Topic: gambling question - changing house edge by altering bet size? - page 2. (Read 2596 times)

Your expected value on any/every flip is 0. No matter whether you are betting 1 or 2, the EV of each flip is exactly the same.  Thus, your long term EV is also the same.  Everything else is just noise and variance.

As long as each event is 50/50 and the amount you win is equal to the amount you can lose, the EV is 0.  You will win half your bets of 2, and lose half your bets of 2.  Same when wagering 1.  You will lose half, and win half.

(07:39:32 PM) deb: You people really get off on this shit don't You?
(07:39:57 PM) Chris: it's not a sexual thing
(07:40:04 PM) deb: i mean no disrespect, i just can't fathom where making charts to work out gambling strategies that don't really matter is a good thing to do
(07:40:04 PM) Chris: it's more of an obsessive thing
(07:40:12 PM) Chris: I couldn't relax before I understood wtf was going on

Yes.  Once I've started playing the 1 game (A?), I only play the 2 game (B?) if I've won more than I've lost at A.  So when my wins and losses eventually match, A will have more wins than losses, and B will have more losses than wins, and so I'll be making a loss overall.

It's OK that the most common result (which is #wins == #losses) is a net loss, because the whole curve is skewed to the left.  The most likely final balance is a little less than the starting 1000, but there's a good sized chunk of significant winnings to the right.  Overall, the average is 1000:



Edit: Notice how the likely range of balances when I lose (600-1000) is half as wide as the likely range of balances when I win (1000-1800).

The red (losing) area is twice the size of the green (winning) area, so we're twice as likely to lose as to win.  But it's much narrower, so if we lose, we don't lose much, but if we win, we're likely to win a decent amount.

I think this is incorrectly modeled as a "single game". Your overall expectation is to make it out even on all games where you bet 1 coin, and on all games where you bet 2 coins. Overall, they should both even out and so their sum should also even out.

However, if you do something like 50 games A lost + 50 games B lost + 40 games A won + 60 games B won, the situation isn't correctly described as "100 won, 100 lost".

Yes, I realise.  That's what I was getting at with:


But I was wondering how it was possible that the expected return wasn't at least slightly lower than zero, given that crossing from 1000 to 998 effectively "wastes" a loss.

I think I have a reasonable explanation of that now:

The losses incurred each time we cross from 1000 to 998 are offset by the fact that winnings are slightly skewed upwards.  For example, if we win our first 5 bets, we win 10 coins, but if we lose our first five bets, we lose only 6 coins.  So the expected return can be thought of in a hand-waving way as:

  (p(lots more wins than losses) * big_win) +
  (p(lots more losses than wins) * smaller_loss) +
  (p(similar wins and losses) * loss_due_to_crossings_from_1000_to_998)

where the first term is cancelled out by the other two.

It's very rough and ready, but I think that's enough to let me stop thinking about it!

Reversing the original plan (i.e. betting 2 when below and only 1 when above)
can even be extended further (and often is):

 - When I lose once, I repeat the bet with double value.
 - When I lose again, I double again
 - ...
 - When I win, I've compensated the previous losses.

Assuming the broke-free environment, then at some point every bad luck
streak would break, and you'd end up debt-free with just one lucky throw.

How does it work? Mathematics trolled? Nope.

It's the assumed broke-free environment that's crucial to the reverse "strategies".
(Also, in reality there's typically a max-limit on bets - primarily to cap such cascades.)

By increasing the bets for lower budget, you improve your payout-performance
  but also boost your crash&burn broke-likeliness.

By lowering your bets for lower budget, you can avert broke-ness for somewhat
  longer, but the average luck will give you less money.

What you found on the net about variable betting is most likely for random
variable betting-amounts.  It's not true for such systematically chosen betting
amounts.

You do realize that +16,000 is within the margin of error right.   16,000 deviation on 2 billion flips is 0.0008% of expected.   Trying running it 10 billion, or a trillion, or a quadrillion you will notice the deviation will continue to shrink as a % of the expected.  In other words as there are more and more trials the actual outcome approaches the expected outcome.

Expected Outcome is called that because it is EXPECTED but you rarely will ever hit it EXACTLY (i.e. actual outcome = expected outcome) as an ironic twist of math as the same size gets larger the actual outcome will approach the expected outcome HOWEVER the odds that it will be exactly equal to the expected outcome decrease.

This is easy to observe:
2 flips.  expected outcome = 1 heads.   Odds of actual outcome matching expected outcome = 2 out of 4 combinations (50%)
4 flips. expected outcome = 2 heads.   Odds of actual outcome matching expected outcome = 4 out of 16 combinations (25%)

I started with 1000 chips, and tossed the coin 1 billion times.

I got 499985225 wins and 500014775 losses
29550 more losses than wins
and a final balance of -36034

Each of the 29550 excess losses should have lost me 1 chip, so I should be at -28550, but I'm 7484 chips short of that.

I got the simulation to print out each time the number of wins and losses were the same.  The last time that happened before the billionth turn was after 612,215,220 turns, at which point there had been 306107610 wins and 306107610 losses, and the balance was -6484.  ie. 7484 below where I started.  This suggests to me that the balance crossed from 1000 to 998 7484 times, 'wasting' 1 win each time.

