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# Probability Question Watch

1. Three standard dice are thrown

What is the probability that at least two consecutive numbers show up?
2. how far have you got?

Id start by considering how many sequences are available that fit your conditions. given its non-ordered, its simple to do that way if nothing else.
3. 3rd attempt!

Consider the complementary event, and break it down into three cases:

3 numbers the same.
2 numbers the same.
All different.

I now make it
Spoiler:
Show
7/12
4. well, sort of. Although the complement has a bigger sample space so its probably complicating matters to do it that way.

if you have a probability vector, counting the number value of each dice, the conditions you care about are:
<1,1,1,0,0,0> <0,1,1,1,0,0> etc
and <1,1,0,0,0,0>, <0,1,1,0,0,0> etc etc

Spoiler:
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my reasoning to count the events.

Dealing with three consecutive

you can see there are only 4 events in this space.

if A=2 consec & b = 3 consec
P(AuB) = P(a) + p(b) - P(anb)

consider writing it out by hand, it will aid your understanding loads to go through the possibilities...

5. Ghost walker did exactly the way I did it I got 7/12 .
6. (Original post by ghostwalker)
3rd attempt!

Consider the complementary event, and break it down into three cases:

3 numbers the same.
2 numbers the same.
All different.

I now make it
Spoiler:
Show
7/12
its getting late so maybe my brain is giving up but this is my method.

Spoiler:
Show

notation:
probability vector on a dice <0,0,0,0,0,0> where the nth position represents the side with n spots.
so <1,1,0,0,0,0> is vector representation of one 1 and one 2, while
<1,2,3,0,0,0> is one one, two 2, three 3's etc

2 consc.

<1,1,0,0,0,0> {1}
<0,1,1,0,0,0> {2}
<0,0,1,1,0,0> {3}
<0,0,0,1,1,0> {4}
<0,0,0,0,1,1> {5}

I can populate each vector with an additional entry in 6 ways per vector.

3. Consec
<0,0,0,1,1,1>
<0,0,1,1,1,0>
<0,1,1,1,0,0>
<1,1,1,0,0,0>

Clearly 3 dice, 3 entries, so this has 4 events in it.

So i have (5x6) + 4 events in my space, but must remove the double counts that occur.

If i populate the first set with an additional 1 in each space, i will double count some results. eg.
for this entry <1,1,0,0,0,0>

<2,1,0,0,0,0>
<1,2,0,0,0,0>
<1,1,1,0,0,0>

Clearly the first two are valid (2 1's and one 2 and vice versa), but the third is a triple consecutive, accounted for already in the 3 consec.

so in the two dice vectors, the first and last vectors will give 1 event each which is double counted, and the others will give two.
so my total is 34 - [2+(2*3)] = 34 - 8 = 26 events.

using the stars and bars method placing 3 objects into 6 bins, allowing for zero entries is the same as 9 objects into 6 bins.
so to divide into 6 groups, i must place 5 bars.
9 objects:
x x x x x x x x x

8 positions to place
x| x x x x x x x x through to x x x x x x x x| x

so my total events are 8C5 = 56.

so that is a p of 26/56

edit to add: in general i find it easiest to reduce to vector based approach when possible. especially with dice problems, playing card problems etc. Plus it is by far the easiest way to tackle more open problems like "if i roll a dice N times, what is the probability i get a monotonic sequence?"
7. I'm not entirely sure, but I know a guy who can help.

8. (Original post by c471)
so my total is 34 - [2+(2*3)] = 34 - 8 = 26 events.
so my total events are 8C5 = 56.

so that is a p of 26/56
Thanks for posting your methodology, as I didn't understand what you were getting at.

However, all events do not have equal likelihood of occuring.

e.g. <1,1,1,0,0,0> can arise in 3! ways.

whereas <2,1,0,0,0,0> can arise in only 3 ways.

And pushing it to the extreme <3,0,0,0,0,0> can only arise in 1 way.
9. (Original post by ghostwalker)
Thanks for posting your methodology, as I didn't understand what you were getting at.

However, all events do not have equal likelihood of occuring.

e.g. <1,1,1,0,0,0> can arise in 3! ways.

whereas <2,1,0,0,0,0> can arise in only 3 ways.

