May 13th, 2015 at 11:44:13 AM
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Coin Toss Experiment..

Rules...

Prior to a coin being tossed the player makes one of the following decisions:-

1. To bet 0 points i.e. no bet.= (no prediction)

2.To bet 1 point on heads or tails.= (prediction)

3.To bet 2 points on heads or tails.= (prediction)

4. To bet 3 points on heads or tails.= (prediction)

5. To bet 4 points on heads or tails.= (prediction)

The coin is then tossed. Being a fair game, the pay out is "even". Bet 1 point win 1 point or loose 1 point etc. Bet 0 points neither win or loose points but the coin is still tossed.

The same rule is used throughout the experiment to determine a 0, 1, 2 , 3 or 4 points decision prior to each toss.

The game continues until the player has made 3000 "predictions".

The correct ( prediction points) - (incorrect prediction points) is then calculated as the final score of the experiment.

The above rules seem to be fair.

Assume that after 3000 predictions the player is ahead by +224 points.

So a final score of +224 would suggest correct 1612 and 1388 incorrect predictions ( in a points sense)

sqr root 3000 x .5 x.5 = 27.38

1612 - 1500 = 112

112 / 27.38 = 4.09 standard deviations above the mean.

That seems a much higher value than one should ever expect in something that is normally distributed and should exist within about 3 standard distributions in a practical sense all of the time.

I think the rules of the experiment are fair, but perhaps , as the points bet are "variable" then I have an incorrect view of the results of the experiment in a standard deviation sense. And am missing a mathematical fundamental.

My counter point is for example: that bet 2 points win/loose 2 points etc. should take care of this concern. And 3000 predictions is more than sufficient for normal distribution to occur and not allow such a result that is basically derived from the gamblers fallacy with a limited step martingale money management strategy attached.

I am at a loss to quantify a result of +224 points in terms of standard deviation when a variable bet amount is applied. In essence my question is how should one correctly, in terms of standard deviation, describe this result?

Thank you so so much in advance for your advice and comments.

regards

Charlie

Report

Rules...

Prior to a coin being tossed the player makes one of the following decisions:-

1. To bet 0 points i.e. no bet.= (no prediction)

2.To bet 1 point on heads or tails.= (prediction)

3.To bet 2 points on heads or tails.= (prediction)

4. To bet 3 points on heads or tails.= (prediction)

5. To bet 4 points on heads or tails.= (prediction)

The coin is then tossed. Being a fair game, the pay out is "even". Bet 1 point win 1 point or loose 1 point etc. Bet 0 points neither win or loose points but the coin is still tossed.

The same rule is used throughout the experiment to determine a 0, 1, 2 , 3 or 4 points decision prior to each toss.

The game continues until the player has made 3000 "predictions".

The correct ( prediction points) - (incorrect prediction points) is then calculated as the final score of the experiment.

The above rules seem to be fair.

Assume that after 3000 predictions the player is ahead by +224 points.

So a final score of +224 would suggest correct 1612 and 1388 incorrect predictions ( in a points sense)

sqr root 3000 x .5 x.5 = 27.38

1612 - 1500 = 112

112 / 27.38 = 4.09 standard deviations above the mean.

That seems a much higher value than one should ever expect in something that is normally distributed and should exist within about 3 standard distributions in a practical sense all of the time.

I think the rules of the experiment are fair, but perhaps , as the points bet are "variable" then I have an incorrect view of the results of the experiment in a standard deviation sense. And am missing a mathematical fundamental.

My counter point is for example: that bet 2 points win/loose 2 points etc. should take care of this concern. And 3000 predictions is more than sufficient for normal distribution to occur and not allow such a result that is basically derived from the gamblers fallacy with a limited step martingale money management strategy attached.

I am at a loss to quantify a result of +224 points in terms of standard deviation when a variable bet amount is applied. In essence my question is how should one correctly, in terms of standard deviation, describe this result?

Thank you so so much in advance for your advice and comments.

regards

Charlie

Report

May 13th, 2015 at 11:50:28 AM
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The answer depends heavily on the average bet, and to some extent the distribution of the average bet. If you bet 4 on every decision, then 224/4 = 56 decision spread, which is 1528 - 1472.

28 decisions is only 1 st dev above expectations, which is perfectly reasonable.

Even if the average bet is smaller, if you are using a progression the distribution could include many choices of 4 bets, and the result would still be reasonable.

28 decisions is only 1 st dev above expectations, which is perfectly reasonable.

Even if the average bet is smaller, if you are using a progression the distribution could include many choices of 4 bets, and the result would still be reasonable.

Wisdom is the quality that keeps you out of situations where you would otherwise need it

May 13th, 2015 at 12:05:03 PM
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Oh ohQuote:charlie123112 / 27.38 = 4.09 standard deviations above the mean.

That seems a much higher value than one should ever expect

1 in abouts 16,000 is no such a small value, imo

and to (N)EVER expect it...

ever

where did you learn that false stuff

all of the time, you saysQuote:charlie123in something that is normally distributed and should exist within about 3 standard distributions in a practical sense all of the time.

