I sometimes like to play the dark side in craps. My question is, if I want to wager a fixed amount of dollars, am I better off betting it all on don't pass on the come out roll, or am I better off putting the minimum on the come out roll and then betting the balance as odds once a point is established.

I know if I had to decide before the come out roll the minimum plus odds is the better bet. However, after surviving the come out roll, my edge - now in my favor - gets worse with odds. After surviving the come out, I think my edge is maximized by not adding odds. Yet, the minimum bet is normally less than I want to play.

Any advice would be appreciated.

Silverchip

yes on the latterQuote:silverchipI am new to the Forum. If my question has recently been asked, please point me to the correct post.

I sometimes like to play the dark side in craps. My question is, if I want to wager a fixed amount of dollars, am I better off betting it all on don't pass on the come out roll, or am I better off putting the minimum on the come out roll and then betting the balance as odds once a point is established.

uh ohQuote:I know if I had to decide before the come out roll the minimum plus odds is the better bet. However, after surviving the come out roll,

incorrect.Quote:... my edge - now in my favor - gets worse with odds.

Your confusion probably comes from other players and the dealers, who will point out that the free odds on the darkside pay less than even money, and that's automatically bad, they hint. Meanwhile those same dealers are hawking middle table bets, that the house is getting wealthy on, that all pay less than even money from the house perspectiveQuote:After surviving the come out, I think my edge is maximized by not adding odds. Yet, the minimum bet is normally less than I want to play.

Any advice would be appreciated.

Silverchip

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the reason to play free odds is the same reason rightside or darkside. And the same caveat applies.

* the free odds allow you to bet more against a lower overall house edge

* seemingly contradicting caveat: the free odds don't help you win more money

obviously this can be hard to wrap your mind around. The reason the caveat is true is because you are adding a bet that has no edge for player or the house. If you are not comfortable with that size of a bet, then there is no point in making it. However, you say you want a bigger bet. Well in that case it is plain stupid not to make that bigger amount be the free odds. I see darkside players at 10x odds tables put $100 on Don't Pass, with no intention of adding the free odds, when the minimum bet is $10. The expected value of that $10 is to lose about 14 cents, but $1.40 on the $100 bet. If he adds $100 in free odds behind a $10 bet instead, the expected loss is still 14 cents. This is undeniable math

One thing the naysayers are right about: you need to win those darkside bets. If you see that you are winning about half of the time on the darkside odds bets, you are probably feeling good about it while losing money. You need to win about two thirds of the time on them to actually be winning and not losing or just staying even

The overall house edge will be much lower if you bet, for instance, $25 plus $75 odds instead of $100 on the line because the 1.4% edge applies to $25 in the former and $100 in the latter

Incidentally, craps is a game of chance so there is no “strategy”

I may prefer 3X odds on the Don't then raise my line bet for that if I want to bet more, like raising from a $60 line & $180 odds to $90 line & $270 odds.

Quote:Ace2The issue is that you view the line + odds bet as two separate bets with different house edges at each stage. You have no control of when the bet is resolved (comeout or or point stage) so you must look at the overall edge

The overall house edge will be much lower if you bet, for instance, $25 plus $75 odds instead of $100 on the line because the 1.4% edge applies to $25 in the former and $100 in the latter

Incidentally, craps is a game of chance so there is no “strategy”

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So, if you play for about two hours, have you or anybody else ever seen a player walk away from the table just losing 1.41% of his bankroll?

tuttigym

Yeah, it’s hard to wrap your head around when you’re a dark sider, but do the math. Zero house edge is zero house edge. And that’s a beautiful thing.

