<@spoonitnow> say you're heads up with no blinds or antes
<Hoopy> ok
<@spoonitnow> would you rather go all-in pre-flop for $100 with 55% equity, or $50 with 60% equity
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<@spoonitnow> say you're heads up with no blinds or antes
<Hoopy> ok
<@spoonitnow> would you rather go all-in pre-flop for $100 with 55% equity, or $50 with 60% equity
hmm
well it seems like for the $100 bet you are making $5 every bet (assuming you can play out this scenario an infinite amount of times and completely negate variance)
for the $50 bet you are also making $5 every bet.
So when you are only considering the size of the bet and the equity and nothing else, they are exactly the same. When you bring in other considerations one might be favored over another. For example, if your bankroll is small relative to these stakes you would want the smaller bet because you make the same amount of money but you risk a smaller chunk at one time.
This right?
Nope.Quote:
Originally Posted by kfaess
AIPF 100$ -> Cuz i always run hot
I'd chose the $50 bet. The ev is the same but the variance would be lower.
correct me if im wrong but ev = 10$ in both cases, id still do 100$ for baller tho
$50 @ 60%, 1/2 the risk for the same payout.
Prove it!Quote:
Originally Posted by DanAronG
in before spoon posts http://www.flopturnriver.com/pokerfo...ll-173396.html
ok found the EV equation:
In the EV equation the required pieces of info are our equity, his equity, size of pot, size of bet. Soooo I think in each case we need to know the actions that happened previously/what kind of odds we're getting. If we call off our last $30 in the $100 example then this would be better than calling off $45 of our $50 stack.
Good try, but we're not talking about having a choice between two calls, just the two situations given. It doesn't have be dealing out poker hands to find the equity, it could just be rolling dice or using a random number generator.Quote:
Originally Posted by kfaess
<kfaess> alright so basically its like we're making one bet with either 55% or 60% odds?
<@spoonitnow> yes
<@spoonitnow> think of it as a weighted coin flip
<@spoonitnow> would you rather bet $50 with a 60% chance to win, or $100 with a 55% chance to win
No?Quote:
Originally Posted by supahaole
Also this might be helpful http://www.flopturnriver.com/pokerfo...ad-180192.html
lol ok how bout this:
($200-$100)*0.55-$100*0.45 = $10
This is the money we win times % we win minus money we lost times % we lose.
($100-$50)*0.6-$50*0.4 = $10
I think I got it this time
Prove it!Quote:
Originally Posted by supahaole
Either or?
EV of each:
($50*.6)-($50*.4)=$10
($100*.55)-($100*.45)=$10
I can't cuz I'm wrong.:(Quote:
Originally Posted by spoonitnow
Why do you think you're wrong?Quote:
Originally Posted by supahaole
Because it's not the same payout. The payout is also 1/2.Quote:
Originally Posted by AvatarKava
depends on the rake structure dudeQuote:
Originally Posted by spoonitnow
ez game
Quote:
Originally Posted by spoonitnow
Bankroll: $100
option 1:
bet $100, 55% equity:
potential outcomes:
1. win $100 - probability : 55%
2. lose $100 - probability : 45%
ie 45% chance of going busto.
Option 2:
bet $50 with equity of 60% and bet $50 with equity of 60%.
Potential outcomes:
1. win $100 - probabillity : 36%
2. break even - probability : 48%
3. lose $100 - probability : 16%
ie 16% chance of going bust
Wat.Quote:
Originally Posted by supahaole
This is true, and there are other factors that could influence your decision as well.Quote:
Originally Posted by daven
how big's the bbj?
Uhg, this is wrong because in option 1 we're risking $100 to win $100 and in option 2 we're risking $50 to win $50.Quote:
Originally Posted by supahaole
The EV is the same so...
If we only flip one coin one time it's better to take the $50 @ 60% option because we have a better chance of winning.
If we flip 1000 times then $100 @ %55 we make more actual monies.
