Example: pass line has a 1.41% house edge but with 10x odds it has a .18% house edge.

This is where I’m confused by the Wizard’s advice. If I bet $10 on the on the pass line only, my theoretical loss is 1.41% of $10 = $0.141, about 14 cents. If I take 10x odds in addition to my pass line bet, my total wager is $110. My theoretical loss would be 0.18% of $110 = $0.198, about 20 cents. I would lose more money over the long run.

I wouldn’t think the odds bets would make any difference what so ever if it pays true odds and has no house edge. Why does the Wizard recommend taking odds and why do my calculations show that I would lose more money taking odds?

Quote:MrVegasThe Wizard’s strategy for craps recommends playing pass/come or don’t pass/don’t come with odds. I play the pass line as to avoid the dark side. I understand odds have no house edge. Therefore the EV is zero. Over the long run those bets break even. So what’s the point in betting them? They don’t count toward your minimum wager for rating purposes. The Wizard has a chart showing how taking odds lowers the house edge on what I assume he means your total bet after taking odds.

Example: pass line has a 1.41% house edge but with 10x odds it has a .18% house edge.

This is where I’m confused by the Wizard’s advice. If I bet $10 on the on the pass line only, my theoretical loss is 1.41% of $10 = $0.141, about 14 cents. If I take 10x odds in addition to my pass line bet, my total wager is $110. My theoretical loss would be 0.18% of $110 = $0.198, about 20 cents. I would lose more money over the long run.

I wouldn’t think the odds bets would make any difference what so ever if it pays true odds and has no house edge. Why does the Wizard recommend taking odds and why do my calculations show that I would lose more money taking odds?

1.41%

0.18%

https://wizardofodds.com/games/craps/appendix/1/

as to why bet the odds, the simple answer is this:

if you are a player who wants to bet more than $10 , you can add a bet without a house edge to get more in action

if you are a player who is happy with $10, it is fairly pointless to add the free odds. However, there is another way to approach this and still play free odds

My thought is I am right at my EV for odds before I ever place the bet. I don’t want to find myself a standard deviation or two low.

But it obviously depends on what you want to do. Betting $10 on the pass line, you’ll lose the same amount either way (EV).

Quote:MrVegasThe Wizard’s strategy for craps recommends playing pass/come or don’t pass/don’t come with odds. I play the pass line as to avoid the dark side. I understand odds have no house edge. Therefore the EV is zero. Over the long run those bets break even. So what’s the point in betting them? They don’t count toward your minimum wager for rating purposes. The Wizard has a chart showing how taking odds lowers the house edge on what I assume he means your total bet after taking odds.

Example: pass line has a 1.41% house edge but with 10x odds it has a .18% house edge.

This is where I’m confused by the Wizard’s advice. If I bet $10 on the on the pass line only, my theoretical loss is 1.41% of $10 = $0.141, about 14 cents. If I take 10x odds in addition to my pass line bet, my total wager is $110. My theoretical loss would be 0.18% of $110 = $0.198, about 20 cents. I would lose more money over the long run.

I wouldn’t think the odds bets would make any difference what so ever if it pays true odds and has no house edge. Why does the Wizard recommend taking odds and why do my calculations show that I would lose more money taking odds?

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.

Quote:MrVegasThere is no reason to recommend odds. Mathematically is makes no difference. The advice should be to play the pass/come or don’t pass/don’t come and playing odds or not makes no difference. The best advice mathmatically is to play the don’t pass/don’t come but we all know that will make you public enemy number 1 at the table. Also, slow the game down as much as you can.

I consider your position debatable.

The first thing that needs to be considered is that there is no Pass Line + Odds bet, there is a Pass Line bet and there is an Odds bet, but they are two separate bets. When you bet the Pass Line, you are betting that the initial roll will be a 7 or 11, or in the alternative, that your initial roll will be a 4, 5, 6, 8, 9 or 10 and that said number will repeat before a seven comes up. With the Odds bet, all you're doing is betting that a particular number will show up before the seven.

You knew that of course, but I wanted to fully flesh it out to illustrate that they are separate bets. The only common thread between the two bets is that you must make the Pass Line bet (house edge) to get the privilege of making the zero house edge Odds bet...unless you're permitted to piggyback Odds with a different player.

Anyway, if you make a $10 Pass Line bet and take $100 in odds (let's assume a point number for simplicity) then your expected loss is 0.141 and your exposure is $110. If you make eleven separate $10 Pass Line bets, then your exposure ($110) is the same, but your expected loss on those bets is now $1.41.

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Another way of looking at it is let's say you had a bankroll of $500 and you set a goal to either double it or go El Busto. Would you double it more often betting the $10 and Taking Max Odds whenever possible, or would you double it more often flat betting $10?

Now, if your primary objective is just to play as long as possible, (or until bored) then that makes a very good case for just playing the Pass Line (or Don't Pass). However, it doesn't change the fact that the combined House Edge is less for total dollars exposed by Taking Odds.

I guess Wizard's advice assumes the person is there wanting to gamble, given the choice between no house edge and a house edge, the player would do well to take the former as often as possible and for as much as possible.

Myself, I don't look at it either way, to be honest. I just consider them strictly as separate bets. Combined edge is meaningless to me. One bet has a house edge of 1.41%, the other, 0%.

Quote:Mission1461st 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.

Thanks! That’s exactly what I thought. It makes no difference what so ever whether or not you take odds unless it is factored into your avaerage bet for player rating.

Quote:Mission146I consider your position debatable.

The first thing that needs to be considered is that there is no Pass Line + Odds bet, there is a Pass Line bet and there is an Odds bet, but they are two separate bets. When you bet the Pass Line, you are betting that the initial roll will be a 7 or 11, or in the alternative, that your initial roll will be a 4, 5, 6, 8, 9 or 10 and that said number will repeat before a seven comes up. With the Odds bet, all you're doing is betting that a particular number will show up before the seven.

You knew that of course, but I wanted to fully flesh it out to illustrate that they are separate bets. The only common thread between the two bets is that you must make the Pass Line bet (house edge) to get the privilege of making the zero house edge Odds bet...unless you're permitted to piggyback Odds with a different player.

Anyway, if you make a $10 Pass Line bet and take $100 in odds (let's assume a point number for simplicity) then your expected loss is 0.141 and your exposure is $110. If you make eleven separate $10 Pass Line bets, then your exposure ($110) is the same, but your expected loss on those bets is now $1.41.

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Another way of looking at it is let's say you had a bankroll of $500 and you set a goal to either double it or go El Busto. Would you double it more often betting the $10 and Taking Max Odds whenever possible, or would you double it more often flat betting $10?

Now, if your primary objective is just to play as long as possible, (or until bored) then that makes a very good case for just playing the Pass Line (or Don't Pass). However, it doesn't change the fact that the combined House Edge is less for total dollars exposed by Taking Odds.

I guess Wizard's advice assumes the person is there wanting to gamble, given the choice between no house edge and a house edge, the player would do well to take the former as often as possible and for as much as possible.

Myself, I don't look at it either way, to be honest. I just consider them strictly as separate bets. Combined edge is meaningless to me. One bet has a house edge of 1.41%, the other, 0%.

Thanks for the great explanation. Your last sentence is exactly how I look at it too. I like your explanation of trying to double a $500 bankroll. That’s a good scenario where you might want to take odds.

I just bought Gambling 102 the other day. I’m really enjoying it. I’m backing it up with the website when I want more detail. You’ll probably see me posting on the boards as I work my way through the book. I skipped ahead to the craps section because of a debate I had with a gambler.