# Coin Toss net gain standard deviation

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1. Apr 30, 2015

### charlie !

When tossing a fair coin 1000 times a player correctly predicts 532 outcomes. Then I think I am right in saying that result is about + 2 Standard Deviations from the mean.

sqr root 1000 x .5 x .5 = 15.81.

32/15.81= 2.02.

However, if the results of the coin toss are given by a Net Gain/ Loss value i.e. wins - losses. So after player A makes 1000 predictions the net score = +32. Would it be correct to state that the result is +2 Standard Deviations from the mean.

As in the toss of a fair coin the player would expect to correctly predict 500 tosses and incorrectly predict 500 tosses. 500-500 = 0.

2. Apr 30, 2015

### mathman

Your description is confusing. Net score of 532 seems to be equivalent to 516 correct.

3. Apr 30, 2015

### charlie !

?? The net score , correct predictions - incorrect predictions by the player after 1000 predictions is + 32. i.e. the player made 32 more correct predictions than incorrect predictions.

4. May 1, 2015

### mathman

516-484=32 ????????????????????????????

5. May 2, 2015

### charlie !

Understand,

So then the net score of +32, in 1000 tosses is actually 1 Standard Deviation ?

Hope I'm right this time, and thanx for your help.

6. May 2, 2015

Yes.

7. May 2, 2015

### charlie !

Thank you for your input and time to reply. Appreciate it very much. It is good to learn and with your help understand..

Thank you once again.

charlie

8. May 2, 2015

### WWGD

But wait, if I understood correctly, the player predicted 532 outcomes out of 1000 correctly, with standard deviation ~ 16 , then 532 has a z-value $z= \frac {532-500}{16} =2$? And expecting exactly 500 correct guesses is a bit flexible. This is why you use confidence intervals.

9. May 3, 2015

### Staff: Mentor

500 is the exact expectation value. That does not mean you would be surprised to see other values nearby, of course, that's why we have the standard deviation as additional value.

10. May 3, 2015

### WWGD

I was addressing the statement that " the player would expect to correctly predict 500 values" . I don't know if s/he meant it in a literal sense or in the (correct) sense that np =1000(0.5)=500. I replied in case it was the literal sense.

11. May 3, 2015

### charlie !

hello, and thanx to everyone I am a she by the way not a he. , and by the look of things not the brightest at the moment.

"However, if the results of the coin toss are given by a Net Gain/ Loss value i.e. wins - losses. So after player A makes 1000 predictions the net score = +32. Would it be correct to state that the result is +2 Standard Deviations from the mean".

Mathman is correct I think, in saying a net gain of +32 points in 1000 tosses would be represented by winning 516 tosses and loosing 484 tosses.

516 + 484 = 1000.

So if the standard deviation is worked out as follows:-

sqr root 1000 x .5x.5= 15.81.

Therefore the result is about 1 standard deviation above the expected mean when tossing a fair coin. 516 which is +16 above the mean. But in actual fact one has won 516 tosses and lost 484. And so has won and is ahead by (net +32 tosses).

Looking at the same figures in a different way. If one is given the net gain amount, in this case +32.

Then one could argue that:-

In tossing a fair coin 1000 times. You should in absolute terms should expect to win 500 tosses but also loose 500 tosses. And so be ahead by zero after the 1000 toss.

But being ahead, in this case by +32 seems to suggest being ahead by a standard deviation value of 2. Which of course is very different to a value of 1 S/D.

This is what I am having difficulty with understanding.

And many thanx to all that have taken the time to help. Your patience is very much appreciated.

Charlie.

12. May 3, 2015

### WWGD

You mean you are not a s/he ? :) .

Then if the score is actually 516, then your z is 1; formally, you're using a normal aprox to a binomial, with mean 1000(0.5)=500
and standard deviation 16 , as you computed. Then your z-value for 516 is $z(500)= \frac {516-500}{16} =1$. If the number of correct
is 532, then, computing z(532), you would get a z-value of 2. So it is just a matter of figuring out the number of correct guesses.

13. May 3, 2015

### Staff: Mentor

The standard deviation of correct predictions is 16, the standard deviation of the difference between correct and wrong predictions ("score") is 32.
Those two are different quantities, they have different standard deviations (and different means, obviously).

14. May 3, 2015

### WWGD

Well, in my experience, the usual questions in this context are about the distribution of correct predictions , as a normal approximation
to a binomial, so that is what I am assuming in my posts.

15. May 4, 2015

### charlie !

Thank you mfb, that is my humble but critical point.

16. May 4, 2015

### WWGD

So , Charlie! , you're not clear on what distribution to use?

17. May 5, 2015

### charlie !

18. May 5, 2015

### Staff: Mentor

I don't understand the problem. Both ways to describe the same number of successful of predictions are valid and both agree with their answer.

19. May 5, 2015

### charlie !

Hello mfb,

the problem as I see it ..

Is one correct to define the results as 1 Standard Deviation above the mean or 2.

This is my concern.

Charlie

20. May 5, 2015

### Staff: Mentor

Which result? You had two different results in the thread so far.

532 correct predictions is 2 standard deviations above the mean.
A score of 64, which comes from 532 correct predictions, is 2 standard deviations above the mean.

516 correct predictions is 1 standard deviation above the mean.
A score of 32, which comes from 516 correct predictions, is 1 standard deviation above the mean.

mathman highlighted that difference in post 2.