Streetwiseguy
Streetwiseguy PowerDork
11/25/15 4:03 p.m.

4 wheels, 4 directional tires. Without looking, you mount the 4 tires. What are the chances of getting it right?

I say 1 in 5, or 20%, because there are 5 possible outcomes- 4 lefts, 4 rights, 3 lefts, 3 rights, or two and two.

Co worker says the first two can't be wrong, and both following tires are 50/50 of being a left or right, and twice 50/50 would be a 25% chance of getting it right.

We both kinda think I'm right, but what's the math to prove it?

El Cheapo
El Cheapo HalfDork
11/25/15 5:26 p.m.

A coin toss works out to be 66% and not 50/50. That's all I got.

Karacticus
Karacticus Reader
11/25/15 5:41 p.m.

I think your co-worker has the answer right.

Though there are only five possible outcomes, the odds for each aren't the same-- odds for 4 the same per my likely incorrect math is 12.5%. I'd like to see some fancy math too!

EastCoastMojo
EastCoastMojo Mod Squad
11/25/15 8:30 p.m.

I think you have a 37.5% chance of mounting the tires in a 2L2R configuration, using the first counting rule of probability:

Rule 1: If any one of K mutually exclusive and exhaustive events can occur on each of N trials, there are K^N different sequences that may result from a set of such trials. Example: Flip a coin three times, finding the number of possible sequences. N=3, K=2, therefore, K^N =2^3=8 .

The number of mutually exclusive events is two, you can either mount the tire on the wheel facing left or facing right. The number of trials is four, as you have four wheels. K^N = 2^4 = 16 possible outcomes. Here are the 16 possible ways you can mount the tires, it helps to think of this as the order in which they are mounted.

RRRR ... LLLL
RRRL ... LLLR
RRLL ... LLRR
RLLL ... LRRR
RLRR ... LRLL
RRLR ... LLRL
RLLR ... LRRL
RLRL ... LRLR

6 of these 16 possible outcomes will give you two lefts and two rights. That works out to about 37.5% of the total possible outcomes in your favor. I think.

Streetwiseguy
Streetwiseguy PowerDork
11/25/15 8:33 p.m.

Oh dear.

mith612
mith612 Reader
11/25/15 9:46 p.m.

There is 100% chance the first tire will be mounted "A" way.

There is a 50% chance each that the second and third tires will be mounted "A" way.

Hence there is a 25% chance (0.50 * 0.50) that both tires will be mounted "A" way, resulting in a 75% chance they will be mounted one each way or both "B" way.

So there is 75% chance after mounting three tires there will be either AAB or ABB.

The fourth tire has a 50% chance of being mounted correctly to complete the set.

75% of 50% is 37.5%, or 3 in 8 odds.

egnorant
egnorant SuperDork
11/25/15 10:20 p.m.

Are you saying mounted on the car?...or mounted on the rim?

If you mean the rim then there is no way to avoid getting 2 of them correct. The first 2 tires can be put on the car no matter what. Then you only have 4 options! 4 correct, 2 incorrect or 1 incorrect(in 2 different ways). Correct is one of these 4 permutations or 25%.

Unless I am they guy mounting the tire...near 100% then!

Bruce

JamesMcD
JamesMcD Dork
11/25/15 10:28 p.m.

Now what if the wheels are directional too?

egnorant
egnorant SuperDork
11/25/15 10:41 p.m.
JamesMcD wrote: Now what if the wheels are directional too?

With true directional wheels (2 left and 2 right) I would claim I did it wrong on purpose to see who noticed and then buy them a beer!

Bruce

chandlerGTi
chandlerGTi UltraDork
11/26/15 8:42 a.m.

The first two are gimmes, the third is 66% and the forth is 50%.

Woody
Woody MegaDork
11/26/15 8:44 a.m.

Just for fun, let's throw staggered directional wheels into the mix...

EastCoastMojo
EastCoastMojo Mod Squad
11/26/15 8:25 p.m.

I'll take a stab at it, though math after beer is not always a good idea

For the first wheel, you have essentially eight possible tire options (four different tires with two mounting directions possible for each). Once the first tire has been mounted, the tire choices for the remaining wheels has been reduced. The second wheel has six tire options, and so on. The calculation for the dependent events in this example is (8)(6)(4)(2)=384 possible outcomes, only one of which is correct (correct tire on the correct wheel, facing the correct direction. 1/384 = 0.26% chance of getting this outcome.

Mounting the wheels on the car correctly is the same dependent event formula, as each wheel mounted reduces the available options for the remaining positions. (4)(3)(2)(1)=24 possible wheel mounting outcomes, only one of which is correct. 1/24= 4.16% chance of mounting wheels in their correct positions.

4.16% of 0.26% = .010816% chance of getting the tires mounted on the rims correctly, and the rims mounted on the car correctly.

Knurled
Knurled MegaDork
11/28/15 8:35 a.m.

NEVER TELL ME THE ODDS

Really though your co-worker was right.

After all, if you mount two right, then you have two left...

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