Probability. Oy.

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Ok, so I haven't posted in ages. But anyhow, I was having a little discussion with my Geometry teacher after class today and we somehow got into a conversation on probability. Now, I didn't have much time to ask him about what he was telling me because I was already late for my last class, but I'm hoping you can explain it to me so the same thing doesn't happen tomorrow. Laughing

Can you explain to me why if you have, say, 3 coins, each possible combination ( HHH, HHT, HTT, TTT ) doesn't have a 50/50 chance of occurring? .I figured it would because each coin has a 50/50 chance of landing on heads or tails and each occurrence wouldn't affect the next outcome of the coin. Apparently, I had made the wrong assumption. I remember reading something about it in Martin Gardners book Aha! Gotcha about four cats or something, but I can't find it.

Also, say, you have an average height of a person. You would have a graph like this ( right?):

http://img.photobucket.com/albums/v290/BonelessConnus/graphaverage.bmp

What would your chances of being average be? 0%? I wouldn't think so, but, I'm just confused.

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I have very little time on my hands so I'll only answer the first question. Your assumption is correct if you consider the order (such that the possibilities are HHH HHT HTH HTT THH THT TTH TTT). That, however, is not the case. The probability that you will have HHH is 1/8. The possibility the you will have THH is 3/8 because HHT HTH THH (all with probability 1/8) are all acceptable.

I hope this helps,

Rara Avis.

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Name the coins 1,2,3, say. Then draw a "tree."

1 H (HHH)
` H /
2 / \ T (HHT)
3 H H (HTH)
` \ T /
4 \ T (HTT)

5 H (THH)
` H /
6 / \ T (THT)
7 T H (TTH)
` \ T /
8 \ T (TTT)

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Re: Probability. Oy.

ScissorsRunWithYou wrote:

Also, say, you have an average height of a person. You would have a graph like this ( right?):

What would your chances of being average be? 0%? I wouldn't think so, but, I'm just confused.

Your odds of being an exact "average height" depends on how many people are used to decide the average and the actual heights of the people involved.

Two examples:

You have 3 people - 1 5' tall, 1 6' tall and one 5' 6" tall your odds of being "average" are 1 out of 3 (33% probability) because the average is 5' 6" and one of the people measured is exactly that height.

You have 3 people and they are 5' 3", 5' 7" and 5' 9.5" then your odds are 0. The "average" of those 3 is 5' 6.5" which matches none of the 3.

That's why doctors, etc.. usually use an range for "average".

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satt_focusable wrote:

Name the coins 1,2,3, say. Then draw a "tree."

1 H (HHH)
` H /
2 / \ T (HHT)
3 H H (HTH)
` \ T /
4 \ T (HTT)

5 H (THH)
` H /
6 / \ T (THT)
7 T H (TTH)
` \ T /
8 \ T (TTT)

Erm, your text tree is just a little hard to follow. I'm not sure I understand. Mad

Perhaps you could draw it and post an image of it to show it more clearly? Smile

Even if I think I can read the tree, I don't think, or maybe I just don't understand, why the certain onces have a more probable outcome.

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Re: Probability. Oy.

fishin' wrote:

Your odds of being an exact "average height" depends on how many people are used to decide the average and the actual heights of the people involved.

Two examples:

You have 3 people - 1 5' tall, 1 6' tall and one 5' 6" tall your odds of being "average" are 1 out of 3 (33% probability) because the average is 5' 6" and one of the people measured is exactly that height.

You have 3 people and they are 5' 3", 5' 7" and 5' 9.5" then your odds are 0. The "average" of those 3 is 5' 6.5" which matches none of the 3.

That's why doctors, etc.. usually use an range for "average".

So, you can only have any chance of being average if the average is one of the numbers used? And if that happens, then the probability would just be (number of times that one number occurred) / ( The total number of..numbers)?

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This will be better.

Code:



               H (HHH) 

       H  / 

      /     \ T (HHT) 

 H             

                    H (HTH) 

       \  T    / 

                  \ T (HTT) 





                H (THH) 

`         H / 

       /      \ T (THT) 

 T           

                   H (TTH) 

        \ T / 

                \ T (TTT)

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The amazing thing about this way of thinking is that if you know a girl and you know she has a sibling, there is a 66.66% the sibling is male.

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Sorry, I still can't read it well. :/ Maybe my computer just doesn't work with the script. I don't mean to be annoying. Sad

Rara Avis wrote:

The amazing thing about this way of thinking is that if you know a girl and you know she has a sibling, there is a 66.66% the sibling is male.

I've heard sopmething like this before, but. I ..just don't understand why. :/

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Another nice probability fact (although somewhat out of place) is that if you have 23 random people there is more than 50% that two of them have the same birthday.

Most people just say 23/365 is less than 1/2 and so it's not true. But the secret is that the number of people isn't the thing that matters. It's the number of pairs of people that matters.

Basic probability is cool.

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Yes, I've heard that from Ray Smullyan. I've heard a few times of a story about a time when Smullyan was in a classroom and was telling this to a class of students, but that this specific class only had a certain amount of people, so the chance of that is much LESS likely. One of the students bet him that there was a 100% chance of at least two people in the classroom had the same birthday, when he, in fact, did not know anyone's but his own. Turns out there were two identical twins in the room and Smullyan didn't even notice! Laughing

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Re: Probability. Oy.

ScissorsRunWithYou wrote:

Can you explain to me why if you have, say, 3 coins, each possible combination ( HHH, HHT, HTT, TTT ) doesn't have a 50/50 chance of occurring?

Someone else stated this slightly differently.

If you toss three coins, you have eight possible outcomes:

HHH
HHT
HTH
THH
HTT
THT
TTH
TTT

The chances, then, of all heads is 1/8, two heads one tails is 3/8, one heads two tails is 3/8, and all tails is 1/8.

Therefore, while the chance of each permutation is equal, the combinations clump and do not have equal probabilities.

(I ignore the chances of a coin landing on edge....)

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Re: Probability. Oy.

MerlinsGodson wrote:

The chances, then, of all heads is 1/8, two heads one tails is 3/8, one heads two tails is 3/8, and all tails is 1/8.

Therefore, while the chance of each permutation is equal, the combinations clump and do not have equal probabilities.

(I ignore the chances of a coin landing on edge....)

Aaah thanks so much,I understand now that I realize that he meant all possible outcomes, including the order in which they land. Thanks again. Smile

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Probability. Oy.

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