dice odds help

I have an odds question we can't figure out, limited intellectual capacity
> I suppose.
> 
> I am a bartender, we have a "shift shake" in which a patron may shake one
> time per bartender shift...one flop of the dice to match the numbers that
> the last winner rolled. 
> 
> I argue at any given roll regardless of who is rolling, there is a 7776 to
> one chance of the numbers matching the last winners numbers. Order does not
> matter, only that they match. 
> 
> There are several people who.disagree, saying the odds arenlower that that
> or the more frequently there is a roll the more likely someone is to
> rolling the numbers.
> 
> 
> 
> Can you illuminate?
> 
> Thanks for your time!
> 
> Thomas

Quote: atpeace27

I have an odds question we can't figure out, limited intellectual capacity
> I suppose.
> 
> I am a bartender, we have a "shift shake" in which a patron may shake one
> time per bartender shift...one flop of the dice to match the numbers that
> the last winner rolled. 
> 
> I argue at any given roll regardless of who is rolling, there is a 7776 to
> one chance of the numbers matching the last winners numbers. Order does not
> matter, only that they match. 
> 
> There are several people who.disagree, saying the odds arenlower that that
> or the more frequently there is a roll the more likely someone is to
> rolling the numbers.
> 
> 
> 
> Can you illuminate?
> 
> Thanks for your time!
> 
> Thomas

Welcome to the forums, Thomas. Can you give some more detail? How many dice are there? Do the "dice totals" need to match (i.e. if I roll 4-3 and you roll 5-2, is that a winner because we both rolled 7)?

I can answer better when I know the rest of that information, but off the top my educated guess is that it's much more likely than you believe. On 2 dice there's only 36 possible outcomes. Thus, no matter what you're way off. Even if they roll 1-1, which is a 1-36, then the next person whom attempts has the worst odds possible, but is still going to win 1/36 times (on average)... Definitely less than 1 in 7776.

It would be 7775-1 if the dice were five different colors - say, a red 1, an orange 3, a yellow 4, a green 5, and a blue 6 - or if all five numbers were the same.

However, if all of the dice are the same color, the odds depend on what was previously rolled.

Matching a five of a kind: 1 / 7776

Matching five different numbers: 120 / 7776, or 1 / 64.8

Matching four of a kind: 5 / 7776 (because any of the five dice can be the one not in the four), or 1 / 1555.2

Matching a full house: 10 / 7776, or 1 / 777.6

Matching three of a kind: 20 / 7776 (any of the five can be the higher of the two numbers not in the three, and any of the remaining four can be the lower of the two), or 1 / 3110.4

Matching two pair: 30 / 7776 (any of the five dice can be the number not in either pair, and of the remaining four, each of the six pairs can be the higher pair of the two pair), or 259.2

Matching one pair: 60 / 7776 (any of the five can be the highest of the three numbers not in the pair, any of the remaining four can be the middle number of the three, and any of the remaining three can be the lowest of the three), or 1 / 129.6.


The overall probability that two consecutive rolls will match (before knowing what the first roll is) is about 1 / 157.

5 dice, randomly rolled, all same color, yes the totals need to match because you have to match the exact previous roll....
For instance, the current roll is 1, 2, 4,4,5

These are 5 black and white bar dice..... Ramdomly rolled. Each patron pays a quarter for their one flop of the dice to try to match the numbers.

Quote: atpeace27

These are 5 black and white bar dice..... Ramdomly rolled. Each patron pays a quarter for their one flop of the dice to try to match the numbers.

What does the patron get for a win?

If the bet is made after knowing what the needed numbers are, then, as I said, the probability of winning depends on the numbers needed (five of a kind, four of a kind, full house, three of a kind, two pair, one pair, five different numbers).

If the bet is made before knowing what you need to roll, then it is about 1/157.

deleted... more thought needed.

Quote: indignant99

Quote: ThatDonGuy

Matching five different numbers: 120 / 7776, or 1 / 64.8

followed by more nonsense, irrelevant to the problem

You over-thought this problem, to the point of botching it badly.
Say the previous roll result was A-B-C-D-E. It does not matter what the number of pips are, as represented by A, B, C, D, and E. You have to repeat A-B-C-D-E, in any order, of which there are 120 orderings. There are exactly 120 roll results that match the desired result, out of 7776 all possible rolls.

Your 1-in-64.8 is the correct answer.

Only if all five numbers are different.

