Ask the Wizard: Probability - Dice
If you are rolling 6 six-sided standard dice what are the odds of rolling six of a kind?
|Probability of a Pair or More|
What is the chance of getting a sum over 100, when rolling 20 dice? Kind regards
— Terje from Stockholm
Wizard, could you please describe the equivalent odds of the California SuperLotto Plus (1 in 41.4 Million), in terms of number of consecutive times of rolling 7 or 11? I heard it somewhere before. Most people cannot comprehend the lottery odds. But, the rolling of dice -- they can relate.
— Tim from Belmont, CA
(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.
We are in a disagreement between workers. there is a bar down the street that has a shake a day. which is you must throw five dice at once and all five must end up being the same "like yahtzee" but he gives you three chance at it. but you must pick up all the dice all three times. so the questions is what’s the odds to do it in one shake and what’s the odds to do it in the three shakes allowed. Thanks , if you already answered this before i am sorry but i couldt find it.
— Dan and co workers at maple island from Forest Lake
What is the expected number of tosses required in order to obtain at least one of each of the possible outcomes on an unbiased 6 sided dice?
— Michael from Melbourne
What is the classical probability of getting a total of 12 when 5 balanced dice are rolled?
— Kennith H. from Winters
Somebody bet he that he could roll a total of 6 and 8, with two dice, before I could roll a total of seven twice. This seemed like a good bet because seven is the most common total. However I lost $2500 doing this over and over. What are the odds?
— Anthony from Indiana
|Two Sevens before Six and Eight Bet|
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.
Two questions, please: 1) What is the probability of rolling 6,6,6,6,6,6 @ one time, with (6)6-sided die? 2)What is the probability of rolling 1,2,3,4,5,6 @ one time with (6) 6-sided die? Thanks! It’s killing me!
— Heather from Petaluma
What is the probability of getting any given number more than once if you roll a die ten times?
— Ryan from Silay
In a game called Taxation and Evasion, a player rolls a pair of dice. On any roll if the sum is 7,11, or 12, the player gets audited; any other sum they avoid taxes. If a player rolls the pair 5 times, what is the probability that he avoids taxes?
— George P. from Stevens Point, WI
How many throws of a die does it take before it is likely that you have thrown a 1, 2, 3, 4, 5, and 6 at least once each? Any ideas on generalizing this for an n-sided die?
— Asif from Columbia, SC
I know you’re skeptical of dice control. I have been practicing dice setting and controlled shooting for 3 months. What is the probability of throwing 78 sevens over 655 throws randomly? Thanks for the help :)
— Eric B. from Boston, MA
This is about controlling the dice at Craps. You previously discussed the Stanford Wong Experiment, stating, "The terms of the bet were whether precision shooters could roll fewer than 79.5 sevens in 500 rolls of the dice. The expected number in a random game would be 83.33. The probability of rolling 79 or fewer sevens in 500 random rolls is 32.66%.... The probability of rolling 74 or fewer sevens in 500 random rolls is 14.41%."
The question I have about this bet is that 14.41% still isn’t "statistically significant" [ i.e. p < 0.05 ] , which is usually taken to mean greater than two Standard Deviations from the Mean -- or a probability of less than a *combined* 5% of the event happening randomly on EITHER end of the series.
How many Sevens would have to be rolled in 500 rolls before you could say that there is a less than 2.5% chance that the outcome was entirely random (i.e. that the outcome was statistically significant) ?
Many Thanks & BTW , yours is ABSOLUTELY the BEST web site on the subject of gambling odds & probabilities that I’ve found .... keep up the good work !!!
— Plexus from Warwick, Rhode Island
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%.
What is the expected number of rolls needed to get a Yahtzee?
— Ian F. from Provo
Consider a hypothetical game based on the roll of a die. If die lands on 1, the player loses $1 and the game ends. If the die lands on anything else, the player’s wins $1. At this point the player may let it ride, or quit. The player may keep playing, doubling each bet, until he loses or quits. What is the best strategy?
— Byron P. from Newington, CT
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.
Players A and B throw a pair of dice. Player A wins if he throws a total of 6 before B throws a toal of 7, and B wins if he throws 7 before A throws 6. If A begins, show that A’s chances of winning are 30/61.
— Sangeeta from Mumbai, India
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.
How many ways are there to roll n six-sided, non-distinct dice? As stated, the dice are non-distinct, so with five dice, for example, 1-1-3-5-6 and 1-6-5-1-3 would be considered the same roll. With two dice, it’s trivial to determine that the answer is 21, but I can’t figure out an elegant, generalized solution.
— Don from New York
|Non-Distinct Dice Combinations|
Credit to Alan Tucker, author of Applied Combinatorics.
Can you calculate what the probability is of two numbers coming up behind each other in a roll of the dice? Meaning what is the probability of two 4’s or two 6’s or two 7’s back to back? I realize that the past cannot predict the future but is there a way to calculate 7/36 X 7/36 happening back to back? I hope that makes sence.
— James from Birmingham
In 180 consecutive rolls of the dice, how many times can I expect to see the following:
Two sevens in a row?
Three sevens in a row?
Four sevens in a row?
Thanks for your time :-).
— Melanie D. from Elizabeth City, NC
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.
|Expected Runs of Sevens in 180 Rolls|
Two dice are rolled until either a total of 12 or two consecutive totals of 7. What is the probability the 12 is rolled first?
I’m a tabletop gamer, and was having some discussion with my friends about non-cubical platonic solid dice (If you’re a big enough nerd, that means d4, d8, d12, and d20). They argued that those would be the only ones that would be demonstratively fair. I argued that manufacturing them to be fair would be entirely too difficult. Also, the only games would be craps variants rendered overly cumbersome due to the number of extra outcomes. Has any casino ever had a game that used non-traditional six sided dice?
— Bayani from Carnagie, PA
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.
If I roll three six-sided dice, what are the odds of rolling a straight and, also, what are the odds of rolling a three of a kind?
— Mark from Fargo, ND
What is the average sum when rolling four six-sided dice after subtracting the lowest result (known as 4d6-L)? What is the standard deviation for this roll?
— Aaron from New York
|Combinations in 4d6-L|
The mean result is 12.2446, and the standard deviation is 2.8468.
My question is based on dice odds. I know that there are six ways to get 7 and one way to get 12, but what are the chances of getting six 7’s before one 12? Are they even, and if not, how many twelves should be added to the equation to make it an even proposition?
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 .
Is there an easy way to calculate the probability of throwing a total of t with d 6-sided dice?
— Anon E. Mouse
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.
|Two Dice Probabilities|
For three dice, just repeat. This will represent the number of combinations of 3 to 18.
|Three Dice Probabilities|
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.
How many rolls of two dice would it take to have a 50/50 chance of rolling at least one 12?
(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.
In Dice Wars , what is the probability of success for any given number of attacking and defending dice? As an attacker, what ratio has the greatest expected gain?
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.
|Probability of Attacker Win|
|1 Army||2 Armies||3 Armies||4 Armies||5 Armies||6 Armies||7 Armies||8 Armies|
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.
|Net Gain of Attacker Win|
|1 Army||2 Armies||3 Armies||4 Armies||5 Armies||6 Armies||7 Armies||8 Armies|
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