Irish,
OK, I’m up for a little fun
The 6 or 8 can win 5 ways and lose 6 ways making 11 events that can affect it. If we bet either the 6 or 8 for $6 there is $66 invested in the number of possible events. 5 winning events = $65 on the 5 wins (5x (the $7 won & $6 bet returned)). $66-$65 = $1 & 1/66 = 1.515%
Bet 6 & 8. 10 ways to win and 6 ways to lose making 16 events that can affect it. Investment is $12 ($6 + $6) x 16 = 192. 10 winning events = $190 = (($7 won + $12 bet returned) x 10). $192 - $190 = $2 & 2/192 = 1.042%
Bet Inside. 18 ways to win and 6 ways to lose making 24 events that can affect it. Investment is $528 ($5 + $6 + $6 + $5) x 24. 18 winning events = $522 = ($7 won + $22 bet returned) x 18). $528 - $522 = $6 & 6/528 = 1.136%
Bet Across. 24 ways to win and 6 ways to lose making 30 events that can affect it. Investment is $960 ($5 + $5 + $6 + $6 + $5 + $5) x 30. 24 winning events = $948 = $9 won 6 times ($54) + $7 won 18 times ($126) + ($32 bet returned) x 24) ($768). $960 - $948 = $12 & 12/960 = 1.25%
This is using the old Zeke Feinberg math and as you can see while the 6 & 8 match yours the Inside and Across do not. Just looking at the Inside and Across percentages in your post has me scratching my head as to how the Inside would have a higher edge in any math.
I agree that volatility is increased. There is always a price for more ways to win.
Kelph
How are your math skills?
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shunkaha
Re: How are your math skills?
I'm reasonably sure I've seen one of those math challenged posters in action on a forum as well... it immediately brought a Val Kilmer quote to mind from the movie "Kiss Kiss Bang Bang"... "who taught you math?"
The upshot was the guy that posted the statement, presumably after being dropped on his head as a child, stated that the house edge on a buy 4 and 10 for $25 each was 1% on a 1 hit and down scenario. His explanation involved the fact you had $50 in play and won $49 at the cost of $1. Thus to him it was 1/100 or 1%, yet he claimed a higher house edge on either a buy 4 or buy 10. In his case he stated a 1.33% on either, yet a 1% if you do both.
Which brought me to the quote, because I want to know who taught this guy math.
The upshot was the guy that posted the statement, presumably after being dropped on his head as a child, stated that the house edge on a buy 4 and 10 for $25 each was 1% on a 1 hit and down scenario. His explanation involved the fact you had $50 in play and won $49 at the cost of $1. Thus to him it was 1/100 or 1%, yet he claimed a higher house edge on either a buy 4 or buy 10. In his case he stated a 1.33% on either, yet a 1% if you do both.
Which brought me to the quote, because I want to know who taught this guy math.
Re: How are your math skills?
irish,
Can you PM me the site and name of thread? I'm curious to understand what was done to get to those other two numbers.
Kelph
Can you PM me the site and name of thread? I'm curious to understand what was done to get to those other two numbers.
Kelph
Re: How are your math skills?
When you read this calculation it seems to make sense. The calculation for the 6 OR 8 comes with the correct answer and so it seems to be a proof that the second calculation must also be correct.Kelph wrote:...
The 6 or 8 can win 5 ways and lose 6 ways making 11 events that can affect it. If we bet either the 6 or 8 for $6 there is $66 invested in the number of possible events. 5 winning events = $65 on the 5 wins (5x (the $7 won & $6 bet returned)). $66-$65 = $1 & 1/66 = 1.515%
Bet 6 & 8. 10 ways to win and 6 ways to lose making 16 events that can affect it. Investment is $12 ($6 + $6) x 16 = 192. 10 winning events = $190 = (($7 won + $12 bet returned) x 10). $192 - $190 = $2 & 2/192 = 1.042%...
Before we get into it, take a step back. Let's first look at this from a holistic angle. If combining bets on the craps table reduces the house edge overall, would that not lead to a conclusion that if you found the correct combination of bets you could reduce the house edge to near zero? or zero? or even reverse it to be positive for the player?
I think we all realize that NO combination of bets can reduce the house edge. If that were the case then there would be many sufficiently bankrolled players that would take advantage.
So if intuitively we can surmise that combining bets should not be reducing the house edge, then why does that second calculation seem to show that it absolutly can?
The answer is that the second calculation is working on an incorrect assumption.... What is wrong you ask? How about if I give you a hint and see if you can figure out what is wrong?
HINT: When a six (6) rolls, what is the impact on the 8? How is that impact any different then if a 5 or a 4 or a 2 rolled? The $2 lost should not be divided by 192, but instead by 132.... why?
What do you come up with?