I suspect that the 7484 crossings is insignificant compared to the amounts won and lost, and that's why the expected return is still 0.  Could that be the case?

Edit: I left the simulation running.  After 2 billion tosses I was back in profit:

2000000000 turns; 1000008100 wins; 999991900 losses; 16200 more wins than losses; 6002 balance

the last time the numbers of wins and losses were the same:

1821704084 turns; 910852042 wins and 910852042 losses; -8477 balance

so now there have been 9477 crossings from 1000 to 998, and my balance is only positive because I have been lucky; even though I've have had 16k more wins than losses, I'm only 5k up.

Here's the Python script I used by the way: http://pastie.org/5439806

Yes, it sounds stupid, but probably is just as valid as my argument.  With your scheme it takes 2 losses to fall to 999 from 1001, but only one win to recover.  So it is a winning strategy.

Now that's got to be wrong...  Just wish I could see why!

You can't start after an event and consider that part of the random set.

i.e.
Either starting at 1,000 he has equally chance of winning or losing.  Expected outcome is 1,000 for an any number of flips going forward.

If he ALREADY (as in the past) lost 2 and only has 998 chips then 998 chips becomes the starting point and he has an equal chance of winning or losing and 998 is the expected outcome for any number of flips going forward. 

if he is starting with 998 and wants to acheive 1,000 coins he is looking to gain two.  then the odds of that ARE going to be less than zero.   Lets take this to the logical extreme.  Say he played his game until he had lost all but 2 chips.   Now say from this point he plays until he either loses all chips or gets back to 1,000.  The odds of that are <1%.  Would you say the odds of having break even or better on a coin flip are <1%?  Of course not.  The odds are only so low because he is trying to turn 2 chips into 1,000 chips.  What happened BEFORE he ended up at 2 chips is irrelivent.  The coins don't "know" he originally started at 1,000 and lost it down to 2.  The odds going forward are exactly the same as any other fair coin flip.

I'm thinking that since it's a fair coin, my path will be essentially a random walk through profit and loss.  So if I do win the first trial, at some point in the future I'll be back to 1000 coins, and can effectively discount all the coin tosses in between, since the wins and the losses will be equal.

If I lose that first trial, or any trial when my balance is 1000, I'll lose 2 coins and 'waste' 2 wins getting back to even.  Other than that everything seems equal, so it appears to me that this case, which will keep happening each time I cross from 1000 to 998, will result in me losing in the long run.

I agree that this is probably incorrect, but can't see the hole in my reasoning.


I will do.  Here's a worked example of how I get 22 wins and only 20 losses, but break even:

Actually it is.   Crossing the borderline downwards means he lost 2 in one game, while
crossing it upwards means he won one in one win-game.  If you then pair up the total
wins and losses, then those crossing the border will be (almost) equal (as you can't cross
the line twice in same direction without at least once the other direction inbetween),
but the upward crosses will have earned less than the downward crosses lost.

It isn't.  


You are merely observing the effect of a small sample size.  First your number of wins shouldn't be exactly half.  Due to the house odds it should be slightly less than half (the 0 & 00 numbers).  The fact that it was exactly half was just conicidental.  You simply won less of the larger wagers and more of the smaller wagers.  Had the reverse been true would you believe you found a method to always beat the house?  If you played for quadrillions upon quadrillions of spins and had infinite amount of money to lose your expected loss =  (house odds) x (total amount bet).  The more you play the more it aproaches this value, in the short term there can be anomoloies.  Note: a hundred spins isn't statistically valid.  Try doing 100,000 spins or more.


Your penny scenario is equally flawed because you assume you will lose on the first trial.  What if you win, or win 30 times in a row.  By simply looking at only the scenario's which begin below break even obviously your expected outcome will always be below breakeven.  Once you lost to 998 going forward your expected gain is 0 and thus 998 is the break even.  The coin has no memory that in the past you lost a bet and thus are "owed" one extra win.

You can simulate your penny game rather easily in software using a random number generator.  You have it flip trillions of times and try every possible betting combination you can think of.  In the end total return would be 0 (+/- margin of error).

Just play it reverse:  When you're below the 1000-line, play two coins, and when above you play only one coin per toss. -> winning strategy *lol* *JK!*

I've always thought that it's not possible to change the expected return of a game with fixed odds by altering your bet size.  I've seen this stated all over the place.

For the last hour or two I've been playing roulette at luckybitcoincasino, and seem to have found a way of reducing my expected return.

I was betting the 3-36 line over and over, starting with 100 chips.  I was betting 2 chips each spin, but whenever I fell below 70 chips I was reducing my bet to 1 chip per spin, and increasing it back to 2 chips per spin when I got back to 70 chips.

I counted my wins and losses.  Each time my number of wins was exactly equal to half my number of losses, I noticed my balance was a little lower than the previous time it happened.

I would expect that since a win pays out twice as much as a loss loses, having N wins and 2N wins losses should mean I was breaking even, but that wasn't the case.

I tried to understand why, and came up with this thought experiment:


So what's wrong with the above argument?  It looks to me like I've found a strategy that reduces my expected return, making it negative in a zero-expectation coin toss game.  I must be missing something.  Mustn't I?