When you play craps for a few hours, what’s your expectation and parameters around it?Quote:tuttigymQuote:Ace2The issue is that you view the line + odds bet as two separate bets with different house edges at each stage. You have no control of when the bet is resolved (comeout or or point stage) so you must look at the overall edge

The overall house edge will be much lower if you bet, for instance, $25 plus $75 odds instead of $100 on the line because the 1.4% edge applies to $25 in the former and $100 in the latter

Incidentally, craps is a game of chance so there is no “strategy”

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So, if you play for about two hours, have you or anybody else ever seen a player walk away from the table just losing 1.41% of his bankroll?

tuttigym

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Btw, house edge is based on units wagered not bankroll

Quote:ChumpChangeIf you make 50X $30 DP bets over 400 rolls or something, 1.4% of that is about $21. You could have made odds bets well in excess of your line bets and they won't add to the HA, but the variance will drown out any notion of the HA. You can compare what you estimate your HA will be and see how variance treated you compared to the HA. If you won $279, you could point to the $21 HA on the line bet as to why you didn't win $300.

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So what, the question before you is: Have you seen it done? The scenario of 50X30 DP bets over 400 rolls just doesn't happen in reality at the tables let alone provide that outcome shown.

tuttigym

Quote:Ace2When you play craps for a few hours, what’s your expectation and parameters around it?Quote:tuttigymQuote:Ace2The issue is that you view the line + odds bet as two separate bets with different house edges at each stage. You have no control of when the bet is resolved (comeout or or point stage) so you must look at the overall edge

The overall house edge will be much lower if you bet, for instance, $25 plus $75 odds instead of $100 on the line because the 1.4% edge applies to $25 in the former and $100 in the latter

Incidentally, craps is a game of chance so there is no “strategy”

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So, if you play for about two hours, have you or anybody else ever seen a player walk away from the table just losing 1.41% of his bankroll?

tuttigym

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Btw, house edge is based on units wagered not bankroll

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My expectations are irrelevant.

You among many others continue to say one cannot win at the game of craps because of the house edge. What is the "house edge" for the GAME of craps? Units wagered? At what? Everyone who plays should understand that each bet regardless of size has a risk. My question stands.

tuttigym

Quote:tuttigymQuote:ChumpChangeIf you make 50X $30 DP bets over 400 rolls or something, 1.4% of that is about $21. You could have made odds bets well in excess of your line bets and they won't add to the HA, but the variance will drown out any notion of the HA. You can compare what you estimate your HA will be and see how variance treated you compared to the HA. If you won $279, you could point to the $21 HA on the line bet as to why you didn't win $300.

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So what, the question before you is: Have you seen it done? The scenario of 50X30 DP bets over 400 rolls just doesn't happen in reality at the tables let alone provide that outcome shown.

tuttigym

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Taking a page out of my Roku BJ game that has me losing like never before, my total bet is just over $9 million and my total loss is just about $37K for an average loss of 0.411%. So what is the actual HA on this BJ game? Is it 0.75%? If it is, I'm still doing better than the HA. If the actual HA is 0.8222%, then I've cut my losses in half from the HA of $74K.

Odds bets in craps are mathematically simulated to lower the HA on the total bet amount. The more odds you bet, the lower the HA in relation to your total bet and you get figures below 1% and even down to 0.5%.

Quote:tuttigymQuote:Ace2

The overall house edge will be much lower if you bet, for instance, $25 plus $75 odds instead of $100 on the line because the 1.4% edge applies to $25 in the former and $100 in the latter

Incidentally, craps is a game of chance so there is no “strategy”

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So, if you play for about two hours, have you or anybody else ever seen a player walk away from the table just losing 1.41% of his bankroll?

tuttigym

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The only way that statement makes any sense is if the player stops playing after the sum of their bets equals their bankroll.

Part of the problem is, house edge - or, rather, a player's expected loss - is being thought of in terms of a percentage rather than an actual amount.

Betting 25 on DP and then planning on betting 75 on odds (well, 30 and 90 might be a better example, as DP odds only pay out correctly every time if the bet is a multiple of 6) has an expected loss of 25 x 1.414% + 75 x 0% whether you planned on betting odds before the comeout roll or decided to wait until after the roll before doing so.

We have had this discussion before, but usually with "Martingale" and "roulette" thrown in.

