Homework 3: Analyzing the Casino Game of Craps
Due Monday April 18 (send by email to pasquale at cs.ucsd.edu)
1. The house edge for a pass line bet in Craps is 1.41%. Derive this value and show your work. Here are some hints to guide you:
a. On the opening roll, a 7 or 11 wins: what is the probability of a win on the first roll? Call this probability A.
Since there are 6 (out of 36) ways for a 7 to appear, and 2 ways for 11, there are 8 = 6 + 2 out of 36 ways for a 7 or 11 to appear, so the probability is:
A = 8/36
b. On the opening roll, a 2, 3, or 12 loses: what is the probability of a loss on the first roll? Call this probability B.
There is 1 way for a 2 to appear, 2 ways for a 3, and 1 way for a 12, summing to 4, so the probability is:
B = 4/36
c. What is the probability of a 4 or 10 "point" being established on the first roll? Call this probability C.
There are 3 ways for a 4 to appear and 3 for a 10, summing to 6, so the probability is:
C= 6/36
d. What is the probability of a 5 or 9 "point" being established on the first roll? Call this probability D.
There are 4 ways for a 5 to appear and 4 for a 9, summing to 8, so the probability is:
D = 8/36
e. What is the probability of a 6 or 8 "point" being established on the first roll? Call this probability E.
There are 5 ways for a 6 to appear and 5 for an 8, summing to 10, so the probability is:
E = 10/36
We now have probabilities for all possible first rolls. As a check, they should all add up to 1. If they don't, you made some error above.
8/36 + 4/36 + 6/36 + 8/36 + 10/36 = 36/36 = 1
f. Assuming a 4 or 10 "point" was established on the first roll, what is the probability that this point will come out BEFORE a 7? Call this probability F.
On the next roll, there are 3 ways for the point to appear, 6 ways for a 7, and 27 ways for some other number. So, out of the 36 possibilities for rolling two dice, only 9 matter, with 3 of those being a winner (the point comes out), and 6 being losers (7 comes out). So, the probability that the point comes out before 7 is:
F = 3/9
g. Assuming a 5 or 9 "point" was established on the first roll, what is the probability that this point will come out BEFORE a 7? Call this probability G.
Since there are 4 ways for the point to appear, and 6 ways for a 7 to appear, the probability that the point comes out before 7 is:
G = 4/10
h. Assuming a 6 or 8 "point" was established on the first roll, what is the probability that this point will come out BEFORE a 7? Call this probability H.
Since there are 5 ways for the point to appear, and 6 ways for a 7 to appear, the probability that the point comes out before 7 is:
H = 5/11
You can now compute the probability P of a win: A + C*F + D*G + E*H.
P = 8/36 + (6/36 * 3/9) + (8/36 * 4/10) + (10/36 * 5/11)
= 0.2222... + 0.05555... + 0.08888... + 0.12626... = 0.492929...
i. What are the fair odds for a pass line bet?
(1-P)/P = 0.507/0.493 = 1.029
Fair odds = 1.029 to 1
j. What is the house edge (given that the casino pays 1 to 1 for a pass line bet)?
House edge = (1.029 - 1)/(1.029 + 1) = 0.029/2.029 = 1.41%
2. Read the FAQ on Craps located at http://www.faqs.org/faqs/gambling-faq/craps/. Learn how to place all the bets below so that you are ready to play at our next meeting. For each of the bets below, what are the fair odds (you need to compute these), casino odds (look them up in the FAQ) and house edges (look them up in the FAQ, and if you have the time, compute and check that they match what are in the FAQ):
An "odds" bet can be made after a point is established. If the point comes out, you get more than the 1 to 1 that your pass line bet gets.
a. "Odds" on 4 or 10 point:
Fair: 2 to 1
Casino: 2 to 1
House edge: 0%
b. "Odds" on 5 or 9 point:
Fair: 3 to 2
Casino: 3 to 2
House edge: 0%
c. "Odds" on 6 or 8 point:
Fair: 6 to 5
Casino: 6 to 5
House edge: 0%
d. "Place" on 4 or 10 point:
A place bet is that a number you select from {4, 5, 6, 8, 9, and 10}
will be rolled before a 7.
