May 18th, 2017 at 8:20:35 PM
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So I want to play Switzerland. $5 don't pass and $5 pass. Then playing the odds. How would you go about calculating the house edge?

May 18th, 2017 at 8:43:01 PM
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Pass and don't pass: 0.5/36

Odds: 0

Odds: 0

"should of played 'Go Fish' today ya peasant" -typoontrav

May 19th, 2017 at 3:23:09 AM
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Each bet has a house edge, they do not cancel each other out. You are just doubling your expected loss by betting twice as much.Quote:alphastormSo I want to play Switzerland. $5 don't pass and $5 pass.

The expected loss is not changed by playing the odds. Maybe since the whole thing is wrong-minded I've never seen anyone try to calculate house edge - it's one of those times it's better to look at EV instead of HE. BTW it is hard to find what you mean by "switzerland" by using google, so I'm guessing it's the usual scheme.Quote:Then playing the odds. How would you go about calculating the house edge?

It's wrong-minded because of point one above, not canceling out, but also because it doesn't do what is usually wanted with hedging - reducing variance. Sure, the variance is reduced on the line bets - to a guaranteed win for the casino each time a 12 is rolled [otherwise push]. But the variance is very low already for these bets, mind-numbingly low in fact. But for the odds bet [you have to do one side only] which is where all the variance is, it is unaffected.

Just don't fail to realize this is just hedging, a dumb way to gamble.

"Baccarat is a game whereby the croupier gathers in money with a flexible sculling oar, then rakes it home. If I could have borrowed his oar I would have stayed." .......... Mark Twain

May 19th, 2017 at 5:40:52 AM
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Quote:odiousgambit

Thanks for your reply. I wasn't asking if it's a good play. Was asking a math question.

Let's say I play $5 pass and $5 don't pass. The point is 6. I then add $25 odds bet to the pass line. What would the house edge be? Would you just add the 1.41 percent x 2 and then subtract the zero edge from odds bet? So the house edge is ~2.8%. I know that is somehow wrong. Just wanting an explanation.

May 19th, 2017 at 5:59:01 AM
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Quote:alphastormThanks for your reply. I wasn't asking if it's a good play. Was asking a math question.

Let's say I play $5 pass and $5 don't pass. The point is 6. I then add $25 odds bet to the pass line. What would the house edge be? Would you just add the 1.41 percent x 2 and then subtract the zero edge from odds bet? So the house edge is ~2.8%. I know that is somehow wrong. Just wanting an explanation.

I'm sure I'll get corrected if I have this wrong, and i would appreciate that if necessary.

12 rolled is 1 in 36 chances. This roll loses on the pass line, pushes ("bar 12" on the felt) on the don't pass. All other rolls pay one or the other at even money, for a wash. So, as RS says above, the HE IS 0.5/36, or 1.3889% no matter what you do, as long as both bets are equal and played to resolution.

The odds don't figure into the HE calculation. They increase the volatility of your bet, by exposing more money to the variance of the dice, but since they pay at true odds, their long term effect on the HE (as HE is defined) is zero.

So, I'm repeating what RS said above, but with a bit more context. Hope that helps.

"If the house lost every hand, they wouldn't deal the game."

May 19th, 2017 at 11:34:26 AM
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Amazing how many people don't get the ".5" part. And try to tell us that the doey/don't has a %2.8+ advantage for the house. A smart house will rate the player at $0, and many of them do.

May 19th, 2017 at 11:39:23 AM
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While they are independent bets and don't exactly "cancel" each other out... You're only losing money on the come out throw to a 12.

Thus, EV(come out roll) = Bet*(-1/36)... so if your bet is $5, then your EV on the come out roll = 5*(-1/36) = -14 cents. (don't forget to compare this to a regular line bet... say DP... 5*(-.0136) = -.07 cents... so you're almost doubling your expected loss on the line bet to do this).

After the come out your odds bets have zero house edge, and thus zero expected value in the long run.

Thus, EV(come out roll) = Bet*(-1/36)... so if your bet is $5, then your EV on the come out roll = 5*(-1/36) = -14 cents. (don't forget to compare this to a regular line bet... say DP... 5*(-.0136) = -.07 cents... so you're almost doubling your expected loss on the line bet to do this).

After the come out your odds bets have zero house edge, and thus zero expected value in the long run.

Playing it correctly means you've already won.