Quote:ThatDonGuyQuote:tuttigymQuote:Ace2

The overall house edge will be much lower if you bet, for instance, $25 plus $75 odds instead of $100 on the line because the 1.4% edge applies to $25 in the former and $100 in the latter

Incidentally, craps is a game of chance so there is no “strategy”

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So, if you play for about two hours, have you or anybody else ever seen a player walk away from the table just losing 1.41% of his bankroll?

tuttigym

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The only way that statement makes any sense is if the player stops playing after the sum of their bets equals their bankroll.

Part of the problem is, house edge - or, rather, a player's expected loss - is being thought of in terms of a percentage rather than an actual amount.

Betting 25 on DP and then planning on betting 75 on odds (well, 30 and 90 might be a better example, as DP odds only pay out correctly every time if the bet is a multiple of 6) has an expected loss of 25 x 1.414% + 75 x 0% whether you planned on betting odds before the comeout roll or decided to wait until after the roll before doing so.

We have had this discussion before, but usually with "Martingale" and "roulette" thrown in.

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Don: I really respect your mathematical abilities. They far exceed my own by light years. The continuous touting of a miniscule unapproachable "house edge" at the table does not equate with reality, and somehow because of those preachings, players believe bankrolls can last a long time or they can actually overcome such a small margin.

tuttigym

Quote:ChumpChangeQuote:tuttigymQuote:ChumpChangeIf you make 50X $30 DP bets over 400 rolls or something, 1.4% of that is about $21. You could have made odds bets well in excess of your line bets and they won't add to the HA, but the variance will drown out any notion of the HA. You can compare what you estimate your HA will be and see how variance treated you compared to the HA. If you won $279, you could point to the $21 HA on the line bet as to why you didn't win $300.

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So what, the question before you is: Have you seen it done? The scenario of 50X30 DP bets over 400 rolls just doesn't happen in reality at the tables let alone provide that outcome shown.

tuttigym

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Taking a page out of my Roku BJ game that has me losing like never before, my total bet is just over $9 million and my total loss is just about $37K for an average loss of 0.411%. So what is the actual HA on this BJ game? Is it 0.75%? If it is, I'm still doing better than the HA. If the actual HA is 0.8222%, then I've cut my losses in half from the HA of $74K.

Odds bets in craps are mathematically simulated to lower the HA on the total bet amount. The more odds you bet, the lower the HA in relation to your total bet and you get figures below 1% and even down to 0.5%.

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So, is that what you do at the real table? You wager large sums only to lose 1% or less? Somehow, I really have a hard time finding that factual. Now playing with your ROKU with a phantom monopoly bankroll might work out for you, but it hardly represents what most might do with $9 mil.

tuttigym

Quote:ChumpChangeI was over $100K ahead on the Roku game previously, and if I made a comeback tonight from a $2K bankroll to $100K and tacked on another $1 million in bets, I'd be up 1% on $10 million.

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ROKU for the ages. You go dude! Don't spend it all in one place.

tuttigym

So far so good. Does that mean that, if I am playing a single dark side bet, I should just bet the minimum and let it go until it resolves? Is that also true if I play the same thing over and over until leaving?

Siverchip

I believe you’re Tuttigym trolling. Who else would ask the same dumb questions repeatedly and sign his name at the end of every postQuote:silverchipWell to clarify, as far as I can tell, the house edge on don't bets arises from the come out rolls. Just on the come out roll, the house edge, I think, is about 14%. However, once the come out roll is survived, the edge shifts to me, about 12.5% I think. So what that means is any don't odds bet placed - even at fair odds - reduces my conditional edge towards zero. The implication is that given I survive the come-out, I shouldn't place an odds bet - just the opposite of pass line odds plays.

So far so good. Does that mean that, if I am playing a single dark side bet, I should just bet the minimum and let it go until it resolves? Is that also true if I play the same thing over and over until leaving?

Siverchip

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Quote:Ace2I believe you’re Tuttigym trolling. Who else would ask the same dumb questions repeatedly and sign his name at the end of every postQuote:silverchipWell to clarify, as far as I can tell, the house edge on don't bets arises from the come out rolls. Just on the come out roll, the house edge, I think, is about 14%. However, once the come out roll is survived, the edge shifts to me, about 12.5% I think. So what that means is any don't odds bet placed - even at fair odds - reduces my conditional edge towards zero. The implication is that given I survive the come-out, I shouldn't place an odds bet - just the opposite of pass line odds plays.