Fair: 2 to 1
Casino: 9 to 5
House edge: 6.7%
To calculate this:
1. Express both fair and casino odds in the form X to 1
Fair = 2 to 1, Casino = 1.8 to 1
2. House edge is (F-C)/(F+1), where Fair = F to 1, Casino = C to 1
House edge = (2-1.8)/(2+1) = 0.2/3 = 0.067 = 6.7%
e. "Place" on 5 or 9 point:
Fair: 3 to 2 (1.5 to 1)
Casino: 7 to 5 (1.4 to 1)
House edge: 4.0%
f. "Place" on 6 or 8 point:
Fair: 6 to 5 (1.2 to 1)
Casino: 7 to 6 (1.167 to 1)
House edge: 1.5%
g. "Field" (with 2 and 12 paying 2 to 1):
A field bet is a single-roll bet that one of the numbers 2, 3, 4, 9, 10, 11, 12 will be rolled; any other number is a loser.
To determine fair odds, count the number of winning rolls and number of losing rolls. The winning rolls are 2, 3, 4, 9, 10, 11, and 12, and there are 16 ways of rolling them. The losing rolls are 5, 6, 7, and 8, and there are 20 ways of rolling them.
Fair odds = 20 to 16, or 5 to 4 (1.25 to 1)
The casino expresses casino odds as 1 to 1 for a 3, 4, 9, 10, or 11, and 2 to 1 for a 2 or 12. Since the odds are not stated in simple terms of a single X to Y, we will need to figure them out. We do this by figuring out the house edge, and then work backwards to get casino odds. Recall that we can calculate the house edge directly from expected gain as follows.
If you bet $1, there is a 14/36 probability (for rolls of 3, 4, 9, 10, or 11) that you will win $1, a 2/36 probability (for rolls of 2 and 12) that you win $2, and 18/36 that you lose $1. Computing the expected gain:
EG = (14/36)*1 + (2/36)*2 + (20/36)*(-1) = -0.056
This means that for every $1 bet, you will lose 5.6 cents, and so:
House edge = 5.6%
Now we can compute casino odds, since (F-C)/(F+1) = House edge
(1.25-C)/2.25 = 0.056
Solving for C,
C = 1.125 to 1 (or 9 to 8)
Since fair odds are 1.25 to 1, or 10 to 8, we see the casino's odds of 9 to 8 are a bit less, as expected.
h. "Big 6 or 8":
A Big 6 (or Big 8) is a bet that a 6 (or an 8) will come out before a 7. Since there are 5 winning rolls to 6 losing rolls:
Fair odds: 6 to 5 (or 1.2 to 1)
Casino odds: 1 to 1
House edge: 0.2/2.2 = 9.1% (not a good bet at all!)
i. "Proposition" on 2 (or 12):
[This was a bit confusing (sorry) because there was a typo - it originally said "2, 3, or 12". Also, I meant proposition on these separately.]
A "proposition" bet is a single-roll bet that a certain number will be rolled. If that number comes out, you win, else you lose.
Proposition on 2 (or Proposition on 12)
Fair odds = 35 to 1
Casino odds = 30 to 1
House edge = 5/36 = 13.9% (very bad bet)
Often the casino will say "30 FOR 1" rather than "30 to 1". The former means that they'll give you $30 "for" your $1, i.e., you lose your original $1 even if you win! In this case:
Casino odds = 29 to 1
House edge = 6/36 = 16.7% (very very bad bet!)
