a positive expectation of 91 cents on the $10 bet at this point, which is an advantage of 9.09091%, and should ABSOLUTELY NOT remove the bet.

Some, “Dark Side,” players do not like Laying Odds when the point is six or eight, and that’s just fine, as odds bets carry an expectation of +/-$0.00, anyway. However, to remove a Don’t Pass bet after ANY point is established is just foolhardy.

Answer: Given that only the Pass Line bets and the Come Bets have a House Edge working against you, the, 'Better choice,' is to make as few of those as possible. In terms of the House Edge relative to the total $$$ exposed, then again, you're better off to make as few Pass Line and Come Bets as possible.

Lesson: No House Edge is better than a House Edge. 100% RTP is better than less than 100% RTP. Pretty simple stuff.

Keno and Slot Assumptions? (Prozema at WoV)

Question: Is there any general assumptions one can make on the allocation of that return between free spin bonus games, progressives, and the base game?

Answer: A. The contribution to a player from a Progressive(s) can vary pretty wildly. There are several factors, such as:

• Meter Move
• Progressive Start Amount Relative to Probability of Hitting
• Probability of Hitting
• Can the Progressive be won w/o Max Bet? (Assuming no change to probability AND not a must-hit, a lower bet would be better and the Progressive, as a result, would constitute a greater percentage of the return.)

B. Free Spins can also vary dramatically based on their average value and probability of hitting. Free Games often make up as much as 40% of a game's return, but it is also sometimes less than that. You can even look at the same game. Not a slot, but:

As you can see, the Bonus Return on Cleopatra Keno can run from under 25%, to as high as 47.4%, just based on how many spots you're playing.

Lesson: Do not make assumptions! You either know the math behind a game or you do not know the math behind a game. Assumptions are what get recreational players chasing disadvantageous Progressives that were never positive to begin with.

Fun with Flights! (OneNickelMiracle at WoV)

Question: There are three siblings, two of them are a taking a flight, one is staying home. Should the two siblings take the same flight together, risking the third sibling being alone should a flight tragedy occur, or should the two siblings take separate flights ensuring the third sibling is not left alone.

Answer: Let's get morbid and say planes have a 90% probability of going down on any individual flight, even though it's nowhere near that:

One flight, both siblings: .1 * 2 = 0.2 Siblings Dead on Average

Two flights land safely: .81 * 0 = 0

One flight crashes, the other doesn't: (.9 * .1) + (.1 * .9) = .18, siblings dead, also .18

Both flights crash: (.1 * .1 * 2) = .02

TWO FLIGHTS, AVERAGE SIBLINGS DEAD: 0.2

So, the probability of at least one being involved in a crash increases with two flights, but the average number of siblings who die remains the same and the probability of both siblings dying (which seems the focal point of the OP) goes down relatively dramatically.

It doesn't double the chance that one of them will, though. It doubles the chance that AT LEAST one of them will.

Lesson: Don’t Fly Mission Airlines! 90% of our planes apparently crash.