So far so good. Does that mean that, if I am playing a single dark side bet, I should just bet the minimum and let it go until it resolves? Is that also true if I play the same thing over and over until leaving?

Siverchip

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Wrong again, Ace. Check with the Wizard. I am sure he could tell you that the thread and post are not mine.

Hey Ace, you want to tell us what the HE/HA is for the entire game of craps? Does that number even exist?

tuttigym

Quote:silverchipWell to clarify, as far as I can tell, the house edge on don't bets arises from the come out rolls. Just on the come out roll, the house edge, I think, is about 14%. However, once the come out roll is survived, the edge shifts to me, about 12.5% I think. So what that means is any don't odds bet placed - even at fair odds - reduces my conditional edge towards zero. The implication is that given I survive the come-out, I shouldn't place an odds bet - just the opposite of pass line odds plays.

So far so good. Does that mean that, if I am playing a single dark side bet, I should just bet the minimum and let it go until it resolves? Is that also true if I play the same thing over and over until leaving?

Siverchip

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No, and here's why.

Suppose the point is 4, and, to simplify matters, let's say your initial bet is 10 and your odds bet is 30.

Taking just the DC bet into account, you now have a 2/3 chance of winning 10 and a 1/3 chance of losing 10, so the player edge = (10 x 1/3) / 10 = 1/3, or 33.33%.

Taking both bets into account, you have a 2/3 chance of winning 10 + 15 = 25 and a 1/3 chance of losing 10 + 30 = 40, so the player edge = (2/3 x 25 - 1/3 x 40) / 40 = 1/12, or about 8.33%.

However, while the player edge on the DC bet has reduced from 33.33% to 8.33%, the player edge on the odds bets has increased from 0% to 8.33% when you combine them.

Notice that 25 x 8 1/3 % + 75 x 8 1/3 % = 25 x 33 1/3 %. How about that!

Quote:ThatDonGuyQuote:silverchip

So far so good. Does that mean that, if I am playing a single dark side bet, I should just bet the minimum and let it go until it resolves? Is that also true if I play the same thing over and over until leaving?

Siverchip

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No, and here's why.

Suppose the point is 4, and, to simplify matters, let's say your initial bet is 10 and your odds bet is 30.

Taking just the DC bet into account, you now have a 2/3 chance of winning 10 and a 1/3 chance of losing 10, so the player edge = (10 x 1/3) / 10 = 1/3, or 33.33%.

Taking both bets into account, you have a 2/3 chance of winning 10 + 15 = 25 and a 1/3 chance of losing 10 + 30 = 40, so the player edge = (2/3 x 25 - 1/3 x 40) / 40 = 1/12, or about 8.33%.

However, while the player edge on the DC bet has reduced from 33.33% to 8.33%, the player edge on the odds bets has increased from 0% to 8.33% when you combine them.

Notice that 25 x 8 1/3 % + 75 x 8 1/3 % = 25 x 33 1/3 %. How about that!

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Don, Help me with this, please. I cannot get past the fact that there are 6 ways to lose the 4 + odds and only 3 ways to win those bets. How is there only a 1/3 chance of winning when it is clearly 2 to 1 against? The PL is 8 ways to win and 4 ways to lose or a 2 to 1 PL advantage, right? That is how my mind works.

tuttigym

Quote:tuttigymQuote:ThatDonGuyQuote:silverchip

So far so good. Does that mean that, if I am playing a single dark side bet, I should just bet the minimum and let it go until it resolves? Is that also true if I play the same thing over and over until leaving?

Siverchip

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No, and here's why.

Suppose the point is 4, and, to simplify matters, let's say your initial bet is 10 and your odds bet is 30.

Taking just the DC bet into account, you now have a 2/3 chance of winning 10 and a 1/3 chance of losing 10, so the player edge = (10 x 1/3) / 10 = 1/3, or 33.33%.