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Re: How are your math skills?
"The $2 lost should not be divided by 192, but instead by 132.... why?"
If you use this assumption, "(($7 won + $12 bet returned)" if either the 6 or the 8 hit you have to take both bets down.
If you use this assumption, "(($7 won + $12 bet returned)" if either the 6 or the 8 hit you have to take both bets down.
Re: How are your math skills?
Maddog,
As I pointed out I used the old Zeke Feinberg math from his book because it caused a lot of talk back then and I was trying to see if it would arrive at the figures irish posted.
The problem with Zeke's seductive math is that the numbers are being combined as a single bet when in fact they are independent. One number can win without any affect on the other even though both can lose. Each number needs to be calculated for an independent win even though a loss wipes out all bets. Investment on each number is $6 on each of 5 ways to win ($30) plus 6 ways to lose ($36) or $66 on 6 and $66 on 8 for a total of $132.
The number of events really should be 11 and not 16 since only one of the two numbers can actually win while the other is unaffected ($12 x 11 = $132). But I must says Zeke's math does look so logical.
Kelph
As I pointed out I used the old Zeke Feinberg math from his book because it caused a lot of talk back then and I was trying to see if it would arrive at the figures irish posted.
The problem with Zeke's seductive math is that the numbers are being combined as a single bet when in fact they are independent. One number can win without any affect on the other even though both can lose. Each number needs to be calculated for an independent win even though a loss wipes out all bets. Investment on each number is $6 on each of 5 ways to win ($30) plus 6 ways to lose ($36) or $66 on 6 and $66 on 8 for a total of $132.
The number of events really should be 11 and not 16 since only one of the two numbers can actually win while the other is unaffected ($12 x 11 = $132). But I must says Zeke's math does look so logical.
Kelph
Re: How are your math skills?
But you gotta admit, Zeke had a way with words when it came to book titles.
"Get in, get up, and get gone."
- Heavy
- Heavy
Re: How are your math skills?
I've been preoccupied with the thread on math skills and multiple Place Bet calculations. I’ve done a little digging and have come up with some interesting bits that I’d like to share at the risk of being assigned a seat on the short bus. To avoid that I’ll try my best to offer somewhat sound reasoning and support from sources far more skilled in math than me. I cannot claim I’m correct but I have found enough to make me wonder. This is really a mental exercise for anyone interested.
The way we calculate HA on Place Bets. The 6 or 8 can win 5 ways and lose 6 ways making 11 events that can affect it. If we bet either the 6 or 8 for $6 there is $66 invested in the number of possible events. 5 winning events = $65 on the 5 wins (5x (the $7 won & $6 bet returned)). $66-$65 = $1 & 1/66 = 1.515%.
To start and recap I’ll repeat what I said regarding Zeke Feinberg’s multiple Place Bets calculations. Zeke said that when Place Bets were combined the House Advantage lessened based on:
Bet 6 & 8. 10 ways to win and 6 ways to lose making 16 events that can affect it. Investment is $12 ($6 + $6) x 16 = 192. 10 winning events = $190 = (($7 won + $12 bet returned) x 10). $192 - $190 = $2 & 2/192 = 1.042%
Bet Inside. 18 ways to win and 6 ways to lose making 24 events that can affect it. Investment is $528 ($5 + $6 + $6 + $5) x 24. 18 winning events = $522 = ($7 won + $22 bet returned) x 18). $528 - $522 = $6 & 6/528 = 1.136%
Bet Across. 24 ways to win and 6 ways to lose making 30 events that can affect it. Investment is $960 ($5 + $5 + $6 + $6 + $5 + $5) x 30. 24 winning events = $948 = $9 won 6 times ($54) + $7 won 18 times ($126) + ($32 bet returned) x 24) ($768). $960 - $948 = $12 & 12/960 = 1.25%
My response was: “The problem with Zeke's seductive math is that the numbers are being combined as a single bet when in fact they are independent. One number can win without any affect on the other even though both can lose. Each number needs to be calculated for an independent win even though a loss wipes out all bets. Investment on each number is $6 on each of 5 ways to win ($30) plus 6 ways to lose ($36) or $66 on 6 and $66 on 8 for a total of $132.
The number of events really should be 11 and not 16 since only one of the two numbers can actually win while the other is unaffected ($12 x 11 = $132). But I must say Zeke's math does look so logical.”
In spite of my answer this continued to gnaw at me. I found a 7/25/15 article by Jerry Stickman at
http://stickman.casinocitytimes.com/art ... side-64374
that pretty much backs up my response. I also found another Stickmen article from 5/18/07 at
http://stickman.casinocitytimes.com/art ... raps-34389
that provides a nifty ways to calculate combined bets while maintaining their independence. Using that method Inside is 2.597% and Across is 3.75%.
