Is probability so hard to understand?

[Agathon]

People disagree over probability because it isn't easy to understand. I mean which of these is correct?

1. Probability is a priori. e.g. Given equal chances of a die side turning up the probability of a 6 is 1/6 (no empirical calculation is required as this is a logical truth).

2. Subjective (Bayesian) probability. Probability is measured by our degree of belief in a particular outcome rather than anything having to do with the events themselves.

3. Frequency: Probability is measured by the proportion of the desired results over the total results in the long run (however one is supposed to measure that).

4. Realist probability: God plays dice. There is randomness built in to the very fabric of the world.

All of these have compelling reasons for themselves and against the others - it just doesn't seem a well understood concept.

[orange]

Not entirely true, orange. If starting position was chosen randomly and there were no Chance or Community Chest cards, you'd be correct. However, the fact that you always start in one spot does skew Monopoly a bit, as do (obviously) cards that say things like "Go to St. James Place." Consider that Mediterranean & Baltic Ave. are screwed over anyway not just because they have only 2 slots, but because you're less likely to hit them in your initial pass around the board. Not that this matters much, considering that the first pass around the board- when the probabilities are most skewed- is also when nobody owns anything, and therefore arguments about what to buy become more moot.

Originally posted by LulThyme
Yes Orange, computer simulations proved not long ago taht Kentucky is the most landed on property in a game of Monopoly, not long ago.
Although your basic argument is correct, the difference in probability is very thin between each, but the effects that teleport you affect it a bit, as the starting position but almost imperceptibly. The oranges ARE often landed upon while coming out of prison which is the most landed on space so he wasnt totally wrong...

Ignoring all chance and community chest...we're talking straight dice rolls...and the only one that can't be landed on is the first one.

[yavoon]

Originally posted by orange
Good timing.

I didn't want to take Calc242 as my math req., which I had placed into with that test during orientation, so I decided to take Math114. I'm a Hist/Poly Sci major, so it didn't need to be anything really great...just a math class. Math114 is Basic College Math and Statistics.

He was trying to explain normal distribution curves and standard deviation, and it took him the whole damn class. I understood the material as he was drawing it, and it took 50 minutes to explain it because people just weren't getting it. Unreal. I almost wish I had taken the higher course.

Oh, and then during a game of monopoly one of my (slower...) friends was trying to argue that it was best to build houses on orange because "it gets landed on more than any other set" to which we all tried to explain to him that it had the same probability as every other set (minus the ones that only have two tiles), to which he said "yeah, but it still gets landed on more"

Needless to say, he lost quite badly as usual.

u land on orange the most because of jail. u land on it more than purple because higher 2d #'s are easier to attain than lower 2d #'s.

[Curiosity]

Try this problem -

A man on a gameshow chooses from three boxes, one of which contains a prize. The gameshow host then opens one of the other two boxes, showing that it's empty. The man is offered the chance to change his choice to the remaining box, or stick with the one he chose in the first place.

What should he do, and does it make any difference?

Quite a basic problem in probability, but many mathematicians have trouble with it.

[Immortal Wombat]

Old one. Swap boxes.

[yavoon]

Originally posted by Curiosity
Try this problem -

A man on a gameshow chooses from three boxes, one of which contains a prize. The gameshow host then opens one of the other two boxes, showing that it's empty. The man is offered the chance to change his choice to the remaining box, or stick with the one he chose in the first place.

What should he do, and does it make any difference?

Quite a basic problem in probability, but many mathematicians have trouble with it.

always switch. switching is 2/3rds. many ways to explain it. but the way I like is that.

If u pick the right box initially then the host removes half the wrong boxes. if u pick the wrong box initially then the host removes all remaining incorrect boxes.

[chequita guevara]

Originally posted by yavoon
u land on orange the most because of jail. u land on it more than purple because higher 2d #'s are easier to attain than lower 2d #'s.

Irrelevent. You need doubles to get out of jail, so the probabilities are evenly distributed.

[yavoon]

Originally posted by chegitz guevara


Irrelevent. You need doubles to get out of jail, so the probabilities are evenly distributed.

incorrect. u need doubles, 50$ or a get out of jail free card.

[MrFun]

Originally posted by chegitz guevara


Irrelevent. You need doubles to get out of jail, so the probabilities are evenly distributed.

But in the game, if a commie player lands in jail, there's no way he will ever get back out . . .

[orange]

u land on it more than purple because higher 2d #'s are easier to attain than lower 2d #'s.

Huh? Equal chance of rolling a 2, 3, 4, 5, or 8 (non-orange as it is of roling a 7, 9, or 10 (orange) from Jail.

Also - there are 9 and 9 types of rolls that are greater and less than 7, and 3 types of rolls that are equal to 7

1,1 < 7
1,2 < 7
1,3 < 7
1,4 < 7
1,5 < 7
2,2 < 7
2,3 < 7
2,4 < 7
3,3 < 7

1,6 = 7
2,5 = 7
3,4 = 7

2,6 > 7
3,5 > 7
3,6 > 7
4,4 > 7
4,5 > 7
4,6 > 7
5,5 > 7
5,6 > 7
6,6 > 7

[orange]

Originally posted by yavoon
always switch. switching is 2/3rds. many ways to explain it. but the way I like is that.

If u pick the right box initially then the host removes half the wrong boxes. if u pick the wrong box initially then the host removes all remaining incorrect boxes.

but statistically does it really improve your chances? Either way you'll end up making a choice between two boxes, one of which is correct and one of which isn't.

IE- as you explained, it's going to come down to a 50/50 chance whether you swap or do not swap...no?

Edit: Hmmm...statistically the first pick is more likely to be wrong...in which case I could see why the removal and swap would give you an advantage.

[tinyp3nis]

Urbans problem has nothing to do with math. It's just common sense, but remember that there are bad students and also bad teachers. Maybe she didn't get it because she isn't quite smart, or she didn't consentrate or you can't teach.

[tinyp3nis]

Edit: Hmmm...statistically the first pick is more likely to be wrong...in which case I could see why the removal and swap would give you an advantage.

Because the host allways chooses one from the _other two boxes_, you have 50% from the start...

[Nacht]

Originally posted by JellyBean
The probability of rolling at least one six on n dice is:

1 - (probability of not rolling any sixes on n dice)

The chance of not rolling a six on one dice is 5/6, so the chance of not rolling any sixes on n dice is (5/6)^n. So, the expression given above is:

1 - (5/6)^n

Or, for n = 3, 42%.

Yes JellyBean, that was the simple solution I had in mind.

[Dis]

I live in Vegas. I like to think I know about probablility. One main reason why I don't gamble