Taking both bets into account, you have a 2/3 chance of winning 10 + 15 = 25 and a 1/3 chance of losing 10 + 30 = 40, so the player edge = (2/3 x 25 - 1/3 x 40) / 40 = 1/12, or about 8.33%.

However, while the player edge on the DC bet has reduced from 33.33% to 8.33%, the player edge on the odds bets has increased from 0% to 8.33% when you combine them.

Notice that 25 x 8 1/3 % + 75 x 8 1/3 % = 25 x 33 1/3 %. How about that!

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Don, Help me with this, please. I cannot get past the fact that there are 6 ways to lose the 4 + odds and only 3 ways to win those bets. How is there only a 1/3 chance of winning when it is clearly 2 to 1 against? The PL is 8 ways to win and 4 ways to lose or a 2 to 1 PL advantage, right? That is how my mind works.

tuttigym

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Easy. He's betting Don't, so the odds being placed are against making the point.

The percentages seem to be wrong, just for the recordQuote:silverchipWell to clarify, as far as I can tell, the house edge on don't bets arises from the come out rolls. Just on the come out roll, the house edge, I think, is about 14%. However, once the come out roll is survived, the edge shifts to me, about 12.5% I think.

But let's put this to rest. Of course if there was a way to decrease the overall edge against you at the second stage it would be right to do. Or, not make it worse, it would be right to do. But the fact is there is nothing you can do to take you to either direction in terms of expected value of the bet. So you are off on the wrong track here.

This is where right thinking is to say the edge as a percent doesn't matter. As a Craps player, you can use the free odds to change the edge percent of the new overall size of the combined bet, but you can't change the expected value. So for the type of evaluation you are trying to make, the percentages just mislead.Quote:So what that means is any don't odds bet placed - even at fair odds - reduces my conditional edge towards zero.

I think you are listening to the other players and the dealers and trying to make sense of what they are saying.Quote:The implication is that given I survive the come-out, I shouldn't place an odds bet - just the opposite of pass line odds plays.

I can tell you are not going to come out of this until you understand it about the expected value of a bet, which is bet times edge, and how that doesn't change. You can look at percentages of edge change when you add free odds and get totally confused, as you indeed have. Others have decided the free odds are a waste of time, and they are not necessarily wrong. In the long run, a player making odds bets has an expected value of zero on themQuote:So far so good. Does that mean that, if I am playing a single dark side bet, I should just bet the minimum and let it go until it resolves? Is that also true if I play the same thing over and over until leaving?

Siverchip

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Take this to the bank, though, as I have seen this many times. The darkside player is losing and he asks why, and a trusted dealer tells him 'you can't win because odds pay less than even money'. So that player starts putting down the size bet he likes to make, and it's, say, $100 on a $25 table, and he adds no odds when it travels.

This is going to be you. Can you see that? The logic of "don't add free odds" leads to this. Can you see this is idiotic? It's not idiotic because he didn't add odds to a $100 bet, it's idiotic because he didn't bet $25 at first and then add odds to make up the $100 he wants to bet.

Quote:tuttigymSo, if you play for about two hours, have you or anybody else ever seen a player walk away from the table just losing 1.41% of his bankroll?

Betting only the Pass Line and Come, they are expected to lose 1.41% of the total amount bet. Not 1.41% of their bankroll.

OK, I think I get it. Assume a $25 minimum. Adding odds to my $25 don't line bet lowers my edge (having survived the come-out), but does not change the expected value - who cares about the edge. So, from an expected value point of view, once the come-out is survived, it makes no difference whether I add odds. Adding odds does increase the variance. If I don't want to add variance then I shouldn't add odds. That's the reason, not because it changes the edge. Of course, if I don't want variance, why am I playing craps at all since it is a negative expected value game.

It is also now apparent that adding to the line (betting $100 instead of $25) is an inferior play. It is making "4" negative expected value bets instead of 1 negative expected value bets and 3 zero expected value bets. If for some reason I want to make $100 bets each resolution it is better to bet $25 on the line with 3x odds rather than to multiply the initial bet. But, another place where I was thinking wrong is the idea that I have to make $100 bets. If everything is random, it is better just to make $25 bets.