Just so we are all looking at this the same way the combined bet must be treated as a single bet. You cannot add to or regress any individual number unless you do so equally to all those numbers. If the bet is removed you do so to all the numbers in the combined bet. Under these conditions the player certainly has acquired more ways to win than lose.
I understand that doesn’t cut it for many math people but then I came across 12/6/10 article by Al Krigman at http://krigman.casinocitytimes.com/arti ... raps-59413
That pretty much lines up with Zeke’s take. So even the experts can’t agree?
Then I came across the Place Bet section in Scarne’s New Complete Guide To Gambling, page 332 – 333 that addresses this very question. In addition there is the 10/2/2000 Al Krigman article at
http://krigman.casinocitytimes.com/arti ... guise-5541
that arrived at the same conclusion as Scarne. Combined Place Bets are actually Lay Bets in disguise. Someone who has only had vanilla ice cream their entire life might believe ice cream can only be vanilla. Many are going to say that a Lay is betting that a 7 will show before a specific chosen number. That’s what the game tells us but perhaps the math doesn’t care about the game, just the math. In the simplest terms a Lay is betting with the odds in your favor that another event will not occur. This friendly situation means betting more than you’ll win and the accepted casino version of a Lay means tacking on vig for casino profit.
If you work out the combined Place bets as Lay Bets the percentages line up as Zeke calculated them but for a different reason. They provide more ways to win than lose; you risk more to win less but the 7 is still not your friend.
Regular Lays as calculated at Wizard of Odds site.
Lay bet to lose on 6 or 8: [(6/11)×19 + (5/11)×(-25)]/25 = (-11/11)/25 = -1/25 = -4.000%
Lay bet to lose on 5 or 9: [(6/10)×19 + (4/10)×(-31)]/31 = (-10/10)/31 = -1/31 = -3.226%
Lay bet to lose on 4 or 10: [(6/9)×19 + (3/9)×(-41)]/41 = (-9/9)/41 = -1/41 = -2.439%
Combined Place Bets Calculated As Lays
Lay bet on 6 & 8: {(10/16)X7+(6/16)X-12]/12 = -.125/12 = -1.04%
Lay bet on 5 & 9: {(8/14)X7+(6/14)X-10]/10 = -.2857/10 = -2.857%
Lay bet on 4 & 10: {(6/12)X9+(6/12)X-10]/10 = -.5/10 = -5%
Inside & Across Combined Place Bets Calculated As Lays
Lay bet on Inside: {(18/24)x7+(6/24)X-22]/22 = -.25/22 = -1.136%
Lay bet on Across: {(6/30)x9+(8/30)x7+(10/30)x7+(6/30)X-32]/32 = .401/32 = 1.25%
As far as combining the Lay bets I point to the 7/28/97 Al Krigman article at
http://krigman.casinocitytimes.com/arti ... craps-5367
Krigman says that combining casino accepted Lay Bets moves the advantage back to the casino as the player accepts more ways to lose. I find it hard to believe that combining casino Lay Bets benefits the casino while no such benefit is derived by the player for combining Place Bets. That just doesn't seem logical if the conditions I laid out hold true.
I found this to be interesting but maybe that’s just me.
Kelph
The way we calculate HA on Place Bets. The 6 or 8 can win 5 ways and lose 6 ways making 11 events that can affect it. If we bet either the 6 or 8 for $6 there is $66 invested in the number of possible events. 5 winning events = $65 on the 5 wins (5x (the $7 won & $6 bet returned)). $66-$65 = $1 & 1/66 = 1.515%.
To start and recap I’ll repeat what I said regarding Zeke Feinberg’s multiple Place Bets calculations. Zeke said that when Place Bets were combined the House Advantage lessened based on:
Bet 6 & 8. 10 ways to win and 6 ways to lose making 16 events that can affect it. Investment is $12 ($6 + $6) x 16 = 192. 10 winning events = $190 = (($7 won + $12 bet returned) x 10). $192 - $190 = $2 & 2/192 = 1.042%
Bet Inside. 18 ways to win and 6 ways to lose making 24 events that can affect it. Investment is $528 ($5 + $6 + $6 + $5) x 24. 18 winning events = $522 = ($7 won + $22 bet returned) x 18). $528 - $522 = $6 & 6/528 = 1.136%
Bet Across. 24 ways to win and 6 ways to lose making 30 events that can affect it. Investment is $960 ($5 + $5 + $6 + $6 + $5 + $5) x 30. 24 winning events = $948 = $9 won 6 times ($54) + $7 won 18 times ($126) + ($32 bet returned) x 24) ($768). $960 - $948 = $12 & 12/960 = 1.25%
My response was: “The problem with Zeke's seductive math is that the numbers are being combined as a single bet when in fact they are independent. One number can win without any affect on the other even though both can lose. Each number needs to be calculated for an independent win even though a loss wipes out all bets. Investment on each number is $6 on each of 5 ways to win ($30) plus 6 ways to lose ($36) or $66 on 6 and $66 on 8 for a total of $132.
