simple Dice Game Probability
February 8th, 2018 at 1:42:33 PM
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My friends and I roll dice in a simple game.
1st we flip a fair coin and IF Heads:
we select 3 of the faces of 1 die and try to roll the 3 other faces not selected in 7 rolls.
and IF Tails:
we select 4 of the faces of 1 die and try to roll the 2 other faces not selected in 7 rolls.
7 rolls can be:
1 die 7 times
or
4 dice 1 time followed by 3 dice the next time
or other combos of dice as long as max dice rolls = 7
The question (2 parts) is
(1)the probability of getting all 6 faces in at most 7 rolls
starting with 3 faces and starting with 4 faces.
or
(2)the probability of getting exactly 5 faces in at most 7 rolls
starting with 3 faces and starting with 4 faces.
I think that part is easy and I solved it with a transition matrix in Excel
not showing the results yet
someone asked me if there was an easy way to do the math where most could understand the chances of success.
I used a recursion method in Excel that only uses adding and multiplication
but it still takes time to set up.
I also
did the select 4 faces 1st and crunched numbers til I got it right
for staying at 4 faces or ending at 5 faces, not getting that 6th face.
It hurt my head and was really messy looking until the last try.
I would like to see how others would approach this.
Simulation is fine too! (I also did that 1st)
Sally
1st we flip a fair coin and IF Heads:
we select 3 of the faces of 1 die and try to roll the 3 other faces not selected in 7 rolls.
and IF Tails:
we select 4 of the faces of 1 die and try to roll the 2 other faces not selected in 7 rolls.
7 rolls can be:
1 die 7 times
or
4 dice 1 time followed by 3 dice the next time
or other combos of dice as long as max dice rolls = 7
The question (2 parts) is
(1)the probability of getting all 6 faces in at most 7 rolls
starting with 3 faces and starting with 4 faces.
or
(2)the probability of getting exactly 5 faces in at most 7 rolls
starting with 3 faces and starting with 4 faces.
I think that part is easy and I solved it with a transition matrix in Excel
not showing the results yet
results after 7 rolls
someone asked me if there was an easy way to do the math where most could understand the chances of success.
I used a recursion method in Excel that only uses adding and multiplication
but it still takes time to set up.
I also
did the select 4 faces 1st and crunched numbers til I got it right
for staying at 4 faces or ending at 5 faces, not getting that 6th face.
It hurt my head and was really messy looking until the last try.
I would like to see how others would approach this.
Simulation is fine too! (I also did that 1st)
Sally
I Heart Vi Hart
February 8th, 2018 at 2:55:07 PM
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Note that the physical number of dice is irrelevant.
From 4 faces (assume 3, 4, 5, 6 were chosen):
The probability of not rolling at least one 1 or at least one 2:
at least one 1: (5/6)^7
at least one 2: (5/6)^7
but that includes these twice:
any 1s or 2s: (4/6)^7
The total is (2 x 5^7 - 4^7) / 6^7
The probability of doing it is 1 - this value = 140070/279936
From 3 faces (assume 4, 5, 6 were chosen):
The probability of not rolling at least one 1 or at least one 2 or at least one 3:
at least one 1: (5/6)^7
at least one 2: (5/6)^7
at least one 3: (5/6)^7
but that includes each of these twice:
any 1s or 2s: (4/6)^7
any 1s or 3s: (4/6)^7
any 2s or 2s: (4/6)^7
but removing those removes all references to this:
any 1s, 2s, or 3s: (3/6)^7
The total is (3 x 5^7 - 3 x 4^7 + 3^7) / 6^7
The probability of doing it is 1 - this value = 92526/279936
From 4 faces (assume 3, 4, 5, 6 were chosen):
The probability of not rolling at least one 1 or at least one 2:
at least one 1: (5/6)^7
at least one 2: (5/6)^7
but that includes these twice:
any 1s or 2s: (4/6)^7
The total is (2 x 5^7 - 4^7) / 6^7
The probability of doing it is 1 - this value = 140070/279936
From 3 faces (assume 4, 5, 6 were chosen):
The probability of not rolling at least one 1 or at least one 2 or at least one 3:
at least one 1: (5/6)^7
at least one 2: (5/6)^7
at least one 3: (5/6)^7
but that includes each of these twice:
any 1s or 2s: (4/6)^7
any 1s or 3s: (4/6)^7
any 2s or 2s: (4/6)^7
but removing those removes all references to this:
any 1s, 2s, or 3s: (3/6)^7
The total is (3 x 5^7 - 3 x 4^7 + 3^7) / 6^7
The probability of doing it is 1 - this value = 92526/279936
February 9th, 2018 at 10:33:26 AM
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just thinking about each roll...Quote: mustangsally(2)the probability of getting exactly 5 faces in at most 7 rolls
starting with (3 faces later) starting with 4 faces.