Thank you for the great answer.

Silverchip

EV = 1 - edge. You can’t change edge without changing EV.Quote:silverchipAdding odds to my $25 don't line bet lowers my edge (having survived the come-out), but does not change the expected value - who cares about the edge.

Betting $25 flat + $150 odds, you’ll have lower edge and higher EV than betting $175 flat. Much lower edge…81% lower relatively

The standard deviation of a $175 flat bet is $175. The standard deviation of $25 flat plus $150 odds is $123. So playing DP plus 6x odds decreases variance significantly.Quote:silverchip

OK, I think I get it.

Adding odds does increase the variance. If I don't want to add variance then I shouldn't add odds.]

Quote:Ace2The standard deviation of a $175 flat bet is $175. The standard deviation of $25 flat plus $150 odds is $123. So playing DP plus 6x odds decreases variance significantly.Quote:silverchip

OK, I think I get it.

Adding odds does increase the variance. If I don't want to add variance then I shouldn't add odds.]

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The comparison is $25 flat bet vs $25 + $150 odds. Same EV, greater variance.

You are welcome. And you don't have to make $100 bets if you are happy with $25, certainly not!Quote:silverchipOdiousgambit:

OK, I think I get it. Assume a $25 minimum. Adding odds to my $25 don't line bet lowers my edge (having survived the come-out), but does not change the expected value - who cares about the edge. So, from an expected value point of view, once the come-out is survived, it makes no difference whether I add odds. Adding odds does increase the variance. If I don't want to add variance then I shouldn't add odds. That's the reason, not because it changes the edge. Of course, if I don't want variance, why am I playing craps at all since it is a negative expected value game.

It is also now apparent that adding to the line (betting $100 instead of $25) is an inferior play. It is making "4" negative expected value bets instead of 1 negative expected value bets and 3 zero expected value bets. If for some reason I want to make $100 bets each resolution it is better to bet $25 on the line with 3x odds rather than to multiply the initial bet. But, another place where I was thinking wrong is the idea that I have to make $100 bets. If everything is random, it is better just to make $25 bets.

Thank you for the great answer.

Silverchip

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I see you know about variance, I was reluctant to bring that up. That's one problem with the $25 bet with no odds. The variance is quite low, it's boring really

but it is pointed out all the time that the edge *percentage* goes down from @1.4% to @0.37% when maxing 3x4x5x free odds [rightside]. So I think you are adding confusion to things here.Quote:Ace2EV = 1 - edge. You can’t change edge without changing EV.Quote:silverchipAdding odds to my $25 don't line bet lowers my edge (having survived the come-out), but does not change the expected value - who cares about the edge.

with the latter remark it makes it seem you are contradicting yourself.Quote:Betting $25 flat + $150 odds, you’ll have lower edge and higher EV than betting $175 flat. Much lower edge…81% lower relatively

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Silverchip merely said it slightly wrong, I think he has the idea right now

Not comparable. Would be similar to comparing the results of a $100 blackjack player to those of a $500 player. Betting volume and variance differ by factor of 5. Different leaguesQuote:unJonQuote:Ace2The standard deviation of a $175 flat bet is $175. The standard deviation of $25 flat plus $150 odds is $123. So playing DP plus 6x odds decreases variance significantly.Quote:silverchip

OK, I think I get it.

Adding odds does increase the variance. If I don't want to add variance then I shouldn't add odds.]

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The comparison is $25 flat bet vs $25 + $150 odds. Same EV, greater variance.

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how so ?Quote:odiousgambitwith the latter remark it makes it seem you are contradicting yourself.

well sir, that's another reason not to bet $175 on the DP when the lower min and the odds are available ... hadn't thought of thatQuote:Ace2Quote:silverchip

OK, I think I get it.

Adding odds does increase the variance. If I don't want to add variance then I shouldn't add odds.]