This also assume there's no rake or any other factors that might change the maths.
In w/ wat b4 spoon
bankroll?Quote:
Originally Posted by spoonitnow
etc
Wat?Quote:
Originally Posted by supahaole
If we flip the coin 1000 times, we make 10k either way...
($100*.55)-($100*.45)(1000)=$10,000
($50*.6)-($50*.4)(1000)=$10,000
We need slightly more information to make a definitive statement. I will assume no rake.
ROI on case one is 10% (Case 1)
ROI on case two is 20% (Case 2)
EV is identical.
To truly answer this question we need to know the size of the bankroll. If the bankroll isn't constant for both scenarios and it is a one shot deal. Meaning you can either bet 50 to win 10 OR bet 100 to win 10, the 20% return case is the best choice (less risk more reward, risk defined as variance). If your bankroll is infinite and so is the number of times you are allowed to run this scenario than both seniors will result in infinite profit. If however you are limited by total amount wagered, with an infinite bankroll more calculations need to be done.
If you are limited to wagering 10,000 (another words 1000 trials for Case 1 and 2000 trials for case two) than once again case 2 is superior, since it will converge to a grand total return of 20%, while case 1 will return 10%.
Lastly, if you are limited by trials, once again the 20% ROI case will be supreme. I am trying to think of a situation where Case 1 is optimal. The best I can do is find where it is equivalent in dollar terms, which is at infinity.
In all cases that I can think of the ROI case of 20% is better, same reward for small capital risked. Only rake can change this for the ROI to be the same rake has to be 15.38%, with a maximum rake of 15.38 dollars.
!luck
Are both options all-in situations, or is the $50 option out of a $100 stack?
100$ option if there is rake with no risk of ruin
50$ option even if there is rake but we have a decent risk of ruin using option A, Obviously there is some transition point here where we are going to take option A instead of B.
No rake with high, medium, or low risk of ruin(as low as .0000001%) through option A we obviously take option B also.
If there is no rake and no risk of ruin then it doesn't matter what option we choose.
Tired when I typed this up so could be wrong.
If you have an infinite amount of bets (or very large sample) you never ever ever ever choose the $100 option, unless there is a rake issue, as why would you ever chose to risk more for the same reward?
If you have a finite amount of times to make the bet, then it comes down to risk preference.
Quote:
Originally Posted by DanAronG
Quote:
Originally Posted by spoonitnow
Variance of each:Quote:
Originally Posted by bhaley66
( 50^2 * .6 ) + ( -50^2 * .4 ) - 10^2 = 2400
(100^2 * .55 ) + ( -100^2 * .45 ) - 10^2 = 5568.75
If we call the EV E(X), and the variance V(X), we have the following:
V(X) = E(X^2) - E(X)^2.
As you can see comparing the two computations, we just square the "x" value (the amount of $$ won/loss) and carry out the same EV computation. The last term is the E(X) = 10, but we square before subtraction.
As a final note, we keep the minus sign with the money lost, and square it, so the first two terms are both positive.
The standard deviation in both cases is found by taking the square root of the variance, $48.99 and $74.62 respectively, I believe.
The nice thing about the standard deviation is that it has the same units (dollars) as the random variable in the calculations. Either way, though, this would "prove" the variance in the $100 bet is higher.
As a caveat, someone ought to check my math - I did it off the top of my head and I'm not 100% of anything other than the formulas. I'm not feeling great atm.
_____
To answer spoon's original question, let me first ask this - how sure are we of these reads? If there's any margin of error in these reads (and it's equal for both cases), then the 60% equity choice is the better bet.
I couldn't remember the formula for SD, and couldn't be assed to look it up. But that should push you towards the $50 bet.
Why?Quote:
Originally Posted by DanAronG
Same expected value, lower risk to return ratio.Quote:
Originally Posted by spoonitnow
Expected value may seem the same in $, but it is different when you look at it in terms of percentage...
