# Questions on house advantage and odds

Discussion in 'Beginner Zone' started by jimijazz, Dec 22, 2010.

1. jimijazz, Dec 22, 2010

### jimijazz Member

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Q1: How is the HA calculated on a place bet? The table of odds gives the HA on a 6 or 8 place bet as 1.52%. If the true odds are 6:5 (1.20) and the payout 7:6 (1.167) shouldn't the HA be the difference between those two numbers or 0.033 (3.3%)?

Q2: How does taking odds on the pass line bet decrease the HA, irrespective of the point? Since the decision to take odds is made after the point is established it seems to me it should depend upon the point. If you make a \$5 PL bet and the point is a 4 then taking 2x odds increases the pay out from \$5 for \$5 bet (50% of the true odds) to \$25 for a \$15 dollar bet (83% of the true odds). WHile the player is still at a disadvantage that's a huge improvement in positsion and it seems to me it's a must do. If you take 2x odds on a 6 point the increase is from 83% of true odds to 94% of the true odds. WHile this is a good increase in value it's not as good as for the 4 point. I'm asking because if you're a small stakes player and you want to protect your bankroll doesn't it make more sense to take odds on the 4,5, 9, 10 and consider letting the 6 and 8 go?

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2. splitdoubledestroy, Dec 22, 2010

### splitdoubledestroy Member

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Answer to question 1+ The house advantage of the 2 combined place bets would be 1.3435.

Add each HA together and divide by two to get the mean or average.

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3. The Midnight Skulker, Dec 22, 2010

### The Midnight Skulker Member

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The long hand method for calculating house advantage of any bet is to establish the probability of winning and losing, to multiply those probabilities by the respective amounts won and lost (using negative numbers for losses), to add the two products together, to divide by the size of the bet, and finally to multiply by 100 to convert to a percentage. So, for a place bet on 6 or 8,
p(win)=5/11 and p(loss)=6/11.
(5/11)*\$7 = \$3.182 and (6/11)*(-\$6) = -\$3.273
\$3.182 - \$3.273 = -\$0.091
-\$0.091 / \$6 = -0.0152
-0.0152 * 100 = -1.52% (The minus sign indicates the house has the advantage.)
For bets like place bets, that have only two possible monetary outcomes (i.e. there is not a range of wins available), there is a short hand method available, namely simply taking the product of the number of ways to win times the amount won and subtracting the product of the number of ways to lose times the amount lost, dividing by the total bet handle, and as above finally multiplying by 100 to convert to a percentage. So, for a place bet of 6 or 8,
((5 ways to win) * \$7) - ((6 ways to lose) * \$6) = -\$1
-\$1 / (11 ways to resolve bet) * \$6 = -0.0152
-0.0152 * 100 = -1.52% (The minus sign indicates the house has the advantage.)
If you are going to take this approach then the correct methodology is to calculate what the house should pay for a fair bet, subtract what it does pay (to give the amount you are being shorted every time you win), divide by the amount of the bet, multiply that quotient by the probability of winning, and finally multiply that product by 100 to convert to a percentage. So, for a place bet on 6 or 8,
1.20 * \$6 = \$7.20
\$7.20 - \$7.00 = \$0.20
\$0.20 / \$6 = 0.0333
0.0333 * (5/11) = 0.0152
0.0152 * 100 = 1.52%
This last method actually highlights the mechanism by which the house makes its profits: by "taxing" the players' wins.
When making a statement like, "Taking odds decreases the house advantage," one must pay very close attention to what the house advantage is being decreased on. Taking odds has absolutely no effect on the house advantage on the flat portion of the Pass Line bet: it was 1.41% when the bet was made and remains at 1.41% after the bet has established a point. (Yes, yes; the chances of winning the bet have decreased after it has established a point, but the player is still expected to lose 1.41% of the bet every time it is made, which is before the comeout roll.) Because odds are a fair bet (i.e. the payoff exactly balances the probability of winning the bet) the house advantage on the total amount at risk is less than the house advantage on the flat bet alone.
The flip side is that you are less likely to win your odds bet on points of 4, 5, 9, and 10 than you are on 6 and 8. What odds bets do is increase variance without increasing expected loss. This gives the player a better chance to show a profit, but lamentably also increases the chance of heavier losses because variance is a double edged sword. As a rule of thumb one should allocate as much of one's total bet to odds as the table allows.

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4. goatcabin, Dec 22, 2010

### goatcabin Member

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The reason for this last statement is that the flat bet is the only part that carries a house advantage. Taking odds on a given flat bet increases variance without increasing expected loss, as Midnight says. However, REDUCING the amount of your flat bet and, instead, putting that money on an odds bet REDUCES your expected loss. Of course, that is only possible if your flat bet is not already the table minimum.