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Quote:Ace2Not comparable. Would be similar to comparing the results of a $100 blackjack player to those of a $500 player. Betting volume and variance differ by factor of 5. Different leaguesQuote:unJonQuote:Ace2Quote:silverchip

OK, I think I get it.

Adding odds does increase the variance. If I don't want to add variance then I shouldn't add odds.]

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The comparison is $25 flat bet vs $25 + $150 odds. Same EV, greater variance.

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It’s comparable. We are comparing it. And it’s the right comparison in showing the change of variance with EV fixed.

I dont think I'm going to catch you with a technically incorrect remarkQuote:Ace2how so ?Quote:odiousgambitwith the latter remark it makes it seem you are contradicting yourself.

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what are you trying to argue anyway? that all the players and dealers who say 'don't play the free odds darkside' are right ? I don't think so, but you are nit-picking everything even when the OP is finally "getting it"

Quote:Ace2The standard deviation of $25 flat plus $150 odds is $123. So playing DP plus 6x odds decreases variance significantly.

Looks like the SD for a 3/4/5 odds game. 4.9156 x 25 = $122.89

Using mustangslly's table, 6x SD is 6.817. x 25 = $170.42

Which is only a slight improvement over $175 no odds..

$170.42 is for game that allows 6x odds on the pass and 12/9/7.2 odds on DP. Irrelevant and you’ll never find those odds anywhereQuote:TankoQuote:Ace2The standard deviation of $25 flat plus $150 odds is $123. So playing DP plus 6x odds decreases variance significantly.

Looks like the SD for a 3/4/5 odds game. 4.9156 x 25 = $122.89

Using mustangslly's table, 6x SD is 6.817. x 25 = $170.42

Which is only a slight improvement over $175 no odds..

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3/4/5 on pass is 6x on DP.

Quote:Ace2$170.42 is for game that allows 6x odds on the pass and 12/9/7.2 odds on DP. Irrelevant and you’ll never find those odds anywhereQuote:TankoQuote:Ace2The standard deviation of $25 flat plus $150 odds is $123. So playing DP plus 6x odds decreases variance significantly.

Looks like the SD for a 3/4/5 odds game. 4.9156 x 25 = $122.89

Using mustangslly's table, 6x SD is 6.817. x 25 = $170.42

Which is only a slight improvement over $175 no odds..

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3/4/5 on pass is 6x on DP.

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Hey Ace: Which is the better wager: PL + 3X odds or Place Bet on the point for the same $$$ amount??

tuttigym

I know…house edge is irrelevant

Quote:Ace2The edge of any place bet is higher than the PL bet edge with or without odds

I know…house edge is irrelevant

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Answer the question: Which bet is BETTER?

Being fixated on "edge," then what is the "edge" on these two bets individually?

tuttigym

https://www.goldentouchcraps.com/article.php?p=stickman_015.html

So I've been betting the DC on the 4 count since the point got established (at the tables), trying to skip the 2nd & 3rd rolls where random rollers typically throw a 7 and is a winner for the DP bets. I'm hoping that there will still be a 57% chance they will 7-out during the next 5 rolls after I get my DC point established. But if 4 out of 7 shooters aren't even going to get past 5 rolls, I'll miss a lot of shooters that I could have picked up if I bet the DP. It's just that they may throw a bunch of 7's before the point gets established and make my DP a loser. That's why I would bet lower on the come-out on the Don'ts and then add odds, because I'm disadvantaged on the come-out. I have a 2/9ths chance of losing my DC come-out roll and 1/9th chance of winning my DC come-out roll. Non-tourist rollers may be throwing a lot more 7's on the come-out rolls making my odds of losing a DP bet higher on the come-out.

I don't know that the 57% chance goes down for non-tourist rollers.

Now if I have to bet on the shooter before me in order to qualify to shoot when it's my turn, I'll have to bet the DP because they may not make it to my 4 count to place my DC bet.

The place bet is BETTER because you only have to roll the point ONCE. To win a PL bet you must roll the point number TWICEQuote:tuttigymQuote:Ace2The edge of any place bet is higher than the PL bet edge with or without odds

I know…house edge is irrelevant

link to original post

Answer the question: Which bet is BETTER?

