Probability - Dice

The average number of rolls per shooter is 8.525510. For the probability of exactly 2 to 200 rolls, please see my craps probability of survival page.

My craps appendix shows how to work out the odds for any one bet. There you will see the probability of losing the don’t pass bet is 2928/5940. The probability of losing n bets in a row is (2928/5940)ⁿ. The frequency in 100,000 of losing exactly n can be closely approximated as 100,000 * (2928/5940)ⁿ⁺².

The odds of getting six of the same number with six dice is 6*(1/6)⁶=1/7776 =~ 0.01286%.

Thanks for the compliment. I take it you mean what is the probability of rolling a pair of dice 28 times without getting a 7. The probability of not rolling a 7 on any one roll is 5/6. The probability of not rolling a 7 in 28 rolls is (5/6)²⁸ = 0.006066, or about 1 in 165.

The probability of three matching is 1/216. The probability of two matching is 3*5/216. The probability of one matching is 25*5/216. The probability of 0 matching is 5*5*5/216. So the expected return is 3*(1/216)+2*(15/216)+1*(75/216)-1*(125/216)=-17/216=-7.87%. There is no optimal number of bets, you will give up an expected 7.87% of total money bet no matter what you do.

These bets can be made in both sic bo and chuck a luck.

There are combin(6,2)=15 different sets of pairs possible. There are combin(4,2)=6 ways the dice can roll any specific two pair. There are 6^4=1296 ways to roll four dice. So the probability is 90/1296=6.9444%.

If you rolled x dice the probability of getting at least one 6 is 1-(5/6)². In the case of two dice this is 30.56%.

First there are combin(6,3)=20 ways you can choose three dice out of 6 for the three ones. Then each of the other three can be any of five numbers. So, the total ways are 20×5³=2500. The total number of ways to throw all the dice are 6⁶=46,656, so the probability of rolling exactly three ones is 2500/46656=0.0536. For help with the combin function see my probabilities in poker section.

The probability of rolling exactly one one with three dice is 3*(5/6)²*(1/6) = 75/216 = 34.72%.

The pair can be any one of 6 numbers. The other two singletons can be among the other five. So there are 6*combin(5,2)=60 combinations already. There are combin(4,2)=6 combinations of dice on which the pair can appear. The two singletons can be arrange in two ways. So there are 60*12=720 ways to throw a pair. The total number of all ways to throw the dice is 6⁴=1296. So the probability is 720/1296 =~ 55.56%.

The probability that if you roll 10 dice and exactly 8 numbers are the same is 6*combin(10,8)*(1/6)⁸*(5/6)² = 1/8957.952. The probability of matching at least 8 is 6*[combin(10,8)*(1/6)⁸*(5/6)² + combin(10,9)*(1/6)⁹*(5/6) + (1/6)¹⁰] = 1/8569.469.

With every new roll the probability the next four rolls will be all double sixes is (1/36)⁴ = 1 in 1679616.

There are two possible spans: 1 to 5 and 2 to 6. Each of these spans can be ordered in 5!=120 ways. There are 6⁵ = 7776 ways to roll five dice. So the probability is 2*120/7776 = 3.09%. The probability of this seems to be much higher right after I put mark 0 for large straight during a game of Yahtzee.

The expected number of ones is 30*(1/6) = 5. The probability of exactly 5 ones is combin(30,5)*(1/6)⁵*(5/6)²⁵ = 19.21%.

The probability that all the dice will not be a one is (5/6)ⁿ. So the probability of at least one 1 is 1-(5/6)ⁿ. Let’s take an example of five dice. The answer would be 1-(5/6)⁵ = 59.81%.

1-(5/6)³⁶ = 99.86%

Each roll the expectation is that 5/6 of the dice will remain. So the expected number of dice remaining after n throws would be 36*(5/6)ⁿ. For example after 10 throws you would have 5.81 dice left, on average.

The probability all numbers will be different is (5/6)*(4/6)=20/36. So the probability at least two numbers will be the same is 1-(20/36) = 16/36 = 44.44%.