For example:
\$10 flat bet, single odds, 60 bets, average bet \$16.66 (odds taken only when a point is established)
expected loss: \$8.48, standard deviation \$152

\$5 flat bet, double odds, 60 bets, average bet \$11.67
expected loss: \$4.24, standard deviation \$111

\$5 flat, 3, 4, 5X odds, 60 bets, average bet \$18.89
expected loss: \$4.24, standard deviation \$190

So, you cut your expected loss in half and can choose how much volatility you want by selecting an odds multiple. Most places allow 3, 4, 5X odds, so that the payoff is six units, regardless of the point.
Good luck.
Cheers,
Alan Shank
Woodland, CA

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5. donald, Dec 25, 2010

### donald Member

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How do you calculate standard deviation? Is there a formula for it?
Thanks.

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6. goatcabin, Dec 25, 2010

### goatcabin Member

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ev = ( (x+1) * p) - 1

std dev = (x + 1) * sqrt(p*(1 - p) )

These work for single bets, like pass line, place bets, etc. x is the payoff, so 1 for even-money bets, 1.167 for place 6 etc. p is the probability of winning the bet, so .4929 for place, .4545 for place six, etc.

For example, for place six:

ev = ( 1.167 + 1) * .4545) -1 = (2.167 * .4545) - 1 = -.0151
std dev = (1.167 + 1) * sqrt(.4545 * .5454) = 2.167 * sqrt (.2479) = 1.079 (This is as a percentage of the bet, so for \$6 place six, the SD is \$6.47.

For pass line with odds, this does not work, because there are different payoffs. You have to do it long hand, as the square root of the weighted sum of the squares of the differences from the mean.

I don't have time to illustrate that right now, as it is Christmas Day, and my wife is bugging me to check on dinner!
Merry Christmas,
Alan Shank
Woodland, CA

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7. donald, Dec 26, 2010

### donald Member

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\$5 flat, 3,4,5X odds
At your convenience, can you show how you got \$190 for the standard deviation.
Merry Christmas.

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8. goatcabin, Dec 26, 2010

### goatcabin Member

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Code:
```outcome -    mean       diff   squared    ways   gross variance
5       -  -.0707 =   5.0707    25.712     440   11313.28
-5      -  -.0707 =   4.9293    24.298     220    5345.56
-25     -  -.0707 = 24.9293   621.4700     264  164068.07
-20     -  -.0707 = 19.9293   397.1770     220   87378.94
-30     -  -.0707 = 29.9293   895.7630     300  268728.89
35     -  -.0707 =  35.0707  1229.954     536  659255.34

----------
1196090 / 1980 = 604.086
```
With 3, 4, 5X odds, all winning points yield the same win.
Variance is 607.09, standard deviation is its square root, 24.578.

Variance is weighted sum of squared differences from mean. I weighted the outcomes using the "perfect 1980".

Those figures are for one bet, of course. Now, the ev increases directly along with the number of bets, but the standard deviation increases with the square root of the number of bets. So, for 60 bets:

ev = -.0707 * 60 = -\$4.24
SD = 24.578 * sqrt(60) = \$190

Of course, I have a computer program that does this.
Cheers,
Alan Shank
Woodland, CA

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9. splitdoubledestroy, Dec 26, 2010

### splitdoubledestroy Member

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Goatcabin I thank you for your time. The way you explain things when you are trying to teach someone is far easier to read and understand than the traditional text book explanations. You dumb it down in a way that is simple for even the most inept people. Your explanations have allowed me to understand mathematical proofs a little better thus giving me a better understanding if not an easier understanding, of the various subjects in the field of mathematics. Thank you for your time. Hope you had a good Christmas and hope you have an even better New Year.

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10. goatcabin, Dec 27, 2010

### goatcabin Member

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My pleasure. Happy New Year,
Alan Shank
Woodland, CA

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11. donald, Dec 27, 2010

### donald Member

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Goatcabin: I thank you for showing how to calculate the standard deviation. I have not seen such computations in any craps book. Have you considered writing a book with such calculations and computer programs?

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12. goatcabin, Dec 27, 2010

### goatcabin Member

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I actually have a book, or rather parts of a book, that I started quite a few years ago. I don't know whether I will ever finish it.
Thanks for your kind words.
Cheers,
Alan Shank
Woodland, CA

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13. jimijazz, Dec 30, 2010

### jimijazz Member

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Thanks all for the thoughtful and thorough explanations. It's an exciting game and I enjoy it more than other gambling options.

I'm off to Vegas in mid-January, and plan to play small stakes craps a couple of times.

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14. basicstrategy777, Dec 30, 2010

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