MJ: Standard Deviation

[MJ]

I think I get it. So the probability of winning more or less then $2000 can be calculated. However, the probability of winning EXACTLY $2000 is 0%?

I guess that makes sense intuitively as there are MANY different amounts one can win/lose. Graphically this also makes sense as there would be no area under the curve for an exact win amount. It would be like integrating a function on an interval from a to a. There is no area to compute hence the probability is 0%.

-MJ

> No you don't. Using a normal approximation,
> the probability of being ahead $2000 is
> exactly zero. Bonus marks for answering the
> question he meant to ask though. ;-)

[Don Schlesinger]

> I think I get it. So the probability of
> winning more or less then $2000 can be
> calculated. However, the probability of
> winning EXACTLY $2000 is 0%?

I suppose it has some probability, but it is obviously very, very small.

> I guess that makes sense intuitively as
> there are MANY different amounts one can
> win/lose. Graphically this also makes sense
> as there would be no area under the curve
> for an exact win amount. It would be like
> integrating a function on an interval from a
> to a. There is no area to compute hence the
> probability is 0%.

Exactly. Note, however, that for smaller sample sizes -- even for something like flipping a coin 100 times and charting the number of heads -- getting a 1.77 SD event carries a definite probability of greater than zero. In fact, in this example, one SD above the mean of 50 heads is 5, so 1.77 SDs would be 5*1.77, or (rounded) 9.

Now, the probability of getting precisely 59 heads in 100 tosses is the very non-zero
1.587%.

Don

[MJ]

> I suppose it has some probability, but it is
> obviously very, very small.

I agree. It certainly wouldn't be impossible to win EXACTLY $2000.00 but the chances of that are slim to none. There are just so many possible amounts one win after playing 100 hands of BJ.

Now I ask howcome the bell curve leads one to erroneously conclude the probability of winning precisely $2000.00 is 0%? I suppose the bell curve isn't really used to predict the exact probability of ONE specific event(IE winning $2000.00). Rather it is used to predict the likelyhood of an event occuring within a certain range. To try and do otherwise would constitute a misuse of the bell curve and result in an erroneous conclusion.

> Exactly. Note, however, that for smaller
> sample sizes -- even for something like
> flipping a coin 100 times and charting the
> number of heads -- getting a 1.77 SD event
> carries a definite probability of greater
> than zero. In fact, in this example, one SD
> above the mean of 50 heads is 5, so 1.77 SDs
> would be 5*1.77, or (rounded) 9.

So 100 hands of BJ has no definite probability but 100 coin tosses do??

Well anyway I guess you took the square root of 100(sample size) and acquired SD = 10. You then divide 10 by 2 because there is a 50/50 chance of any toss resulting in heads or tails.

1 SD falls within 45T to 55H or 45H to 55T.
2 SD falls within 40T to 60H or 40H to 60T.
3 SD falls within 35T to 65H or 35H to 65T.

> Now, the probability of getting precisely
> 59 heads in 100 tosses is the very non-zero
> 1.587%.

Ok now I'm lost. I don't think you used the Z Chart for that calculation so how did you arrive at 1.587%?

-MJ

> Don

[ET Fan]

> Now I ask howcome the bell curve leads one
> to erroneously conclude the probability of
> winning precisely $2000.00 is 0%?

The central limit theorem tells us that any infinite set of random variable, if they are independant, sum to a normal curve. If you played an infinite amount of blackjack, then the probability of any one particular result would actually be zero. Hands of blackjack within a shoe aren't completely independant, but they're pretty close. Shoe to shoe results are independant, so if you play enough shoes, the normal curve is guaranteed to be a "useful" approximation.

As a practical matter, you want at least 20 or 30 hands of blackjack before applying the normal curve.

Working out the real exact odds for a range of blackjack results would be horrendously complicated. I think a guy was working on it for a while on bjmath.com.

> I suppose
> the bell curve isn't really used to predict
> the exact probability of ONE specific
> event(IE winning $2000.00). Rather it is
> used to predict the likelyhood of an event
> occuring within a certain range. To try and
> do otherwise would constitute a misuse of
> the bell curve and result in an erroneous
> conclusion.

There's a trick to misusing the bell curve that way. You apply it to the range R-0.5 to R+0.5 for the Result your interested in.

> So 100 hands of BJ has no definite
> probability but 100 coin tosses do??

Sure 100 hands of BJ has a definite probability. You figure it out! (I'm on holiday.)

> Well anyway I guess you took the square root
> of 100(sample size) and acquired SD = 10.
> You then divide 10 by 2 because there is a
> 50/50 chance of any toss resulting in heads
> or tails.

> 1 SD falls within 45T to 55H or 45H to 55T.
> 2 SD falls within 40T to 60H or 40H to 60T.
> 3 SD falls within 35T to 65H or 35H to 65T.

> Ok now I'm lost. I don't think you used the
> Z Chart for that calculation so how did you
> arrive at 1.587%?

He used the binomial distribution, which is exact for coin flips:
(100 C 59)/2^100

ETF

[ET Fan]

I wouldn't use the R +/- .5 trick for blackjack without a few hundred hands in the sample, if you're varying bets. Also, I'm assuming $0.50 payoffs are impossible -- bets are all even numbers -- unless you want to consider $1999.50 a hit. (I think I have this part straight, but don't quote me!)

ETF

> There's a trick to misusing the bell curve
> that way. You apply it to the range R-0.5 to
> R+0.5 for the Result your interested in.