The number of events really should be 11 and not 16 since only one of the two numbers can actually win while the other is unaffected ($12 x 11 = $132). But I must say Zeke's math does look so logical.”
In spite of my answer this continued to gnaw at me. I found a 7/25/15 article by Jerry Stickman at
http://stickman.casinocitytimes.com/art ... side-64374
that pretty much backs up my response. I also found another Stickmen article from 5/18/07 at
http://stickman.casinocitytimes.com/art ... raps-34389
that provides a nifty ways to calculate combined bets while maintaining their independence. Using that method Inside is 2.597% and Across is 3.75%.
Just so we are all looking at this the same way the combined bet must be treated as a single bet. You cannot add to or regress any individual number unless you do so equally to all those numbers. If the bet is removed you do so to all the numbers in the combined bet. Under these conditions the player certainly has acquired more ways to win than lose.
I understand that doesn’t cut it for many math people but then I came across 12/6/10 article by Al Krigman at http://krigman.casinocitytimes.com/arti ... raps-59413
That pretty much lines up with Zeke’s take. So even the experts can’t agree?
Then I came across the Place Bet section in Scarne’s New Complete Guide To Gambling, page 332 – 333 that addresses this very question. In addition there is the 10/2/2000 Al Krigman article at
http://krigman.casinocitytimes.com/arti ... guise-5541
that arrived at the same conclusion as Scarne. Combined Place Bets are actually Lay Bets in disguise. Someone who has only had vanilla ice cream their entire life might believe ice cream can only be vanilla. Many are going to say that a Lay is betting that a 7 will show before a specific chosen number. That’s what the game tells us but perhaps the math doesn’t care about the game, just the math. In the simplest terms a Lay is betting with the odds in your favor that another event will not occur. This friendly situation means betting more than you’ll win and the accepted casino version of a Lay means tacking on vig for casino profit.
If you work out the combined Place bets as Lay Bets the percentages line up as Zeke calculated them but for a different reason. They provide more ways to win than lose; you risk more to win less but the 7 is still not your friend.
Regular Lays as calculated at Wizard of Odds site.
Lay bet to lose on 6 or 8: [(6/11)×19 + (5/11)×(-25)]/25 = (-11/11)/25 = -1/25 = -4.000%
Lay bet to lose on 5 or 9: [(6/10)×19 + (4/10)×(-31)]/31 = (-10/10)/31 = -1/31 = -3.226%
Lay bet to lose on 4 or 10: [(6/9)×19 + (3/9)×(-41)]/41 = (-9/9)/41 = -1/41 = -2.439%
Combined Place Bets Calculated As Lays
Lay bet on 6 & 8: {(10/16)X7+(6/16)X-12]/12 = -.125/12 = -1.04%
Lay bet on 5 & 9: {(8/14)X7+(6/14)X-10]/10 = -.2857/10 = -2.857%
Lay bet on 4 & 10: {(6/12)X9+(6/12)X-10]/10 = -.5/10 = -5%
Inside & Across Combined Place Bets Calculated As Lays
Lay bet on Inside: {(18/24)x7+(6/24)X-22]/22 = -.25/22 = -1.136%
Lay bet on Across: {(6/30)x9+(8/30)x7+(10/30)x7+(6/30)X-32]/32 = .401/32 = 1.25%
As far as combining the Lay bets I point to the 7/28/97 Al Krigman article at
http://krigman.casinocitytimes.com/arti ... craps-5367
Krigman says that combining casino accepted Lay Bets moves the advantage back to the casino as the player accepts more ways to lose. I find it hard to believe that combining casino Lay Bets benefits the casino while no such benefit is derived by the player for combining Place Bets. That just doesn't seem logical if the conditions I laid out hold true.
I found this to be interesting but maybe that’s just me.
Kelph
Re: How are your math skills?
Really interesting approach to think of multi placing as a lay. I need to think more on this. Just wanted to say good thread.
I wanna see the dust...
Re: How are your math skills?
Just a quick followup with a definitive answer to the multiple Place bet HA question that has always bothered me and that I couldn't resolve to my satisfaction.
I went the Wizard of Vegas and asked the math experts to explain and I received what I asked for plus something I found online that's a little less complicated. If interested see at http://wizardofvegas.com/forum/gambling ... post506986
Kelph
I went the Wizard of Vegas and asked the math experts to explain and I received what I asked for plus something I found online that's a little less complicated. If interested see at http://wizardofvegas.com/forum/gambling ... post506986
Kelph