There is only one ordering of A-A-A-A-A, so if you are trying to match a five of a kind, it becomes 1 in 7776.
There are five orderings of A-A-A-A-B (AAAAB, AAABA, AABAA, ABAAA, BAAAA), so if you are trying to match a four of a kind, it becomes 5 in 7776, or 1 in 1555.2.
And so on.

Quote: ThatDonGuy

Quote: indignant99

more nonsense,

I offer you a sincere apology. My post was the actual nonsense. That's why I deleted it. It is incontrovertibly correct that the probability of repeating depends on the preceding result. I realized that the distribution of results throwing 5 dice, is a graph quite similar to the 2-dice "curve" in ordinary craps. Heavy in the middle (17 and 18), and extremely improbable out in the tails (5, 6 , 29, 30). The details I am investigating, are the distinct vs. duplicate roll-outcomes from 17 down to 5, and from 18 up to 30.

Again, Don, I'm sorry.

Quote: ThatDonGuy

one ordering of A-A-A-A-A

At the risk of beating a dead horse, and at the risk of perpetuating a dead subject,
the letters ABCDE did not stand for a pip-count on a face. They stood for each specific die. That's why I said that the pip-counts on A-B-C-D-E did not matter. Given a roll-result, the pips on "A" are invariant, wherever the A-die falls in the 5-member sequence. And in my "poor example" (granted) you could not ever roll any result but A-B-C-D-E - perhaps rearranged - because they are the identities of the five dice.

Quote: indignant99

I offer you a sincere apology. My post was the actual nonsense. That's why I deleted it. It is incontrovertibly correct that the probability of repeating depends on the preceding result. I realized that the distribution of results throwing 5 dice, is a graph quite similar to the 2-dice "curve" in ordinary craps. Heavy in the middle (17 and 18), and extremely improbable out in the tails (5, 6 , 29, 30). The details I am investigating, are the distinct vs. duplicate roll-outcomes from 17 down to 5, and from 18 up to 30.

I think I misread the problem. I thought the player had to match each die separately. Is it that the player only has to match the total?

For example, suppose the "house" roll is 1,2,4,4,5 - does the player have to roll 1,2,4,4,5, or just anything that adds up to 16?

If they're trying to match the total, then here are the odds based on each total:

Total	Ways to roll it	Probability
5,30	1	1 / 7776
6,29	5	1 / 1555.2
7,28	15	1 / 518.4
8,27	35	1 / 222.171
9,26	70	1 / 111.086
10,25	126	1 / 61.714
11,24	205	1 / 37.932
12,23	305	1 / 25.495
13,22	420	1 / 18.514
14,23	540	1 / 14.4
15,20	651	1 / 11.945
16,19	735	1 / 10.58
17,18	780	1 / 9.97

Quote: ThatDonGuy

I think I misread the problem. I thought the player had to match each die separately.

I don't think you misread the problem.

In the first post, the OP doesn't mention the word TOTAL at all. He/she mentions matching the "last winning numbers."

Only in his second post, when clarifying that the roller has match "the EXACT previous roll," does he even mention the word TOTAL at all. But that mention was just to confirm that yes, the exact roll must match. (And when it does, the totals will match too.)

Quote: ThatDonGuy

does the player have to roll 1,2,4,4,5

Yes.
Here is an overview of all the ways to roll a total of 16. Then I'll get to all the dirty little details regarding 1-2-4-4-5.

Sum Pips	The Pips	# of ways	Note
16	1-1-2-6-6	30	Two Pair
16	1-1-3-5-6	60	One Pair
16	1-1-4-4-6	30	Two Pair
16	1-1-4-5-5	30	Two Pair
16	1-2-2-5-6	60	One Pair
16	1-2-3-4-6	120	All Different
16	1-2-3-5-5	60	One Pair
16	1-2-4-4-5	60	One Pair
16	1-3-3-3-6	20	Three of a Kind
16	1-3-3-4-5	60	One Pair
16	1-3-4-4-4	20	Three of a Kind
16	2-2-2-4-6	20	Three of a Kind
16	2-2-2-5-5	10	Full House
16	2-2-3-3-6	30	Two Pair
16	2-2-3-4-5	60	One Pair
16	2-2-4-4-4	10	Full House
16	2-3-3-3-5	20	Three of a Kind
16	2-3-3-4-4	30	Two Pair
16	3-3-3-3-4	5	Four of a Kind

And here are the 60 ways to achieve 1-2-4-4-5.

