Exponential Function: Working Through a Coin Flipping Exercise

[kateman]

hello!

there's this question I am working on and I am mostly through it (well, I hope I am at least), there's just one or two things still annoying me/need confirmation.

Here's the question: 50 coins were flipped. For every coin that landed with the head showing upwards, it was taken away. The remaining amount that landed on tails side up were flip again. This process happened repeatedly until all the coins were taken away. The results are below.

flips: 1 2 3 4 5 6 7
amount remaining: 22 7 2 1 1 1 0
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So, when flips = 0, amount remaining = 50

basically, I can see that it is exponential (I can also prove mathimatically that it isn't linear), I can plot it and show that its exponential but my maths is a bit rough trying to prove it being exponential.

I know i can use the formula b = (y2/y1)^(1/x2-x1) for the separate values of the table to prove that it is exponential. For a perfect expoential, they would all be the same but mine are slightly out (some more so at different places). Do you think it would still be acceptable to so that its only up to chance (50/50) for the coin to land in the same place and its only a slight varience in chance that leads to it not being a perfect exponential function?

The next problem is my function for this data. Following the formula for a above, I guess I would do an average of all the b's to find the b for the function y = ab^x ?

then I just add the a value of 50 to get y = 50b(the average of all the b's)^x
does that seem right?
are there any flaws in my reasoning?

Thank-you very much!

 

[HallsofIvy]

Well, of course, you can't get a specific formula because the number is random. You can get a formula for the expected number of heads each time:

You start with 50 coins and they have equal probability, 1/2, of coming up head or tails. After one flip the expected number of coins left is 50(1/2)= 25. After the second flip the expected number of coins left is (50(1/2))(1/2)= 50(1/2)²= 25.5 (since this is an expected value, we can have a fractional number of coints). After the third flip, 50(1/2)³ and, in general, after the n^th flip, 50(1/2)ⁿ coins. Yes, that is an exponential.

I can't quite follow your reasoning since I do not see where you got b= (y2/y1)^(1/x2-x1)- you did not say what x1, x2, y1, and y2 represent. Once again, since different trials will give results, you cannot get a formula for specific results, only for "expected" or "average" results.

 

[kateman]

I can't quite follow your reasoning since I do not see where you got b= (y2/y1)^(1/x2-x1)- you did not say what x1, x2, y1, and y2 represent. Once again, since different trials will give results, you cannot get a formula for specific results, only for "expected" or "average" results.

I found the formula on this website (on page 4):
http://wcherry.math.unt.edu/math1650/exponential.pdf

I was using it as any y value for y1 and then the y2 value was the one after it (like wise with the corresponding x values)

I'am aware a formula can't be exactly taylored for this situation, but trying to get one as close as possible for these resluts.

cheers

 

[olliemath]

That page seems to be for problems where you have no way of justifying a specific exponential, but the data seems to fit, e.g.

A biologist starts a bacteria culture growing. The culture began with 1000 cells. Three hours later the biologist returned to find 5000 cells in the culture. Assuming the bacteria culture grows exponentially, how many cells will there be in the culture 10 hours after the culture began growing?

In that case finding the constants from the data makes sense, since a priori it's not obvious how fast the average bacterium multiplies/dies. In your case, Halls' solution is much nicer - we know the probability that a coin survives the experiment, so we can work out the distribution without recourse to the method in the link.

If you really want to test if the exponential and the data are significantly different then you can do this, but you might want to use more coins (50 is a very small number from a probability point of view) :p

 

[kateman]

okay, thanks Olliemath and HallsofIvy!