Ask the Wizard: Dice Probability

Questions about Probability, Statistics, & Math

Probability FAQ General probabilities, statistics, & math Probabilities with Dice Probabilities with Cards Probabilities with Coins Math-related puzzles & tricks

NOTE: Probabilities for specific games (e.g., blackjack) are listed in the section for that game, not here. This probabilities section is only for general questions about probability, risk of ruin, and other gambling math issues.

Can you tell me the odds of rolling two of the same number with two dice, three dice, and four dice? I am wondering how many dice would one have to roll at one time so that the odds are on the side of the person rolling the dice. (It does not make any difference which number is doubled.) - Mary from Minneapolis, MN

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

Probability of a Pair or More
Rolls	Probability
2 rolls	16.67%
3 rolls	44.44%
4 rolls	72.22%
5 rolls	90.74%
6 rolls	98.46%

So if you were to book this you should offer the yes with 3 rolls or the no at 4 rolls. Feb. 1, 2006

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. Nov. 9, 2005

There are 3 dice, 2 are proper six sided dice, while one is a die with all sides containing a six. All the dice are in my pocket. I randomly take out a die and throw it. The result is a 6. What is the probability that the die was one of the proper dice with 6 different values? - Annojh from Toronto

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. Nov. 9, 2005

In a recent programming exercise myself and other students were asked to describe a 6-sided die in code, and then use our dice to determine play simple game. The object of the game was to roll the dice until the sum of the tosses reached exactly 100. Any toss that put the total over 100 would not be added and merely added to statistics. Quickly it was determined that 17 throws would be the least amount of throws needed to reach 100. However calculating the odds of that occurring has proved elusive. Calculating the odds of a specific sequence of throws is rather straight forward, but how might one factor in both non-specific ordering of throws, and the different ways of reaching 100 in 17 throws (16*6 + 1*4 and 15*6 + 2*5)? - Björn from Göteborg, Sweden

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)*Ö*(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. Sept. 25, 2005

Suppose we roll three fair six-sided dice. What's the conditional probability that the first dice shows 4, given that the sum of the three numbers showing is 12? -- Shikha from North Ryde

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%. Sept. 18, 2005

What is the probability of rolling 13 or more with 3, 4, and 5 dice, if you are allowed to keep the highest three dice in your roll?

Here are the probabilities:

3 dice: 25.93%
4 dice: 48.77%
5 dice: 66.13%. Jan. 9, 2005

If I roll a single die 6 times, what is the probability of getting a "2" exactly 4 times?

Combin(6,2)*(1/6)⁴*(5/6)² = 0.008037551. Jan. 9, 2005

In the game of Yahtzee if only the Yahtzee itself is left on the card what is the probability of making it?

The following table shows the probability of success on the last roll according to the number of additional dice you need to make a Yahtzee.

Last Roll Yahtzee Probabilities
Needed	Probability of Success
0	1
1	0.166667
2	0.027778
3	0.00463
4	0.000772

The next table shows the probabilities of improvement. The left column shows how many dice you need before any given roll and the top column shows how many you need after the roll. The body shows the probability of the given degree of improvement.

Probabilities of Improvement
Need Before Roll	0	1	2	3	4	Total
0	1	0	0	0	0	1
1	0.166667	0.833333	0	0	0	1
2	0.027778	0.277778	0.694444	0	0	1
3	0.00463	0.069444	0.37037	0.555556	0	1
4	0.000772	0.01929	0.192901	0.694444	0.092593	1

The next table shows the probability on the initial roll of needing 0 to 4 more dice to make a Yahtzee.

First Roll Yahtzee Probabilities
Needed	Probability
0	0.000772
1	0.019290
2	0.192901
3	0.694444
4	0.092593

The next table shows the probability of improvement and then eventual success according to the number needed after the first roll. For example, if the player needs 3 more dice to make a Yahtzee the probability of improving to needing 2 more after the second roll and making the Yahtzee on the third roll is 0.010288066.

Probabilities of Yahtzee after first roll according to number needed before and after second roll
Need Before Roll	0	1	2	3	4	Total
0	1	0	0	0	0	1
1	0.166667	0.138889	0	0	0	0.305556
2	0.027778	0.046296	0.01929	0	0	0.093364
3	0.00463	0.011574	0.010288	0.002572	0	0.029064
4	0.000772	0.003215	0.005358	0.003215	0.000071	0.012631

To get the final answer take the dot product of the number needed after the first roll two tables up and the probability of eventual success in the final column one table up. This is 0.092593*0.012631+ 0.694444*0.029064 + 0.192901*0.093364 + 0.019290*0.305556 + 0.000772*1 = 4.6028643%. To confirm this I did a 100,000,000 game simulation and the simulated probability was 4.60562%. Dec. 13, 2004

What is the probability of rolling 1,2,3,4,5,6 with six dice, six times in a row?

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. Oct. 17, 2004

Hi Wizard, I wanted to know if you can answer this. In a popular gambling game in 17th century France, a player would roll a pair of dice 24 times. He would win his bet if at least one of these rolls was a double six. There was a debate at the time over whether the probability of winning was above or below an even 50%. Can you help me?