And pushing it to the extreme <3,0,0,0,0,0> can only arise in 1 way.
indeed, but is this not dealt with using the counting principle to determine the total number of events?
10. (Original post by c471)
indeed, but is this not dealt with using the counting principle to determine the total number of events?
In a word, No. When working out the final probability, errors in the numerator arising from adding together disparate events, will not be compensated for in the denominator, except by chance sometimes.
11. (Original post by c471)
...
I think the problem with your method is that the events as you have them do not occur with equal probability.

If you consider the 3 dice ordered, then each event, (2,2,4) say, where the first die is 2, the second 2, and third 4, will have equal likelihood of occuring, then a method along the lines you're suggesting should work.
12. The answer is 5/6! Not sure why though!
13. Brute force also says 7/12.

Explicitly, the following combinations work, and there are 126 of them:

Spoiler:
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112 121 122 123 124 125 126 132 134 142
143 145 152 154 156 162 165 211 212 213
214 215 216 221 223 231 232 233 234 235
236 241 243 245 251 253 254 256 261 263
265 312 314 321 322 323 324 325 326 332
334 341 342 343 344 345 346 352 354 356
362 364 365 412 413 415 421 423 425 431
432 433 434 435 436 443 445 451 452 453
454 455 456 463 465 512 514 516 521 523
524 526 532 534 536 541 542 543 544 545
546 554 556 561 562 563 564 565 566 612
615 621 623 625 632 634 635 643 645 651
652 653 654 655 656 665
14. Tree diagrams ftw?

Case 1: The first roll is 1 or 6.
The second roll can be 0, 1, 2, 3, 4, or 5 units away.
Case 1a: The second roll is 1 unit away. Consecutive get! Probability is (1/3)(1/6).
Case 1b: The second roll is b=3,4 units away. Third roll must be 1 unit, b+1, or b-1 units away to get a consecutive. Probability is (1/3)(1/3)(1/2).
Case 1c: The second roll is 2,5 units away. Third roll must be 1 unit or 3,4 units away to get a consecutive. Probability is (1/3)(1/3)(1/3)
Case 1d: The second roll is 0 units away. Third roll must be 1 unit away to get a consecutive. Probability is (1/3)(1/6)(1/6)

Total probability of Case 1 and a consecutive occurring: (1/3)(6/36 + 6/36 + 4/36 + 1/36)=(1/3)(17/36).

Case 2: The first roll is 2 or 5. The second roll can be -1, 0, 1, 2, 3, or 4 units away.
Case 2a: The second roll is ±1 unit away. Consecutive get! Probability is (1/3)(1/3).
Case 2b: The second roll is 2,4 units away. Third roll must be ±1 unit or 3 units away to get a consecutive. Probability is (1/3)(1/3)(1/2).
Case 2c: The second roll is 3 units away. Third roll must be ±1 unit, 2 units, or 4 units away to get a consecutive. Probability is (1/3)(1/6)(2/3).
Case 2d: The second roll is 0 units away. Third roll must be ±1 unit away to get a consecutive. Probability is (1/3)(1/6)(1/3).

Total probability of Case 2 and a consecutive occurring: (1/3)(6/18 + 3/18 + 2/18 + 1/18)=(1/3)(12/18)=(1/3)(2/3).

Case 3: The first roll is 3 or 4. The second roll can be -2, -1, 0, 1, 2, or 3 units away.
Case 3a: The second roll is ±1 unit away. Consecutive get! Probability is (1/3)(1/3).
Case 3b: The second roll is 2,3 units away. Third roll must be ±1 unit or 3,2 units away. Probability is (1/3)(1/3)(1/2).
Case 3c: The second roll is -2,0 units away. The third roll must be ±1 units away. Probability is (1/3)(1/3)(1/3).

Total probability of Case 3 and a consecutive occurring: (1/3)(12/36 + 6/36 + 4/36) = (1/3)(22/36)

Total probability of Case 1, 2, or 3 and a consecutive occurring: (1/3)(17/36 + 22/36 + 2/3) = (1/3)(39/36 + 2/3) = (1/3)(13/12 + 8/12) = (1/3)(21/12) = 7/12.

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Updated: April 29, 2013
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