May 19th, 2017 at 5:56:37 PM
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So what I don't understand is that on wizards site he states that when you combine your bet with the odds bet, it will bring down the overall house edge. So the pass line odds of 3X-4X-5X has a house edge of 0.00111 but 100X is 0.00006. Why would this not factor into a pass + don't pass play when taking odds?

https://wizardofodds.com/games/craps/basics/#toc-TheOdds

https://wizardofodds.com/games/craps/basics/#toc-TheOdds

May 19th, 2017 at 6:47:35 PM
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Quote:alphastormSo what I don't understand is that on wizards site he states that when you combine your bet with the odds bet, it will bring down the overall house edge. So the pass line odds of 3X-4X-5X has a house edge of 0.00111 but 100X is 0.00006. Why would this not factor into a pass + don't pass play when taking odds?

https://wizardofodds.com/games/craps/basics/#toc-TheOdds

We had this discussion at some length a few months ago. I think it's more a question of phrasing than of math. You can't get a zero HE odds bet without placing a p/dp bet that has an HE, and that's also required for place, buy, and come bets.

There is an effective reduction in the casino's edge over you when placing odds, and the reduction is proportionate to the amount of odds you bet. But the HE of any individual bet does not change. The percentage of your bet that is exposed to a HE is.

I think it is mostly for convenience and comprehension that he explains it the way he does on that page. It reinforces the better choice between, say, putting 30 on the pass line vs. 5 on the pass line and 25 behind. If it loses, you lose the same 30. But if you win, the first pays even money; the second pays more.

"If the house lost every hand, they wouldn't deal the game."

May 20th, 2017 at 4:24:07 AM
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0.5/36 is perfectly acceptable if you want to think of passline and don't pass wagers as inseparable entities.

However, if you'd like an answer which better represents the combined expectation of both wagers, allows a fair comparison with other wagers, and doesn't take you down the path of making combined wager calculation errors then I beg to differ.

Everyone acknowledges that Don't Pass pushes on 12 yet they make the error of counting it in the action. This leads to the following EVs:

Passline -28/1980 = -0.0141414

Don't Pass -27/1980 = -0.1363636

Combined = -55/3960 = -0.01388889 (note that -55/3960 = -0.5/36)

However, for every 36 Passline resolutions there are only 35 Don't Pass resolutions. So correctly accounting for the action in the denominator leads to the following EVs:

Passline -28/1980 = -0.0141414

Don't Pass -27/1925 = -0.0140259

Combined -55/3905 = -0.0140845

These are the typical EVs you've likely seen published since you were knee-high to a stickman.

Truth be told, the difference between -0.01388889 and -0.0140845 is too small to worry about, but as I mentioned, it's the idea of correctly accounting for combined bets that matters because doing so incorrectly here can lead to erroneous conclusions about other combinations.

Allow me to quote a passage from my help file where I give an example of combined bets and talk about correctly accounting for the action.

This concept is vital when considering multiple, simultaneous bets. For instance, occasionally you'll come across someone who'll try to convince you that the expected value of some combination of bets is actually lower than any of the included bets individually (a synergistic effect, so to speak). One such argument claims that a combined bet on Place 5, Place 6, Place 8, & Place 9 loses only 1.136% per decision. Using an example of $5 Place 5, $6 Place 6, $6 Place 8, & $5 Place 9 (total of $22 in wagers) the reasoning is this: There are 18 ways to win $7 and 6 ways to lose $22 for a total of 24 decisions. Hence, the EV is [18/24 * $7] + [6/24 * -$22] = -$0.25 per decision. Divide by the amount wagered and the EV% becomes -$0.25/$22 = -1.136% per decision. That's lower than the individual EV of the any of the included bets!

Well, guess what? Craps bet are not synergistic. The erroneous conclusion of the argument above comes from not expressing the EV as a function of the action. When one of the included bets wins, the amount won is solely a function of the amount that was wagered on the winning bet and not on the other three bets. For instance, if a 9 hits then the $7 win is a function of the $5 that was wagered on Place 9 and not the $17 on the other three bets. Yet the argument treats the full amount wagered on all four bets as if it had a bearing on each winning outcome (i.e. by dividing the average loss per decision by the full amount wagered.) This is wrong. Intuitively, since the Place 5 & 9 each have EV's of -4.0% per decision and the Place 6 & 8 each have EV's of -1.52% per decision, we know that the combined wagers should have a weighted EV which lies somewhere between -1.52% and -4.0%. To arrive at this figure we simply need to divide our average loss per decision by the average action per decision. The average action is figured as follows: There's an 8/24 chance of rolling a 5 or 9 and producing $5 worth of action, a 10/24 chance of rolling a 6 or 8 and producing $6 worth of action, and a 6/24 chance of rolling a 7 and producing $22 worth of action. Hence, action = [8/24 *$5] + [10/24 * $6] + [6/24 * $22] = $9.67 per decision. Now divide -$0.25 by $9.67 and the average loss per dollar of action is 2.59%.

Over time I've seen others here make similar claims that certain combined bets would produce lower EVs than any of the included bets alone, yet they all used the same erroneous reasoning.

So, I contend that combined equally sized Passline and Don't Pass wagers have an EV of -1.40845%

Now the question (and never-ending furball) about odds. It boggles my mind to see folks continually claim that odds don't matter because their EV are 0%. But odds DO MATTER!