Yes. You simply run through all total from 2 to 12 and determine the probability of rolling each twice. So the answer would be (1/36)²+(2/36)²+(3/36)²+(4/36)²+(5/36)²+(6/36)²+(5/36)²+(4/36)²+(3/36)²+(2/36)²+(1/36)² = 11.27%.

The probability of throwing seven sixes with seven dice is (1/6)⁷ = 1 in 279,936. So the car would have to have a value of £279,936 or more for this to be a good bet. Even your average Rolls Royce is not worth this much, so I would say that was a terrible bet.

[Bluejay adds: Uh, yeah, but I think the point was that it was for charity. What’s more fun: Donating £1.00 to charity and getting nothing back but the good feeling of helping out, or donating £1.00 and getting the good feeling plus the longshot chance of winning a car?]

• Five of a kind: 6/6⁵ = 0.08% (obvious)
• Four of a kind: 5*6*5 = 1.93% (five possible positions for the singleton * 6 ranks for the four of a kind * 5 ranks for the singleton).
• Full house: combin(5,3)*6*5/6⁵ = 3.86% (combin(5,3) positions for the three of a kind * 6 ranks for the three of a kind * 2 ranks for the pair).
• Three of a kind: COMBIN(5,3)*COMBIN(2,1)*6*COMBIN(5,2) / 6⁵ = 15.43%. (combin(5,3) positions for the three of a kind * combin(2,1) positions for the larger of the singletons * 6 ranks of the three of a kind * combin(5,2) ranks for the two singletons.
• Two pair: COMBIN(5,2)*COMBIN(3,2)*COMBIN(6,2)*4 / 6⁵ = 23.15% (combin(5,2) positions for the higher pair * combin(3,2) positions for the lower pair * combin(6,4) ranks for the two pair * 4 ranks for the singleton.
• Pair: COMBIN(5,2)*fact(3)*6*combin(5,3) / 6⁵ = 46.30% (combin(5,2) positions for the pair * fact(3) positions for the three singletons * 6 ranks for the pair * combin(5,3) ranks for the singletons.
• Straight: 2*fact(5) / 6⁵ = 3.09% (2 spans for the straight {1-5 or 2-6} * fact(5) ways to arrange the order).
• Nothing: ((COMBIN(6,5)-2)*FACT(5)) / 6⁵ = 6.17% (combin(6,5) ways to choose 5 ranks out of six, less 2 for the straights, * fact(5) ways to arrange the order.

Sure, this is easy. The probability of rolling at least one 12 in 24 rolls is 1-(35/36)²⁴ = 49.14%. So the odds favor betting against a 12. This is a clever bet because the expected number of twelves in 24 rolls is 2/3. However that does not mean the probability of a 12 is 2/3, because sometimes there will be more than one 12, and the player betting on 12 doesn’t win any more for extra twelves after the first one. If the probability of winning any given trial is p, the number of trials is n, and the probability of at least one win is w then solving for n in terms of p and w gives us...

w=1-(1-p)ⁿ
1-w = (1-p)ⁿ
log(1-w) = log((1-p)ⁿ)
log(1-w) = n*log(1-p)
n= log(1-w)/log(1-p)

So in your example n = log(1-.5) / log(1-(1/36)) = log(0.5) / log(35/36) = 24.6051. So if the probability of success is 50% in 24.6 rolls it must be slightly less in 24 rolls.

The probability of rolling 123456 with six dice in a single roll can be expressed as prob(second die does not match first die) * prob(third die does not match first or second die) * ... = 1*(5/6)*(4/6)*(3/6)*(2/6)*(1/6) = 0.015432. So the probability of doing this six times in a row is 0.015432⁶ = 1 in 74,037,208,411.

Combin(6,2)*(1/6)⁴*(5/6)² = 0.008037551.

Here are the probabilities:

3 dice: 25.93%
4 dice: 48.77%
5 dice: 66.13%.