Being fixated on "edge," then what is the "edge" on these two bets individually?

tuttigym

link to original post

Quote:Ace2The place bet is BETTER because you only have to roll the point ONCE. To win a PL bet you must roll the point number TWICEQuote:tuttigymQuote:Ace2The edge of any place bet is higher than the PL bet edge with or without odds

I know…house edge is irrelevant

link to original post

Answer the question: Which bet is BETTER?

Being fixated on "edge," then what is the "edge" on these two bets individually?

tuttigym

link to original post

link to original post

Ace, you surprise me. My question was worded improperly in that I was looking for an answer that regarded the two bets as equals. So, while your answer is technically correct, the PL number being rolled twice to win, the real answer is still out there. I will rephrase. After the point is established, I, player B did not play the PL, play the point number as a Place Bet equal in $$ to the PL + 3X odds. Which bet is better?

Does the "house edge" give a player any real knowledge as to the ability to win a bet? I will try to be specific. Big Red is a pretty bad bet. We both agree on that. Can you tell me what the "house edge" is on Big Red? Based on your answer, I will ask a follow-up.

tuttigym

Quote:Ace2Non-tourist rollers make it past the 5-count 81.3% of the time

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I love that answer. With that in mind, can you tell us all how we can make that distinction?

tuttigym

Quote:ChumpChange57% of the rolls by a random roller would not make it past the five-count.

https://www.goldentouchcraps.com/article.php?p=stickman_015.html

So I've been betting the DC on the 4 count since the point got established (at the tables), trying to skip the 2nd & 3rd rolls where random rollers typically throw a 7 and is a winner for the DP bets. I'm hoping that there will still be a 57% chance they will 7-out during the next 5 rolls after I get my DC point established. But if 4 out of 7 shooters aren't even going to get past 5 rolls, I'll miss a lot of shooters that I could have picked up if I bet the DP. It's just that they may throw a bunch of 7's before the point gets established and make my DP a loser. That's why I would bet lower on the come-out on the Don'ts and then add odds, because I'm disadvantaged on the come-out. I have a 2/9ths chance of losing my DC come-out roll and 1/9th chance of winning my DC come-out roll. Non-tourist rollers may be throwing a lot more 7's on the come-out rolls making my odds of losing a DP bet higher on the come-out.

I don't know that the 57% chance goes down for non-tourist rollers.

Now if I have to bet on the shooter before me in order to qualify to shoot when it's my turn, I'll have to bet the DP because they may not make it to my 4 count to place my DC bet.

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Hey Chump, Like almost all the stuff coming from GT, it is pure horse hockey. You should first note that the "research" was done by computer simulations which means to me that the accuracy is less than reliable.

Your play on the ROKU is just about as accurate.

tuttigym

Quote:tuttigymQuote:Ace2The place bet is BETTER because you only have to roll the point ONCE. To win a PL bet you must roll the point number TWICEQuote:tuttigymQuote:Ace2

I know…house edge is irrelevant

link to original post

Answer the question: Which bet is BETTER?

Being fixated on "edge," then what is the "edge" on these two bets individually?

tuttigym

link to original post

link to original post

Ace, you surprise me. My question was worded improperly in that I was looking for an answer that regarded the two bets as equals. So, while your answer is technically correct, the PL number being rolled twice to win, the real answer is still out there. I will rephrase. After the point is established, I, player B did not play the PL, play the point number as a Place Bet equal in $$ to the PL + 3X odds. Which bet is better?

Does the "house edge" give a player any real knowledge as to the ability to win a bet? I will try to be specific. Big Red is a pretty bad bet. We both agree on that. Can you tell me what the "house edge" is on Big Red? Based on your answer, I will ask a follow-up.

tuttigym

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Bold added.

You consistently want to compare a PL bet after the point is established to a PB. It doesn’t work that way, and the odds the PB gives you is not quite as good as the extra value you get from the PL on the come out.

There is no reason to make a place bet since many properties charge vig on win only for buy bets.

And there is no reason to make a buy bet when you can make a PL bet plus odds