Session Bankroll Question

[DSchles]

I don't own BJA3. Sure you can tell me how to calculate that'd be great.

TO avoid clutter here, I was actually trying to PM you but don't see a way to do that.... anyone if there's a way to PM someone on this server?

You went to the right place, but you asked to be a friend, which, normally, I don't accept. You needed to simply send a message (email) by putting in DSchles in the "To" box.

In any event, maybe an example will help you to understand some fundamental concepts.

Suppose you flat bet $5 a hand for 100 hands. What is the overall s.d. for that? Well, the one-hand s.d. is 1.15 x $5, which is $5.75. Since overall s.d. is a square root function, the total s.d. for 100 hands is $5.75 x 100^0.5 = $5.75 x 10 = $57.50. Simple and straightforward, right?

So, now, let's consider someone who bets $1 a hand for 50 hands and $9 a hand for another 50 hands. Clearly, that person has an average bet of the same $5 as above, and so you might be tempted to use that value and conclude that the overall s.d. for the 100 hands is the identical $57.50 as above, but you would be very wrong to do so!

As I mentioned in my previous post, if bet sizes are different (makes no difference if you are spreading as a card counter, playing progression, or whatever), you cannot use the average bet size to calculate s.d. Instead, you need to use a somewhat more complicated process that involves: using the variance (the square of s.d.), multiplying by the square of the bet size, multiplying by the frequency of that particular bet, summing all the horizontal lines vertically, and, finally, taking the square root of that sum.

Wow! All that, huh? Yup. and there are no shortcuts; that's the way to do it. Period. So, instead of getting the $57.50 that you thought might be the s.d. for the second example, what is the correct answer? Well, let's follow my instructions above:

Variance is 1.3225. First bet squared is 1. Second bet squared is 81. Frequency of each bet is 50 (hands). So, ...

[(1.3225 x 1 x 50) + (1.3225 x 81 x 50)]^0.5 = (66.125 + 5,356.125)^0.5 = 5,422.255^0.5 = 73.64. Oh!! So, $73.64 instead of $57.50, despite the same average bet of $5 each time.

Clear?

Don

[BughouseMASTER]

You went to the right place, but you asked to be a friend, which, normally, I don't accept. You needed to simply send a message (email) by putting in DSchles in the "To" box.

In any event, maybe an example will help you to understand some fundamental concepts.

Suppose you flat bet $5 a hand for 100 hands. What is the overall s.d. for that? Well, the one-hand s.d. is 1.15 x $5, which is $5.75. Since overall s.d. is a square root function, the total s.d. for 100 hands is $5.75 x 100^0.5 = $5.75 x 10 = $57.50. Simple and straightforward, right?

So, now, let's consider someone who bets $1 a hand for 50 hands and $9 a hand for another 50 hands. Clearly, that person has an average bet of the same $5 as above, and so you might be tempted to use that value and conclude that the overall s.d. for the 100 hands is the identical $57.50 as above, but you would be very wrong to do so!

As I mentioned in my previous post, if bet sizes are different (makes no difference if you are spreading as a card counter, playing progression, or whatever), you cannot use the average bet size to calculate s.d. Instead, you need to use a somewhat more complicated process that involves: using the variance (the square of s.d.), multiplying by the square of the bet size, multiplying by the frequency of that particular bet, summing all the horizontal lines vertically, and, finally, taking the square root of that sum.

Wow! All that, huh? Yup. and there are no shortcuts; that's the way to do it. Period. So, instead of getting the $57.50 that you thought might be the s.d. for the second example, what is the correct answer? Well, let's follow my instructions above:

Variance is 1.3225. First bet squared is 1. Second bet squared is 81. Frequency of each bet is 50 (hands). So, ...

[(1.3225 x 1 x 50) + (1.3225 x 81 x 50)]^0.5 = (66.125 + 5,356.125)^0.5 = 5,422.255^0.5 = 73.64. Oh!! So, $73.64 instead of $57.50, despite the same average bet of $5 each time.

Clear?

Don

Not exactly. So what would be the RoR on $9/hand for 50 hands and $1/hand for 50 hands? BTW - You gave me the 1st part of your email, but did not say the domain.

[DSchles]

Not exactly. So what would be the RoR on $9/hand for 50 hands and $1/hand for 50 hands? BTW

How can I give you a ROR without a bankroll? In the calculator, just plug in the desired BR, the (negative) win rate per 100 hands, and the s.d. per 100 hands, which I've shown you how to calculate.