The probability of A given B is the probability of A and B divided by the probability of B. In this case the probability of rolling a 4 on the first die and then a total of 8 on the other two is (1/6)*(5/36) = 5/216. The probability of rolling any total of 12 with 3 dice is 25/216, as shown in my sic bo section. So the answer is (5/216)/(25/216) = 5/25 = 20%.

The two ways you mention are the only ways to throw a total of 100 in 17 throws. The probability of throwing 16 sixes and one four is 17*(1/6)¹⁷. There are 17 possible positions of the 4 and each sequence has a probability of (1/6)*(1/6)*...*(1/6) with 17 terms. The number of ways to get 15 sixes and 2 fives is combin(17,2) = 136. So the probability of 15 sixes and 2 fives is 136*(1/6)¹⁷. So the total probability is (17+136)*(1/6)¹⁷. = 1 in 110,631,761,077.

Let A = Choosing the normal die
Let B = Rolling a 6 with randomly chosen die
Answer = Pr(A given B) = Pr(A and B)/pr(B) = ((2/3)*(1/6))/((2/3)*(1/6)+(1/3)*1) = (2/18)/((2/18)+(6/18)) = 1/4.

There are 6!/(4!*1!*1!) = 30 ways to arrange these numbers in any order. Another way to look at it is there are 6 positions to put the 1, and 5 left to put the 4, so 6*5=30. The probability of getting 666614 in exactly that order is 1 in 6⁶ = 1 in 46656. Multiply that by 30 for the 30 possible orders and the answer is 30/46656 = 0.0643%, or 1 in 1552.2.

It is true that for single events if the probability is p then the average wait time is 1/p. However it gets more complicated with consecutive events. Let x be the state that the last roll was not a two. This is also the state at the beginning. Let y be the state that the last roll was a two. After the first roll there is a 5/6 chance we will still be in state x, and 1/6 chance we will be in state y. Let Ex(x) be the expected number of rolls from state x, and Ex(y) the expected number from state y. Then...

Ex(x) = 1 + (5/6)*ex(x) + (1/6)*ex(y), and
Ex(y) = 1 + (5/6)*ex(x)

Solving for these two equations...

Ex(x) = 1 + (5/6)*ex(x) + (1/6)*( 1 + (5/6)*Ex(x))
Ex(x) = 7/6 + (35/36)*Ex(x)
(1/36)*Ex(x) = 7/6
Ex(x) = 36*(7/6) = 42

So the average wait time for two consecutive twos is 42 rolls.

I have the same type of problem, only the expected flips to get two heads, in my site of math problems, see problem 128.

Here is the probability of getting at least one number more than once according to the number of rolls:

I started to use the Normal approximation to solve this, but the probability of over 100 points is too low for that method to be accurate. So I did a random simulation of 8.25 million trials and the number of trails that were 101 points or more was 127. So the probability is about 1 in 65,000.

Let n be your answer. The probability of rolling a 7 or 11 is 8/36. To solve for n:

(8/36)ⁿ = 1/41,400,000

log((8/36)ⁿ) = log(1/41,400,000)

n × log(8/36) = log(1/41,400,000)

n = log(1/41,400,000)/log(8/36)

n = -7.617 / -0.65321

n = 11.6608

So there you go, the probability of hitting the SuperLotto is the same as rolling a seven or eleven 11.66 times in a row. For those who can’t comprehend a partial throw I would rephrase as the probability falls between 11 and 12 consecutive rolls.

The probability of a five-of-a-kind on one throw is 6*(1/6)⁵ = 1/1,296. This is because there are six different five-of-a-kinds (one to six) and the probability each die will be that number is (1/6). The probability of not getting a five-of-a-kind is 1-(1/1,296)=1,295/1,296. The probability of going three attempts without a three of a kind is (1,295/1,296)³=99.77%. So the probability of getting at least one five-of-a-kind in three tries is 100%-99.77% = 0.23%.