[Zenfighter]

Coin tossing: 59 heads out of 100

1) Exact binomial calculation : .0158691

2) Approximation via normal : .0158489

Proof:

9.5/ sqr(100*.5*.5) = 1.9 thus q(z1) = .9712834

8.5/sqr(100*.5* .5) = 1.7 thus q(z2) = .9554345

q(z1) ? q(z2) = .0158489

Here the sample (n = 100) yields a pretty good approximation.

Zenfighter

[ET Fan]

Admit it, man. You saw my post go up attempting the same calculation. Then you saw me delete it, so you did it yourself!

Maaaan, I should have known I made a simple mistake somewhere. Should have had more faith in the normal curve.

ETF

> Coin tossing: 59 heads out of 100

> 1) Exact binomial calculation : .0158691

> 2) Approximation via normal : .0158489

> Proof:

> 9.5/ sqr(100*.5*.5) = 1.9 thus q(z1) =
> .9712834

> 8.5/sqr(100*.5* .5) = 1.7 thus q(z2) =
> .9554345

> q(z1) ? q(z2) = .0158489

> Here the sample (n = 100) yields a pretty
> good approximation.

> Zenfighter

[ET Fan]

[Zenfighter]

[Magician]

> I think I get it. So the probability of
> winning more or less then $2000 can be
> calculated. However, the probability of
> winning EXACTLY $2000 is 0%?

You've got it. However, this is only true when we use a normal approximation to the actual distribution of results. If we were to look at the actual distribution (as Don did in his coin-toss example), then some exact dollar amounts would have a non-zero probability and some would have a zero probability.

For example, if you have $25 riding on the first hand of a typical 6-deck game you could end up with -$50, -$25, $0, $25, $37.50 or $50 (and other possibilities if you consider splits and surrender). Each of these values will have a non-zero probability (and many here could tell you those probabilities to several decimal places). But the probability of ending up with, say, $3.67 is zero.

If you looked at the normal approximation to the results for that same hand, it would tell you that the probability of ending up with exactly $25 is zero but the probability of between, say, $3 and $4 is non-zero. Obviously this is nonsense but that's why we call it an approximation.

With coin tosses, it's easy to use the actual (binomial) distribution. But in blackjack there are more than 2 outcomes in each trial and the trials are not independent, so it becomes much more difficult. Also, normal approximations become more accurate as the number of trials increases and we like to simulate as many hands as possible.

[MJ]

[Zenfighter]

There is no area to compute hence the probability is 0%

That?s what Magician was trying to tell you, but tricks do exist to approximate this probability with the aid of the normal approach.

1947.85/1100 = 1.77

Looking under the Curve:

z = 1.77 = .961636 as the total probability (area) of achieving this result or less.

More difficult is to approximate the probability of a $2000 win, no more nor less.

1) Add here 0.005 to z and find z1
2) Subtract 0.005 from z and find z2
3) Finally subtract z2 from z1 and voila?

1) z1 = 1.775 = .962051

2) z2 = 1.765 = .961218

3) The final area (probability) = .962051 -.961218 = .000833 = 0.08%

Now let?s find the area computed in amount of dollars

1) x1/1100 = 1.775 thus x1 = 1952.5

2) x2/1100 = 1.765 here x2 = 1941.5

Adding to both your EV = 52.15 we have then:

p = 0.08% to end somewhere between [$1993.65 and $2004.65]

There is always room for further refinements. What you will need, are established borders to make the computation, so as a result you?ll have only an approximation. That?s all.

For independence trials and exact probabilities the adequate tool is the binomial formula. (E.g. coin tossing with a moderate number of throws). BJ do not fall under these categories, fortunately. Hence, we look under the Curve more often.

Hope this helps, also.

Zenfighter

[MJ]

I get it. Thats a good idea. It seems as though you took the Lim Z --> 1.77 from both sides(left and right) to establish a boundary. I suppose for an even better estimate you could add .001 to Z to find z1 and subtract .001 to find z2. Good work Zen!

-MJ

> There is no area to compute hence the
> probability is 0% That?s what Magician was
> trying to tell you, but tricks do exist to
> approximate this probability with the aid of
> the normal approach.

> 1947.85/1100 = 1.77

> Looking under the Curve:

> z = 1.77 = .961636 as the total probability
> (area) of achieving this result or less.

> More difficult is to approximate the
> probability of a $2000 win, no more nor
> less.

> 1) Add here 0.005 to z and find z1
> 2) Subtract 0.005 from z and find z2
> 3) Finally subtract z2 from z1 and voila?

> 1) z1 = 1.775 = .962051

> 2) z2 = 1.765 = .961218

> 3) The final area (probability) = .962051
> -.961218 = .000833 = 0.08%

> Now let?s find the area computed in amount
> of dollars

> 1) x1/1100 = 1.775 thus x1 = 1952.5

> 2) x2/1100 = 1.765 here x2 = 1941.5

> Adding to both your EV = 52.15 we have then:

> p = 0.08% to end somewhere between [$1993.65
> and $2004.65]

> There is always room for further
> refinements. What you will need, are
> established borders to make the computation,
> so as a result you?ll have only an
> approximation. That?s all.

> For independence trials and exact
> probabilities the adequate tool is the
> binomial formula. (E.g. coin tossing with a
> moderate number of throws). BJ do not fall
> under these categories, fortunately. Hence,
> we look under the Curve more often.

> Hope this helps, also.

> Zenfighter