Proposition on 11
Fair odds = 17 to 1
Casino odds = 15 to 1
House edge = 2/18 = 11.1%
If casino says 15 for 1
Casino odds = 14 to 1
House edge = 3/18 = 16.7% (very very bad)
j. "Proposition" on 7:
Fair odds = 30 to 6, or 5 to 1
Casino odds = 4 to 1
House edge = 1/6 = 16.7% (very very bad)
k. "Proposition" on 11 (or 3):
Fair odds = 17 to 1
Casino odds = 15 to 1
House edge = 2/18 = 11.1%
House edge = 3/18 = 16.7% if casino odds are 15 for 1
l. "Hard" 4:
A "Hardways" bet is a bet where the number N comes out with N/2 showing on each die (i.e., making N the "hard way") before a 7 or a non-hardway N come out. For example, a hard 8 bet is where the the dice showing a (4,4) is a winner, and any other 8 (e.g., 3 and 5) or a 7 are losers.
Fair odds = 8 to 1 (since there is 1 winning roll, a [2,2] and 8 losing rolls, 6 of which are 7's and 2 are 4's, i.e., [1,3] and [3,1])
Casino odds = 7 to 1
House edge = 1/9 = 11.1%
m. "Hard" 6:
Fair odds: 10 to 1
Casino odds: 9 to 1
House edge: 1/11 = 9.1%
n. "Hard" 8:
Same as Hard 6
Fair odds: 10 to 1
Casino odds: 9 to 1
House edge: 1/11 = 9.1%
o. "Hard" 10:
Same as Hard 4
Fair odds = 8 to 1
Casino odds = 7 to 1
House edge = 1/9 = 11.1%
p. "Any craps":
This is a proposition bet: you're betting that a 2, 3, or 12 will come out on the next roll.
Fair odds = 32 to 4, or 8 to 1 (4 winners to 32 losers)
Casino odds = 7 to 1
House edge = 1/9 = 11.1%
IN SUMMARY, if we order the various bets by house edge, we get
Taking odds 0.0%
Pass line 1.4%
Place on 6 (or 8) 1.5%
Place on 5 (or 9) 4.0%
Field bet 5.6%
Place on 4 (or 10) 6.7%
Hard 6 (or 8) 9.1%
Big 6 (or 8) 9.1%
Any craps 11.1%
Hard 4 (or 10) 11.1%
Proposition on 11 (or 3) 11.1% (paying 15 to 1)
Proposition on 2 (or 12) 13.9% (paying 30 to 1)
Proposition on 2 (or 12) 16.7% (paying 30 FOR 1)
Proposition on 11 (or 3) 16.7% (paying 15 FOR 1)
Proposition on 7 16.7%
Therefore, a good strategy is to play the pass line, take odds when a point is established (assuming you can afford it). If you want "more action" on the table, you can make a bet on every roll by making a "Come" bet, which works just like the pass line, except that it works on all rolls that are not starting rolls (where you place your pass line bets). You can even take odds on come rolls.
There is no reason to ever place any other kind of bet, as they all pay off poorly (except for Place on 6 or 8 - but why do so when the pass line or come bets are a bit better?).
THE FOLLOWING ARE ADVANCED EXERCISES AND ARE NOT REQUIRED. DO THEM IF YOU HAVE TIME, AS THEY WILL SHARPEN YOUR ABILITY TO COMPUTE PROBABILITIES AND HOUSE EDGES.
3. What is the house edge against the gambler who always takes 2x odds (i.e., if a point is established, you bet twice the amount that was placed on the pass line)?
Say you bet $1 on the pass line. There is a 8/36 probability (rolling a 7 or 11) of winning $1, a 4/36 probability (rolling a 2, 3, or 12) of losing $1, and a 24/36 probability (rolling anything else and setting the "point") of winning X. So far, the expected gain is:
EG = (8/36)*1 + (4/36)*(-1) + X = 0.11111... + X
We need to figure out X.