If the probability of something is p then on average it will take 1/p trials for it to happen the first time. Obviously, on the first roll you’ll cross off one number. The probability of rolling one of the other five numbers next is 5/6. So it will take on average 1/(5/6)=6/5=1.2 rolls for that to happen. Following that reasoning to the end, the expected number of rolls is (6/6)+(6/5)+(6/4)+(6/3)+(6/2)+(6/1) = 14.7.

I hope you’re happy, I just added a new section answering questions such as this for 1 to 25 dice. As the five-dice table shows, the probability of rolling a total of 12 is 0.039223251028807.

I’m afraid you had the square side of this bet. The probability of rolling two sevens before a six and eight is 45.44%. Here are all the possible outcomes. The first column is the order of petintent rolls to the outcome of the bet, ignoring all others.

Basically, the reason the 6 and 8 is the better side is you can hit those numbers in either order: 6 then 8, or 8 then 6. With two sevens there is only one order, a 7 and then another 7.

The probability of six sixes is (1/6)⁶ = 1 in 46656. The probability of rolling 1,2,3,4,5,6 with six dice is 6!/6⁶ = 1 in 64.8

1-(5/6)¹⁰-10 × (1/6) × (5/6)⁹ = 51.55%.

The probability of a 7, 11, or 12 is (6+2+1)/36 = 9/36 = 1/4. See my section on dice probability basics for how I arrived at that figure. The probability of rolling anything else is 3/4. The probability of going five rolls without rolling a 7, 11, or 12 is (3/4)⁵ = 23.73%.

Not that you asked, but let me address the mean first. For a six-sided die, the expected number of throws to get each face at least once is (6/6)+(6/5)+(6/4)+(6/3)+(6/2)+(6/1) = 14.7. For an n-sided die the expected throws is (n/n) + (n/(n-1)) + (n/(n-2)) + ... + n. The median number of throws required is 13. The probability of taking 13 rolls or less is 51.4%, and 13 rolls or more is 56.21%.

For large numbers of throws we can use the Gaussian Curve approximation. The expected number of sevens in 655 throws is 655 × (1/6) = 109.1667. The variance is 655 × (1/6) × (5/6) = 90.9722. The standard deviation is sqr(90.9722) = 9.5379. Your 78 sevens is 109.1667 − 78 = 31.1667 less than expectation. This is (31.1667 - 0.5)/9.5379 = 3.22 standard deviations below expectation. The probability of falling 3.22 or more standard deviations south of expectations is 0.000641, or 1 in 1,560. I got this figure in Excel, using the formula, normsdist(-3.22).

Thank you for the kind words. You should not state the probability that the throws were non-random is p. The way it should be phrased is the probability that a random game would produce such a result is p. Nobody expected 500 rolls to prove or disprove anything. It wasn’t I who set the line at 79.5 sevens, but I doubt it was chosen to be statistically significant; but rather, I suspect the it was a point at which both parties would agree to the bet.

The 2.5% level of significance is 1.96 standard deviations from expectations. This can be found with the formula =normsinv(0.025) in Excel. The standard deviation of 500 rolls is sqr(500*(1/6)*(5/6)) = 8.333. So 1.96 standard deviations is 1.96 * 8.333 = 16.333 rolls south of expectations. The expected number of sevens in 500 throws is 500*(1/6) = 83.333. So 1.96 standard deviations south of that is 83.333 − 16.333 = 67. Checking this using the binomial distribution, the exact probability of 67 or fewer sevens is 2.627%.

Assuming the player always holds the most represented number, the average is 11.09. Here is a table showing the distribution of the number of rolls over a random simulation of 82.6 million trials.

Speaking only in terms of maximizing expected value, the player should play forever. While the probability is 1 that the player will eventually lose, at any given decision point the expected value always favors going again. It seems like a paradox. The answer lies in the fact that some events have a probability of 1, but still may not happen. For example, if you threw a dart at a number line from 0 to 10, the probability of not hitting pi exactly is 1, but it still could happen.