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. Oct. 17, 2004

What are the probabilities for a 5 of a kind, 4 of a kind, 3 of a kind, full house, 2 pair, pair, straight, and nothing with the roll of five dice?

• 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. Sept. 30, 2004

I recently attended a hospital fete. There was a new car as a prize if 7 dice produced 7 sixes in one throw. £1.00 a go. Odds on this must be high but how high?

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. September 30, 2004

[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?]

If two people throw a pair of dice, what is the probability that it is the same number? Is there a formula to figure this out?

Yes. You simply run through all totals 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%. September 23, 2004

If I roll three dice, what is the probability of getting at least two numbers the same?

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%. July 11, 2004

If I throw 36 dice what is the probability of getting at least one six?

1-(5/6)³⁶ = 99.86% April 22, 2004

Great site. I refer to it often as a gambler with an interest in probability and statistics, but this question actually pertains to my work. My HR Department insists that I rate my small staff (5 people) on a bell curve--one in the top 5% of all employees, one in the next 20%, one in the next 50%, one in the next 20%, and one in the bottom 5%. The company has approximately 5000 employees. What is the probability of such a small sample size fitting this distribution?

Thanks for the compliment. This is a good problem. The probability that exactly one employee will be in the bottom 5% is 5*(.05)*(.95)⁴ = 0.203627. Given that one employee is in the lowest 5% the probability of exactly one in the next 20% is 4*(.2/.95)*(.75/.95)³ = 0.414361. Given these two underachievers the probability of exactly one in the next 50% out of the remaining 75% is 3*(.5/.75)*(.25/.75)² = 0.222222. The probability that one of the remaining two falls in the lower 20% of 25% is 2*(.2/.25)*(.05/.25) = 0.32. Taking the product of all these probabilities we get 0.006, or 3/5 of 1%. April 4, 2004

If I kept throwing and removed all the sixes each time, how would I predict the theoretical number of dice remaining after a particular number of throws?

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. April 22, 2004

If I have any given number of dice what is the probability that if I roll them all of them at least one will land on a one?

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%. Jan. 4, 2004

A fair sided die is rolled 30 times. What is the expected number of times that number 1 will come up? What is the probability that number 1 will come up it's expected number of times?

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%. Sept. 26, 2003

What is the probability of getting a straight by a single throw of 5 dice? - Teodoro C. Deocares from Dagupan, Philippines

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. Aug. 31, 2002

I recently rolled, during a game of backgammon, double sixes four consecutive times. What are the odds of this happening again? - David from Sunland, USA

With every new roll the probability the next four rolls will be all double sixes is (1/36)⁴ = 1 in 1679616. Aug. 31, 2002

My friend owns a bar and has a "shake of the day" where there are ten dice in a Tupperware container, what are the odds of matching 8 out of the 10 in one shake. Thank you for your time. - August from Oshkosh, USA

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. June 28, 2002

What is the probability of rolling a "pair" when tossing 4 dice? - Anthony from Toronto, Canada

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%. Jan. 2, 2002

What is the probability of any one die showing ONE when 3 dice are throw together. My understanding is it should be 50% chance (1/6+1/6+1/6=1/2 -->50%) But in your odds table it shown 34.72%. Please help. - John C. from Singapore

The probability of rolling exactly one one with three dice is 3*(5/6)²*(1/6) = 75/216 = 34.72%. Nov. 11, 2001

How many different ways are there of rolling 3 ones using 6 dice? - Jamie Cook from Croydon, England

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=46656, so the probability of rolling three ones is 2500/46656=0.0536. For help with the combin function see my probabilities in poker section. June 6, 2001

If I roll a die, my probability of rolling a six is 1\6. If I roll two dice, does my probability of rolling a six on one of them increase, or does it stay at 1\6? - Mike R. from Rosemount.

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%. May 1, 2001

Mr. Wizard, what is the probability of rolling two pair when rolling four dice? - Brian from St. Catharines, Canada

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=.069444. Feb. 10, 2001

I am interested in finding out some specific information on the odds of rolling dice. If you have 6 dice and roll them all at once, the odds of rolling all "1's" is 46,656. My question is what are the odds of rolling (5) 1's, (4) 1's, (3) 1's, (2) 1's, and (1) 1. I am really interested in finding out the formula that should be used to calculate this type of problem. -- Ken from ?

The probability of rolling x ones out of y dice is combin(y,x)*(1/6)^x*(5/6)^y-x. See my section on probabilities in poker for an explanation of the combin(x,y) function. For example the probability of rolling 4 ones is combin(6,4)*(1/6)⁴*(5/6)² = 0.803755%. Dec. 31, 2000

How often can you roll a pair of dice 28 times without getting a 7? How do you figure this? Congratulation on your site, it's great. - Arturo Gomez from Mexico City, Mexico

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. Oct. 5, 2000

Q: What are the odds of rolling the same number with six dice in one roll? ñ Kyle Hill of Colstrip, Montana

A: The odds of getting six of the same number with six dice is 6*(1/6)⁶=1/7776 =~ 0.01286%. Sept. 3, 2000

Q: If you are rolling 6-six sided standard dice what are the odds of rolling six of a kind? - Jeff Blum of Miami, Florida

The answer is 6*(1/6)⁶ = 6/46656 = 1/7776 =~ 0.0001286 . March 4, 2000

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