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Ask the Wizard #278

In craps, what is the general formula for the combined house edge of a put bet on a point of p, and an odds bet of o?

SONBP2

For points of 4 to 6: ((7-p)/(5+p))*(1/(1+o))

For points of 8 to 10: ((p-7)/(19-p))*(1/(1+o))

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

In the Rams vs Cardinals game of November 6, 2011 the Rams scored two safeties in the third quarter. What are the odds of that?

Anonymous

The probability of at least one safety per game is 5.77%, based on historical experience.

The expected number of safeties per game would be -ln(1-0.0577) = 0.0594.

The expected number per quarter per team would be 0.0594/8 = 0.0074.

The probability of exactly two safeties by the same team in a single quarter would be e-0.0074×0.00742/fact(2) = 1 in 36,505.

In an NFL season there are 267 games, and 267×8=2,136 team quarters. So, according to my estimate, this will occur on average once every 36,505/2,136 = 17.1 years.

This should be considered as just a rough guess. There are factors to the game that I'm not taking into account, in the interest of simplicity.

What would happen if the two dice landed stacked in craps? Would it be a valid roll? If so, how would the dealers reveal what number the lower die landed on?

boxman4

Whether or not it is called a valid roll depends on where you are. New Jersey gaming regulation 19:47-1.9(a) states:

A roll of the dice shall be invalid whenever either or both of the dice go off the table or whenever one die comes to rest on top of the other. -- NJ 19:47-1.9(a)

Pennsylvania has the exact same regulation, Section 537.9(a):

A roll of the dice shall be invalid whenever either or both of the dice go off the table or whenever one die comes to rest on top of the other. -- PA 537.9(a)

I asked a Las Vegas dice dealer who said that here it would be called a valid roll, if it was otherwise a proper throw. Although he has never seen it happen, he said if it did the dealers would simply move the top die to see what number the lower die landed on. However, one can determine the outcome of the lower die without touching, or looking through, the top die. Here is how to do it. First, by looking at the four sides you can narrow down the possibilities on top to two. Here is how to tell according to the three possibilities.

  • 1 or 6: Look for the 3. If the high dot is bordering the 5, the 1 is on top. Otherwise, if it is bordering the the 2, the 6 is on top.
  • 2 or 5: Look for the 3. If the high dot is bordering the 6, the 2 is on top. Otherwise, if it is bordering the the 1, the 5 is on top.
  • 3 or 4: Look for the 2. If the high dot is bordering the 6, the 3 is on top. Otherwise, if it is bordering the the 1, the 4 is on top.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

What is the expected number of rolls of two dice for every total from 2 to 12 to occur at least once?

SixHorse

This question was asked at TwoPlusTwo.com, and was answered correctly by BruceZ. The following solution is the same method as that of BruceZ, who deserves proper credit. It is a difficult answer, so pay attention.

  1. First, consider the expected number of rolls to obtain a total of two. The probability of a two is 1/36, so it would take 36 rolls on average to get the first 2.

  2. Next, consider the expected number of rolls to get both a two and three. We already know it will take 36 rolls, on average, to get the two. If the three is obtained while waiting for the two, then no additional rolls will be needed for the 3. However, if not, the dice will have to be rolled more to get the three.

    The probability of a three is 1/18, so it would take on average 18 additional rolls to get the three, if the two came first. Given that there is 1 way to roll the two, and 2 ways to roll the three, the chances of the two being rolled first are 1/(1+2) = 1/3.

    So, there is a 1/3 chance we'll need the extra 18 rolls to get the three. Thus, the expected number of rolls to get both a two and three are 36+(1/3)×18 = 42.

  3. Next, consider how many more rolls you will need for a four as well. By the time you roll the two and three, if you didn't get a four yet, then you will have to roll the dice 12 more times, on average, to get one. This is because the probability of a four is 1/12.

    What is the probability of getting the four before achieving the two and three? First, let's review a common rule of probability for when A and B are not mutually exclusive:

    pr(A or B) = pr(A) + pr(B) - pr(A and B)

    You subtract pr(A and B) because that contingency is double counted in pr(A) + pr(B). So,

    pr(4 before 2 or 3) = pr(4 before 2) + pr(4 before 3) - pr(4 before 2 and 3) = (3/4)+(3/5)-(3/6) = 0.85.

    The probability of not getting the four along the way to the two and three is 1.0 - 0.85 = 0.15. So, there is a 15% chance of needing the extra 12 rolls. Thus, the expected number of rolls to get a two, three, and four is 42 + 0.15*12 = 43.8.

  4. Next, consider how many more rolls you will need for a five as well. By the time you roll the two to four, if you didn't get a five yet, then you will have to roll the dice 9 more times, on average, to get one, because the probability of a five is 4/36 = 1/9.

    What is the probability of getting the five before achieving the two, three, or four? The general rule is:

    pr (A or B or C) = pr(A) + pr(B) + pr(C) - pr(A and B) - pr(A and C) - pr(B and C) + pr(A and B and C)

    So, pr(5 before 2 or 3 or 4) = pr(5 before 2)+pr(5 before 3)+pr(5 before 4)-pr(5 before 2 and 3)-pr(5 before 2 and 4)-pr(5 before 3 and 4)+pr(5 before 2, 3, and 4) = (4/5)+(4/6)+(4/7)-(4/7)-(4/8)-(4/9)+(4/10) = 83/90. The probability of not getting the four along the way to the two to four is 1 - 83/90 = 7/90. So, there is a 7.78% chance of needing the extra 7.2 rolls. Thus, the expected number of rolls to get a two, three, four, and five is 43.8 + (7/90)*9 = 44.5.

  5. Continue with the same logic, for totals of six to twelve. The number of calculations required for finding the probability of getting the next number before it is needed as the last number roughly doubles each time. By the time you get to the twelve, you will have to do 1,023 calculations.

    Here is the general rule for pr(A or B or C or ... or Z)

    pr(A or B or C or ... or Z) =
    pr(A) + pr(B) + ... + pr(Z)
    - pr (A and B) - pr(A and C) - ... - pr(Y and Z) Subtract the probability of every combination of two events
    + pr (A and B and C) + pr(A and B and D) + ... + pr(X and Y and Z) Add the probability of every combination of three events
    - pr (A and B and C and D) - pr(A and B and C and E) - ... - pr(W and X and Y and Z) Subtract the probability of every combination of four events

    Then keep repeating, remembering to add probability for odd number events and to subtract probabilities for an even number of events. This obviously gets tedious for large numbers of possible events, practically necessitating a spreadsheet or computer program.

The following table shows the the expected number for each step along the way. For example, 36 to get a two, 42 to get a two and three. The lower right cell shows the expected number of rolls to get all 11 totals is 61.217385.

Expected Number of Rolls Problem

Highest Number Needed Probability Expected Rolls if Needed Probability not Needed Probability Needed Expected Total Rolls
2 0.027778 36.0 0.000000 1.000000 36.000000
3 0.055556 18.0 0.666667 0.333333 42.000000
4 0.083333 12.0 0.850000 0.150000 43.800000
5 0.111111 9.0 0.922222 0.077778 44.500000
6 0.138889 7.2 0.956044 0.043956 44.816484
7 0.166667 6.0 0.973646 0.026354 44.974607
8 0.138889 7.2 0.962994 0.037006 45.241049
9 0.111111 9.0 0.944827 0.055173 45.737607
10 0.083333 12.0 0.911570 0.088430 46.798765
11 0.055556 18.0 0.843824 0.156176 49.609939
12 0.027778 36.0 0.677571 0.322429 61.217385

This question was raised and discussed in the forum of my companion site Wizard of Vegas.