However, for practical purposes, there is some stopping point. This is because the happiness money brings is not proportional to the amount. While it is commonly accepted that more money brings more happiness, the richer you get, the less happiness each additional dollar brings you.

I believe a good way to answer this question is to apply the Kelly Criterion to the problem. According to Kelly, the player should make every decision with the goal of maximizing the expected log of his bankroll after the wager. To cut to the end of this (I cut out a lot of math), the player should keep doubling until the wager amount exceeds 96.5948% of his total wealth. Wealth should be defined as the sum of the amount won plus whatever money the player had before he made the first wager. For example, if the player had $100,000 to start with, he should keep doubling up to 23 times, to a win of $4,194,304. At that point the player’s total wealth will be $4,294,304. He will be asked to wager 4,194,304/4,294,304 = 96.67% of his total wealth, which is greater than the 96.5948% stopping point, so he should quit.

Let the answer to this question be called p. The probability of rolling a total of six is 5/36, and the probability of rolling a total of seven is 6/36. If you don’t understand why, please see my section on dice probability basics. We can define p as:

p = Prob(6 on first roll) + Prob(no 6 on first roll)*Prob(no 7 on second roll)*p.

This is because, if neither player wins after the first two rolls, the game is back to the original state, and the probability of player A winning remains the same.

So, we have:

p = (5/36) + (31/36)×(30/36)×p
p = 5/36 + (930/1296)×p
p * (1-(930/1296)) = 5/36.
p * (366/1296) = 5/36
p = (5/36)×(1296/366) = 30/61.

The answer can be expressed as combin(n+5,n) = (n+5)!/(120×n!). Here is the answer for 1 to 20 dice.

Credit to Alan Tucker, author of Applied Combinatorics.

Sure. That would be Pr(2)² + Pr(3)² + ... + Pr(12)² = (1/36)² + (2/36)² + (3/36)² + (4/36)² + (5/36)² + (6/36)² + (5/36)² + (4/36)² + (3/36)² + (2/36)² + (1/36)² = 11.27%.

I can’t think of any useful reason to know this information, but I get asked this kind of thing a lot, so I’ll humor you.

It is a little easier getting a specified sequence of sevens starting with the first roll, or ending with the last, because the sequence is bounded on one side. Specifically, the probability of getting a sequence of s sevens, starting with the first roll, or ending with the last, is (1/6)^s × (5/6). The 5/6 term is because you have to get a non-7 at the open end of the sequence.

The probability of starting a sequence of s sevens at any point in the middle of the sequence is (1/6)^s × (5/6)². We square the 5/6 term, because the player must get a non-7 on both ends of the sequence.

If there are r rolls, there will be 2 places for an inside sequence, and r-n-1 places for a run of n sevens. Putting these equations in a table, here is the expected number of runs of sevens, from 1 to 10. The "inside" column is 2*(5/6)*(1/6)^r, and the "outside" column is (179-r)*(5/6)²*(1/6)^r, where r is the number of sevens in the run. So, we can expect 3.46 runs of two sevens, 0.57 runs of three sevens, and 0.10 runs of four sevens.

The answer and solution can be found on my companion site, mathproblems.info, problem 201.

If you limit yourself to the regular polygons, and want every face to have the same probability, then you are limited to the platonic solids. However, if you can lift the regular polygon requirement, then you can add the 13 Catalan solids as well.

To answer your other question, no, I have never seen a game actually in a casino that used any dice other than cubes. About ten years ago I saw a game demonstrated at a gaming show in Atlantic City that I think used a Rhombic triacontahedron, one of the Catalan solids, but I don’t think it ever made it to a casino floor. There is a game I see year after year at the Global Gaming Expo that uses a spinning top (like a dreidel), but alas, I’ve never seen that in a casino either.