Cardinal #	Result	Dice A-B-C-D-E	# of ways
(1)	16	1-2-4-4-5	1
(2)	16	1-2-4-5-4	1
(3)	16	1-2-5-4-4	1
(4)	16	1-4-2-4-5	1
(5)	16	1-4-2-5-4	1
(6)	16	1-4-4-2-5	1
(7)	16	1-4-4-5-2	1
(8)	16	1-4-5-2-4	1
(9)	16	1-4-5-4-2	1
(10)	16	1-5-2-4-4	1
(11)	16	1-5-4-4-2	1
(12)	16	1-5-4-2-4	1
Cardinal #	Result	Dice A-B-C-D-E	# of ways
(13)	16	2-1-4-4-5	1
(14)	16	2-1-4-5-4	1
(15)	16	2-1-5-4-4	1
(16)	16	2-4-1-4-5	1
(17)	16	2-4-1-5-4	1
(18)	16	2-4-4-1-5	1
(19)	16	2-4-4-5-1	1
(20)	16	2-4-5-1-4	1
(21)	16	2-4-5-4-1	1
(22)	16	2-5-1-4-4	1
(23)	16	2-5-4-1-4	1
(24)	16	2-5-4-4-1	1
Cardinal #	Result	Dice A-B-C-D-E	# of ways
(25)	16	4-1-2-4-5	1
(26)	16	4-1-2-5-4	1
(27)	16	4-1-4-2-5	1
(28)	16	4-1-4-5-2	1
(29)	16	4-1-5-2-4	1
(30)	16	4-1-5-4-2	1
(31)	16	4-2-1-4-5	1
(32)	16	4-2-1-5-4	1
(33)	16	4-2-4-1-5	1
(34)	16	4-2-4-5-1	1
(35)	16	4-2-5-1-4	1
(36)	16	4-2-5-4-1	1
Cardinal #	Result	Dice A-B-C-D-E	# of ways
(37)	16	4-4-1-2-5	1
(38)	16	4-4-1-5-2	1
(39)	16	4-4-2-1-5	1
(40)	16	4-4-2-5-1	1
(41)	16	4-4-5-1-2	1
(42)	16	4-4-5-2-1	1
(43)	16	4-5-1-2-4	1
(44)	16	4-5-1-4-2	1
(45)	16	4-5-2-1-4	1
(46)	16	4-5-2-4-1	1
(47)	16	4-5-4-1-2	1
(48)	16	4-5-4-2-1	1
Cardinal #	Result	Dice A-B-C-D-E	# of ways
(49)	16	5-1-2-4-4	1
(50)	16	5-1-4-2-4	1
(51)	16	5-1-4-4-2	1
(52)	16	5-2-1-4-4	1
(53)	16	5-2-4-1-4	1
(54)	16	5-2-4-4-1	1
(55)	16	5-4-1-2-4	1
(56)	16	5-4-1-4-2	1
(57)	16	5-4-2-1-4	1
(58)	16	5-4-2-4-1	1
(59)	16	5-4-4-1-2	1
(60)	16	5-4-4-2-1	1
Cardinal #	Result	Dice A-B-C-D-E	# of ways

Let's drive a stake through this vampire's heart. (Showing work, like teachers demand.)
Here are all 252 outcomes for rolling five dice:

# Targets	Established	Ways to	Probability
Like This	Target	Repeat Any	Fraction Percent
---->		One Target
6	All Different	120	5/ 324 __ 1.54321
60	One Pair	60	5/ 648 __ 0.77160
60	Two Pair	30	5/1296 __ 0.38580
60	Three of a Kind	20	5/1944 __ 0.25720
30	Full House	10	5/3888 __ 0.12860
30	Four of a Kind	5	5/7776 __ 0.06430
6	Five of a Kind	1	1/7776 __ 0.01286
252

If the "target" roll is unknown/concealed, then the weighted-average probability is:

( ( 720 x (120 / 7776) )
+(3600 x ( 60 / 7776) )
+(1800 x ( 30 / 7776) )
+(1200 x ( 20 / 7776) )
+ ( 300 x ( 10 / 7776) )
+ ( 150 x ( 5 / 7776) )
+ ( 6 x ( 1 / 7776) ) ) / 7776

or, equivalently: (86,400+216,000+54,000+24,000+3,000+750+6) / 7,776²

or, equivalently: 384,156 / 60,466,176 = 0.00635323788 = 0.6353 %

Quote: ThatDonGuy

The overall probability that two consecutive rolls will match (before knowing what the first roll is) is about 1 / 157.

The reciprocal of the average-weighted probability is 1 / 0.00635323788 = 157.4...
Don's right!