Necco Odds

[arbuthnot]

I am working through a roll of Necco wafers, and I just had three liquorice Neccos in a row.

This must be rare:

8 flavors.

A wafer of a given flavor exists. The chances of the next wafer being the same flavor are 1 in 8. Same for the next wafer.

1/8 is 0.125.

Therefore, the chances of three wafers of the same flavor, given a single wafer of that flavor to start, are 0.125 x 0.125 = 0.015625, or 1/64, or 1.56%

Now, Neccos are packaged by weight--2.20oz per roll. But that's about 40 wafers. Let's say 40 wafers/roll. We can ignore wafers 39 and 40, as they cannot be followed by two additional wafers.

This means 38 chances to have three wafers of the same flavor in a row. 38 x .015625 = 0.59375, or 19/32, or 59.375%.

That's actually not that rare. You have only a slightly worse chance of getting three Necco wafers of the same flavor in a row, in a given roll of Necco wafers, than Ruben Patterson of the Portland Trail Blazers had of making a given free throw during this year's NBA season.

Unless I'm missing something. Am I missing something?

[LetYouDown]

Say the first 4 wafers are the same. Does this count twice? If you get 5 in a row, does it count 3 times? You're counting them obviously. Odds change if you say 3 and only 3.

[arbuthnot]

4 in a row counts twice. We are interested in any possible combination of three in a row.

[LetYouDown]

Well, that seems to be the reason that your number is so high...

[eephus]

[ QUOTE ]
Say the first 4 wafers are the same. Does this count twice? If you get 5 in a row, does it count 3 times? You're counting them obviously. Odds change if you say 3 and only 3.

[/ QUOTE ]

Hi. arbuthnot was kind enough to post my crazy question here.

I want to know what the odds are of ANY three-wafer set of the same flavor showing up.

I think I have it right. If you have a roll of 40 all-chocolate Necco wafers, after all, you have 38 sets of three chocolate wafers. Right?

The problem is that if wafer 1 is not chocolate, you only lose the first possible trio (wafers 1/2/3). Same w/wafer 40 (38/39/40). But if wafer 2 or wafer 39 is not chocolate, you lose two possibilities per wafer (1/2/3, 2/3/4 and 37/38/39, 38/39/40).

Any other wafer, from 3 through 38, you lose three possible sets per nonchocolate wafer.

This makes me think there is a subtlety to this that I haven't taken into account.

[MikeL05]

Compare your math to rolling a die. There are 6 sides, so chances of rolling a SIX is .166667. Now, if you roll it 6 times, it's not the case that you'll hit 1.0 SIXes. You'll hit on average 1.0 SIXes. Likewise, multiplying your percentage (which I believe is right) by 38, you've hit upon .59. This is not your chance of hitting one string of three in a row; this is the average number of 3-in-a-row sets you'll hit in one package.

What I think you need to do is take your probability and subtract it from 1.00. So 1.00 - .0156 = .984. Now raise this number .984 to the exponent "n", where "n" is the number of wafers.

So .984^38 = probability of NOT getting a set of 3 in a row.

1.00 - that = probability of hitting at least one set in 38 wafers.

[TomCollins]

If you have a roll of 10,000 wafers, does this mean you have a 10,000 x 0.015625 chance of getting 3 in a row?

[Siegmund]

If we imagine a very long roll of wafers, we can say that 1/64 of the wafers will be immediately followed by at least two identical wafers. In a 10,000-wafer roll we can safely conclude there'll be rouughly 150 matches. That's not the question the original poster asked, of course; but we can get a rough estimate by saying that we expect about 38/64 matches in a roll of 40, and then reduce this a bit more for the chance of streaks of four or two sets of three in one roll.


If you want an exact answer, yes, a recurrence relation works to count the total number of stacks of size n that have no threepeats: x[1]=8, y[1] = 0; x(i)= 7*(x[i-1]+y[i-1]); y(i) = x[i-1].

Are you sorry you asked? Out of 1329227995784915872903807060280344576 possible 40-wafer sequences, 774855372983096359078760714768858504 of them have no sequences of three identical wafers. So: if every sequence is equally likely, 51.95% of rolls have no sets, 48.05% have one or more sets.

[pzhon]

[ QUOTE ]
Out of 1329227995784915872903807060280344576 possible 40-wafer sequences, 774855372983096359078760714768858504 of them have no sequences of three identical wafers. So: if every sequence is equally likely, 51.95% of rolls have no sets, 48.05% have one or more sets.

[/ QUOTE ]
Your count agrees with mine, but the quotient should be 58.29% with no sets, and 41.71% with one or more sets.

Here is the distribution of the number of sets in the sense of the original poster, where a streak of 6 in a row counts as 4 sets of 3:

0: .582936
1: .285553
2: .0966622
3: .0266832
...
37: 8.42594 x 10^-35
38: 6.01853 x 10^-36

I used the following transfer matrix T:

7/8 7/8 7/8
1/8 0 0
0 x/8 x/8

The above are the coefficients of x^n in row(1,1,1) T^39 column(1,0,0).

Of course, the expected number of sets is 38/64.

[gaming_mouse]

[ QUOTE ]
x[1]=8, y[1] = 0; x(i)= 7*(x[i-1]+y[i-1])

[/ QUOTE ]

Where does x[1] come from? Are there 3 flavors? (EDIT: That makes no sense. It would be 8 flavors)

Also, I don't follow the rucursion relation. Could you explain the concept? (EDIT: I'm realizing that you were solving a different problem than I was....)