There are 6³=216 ways to roll three dice. Six of those combinations will result in a three of a kind (1-1-1 to 6-6-6). So the probability of a three of a kind is 6/216 = 1/36. There are four possible spans for a straight (1-2-3 to 4-5-6). There are also 3!=6 ways to arrange the three dice in a straight. So, there are 4*6=24 straight combinations. Thus the probability of a straight is 24/216 = 1/9.

The following table shows the number of combinations for all possible totals from 3 to 18.

The mean result is 12.2446, and the standard deviation is 2.8468.

The probability of rolling a 7 is 1/6, and the probability of rolling a 12 is 1/36. The probability of rolling a 7, given that a roll is a 7 or 12 is (1/6)/((1/6)+(1/36)) = 6/7. So the probability that the first six times a 6 or 12 is rolled it is a 6 every time is (6/7)⁶ = 39.66%.

If you rephrase the question to be what is the probability of rolling five 6’s before a 12, then the answer is (6/7)⁵ = 46.27%. With four rolls it is (6/7)⁴ = 53.98%. So there is no number of 7’s before a 12 that is exactly 50/50. If you’re looking for a good sucker bet, suggest you can either roll four 7’s before a 12, or a 12 before five 7’s.

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

Here is a handy trick, courtesy of Robert Goodhand of Somerset, UK. First put on a row six ones surrounded by five zeros on either side, as follows:

This represents the number of combinations for rolling a 1 to 6 with one die. I know, pretty obvious. However, stick with me. For two dice, add another row to the bottom, and for each cell take the sum of the row above and the five cells to the left of it. Then add another five dummy zeros to the right, if you wish to keep going. This represents the combinations of rolling a total of 2 to 12.

For three dice, just repeat. This will represent the number of combinations of 3 to 18.

To get the probability of any given total, divide the number of combinations of that total by the total number of combinations. In the case of three dice, the sum is 216, which also easily found as 6³. For example, the probability of rolling a total of 13 with three dice is 21/216 = 9.72%.

So for d dice, you’ll need to work your way up through 1 to d-1 dice. This is very easily accomplished in any spreadsheet.

That is a classic problem in the history of the field of probability. Many people incorrectly think the answer is 18, because the probability of a 12 is 1 in 36, and 18×(1/36)=50%. However, by that logic, the probability of getting a 12 in 36 rolls would be 100%, which clearly it isn’t. Here is the correct solution. Let r be the number of rolls. The probability a roll isn’t a 12 is 35/36. The probability there are 0 12s in r rolls is (35/36)^r. So we need to solve for r in the following equation:

(35/36)^r = 0.5
log(35/36)^r = log(0.5)
r × log(35/36) = log(0.5)
r = log(0.5)/log(35/36)
r = 24.6051

So there isn’t a round answer. The probability of rolling a 12 in 24 rolls is 1-(35/36)²⁴ = 49.14%. The probability of rolling a 12 in 25 rolls is 1-(35/36)²⁵ = 50.55%.

If you want to make a bet on this, say you can roll a 12 in 25 rolls, or somebody else can’t in 24 rolls. Either way you’ll have an advantage at even money.

For those unfamiliar with the game, both the attacker and defender will roll 1 to 8 dice, according to how many armies they each have at that point in a battle. The higher total shall win. A tie goes to the defender. If the attacker loses, he will still retain one army in the territory where he initiated the attack. For this reason, he must have at least two armies to attack, so if he wins one can inhabit the conquered territory and one can stay behind.

The following table shows the probability of an attacker victory according to all 64 combinations of total dice.

The next table shows the expected gain by the attacker, defined as pr(attacker wins)*(defender dice)+pr(defender wins)*(attacker dice -1). It shows the greatest expected gain is to attack with 8 against an opponent with 5.

For the benefit of other readers, a Yahtzee is a five of a kind with five dice. In the game of Yahtzee the player may hold any dice he wishes and re-roll the rest. He can do this up to three rolls.

The player may re-roll previously held dice, if he wishes. For example, if the player's first roll is 3-3-4-5-6 and he holds the threes and then has 3-3-5-5-5 after the second roll he may keep the fives and re-roll the threes on his third roll.