If 4 (or 10) is the point, and you take 2 times odds, you are betting an additional $2. If you win, you get $5 ($1 for pass line plus $4 since the 4 or 10 point pay 2 to 1), and if you lose, you lose $3 ($1 for pass line plus $2 for odds). Therefore, the contribution to the expected gain is: (6/36) * ((3/9)*5 + (6/9)*(-3)) = -0.055555...
The 6/36 is the probability of 4 or 10 becoming the point.
The 3/9 is the probability that the point comes out before 7.
The 6/9 is the probability that a 7 comes out before the point.
If 5 (or 9) is the point, and you take 2 times odds, you are betting an additional $2. If you win, you get $4 ($1 for pass line plus $3 since the 5 or 9 point pay 3 to 2), and if you lose, you lose $3 ($1 for pass line plus $2 for odds). Therefore, the contribution to the expected gain is: (8/36) * ((4/10)*4 + (6/10)*(-3)) = -0.044444...
If 6 (or 8) is the point, and you take 2 times odds, you are betting an additional $2. If you win, you get $3.4 ($1 for pass line plus $2.4 since the 6 or 8 point pay 6 to 5), and if you lose, you lose $3 ($1 for pass line plus $2 for odds). Note that a casino will generally NOT pay non-integral amounts, and so you'd have to put down a multiple of $5 to get $6 for every $5. But for this computation, we'll assume non-integral amounts to make the calculation easier. Therefore, the contribution to the expected gain is: (10/36) * ((5/11)*3.4 + (6/11)*(-3)) = -0.0252525...
Totaling the contributions to the expected gain for each of the points, we get X = (-0.055...) + (-0.044...) + (-0.025...) = -0.1252525..., therefore,
EG = 0.111... - 0.1252525... = -0.0141414...
Thus, the expected gain is actually a loss of 1.4 cents for an average bet, not of $1, but of $1 if a 2, 3, 7, 11, of 12 are rolled, which happens with probability 12/36, and $3 if a 4, 5, 6, 8, 9, or 10 are rolled, which happens with probability 24/36. Therefore
Average bet: (12/36)*1 + (24/36)*3 = 2.33333...
House edge = 0.61% = 0.01414/2.333
More generally, when you take N times odds, the expected gain of -0.01414... does not change (see that it is this value both when we calculated pass line with no odds and then with 2x odds). What changes is the average bet, which is (12/36) + (24/36)*(N+1) = 12/36 + 24N/36 + 24/36, simplified to
Average bet: 1 + 2N/3
House edge = 0.01414/(1+(2N/3))
We can now produce a table of house edges for pass lines with N times odds:
Bet House Edge
Pass line, no odds 1.41%
1x odds 0.85%
2x odds 0.61%
3x odds 0.47%
5x odds 0.33%
10x odds 0.18%
100x odds 0.02% (some casinos actually offer 100 times odds)
4. Rather than making a pass line bet, you bet on "don't pass", which means you win whenever a pass line bet loses, and vice-versa, except for a 12 on the opening roll, which is a "push", i.e., you neither win nor lose, you simply get back what you bet and no more. What are the fair odds, and what is the house edge (compare this to a pass line bet).
We calculated that the probability for winning for a pass line bet as 0.492929... Therefore the probability of losing with a pass line bet is 0.50707..., but this is NOT the same as winning with a don't pass bet because of the difference in rolling a 12. Since rolling a 12 has no effect (it is a push, you get your money back), we must do two things to calculate the probability P' of winning with a don't pass bet: (1) subtract Prob(rolling a 12) from 0.507, and (2) divide by 1 - Prob(rolling a 12):
P' = (0.50707... - 1/36)/(1 - 1/36) = 0.492987...
From this, we can calculate fair odds as (1-P')/P' to 1, along with the house edge:
Fair odds: 1.02845 to 1
Casino odds: 1 to 1
House edge: 1.40%
Notice that by making 12 a push, the house edge for a don't pass bet (1.40%) is almost the same as a pass line bet (1.41%). And of course, in both cases, the house has the edge!