100% false, it has to be (well MAYbe 99.9% false)

where did you learn this crap?

as craps is not (it is) normally distributed either

not ever outside 3 SDs (or 4)

i want names, places, addresses, urls

thank you for your entertaining 1st post

hope you find lots of opinions here

have fun and be good

Sally

I Heart Vi Hart

May 13th, 2015 at 12:17:43 PM
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Quote:charlie123I am at a loss to quantify a result of +224 points in terms of standard deviation when a variable bet amount is applied. In essence my question is how should one correctly, in terms of standard deviation, describe this result?

Normalize to outcomes rather than bet amounts. If you have information on each bet, simply divide the points by the wager and get either -1 or +1 for each outcome. Then your stats will make sense and you'll be well shy of 4 SD.

Edit: wait, did I just do your homework for you?

"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563

May 13th, 2015 at 12:40:18 PM
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If I understand you correctly. Then this result is +2.34 standard deviations. In real terms.

+224 / 2 = 112

sqr root 3000 x .25 = 27.38.

average bet size =1.75

64 / 27.38 = + 2.337 standard deviations.

Is that what you mean ?

I wish you had done my homework for me but I am a little too old for school. Oh well

Thanx for your swift reply.

+224 / 2 = 112

sqr root 3000 x .25 = 27.38.

average bet size =1.75

64 / 27.38 = + 2.337 standard deviations.

Is that what you mean ?

I wish you had done my homework for me but I am a little too old for school. Oh well

Thanx for your swift reply.

May 13th, 2015 at 12:40:22 PM
permalink

If I understand you correctly. Then this result is +2.34 standard deviations. In real terms.

+224 / 2 = 112

sqr root 3000 x .25 = 27.38.

average bet size =1.75

64 / 27.38 = + 2.337 standard deviations.

Is that what you mean ?

I wish you had done my homework for me but I am a little too old for school. Oh well

Thanx for your swift reply.

+224 / 2 = 112

sqr root 3000 x .25 = 27.38.

average bet size =1.75

64 / 27.38 = + 2.337 standard deviations.

Is that what you mean ?

I wish you had done my homework for me but I am a little too old for school. Oh well

Thanx for your swift reply.

May 13th, 2015 at 1:28:55 PM
permalink

Yes, this looks more reasonable. However, a progression that trades many small wins for a few large losses is probably more likely than flat-betting the same average amount to show a small total win like this, so the result is even more likely. The distribution of sample results for 3000 tosses may not be perfectly normal yet, so I'd say nothing suspicious here.

Wisdom is the quality that keeps you out of situations where you would otherwise need it

May 13th, 2015 at 2:57:35 PM
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I'm in love, I'm all shook up...

Youuuuuu MIGHT be a 'rascal' if.......(nevermind ;-)...2F

June 1st, 2015 at 1:13:07 PM
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coin toss experiment continued......

the rules of the coin toss experiment seem to be fair..

How to determine the standard deviation of the result has provoked some welcome reply's. Which is great.(all replies are good) and I thank everyone for their input..

The measure of standard deviation is off course important. As it is not enough to just be simply ahead or for that matter behind in a points sense. It is the magnitude of being ahead or behind and this is were standard deviation is useful to me. As it suggests the probability of a given result..

So in this experiment where the bet sizes are variable.. And again thanks to all replies, all are welcome, standard deviation of the results can be calculated by:-

Total points won - total points lost = Net points won. (assuming a positive result of course).

Net points won / Total n bets = Average bet size..

Net points won / Average bet size / 2 / sqr root x .25 x n bets = standard deviation value of the experiment result.

Please agree or feel free to correct my thinking.. I will then post the results of my coin toss experiment in a manner that is generally, acceptable to all, in terms of the standard deviation value.

Thanks in advance I think this is a fun and interesting forum..

Charlie

the rules of the coin toss experiment seem to be fair..

How to determine the standard deviation of the result has provoked some welcome reply's. Which is great.(all replies are good) and I thank everyone for their input..

The measure of standard deviation is off course important. As it is not enough to just be simply ahead or for that matter behind in a points sense. It is the magnitude of being ahead or behind and this is were standard deviation is useful to me. As it suggests the probability of a given result..

So in this experiment where the bet sizes are variable.. And again thanks to all replies, all are welcome, standard deviation of the results can be calculated by:-

Total points won - total points lost = Net points won. (assuming a positive result of course).

Net points won / Total n bets = Average bet size..

Net points won / Average bet size / 2 / sqr root x .25 x n bets = standard deviation value of the experiment result.

Please agree or feel free to correct my thinking.. I will then post the results of my coin toss experiment in a manner that is generally, acceptable to all, in terms of the standard deviation value.

Thanks in advance I think this is a fun and interesting forum..

Charlie

June 1st, 2015 at 6:10:05 PM
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Quote:charlie123the rules of the coin toss experiment seem to be fair..

I haven't seen any rules about how the coin will be tossed.

Accordingly, I cannot assume it's a fair toss.

May the cards fall in your favor.