We've just seen here that combining Passline and Don't Pass bets yields an EV (a weighted average) that lies somewhere between the two individual bets (no matter which method you use). And so it is with odds: Combining 0% odds together with a -1.41% Passline bet yields a weighted average that lies somewhere between the values of 0% and -1.41% depending on how much odds you take.

It may be an interesting fact but it's just not a fair comparison to say that a $5 line bettor loses the same actual amount as a $5 line bettor who also plays odds. The two players do not wager the same amount. They don't have the same handle. They don't have the same amount of action. And that's why we need to look at the percentage lost per dollar of action.

Steen

However, if you'd like an answer which better represents the combined expectation of both wagers, allows a fair comparison with other wagers, and doesn't take you down the path of making combined wager calculation errors then I beg to differ.

Everyone acknowledges that Don't Pass pushes on 12 yet they make the error of counting it in the action. This leads to the following EVs:

Passline -28/1980 = -0.0141414

Don't Pass -27/1980 = -0.1363636

Combined = -55/3960 = -0.01388889 (note that -55/3960 = -0.5/36)

However, for every 36 Passline resolutions there are only 35 Don't Pass resolutions. So correctly accounting for the action in the denominator leads to the following EVs:

Passline -28/1980 = -0.0141414

Don't Pass -27/1925 = -0.0140259

Combined -55/3905 = -0.0140845

These are the typical EVs you've likely seen published since you were knee-high to a stickman.

Truth be told, the difference between -0.01388889 and -0.0140845 is too small to worry about, but as I mentioned, it's the idea of correctly accounting for combined bets that matters because doing so incorrectly here can lead to erroneous conclusions about other combinations.

Allow me to quote a passage from my help file where I give an example of combined bets and talk about correctly accounting for the action.

This concept is vital when considering multiple, simultaneous bets. For instance, occasionally you'll come across someone who'll try to convince you that the expected value of some combination of bets is actually lower than any of the included bets individually (a synergistic effect, so to speak). One such argument claims that a combined bet on Place 5, Place 6, Place 8, & Place 9 loses only 1.136% per decision. Using an example of $5 Place 5, $6 Place 6, $6 Place 8, & $5 Place 9 (total of $22 in wagers) the reasoning is this: There are 18 ways to win $7 and 6 ways to lose $22 for a total of 24 decisions. Hence, the EV is [18/24 * $7] + [6/24 * -$22] = -$0.25 per decision. Divide by the amount wagered and the EV% becomes -$0.25/$22 = -1.136% per decision. That's lower than the individual EV of the any of the included bets!

Well, guess what? Craps bet are not synergistic. The erroneous conclusion of the argument above comes from not expressing the EV as a function of the action. When one of the included bets wins, the amount won is solely a function of the amount that was wagered on the winning bet and not on the other three bets. For instance, if a 9 hits then the $7 win is a function of the $5 that was wagered on Place 9 and not the $17 on the other three bets. Yet the argument treats the full amount wagered on all four bets as if it had a bearing on each winning outcome (i.e. by dividing the average loss per decision by the full amount wagered.) This is wrong. Intuitively, since the Place 5 & 9 each have EV's of -4.0% per decision and the Place 6 & 8 each have EV's of -1.52% per decision, we know that the combined wagers should have a weighted EV which lies somewhere between -1.52% and -4.0%. To arrive at this figure we simply need to divide our average loss per decision by the average action per decision. The average action is figured as follows: There's an 8/24 chance of rolling a 5 or 9 and producing $5 worth of action, a 10/24 chance of rolling a 6 or 8 and producing $6 worth of action, and a 6/24 chance of rolling a 7 and producing $22 worth of action. Hence, action = [8/24 *$5] + [10/24 * $6] + [6/24 * $22] = $9.67 per decision. Now divide -$0.25 by $9.67 and the average loss per dollar of action is 2.59%.

Over time I've seen others here make similar claims that certain combined bets would produce lower EVs than any of the included bets alone, yet they all used the same erroneous reasoning.

So, I contend that combined equally sized Passline and Don't Pass wagers have an EV of -1.40845%

Now the question (and never-ending furball) about odds. It boggles my mind to see folks continually claim that odds don't matter because their EV are 0%. But odds DO MATTER!

We've just seen here that combining Passline and Don't Pass bets yields an EV (a weighted average) that lies somewhere between the two individual bets (no matter which method you use). And so it is with odds: Combining 0% odds together with a -1.41% Passline bet yields a weighted average that lies somewhere between the values of 0% and -1.41% depending on how much odds you take.

It may be an interesting fact but it's just not a fair comparison to say that a $5 line bettor loses the same actual amount as a $5 line bettor who also plays odds. The two players do not wager the same amount. They don't have the same handle. They don't have the same amount of action. And that's why we need to look at the percentage lost per dollar of action.

Steen