EV in $100 bet is $10 -> 10%
EV in $50 bet is $10 -> 20%
Say you have 100K,
You can bet 1.000 times $100. EV is 10% and you win $10.000
You can bet 2.000 times $50. EV is 20% and you win $20.000
See, ez choice...
yeah that sums it well quite well. If you had $100, taking the smaller bet twice gives you a higher expected value and lower risk.
However if you simply had a one off choice of either, it would be down to preference.
In the op, it appears we only get to run this once - if that's the case, does variance matter? Heck, if that's the case, do EV's matter?
Not if you only have 1 shot. The return isn't the same for both options. The expected return is. The return for bet 1 is ppotentially $100 where as teh return for bet 2 is potentially $50.Quote:
Originally Posted by !Luck
So one shot means it's down to preferences, which will be related to bank roll. If you have a $100k roll, you'd take option 1. If you had a $500 roll, you'd prob take bet 2. If you had a $100 roll, you shouldn't be in the game.
Doesn't matter how many shots we have. We always have to choose the better EV. In this case its the second option.
Would you rather bet $1 with 90% EV or $100 with 70% EV? Which is the better bet?Quote:
Originally Posted by Belt
I offer a counter example.Quote:
Originally Posted by Belt
We start with only $100
I take the $50 bet. I lose.
I have $50.
I take the $50 bet, i win.
I have $100.
I take it 2 more times, and lose each time.
Thus starting with only $100, i then could take the $50 bet 4 times.
The EV's are exactly the same no mater how many times we run this. The only reason you would care about one bet over the other is due to rake or due to variance. With a higher variance we have a greater risk of ruin , so we need a larger bankroll to justify even taking the bet. With rake, the EV's of the bets will change. Depending on structure, the smaller bet could become more favorable (higher ev), the larger one could be, or they could both be the same.
However, I have absolutely no idea how i would figure this out if the EV's and variance's were different and we cared about the variance at all.
In original example the outcome is the same in dollars. In yours it's not. This brings another variable to the equation.Quote:
Originally Posted by Robb
But good point none the less. :)
I'd say the EV is different while outcome is the same. That is why the variance differs between the two options.Quote:
Originally Posted by JKDS
I was responding to this statement, which in my view is not true. Simply having a higher EV does not mean one bet is better than another.Quote:
Originally Posted by Belt
I think most of the key points about EV and variance have been made.
Perhaps some of y'all should look at the thread title and see if you can guess where spoon's going with all of this.
Youre going to have to prove this one. I showed you your original argument for saying the EV's were different was invalid, so if you still disagree you have to give something to support yourself here.Quote:
Originally Posted by Belt
Edit: Didnt see your change. Different variances doesnt necessarily mean different EV's, and the EVs arent different in this case.
I think you are saying EV and the outcome in dollars are the same thing... I beleive they are not. I'd like to put EV in units of bets.
Let's say we are looking down on AA, and we know villain has KK, our equity is ~80%. We go all-in for 100bb and the opponent who covers us calls (let's say there are no blinds and the total pot is 200bb). In my point of view our EV is 60bb no matter what stakes we are playing. But the outcome in dollars will surely be different in different stakes.
Perhaps if I put it a little different I can clarify my point;
The equity is different while the outcome is the same.
EV is your average expected profit. The percentage your EV is of the amount being wagered is irrelevant to which has the highest EV. EV determines how profitable something is.
A lot of you have figured out that both bets have the same EV, and the $50 bet has a lower variance.
So which one, if either, is the better bet, and why?
Pretty sure I answered the thread like 87 posts ago.Quote:
Originally Posted by Icanhastreebet
Actually I think I might have been wrong the first time I typed that up. If rake caps @ >0$-<199.99$ then option A is better...And if rake caps @ >= 200$ then option B is better assuming no risk of ruin. Ofcourse once again depending when the rake caps there is a transition point where B > A and A becomes > B.
