Poll: Reality check

[Snoogins47]

"If your opponent is underbankrolled, that does not convert a -EV game to a +EV game for you. That is a fact. If a casino were underbankrolled, it would not suddenly become a good idea to play roulette there. "

Nope, but it makes it probably not the best idea for the casino to stay in operation, unless they have large reserves of cash that aren't in this "bankroll." The entire discussion is now complete.

It's not rocket science. I'm a 20 year old who got a D in high school math, and this entire argument is silly. A game can have a high expectation, even if you will typically go broke. This doesn't make it a bad bet unless "going broke" is a very negative outcome.

[Double Down]

Ok, listen carefully.
Yes, being underbankrolled will make you a dog against a larger br even if you have a +EV situation. BUT guess what? pzhon got himself another 200 and then another, which means that his bankroll is NOT 200, but much more than that. A bankroll is defined by how much money you have to throw at a proposition before you will no longer have ANY money. So if someone came up to me with 200 and that offer, I'd take it if I knew that the game ended when one of us went broke. But if he can keep buying in for 200, eventually his +EV will rear its head and he will take back what we won from me and then some.

[wtfsvi]

</font><blockquote><font class="small">Svar på:</font><hr />
So if someone came up to me with 200 and that offer, I'd take it if I knew that the game ended when one of us went broke.

[/ QUOTE ] Why?

You people are hopeless.

[PollyEmory]

I'd like to give this a shot if no one minds. I'm new though, so be kind.

pzhon is right in that every single event of the flip is -EV for the player. The same way any single hand of blackjack is -EV for a player unless there is something like a bonus to offset the expected loss. The limited bankroll is that bonus.

If you think of each coin flip in the progression of coinflips as a decision tree and assume equal bankrolls in either direction, the player will bust in more then 50% of the eventual outcomes. However (!!!) because of the house limited roll, some of the tree will be artifically pruned out. How much depends on how small the casino bankroll is with respect to the bet size and not with respect to the player's bankroll. If the casino has enough BR to cover three losing flips before busting and the number of flips needed to approach EV is 100, it doesn't really matter if the player has BR of 15k or 150k.

Let's see if I can demonstrate my point.

$50 Flips

1st Flip(Casino = 100, Player = 1500)
a. Player won (Casino = 51, Player = 1549)
b. Player lost (Casino = 150, Player = 1450)
a.1 Player won(casino = 2(basically broke), Player = 1598)

Now if we could continue this tree indef. downwards, the number of leaves where the casino makes money would exceed the number of leaves when the player makes money. However, the limited casino bankroll means that the tree will be artifically pruned once the casino goes broke, therefore reducing its edge. By how much is a calculation I can't really undertake at 6am est, but offhand, I'd say it depends on casino BR v betsize. The higher the ratio, the less edge it gives up.

So yes, the any single event of the flip is -EV if EV is computed based on infinite number of tosses and infinite player/casino bankrolls. In the real world, casinos' rolls are basically an approximation of infinity. But if the roll is small, the approximation is bad. That skews results. Therefore situationally, this could be a +EV gamble for the player. I'm sure someone will step up and either back me up with some rigorous analysis or disprove me with the same. I suspect if you plot expectation vs finite bankrolls of various sizes(BR) approximating infinity (INF) in EV calculations, you will find that under certain values for BR (low v betsize), the EV will be positive for the player.

In other words, if we played out this situations infinite amt of times (casino BR $200, bet size $50, player BR $1500) the player will bust the casino more times then casino will make money off this deal. But the higher the BR, the more precise the approximation, the closer actual expectation will approach theoretical expectation.


--Polly

[wtfsvi]

</font><blockquote><font class="small">Svar på:</font><hr />
In other words, if we played out this situations infinite amt of times (casino BR $200, bet size $50, player BR $1500) the player will bust the casino more times then casino will make money off this deal

[/ QUOTE ] This is correct.

</font><blockquote><font class="small">Svar på:</font><hr />
Therefore situationally, this could be a +EV gamble for the player. I'm sure someone will step up and either back me up with some rigorous analysis or disprove me with the same. I suspect if you plot expectation vs finite bankrolls of various sizes(BR) approximating infinity (INF) in EV calculations, you will find that under certain values for BR (low v betsize), the EV will be positive for the player.

[/ QUOTE ] But this is wrong.

I don't need rigorous analysis. There's no question about it: -EV is -EV.

[pzhon]

[ QUOTE ]

pzhon is right in that every single event of the flip is -EV for the player...
this could be a +EV gamble for the player.

[/ QUOTE ]
No. It can never be +EV for player.

Suppose your task is to write down numbers that add up to 100 (or more). The catch is that you can only write down negative numbers. Your method can be very complicated: You can use the day of the week to decide on the next number, you can toss a coin to choose a number, you can use a progression of numbers, etc. Can you write down negative numbers that add up to 100? No.

No collection of negative numbers adds up to a positive number. This is true even if you have a method for choosing them that is so complicated you aren't quite sure what the sum is.