You gave me the 1st part of your email, but did not say the domain.

It isn't an email address! It's the handle to put into the addressee field on the forum website.

Don

[BughouseMASTER]

You went to the right place, but you asked to be a friend, which, normally, I don't accept. You needed to simply send a message (email) by putting in DSchles in the "To" box.

In any event, maybe an example will help you to understand some fundamental concepts.

Suppose you flat bet $5 a hand for 100 hands. What is the overall s.d. for that? Well, the one-hand s.d. is 1.15 x $5, which is $5.75. Since overall s.d. is a square root function, the total s.d. for 100 hands is $5.75 x 100^0.5 = $5.75 x 10 = $57.50. Simple and straightforward, right?

So, now, let's consider someone who bets $1 a hand for 50 hands and $9 a hand for another 50 hands. Clearly, that person has an average bet of the same $5 as above, and so you might be tempted to use that value and conclude that the overall s.d. for the 100 hands is the identical $57.50 as above, but you would be very wrong to do so!

As I mentioned in my previous post, if bet sizes are different (makes no difference if you are spreading as a card counter, playing progression, or whatever), you cannot use the average bet size to calculate s.d. Instead, you need to use a somewhat more complicated process that involves: using the variance (the square of s.d.), multiplying by the square of the bet size, multiplying by the frequency of that particular bet, summing all the horizontal lines vertically, and, finally, taking the square root of that sum.

Wow! All that, huh? Yup. and there are no shortcuts; that's the way to do it. Period. So, instead of getting the $57.50 that you thought might be the s.d. for the second example, what is the correct answer? Well, let's follow my instructions above:

Variance is 1.3225. First bet squared is 1. Second bet squared is 81. Frequency of each bet is 50 (hands). So, ...

[(1.3225 x 1 x 50) + (1.3225 x 81 x 50)]^0.5 = (66.125 + 5,356.125)^0.5 = 5,422.255^0.5 = 73.64. Oh!! So, $73.64 instead of $57.50, despite the same average bet of $5 each time.

Clear?

Don

Anyway what would be the RoR on $9/hand for 50 hands and $1/hand for 50 hands? with say a $200 bankroll?

[DSchles]

Anyway what would be the RoR on $9/hand for 50 hands and $1/hand for 50 hands? with say a $200 bankroll?

What's the disadvantage? I thought you said you had used the calculator. If player edge is, say, -0.5%, then ROR is certain if you play long enough (100%). You need to specify a number of hands to be played, as well. I don't mind answering your questions -- even though you can do it yourself -- but you have to be more precise when you ask.

Don

[BughouseMASTER]

What's the disadvantage? I thought you said you had used the calculator. If player edge is, say, -0.5%, then ROR is certain if you play long enough (100%). You need to specify a number of hands to be played, as well. I don't mind answering your questions -- even though you can do it yourself -- but you have to be more precise when you ask.

Don

Don,

I had already specified clearly on my last message that it's 50 hands each. Don't know how much clearer I can get when 50 + 50 is 100 total hands. HA would be -0.5%, and I said I had problems using the QFIT calculator....

[DSchles]

Don,

I had already specified clearly on my last message that it's 50 hands each. Don't know how much clearer I can get when 50 + 50 is 100 total hands. HA would be -0.5%, and I said I had problems using the QFIT calculator....

ROR = 0.724%. And a suggestion for the future: the last thing you want to do is cop an attitude with me.

Don

[BughouseMASTER]

ROR = 0.724%. And a suggestion for the future: the last thing you want to do is cop an attitude with me.

Don

???

What are you talking about? Are you upset at yourself since you misread what I clearly wrote in my previous message? How is that my fault and why would you imply I'm copping an attitude when, again, I merely pointed out to you how you missed what I had already wrote as clear as day?!

[ZeeBabar]

???

What are you talking about? Are you upset at yourself since you misread what I clearly wrote in my previous message? How is that my fault and why would you imply I'm copping an attitude when, again, I merely pointed out to you how you missed what I had already wrote as clear as day?!

Copping an attitude is when you wrote “don’t know how much clearer I can get”. That was not necessary! You did not “merely” pointed it out. You could have stated it differently, like “Don, perhaps you missed it, I....”

There are some on the forum you can banter with, take your hostilities out on and more and then there are some, Don S or Norm or other math folks you must respect.