Edit- Also assuming we can do this as many times as we want.
So just strictly viewing profitability, they are the same. The only other factors that come into play then are things like variance, rake, risk and bankroll considerations (I realize that these play into profitability).
Variance - As Robb showed, the variance is greater in the $100/55% situation. Less variance is better, so advantage $50/60%.
Rake - Obviously, the higher the bet, the higher the rake (unless there's a cap that will make the two bet sizes produce equal amounts of rake). Less rake is better, so advantage $50/60%.
Risk - To win the same amount, you have to risk more. If you're risk averse, less risk is better, so advantage $50/60%. If you're risk neutral, it doesn't matter. If you're risk preferred, more risk is better, so advantage $100/55%. However, any rational "investor" will know that if the expected winnings are the same, you'll go with the one that costs less to invest into. Usually you incur more risk to win more. But in this case you'll win the same amount, so advantage $50/60%.
Bankroll - You have to play within your bankroll and the $50/60% bet just gives you less risk for a given bankroll compared to the $100/55% bet. Also, let's say you hit a cold streak and at the beginning you constantly lose money. With the $50 bet, you have double the amount of times you can lose before you go broke. Even though each bet is independent of the prior bets, in the long run the expected winnings will converge to $10/bet. Double the amount of bets will allow you to be more in the long run than the short run.
Also, for the same amount of money:
$100 (55%) ... EV = $10
[$50 (60%)] x 2 ... EV = $10 x 2
The $50/60% is a much better bet for multiple reasons.
I wish people would stop suggesting we can run this with $50 twice. That was never mentioned as an option or even something to consider in the OP.
Additionally, I really like xpaand's signature.
So by now you realize that both of these options have the same EV, while the $50 one has a lower variance if that's of concern. The point of the exercise is to make you understand that comparing two bet-sizing options can be difficult if you aren't practiced.
When you're considering two bet sizes for a value bet, there are other factors to consider besides the actual bet size and your equity when called, like how often you'll get raised, how often you'll get called, and what play will be like on future streets (and hands). But it all starts with considering a bet size and then deciding how your opponent plays his/her range.
I was going to try and do the variance calculation but Robb beat me to it. For anyone interested I found a pretty cool site that explains a lot about statistics in poker:
Poker Variance - How to Calculate Variance in Poker
The same formula that Robb used to calculate variance is listed in that article, so I'm going to assume his figures are correct. Also, I thought the part about CoV was interesting and how skill factor measures up to luck factor over a given sample size of hands.
I'm having a hard time wrapping my head around the math. The highest math I took was intro to alg. and that interfered with my surfing time so you can imagine how well I did with it. So I'm gonna take a shot at it at a more basic level and hope it makes sense.
If I bet $50, 1000 times and win 600 (60%) of those I'm going to win...
50*1000*.60= $30,000
If I bet $100, 1000 times and win 550 (55%) of those I'm going to win...
100*1000*.55= $55,000
The math I'm seeing in this thread is...
No offense to bhaley or anyone with the same #s but it looks to me that all you did is subtract 45 from 55 and 20 from 30 and multiply them by 1000. Could someone please show me the error in my thinking? Thanks.Quote:
Originally Posted by bhaley66
@supa:
Your thinking is correct for finding how much we expect to win BEFORE factoring in how much we expect to lose.
For the $50 bet, if we did it 1000 times we'd expect to lose 400 of those times...and when we do we're out $50.
So in the end, we win $50 a total of 600 times, and lose $50 a total of 400 times...so we end up with
$50*600 - $50*400 = $50*(600-400) = $50*200 = $10,000
gotcha, thanks
@haoleQuote:
Originally Posted by kfaess
there are several good "basic math" articles at the site kfaess linked above
they do the math, show the equations, then give examples
if you're interested, send me a pm with any questions, and I would be happy to help with the maths
spoon also has posted several good "basic math" threads on here, but I'm too lazy right now to look them up