The following table shows the maximum number of dice of the same face for 1 to 20 rolls. The table shows the probability of getting a Yahtzee within three rolls is about 4.6%.

The answer is 50/50. This will be true for any number of dice rolled, not just two.

A bit off-topic, but I've always thought an odd/even set of bets would be a good way to replace the dreaded big 6/8 bets in craps. To give the house an advantage, here are my proposed pay tables and analysis.

The average number of rolls is the inverse of the bonus-ending event, which has a probability of 1/6, so the player will roll six times on average. However, the last roll will be the seven, so an average of five winning rolls per bonus.

Next, here is the probability of each total, assuming no seven:

• 2 or 12: 1/30
• 3 or 11: 2/30
• 4 or 10: 3/30
• 5 or 9: 4/30
• 6 or 8: 5/30

Here was the original question posted in column #179: If two dice are rolled over and over, until either of the following events happen, then which is more likely to happen first:

• A total of six and eight is rolled, in either order, with duplicates allowed.
• A total of seven is rolled twice.

Your twist is that the same roll can't help both players. Instead, they take turns rolling and only the one rolling can use the roll.

The answer depends on who rolls first. If the player needing a six and eight rolls first, then he has a probability of winning of 57.487294%. If the player needing two sevens goes first, then the probability the player needing the six and eight wins is 52.671614%. I solved it using a simple Markov Chain process.

This question is asked and discussed in my forum at Wizard of Vegas.

There are 58 different types of sequences on the initial roll. The way I identify each is the number of the face in majority, then the number of dice of the face second in total, and so on. For example, a roll of of 3,3,3,3,6,6,6,5,5,2 would be signified as 4-3-2-1. The following table shows the number of combinations of each sequence, the probability of rolling it, the probability of completing a 12 of a kind in the second roll, and the product of the two. For the probability on the second roll, I assume the player holds the dice that have the greatest total on the initial roll. The lower right cell shows an overall probability of 0.0000037953, which equals 1 in 263,486.

Click the button below for the answer.

Here is my solution. (PDF)

This question is asked and discussed in my forum at Wizard of Vegas.

Click the button below for a couple infinite series formulas that you may find helpful.

Click the button below for the answer.

Click the button below for the solution.

This question is asked and discussed in my forum at Wizard of Vegas.

This question is asked and discussed in my forum at Wizard of Vegas.

This question is asked and discussed in my forum at Wizard of Vegas.

To answer this one as I did, using calculus, I recommend an integral calculator like the one at integral-calculator.com/.

Here is my solution (PDF).

This problem is asked (in slightly different words) and discussed in my forum at Wizard of Vegas.

This question is asked and discussed in my forum at Wizard of Vegas.

This question is asked and discussed in my forum a Wizard of Vegas.

Here is my solution (PDF).

This question is asked and discussed in my forum at Wizard of Vegas.

Here is my solution (PDF).

  [spoiler=Solution]

Let:

• p = Probability the 12 is rolled first from the initial state or anytime the previous roll was not a 7.
• q = Probability the 12 is rolled first when the previous roll was a 7.

This is what is known as a Markov Chain problem.

Before we get to that, recall the probability of rolling a total of 7 is 1/6 and that of a 12 is 1/36.

We can define p and q in terms of each other, as follows:

• (1) p = (1/36) + (6/36)q + (29/36)p
• (2) q = (1/36) + (29/36)p

Let's multiply equation (1) by 36:

36p = 1 + 6q + 29p
(3) 7p = 1 + 6q

Let's substitute the value for q in (2) into (3):

7p = 1 + 6*((1/36) + (29/36)p)
7p = 1 + (1/6) + (29/6)p
42p = 6 + 1 + 29p
13p = 7
q = 7/13

So, the probability of rolling the 12 first is 7/13 =~ 53.85%.

The probability of rolling two consecutive 7's first is thus 46.15%.

Thus, it's more likely the total of 12 is rolled first.