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Probability Thread

Not cubing related but interesting. What are the odds of 3 baby buncos in a row? Best guess I have is 6^(9)(5/6) = 8398080 but I'm pretty sure I'm wrong as I have next to no experience with probability.
 
Oh, I guess it's a Wisconsin thing. It's a dice game where the player player rolls three dice at once. If you're on the 3rd of 6 games, and you roll all 3s, thats a bunco worth 21 points. A baby bunco is essentially the same thing but the number rolled doesn't match the game you're on. So if you're on the 3rd game still and all 3 dice land on 1,2,4,5, or 6, that's a baby bunco worth 5 points.

Three baby buncos in a row happened to a friend of mine and I'm curious.
 
JackJ said:
Oh, I guess it's a Wisconsin thing. It's a dice game where the player player rolls three dice at once. If you're on the 3rd of 6 games, and you roll all 3s, thats a bunco worth 21 points. A baby bunco is essentially the same thing but the number rolled doesn't match the game you're on. So if you're on the 3rd game still and all 3 dice land on 1,2,4,5, or 6, that's a baby bunco worth 5 points.

Three baby buncos in a row happened to a friend of mine and I'm curious.
Click to expand...

OK.
So you roll three dices and all have the same number. Is that right?

And this happens 3 times in a row...

Let's first calculate the propability of one babu bunco:
First dice can be anything. so P=1
Second must be the same as first so P=1*(1/6)=1/6
Third must be the same as first two so P=(1/6)*(1/6)=1/36
1/36 for one baby bunco
for three in a row it would then be (1/36)³= 1/46656~0.0021%
 
tseitsei said:
OK.
So you roll three dices and all have the same number. Is that right?

And this happens 3 times in a row...

Let's first calculate the propability of one babu bunco:
First dice can be anything. so P=1
Second must be the same as first so P=1*(1/6)=1/6
Third must be the same as first two so P=(1/6)*(1/6)=1/36
1/36 for one baby bunco
for three in a row it would then be (1/36)³= 1/46656~0.0021%
Click to expand...

Instead of 1 it's 5/6 because rolling all 3s is not a baby.
 
RageCuber said:
Just thought of something stupid but, What are the chances of solving the cross then getting a f2l skip? (no OLL or PLL skip)
Click to expand...

corner permutations: 1/ 8*7*6*5
corner orientations: 1 / 3^4
edge permutations: 1 / 8*7*6*5
edge orientations: 1 / 2^4

All of them at once (multiplication) = 1 / 3657830400. You have a much higher chance for winning a lottery with a single ticket (5 numbers out of 90, chance for winning is 1 / 43949268 ).

Note that these numbers are only valid if you solve the cross "blindly", so you don't look at other edges, corners, or already matched edge-corner pairs. In a similar lucky case you probably would see and try to solve xx-cross or xxx-cross intentionally (and get the last one or two corner-edge pairs by chance, hence getting an "F2L skip"). So, in practice, the chance for an F2L skip is much higher.
 
Tried calculating the probability of a LL skip on 4x4 while using K4.

OLL skip: 3^(3)*2^(3)*2^(3)= 1/1728
PLL skip: 4!*4^(2)*3^(2)= 1/3456. That's an AUFless PLL skip so it's really 1/864.
LL skip: 1728*864= 1/1492992

Nearly 1 in 1.5 million?! Surely that's not right. It's probably something to do with parity but I'm not sure how to implement it. Help anyone?
 
Last edited:
JackJ said:
Tried calculating the probability of a LL skip on 4x4 while using K4.

OLL skip: 3^(3)*2^(3)*2^(3)= 1/1728
PLL skip: 4!*4^(2)*3^(2)= 1/3456. That's an AUFless PLL skip so it's really 1/864.
LL skip: 1728*864= 1/1492992

Nearly 1 in 1.5 million?! Surely that's not right. It's probably something to do with parity but I'm not sure how to implement it. Help anyone?
Click to expand...
When solving the 4x4x4 K4 style, all steps through F3L will solve all pieces except for:

4 Corners, which have (4!)(3)(3)(3) positions.
8 Wing Edges, which have 8! positions.

The parity of corners and wing edges are independent of one another (and thus all permutations of each are allowed independently), and so we just multiply these two numbers together. We also divide by 4 for AUF since all unsolved pieces are in the same layer and thus we can apply this rotational symmetry reduction.

1/6,531,840.
 
CyanSandwich said:
What's the probability of getting at least 1 LL skip in 15522 solves?
Click to expand...

I'm going to assume you meant 15552, not 15522. Also assuming no edge/OLL control of any sort.

Simpler question - what's the chance of none happening? In one solve, 15551/15552. Take that to the 15552nd power to get the probability of no LL skips in that many solves.

(15551/15552)^15552 = .368 or 36.8%

Thus, the probability of at least one happening is the complement of that.

1 - .368 = .632 or 63.2%
 
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