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**goldfingerfif****Member**- Registered: 2006-01-25
- Posts: 2

A person sets out to toss a coin 20 times. On average, they expect to get 10 heads and 10 tails.

Their first four tosses hapen to be heads. What do you expect to happen in the next 16 tosses?

Student A: I still expect 10 heads and 10 tails, and since I've already got 4 heads, I now expect 10 tails and 6 heads from the remaining 16 tosses. So in the next few tosses I expect more tails than heads.

Student B: There are 16 tosses to go. For these 16 tosses, I expect 8 head and 8 tails. This means I now expect 12 heads and 8 tails from my original 20 throws.

Which argument is right, and why?

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**darthradius****Member**- Registered: 2005-11-28
- Posts: 97

I would agree with Student B because the probability of tossing heads is always 50%...just because you have already seen the outcome of the first four tosses does not change the likelihood of the next outcome.

Student A's answer implies that once you toss the first four coins, the probability of seeing tails on your next tosses increases...which is obviously incorrect because we know that the coin always has two sides...

The greatest challenge to any thinker is stating the problem in a way that will allow a solution.

-Bertrand Russell

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**goldfingerfif****Member**- Registered: 2006-01-25
- Posts: 2

True. I guessed student A because I guess I didnt read it real good. I didn't see the expect part. I was taking it for the whole instead of the next flip.

Student A also has a good response. If you say it is a 10:10 average then that would be right. Student B is saying this particular flip will be a 12:8 = 3:2 So for every 5 flips 3 will be heads. That is off setting the odds. The odds go up with each flip staying the same. So it is a 1 out of 2(2^1) chance to be heads then the second time is 1 out of 4(2^2) and the 3rd time is 1 out of 8(2^3) and 4th time is 1 out of 16 (2^4). Within the bounds of this single flip, of course the odds are 1/2. But within the context of the 5th flip, the odds 1/32 (2^5). Therefore if you look at it through an average of all flips then Student A is right. If you are looking at it from a perspective where you examine the context of the next flip only then Student B is right. For all we know the next 4 flips will be all tails, giving us our 1:1 odds in which case it will be Student A.

Which is right for homework though? I am now guessing student B.

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**darthradius****Member**- Registered: 2005-11-28
- Posts: 97

I'm a little rusty on my probability theory and I don't want to give you the wrong answer, so you may want to see what someone else says...

but I think that the conditional probability would only come into play if the previous events have some sort of impact on the upcoming ones...That is, I think you ought to be able to look at the next 16 tosses as a totally new series of events and so that each toss still has the very same 50/50 probability.

The greatest challenge to any thinker is stating the problem in a way that will allow a solution.

-Bertrand Russell

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I think my biology professor said it best:

"If you think that the coin will have anything but a 50/50 chance on the next flip, then you would have to think that the coin a) has the ability to realize what has happened in the past and b) has the ability to change the future. If you still want to stick to your answer, there are paramedics standing by."

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**RickyOswaldIOW****Member**- Registered: 2005-11-18
- Posts: 212

I guess that it'll land in akward position between two objects and thus land neither on heads or tails

*Last edited by rickyoswaldiow (2006-01-26 04:37:35)*

Aloha Nui means Goodbye.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Student B is correct. Although you would expect 10 heads and tails from 20 coin tosses, this number changes with the additional information of the 4 heads.

The probability is *always* 50% each way. Well, technically it's a bit less because of the possibility that it would land on its edge, or the slight extra weight of the head side might make it more likely to land that way, but those things are negligible.

Interestingly, although 10 heads and tails is the most likely outcome, there's still only an 18% chance of it actually happening.

Why did the vector cross the road?

It wanted to be normal.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

rickyoswaldiow wrote:

I guess that it'll land in akward position between two objects and thus land neither on heads or tails

The quantum physics joke is that the coin lands and stays on it's side, and is both heads and tails at the same time. Wave/Particle duality rocks!

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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