1st to warm up
to stay at 4 faces - I select 1,2,3,4 and need to roll at least one 5 and one 6
in 7 rolls should be easy like this: (4/6)^7 = about 0.058527663 or about 1 in 17
to end at 5 faces not get that 6th face
| roll | A = (4/6)^(roll-1)*(2/6) | B = (5/6)^(7-roll) | A*B |
|---|---|---|---|
| 0 | getting 5th face on roll X | NOT getting 6th face after Roll X | getting 5th face not getting 6th |
| 1 | 0.333333333 | 0.334897977 | 0.111632659 |
| 2 | 0.222222222 | 0.401877572 | 0.089306127 |
| 3 | 0.148148148 | 0.482253086 | 0.071444902 |
| 4 | 0.098765432 | 0.578703704 | 0.057155921 |
| 5 | 0.065843621 | 0.694444444 | 0.045724737 |
| 6 | 0.043895748 | 0.833333333 | 0.03657979 |
| 7 | 0.029263832 | 1 | 0.029263832 |
| . | . | ending with 5 faces after roll 7 >> | 0.441107968 |
0.441107968
thinking out loud
1 - (stay at 4 + end at 5) = about 50% getting all 6 faces
that I find amazing
Sally
I Heart Vi Hart
February 9th, 2018 at 10:38:52 AM
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yes it is. but some like to roll 2 dice at a time instead of 1 at a time.Quote: ThatDonGuyNote that the physical number of dice is irrelevant.
You really have that exclusion-inclusion principle down pat. great work!
I Heart Vi Hart
February 9th, 2018 at 11:13:57 AM
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The inclusion-exclusion method that Thatdonguy presented is the most efficient way.
Since there are only 6^7 permutations you can also solve this quite easily by brute force in Excel. List out the 279,936 perms (use base 6, length 7) and count the ones that match your criteria.
Since there are only 6^7 permutations you can also solve this quite easily by brute force in Excel. List out the 279,936 perms (use base 6, length 7) and count the ones that match your criteria.
It’s all about making that GTA
February 9th, 2018 at 11:34:55 AM
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agree on efficient.Quote: Ace2The inclusion-exclusion method that Thatdonguy presented is the most efficient way.
Here is my recursion solution in Excel for those that want to see
(I always do this after making my transition matrix solution)
https://goo.gl/A6qAyG
cell K1 is a drop-drown list
that way one can start with any number of faces (from 0 to 6. 0 takes time and starting with 6 is funny!)
yes, another possible solution.Quote: Ace2Since there are only 6^7 permutations you can also solve this quite easily by brute force in Excel. List out the 279,936 perms (use base 6, length 7) and count the ones that match your criteria.
had to add:
this proved to be the best way to show others
how to figure this out and can understand
*******
btw, my friends and I play this dice game
and score a win as 1 point for 4 face game and 2 points for 3 face game.
A few friends asked why 2 points for the 3 face game and I tried to explain it is close to true odds and lost them all.
The winner is first to 7 points (sometimes 11 if lots of food and drinks are handy)
Sally
Last edited by: mustangsally on Feb 9, 2018
I Heart Vi Hart
February 9th, 2018 at 3:25:17 PM
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Well, counting in Excel is not only super easy and takes about a minute to do, most can easily grasp the concept and they get it!Quote: mustangsally(2)the probability of getting exactly 5 faces in at most 7 rolls
starting with 3 faces.
here is for 3 faces
142632/279936 = 0.509516461
the distribution in photo
it answers the probability of ending at 5 faces after 7 rolls
interesting about starting at 2 faces
almost same probability to end at exactly 5 faces
0 128 0.000457247
1 8236 0.029421011
2 72828 0.260159465
3 141624 0.505915638 <<< 5 faces ending
4 57120 0.204046639
game on in 30 minutes!
Sally
I Heart Vi Hart
