
blackjack double up question
here's the question. the table limit is £50 (although it is multihand) and i want to double up my balance (which will be anywhere between £180 and £325). although i can't put the whole lot on one hand i assume the chances are the same as if i could put the lot on one hand. it doesn't matter how much wagering it takes as it all counts towards the WR. anyone see an error in this?

The EV of a game is independent of wager size.
But for doubling up (since blackjack as any other game is EV) you need high variance.
Variance is highly sensible to the way the game is played. The smaller your wager per hand is, the smaller is your variance. However the more hands you play the higher is your variance.
Given a constant wagerperhand * numberofhands (your WR), the "battle" of variance from the two opposing effects is clearly dominated by the wagersize.
If you must double up, play with maximum bet size. Or better switch to another game with higher max bet and even odds.

thanks for taking the time out to reply although you drifted into your variance theory a little there and perhaps missed the point.
if i had for example £5,000,000 and wanted to double to £10,000,000 playing blackjack with a max bet size of £50 it would never happen. if i only wanted to double from £50 to £100 then my chances would be almost 50/50. my proposed balance of between £180 and £235 is of course somewhere between £50 and £5,000,0000. but am i so closed to £50 that i need not worry or is there a percent or 2 more chance i will bust than double? a percent or 2 in my situation is important to take into account (as i am planning world domination with my new super strategy).

Premium Member
With a starting bank of £200 (for example) and bets of £50 I calculate you will bust 51  53% of the time if you are trying to get to a £400 target.

Originally Posted by fraser
With a starting bank of £200 (for example) and bets of £50 I calculate you will bust 51  53% of the time if you are trying to get to a £400 target.
thats about the region i was guesstimating  thanks.

Thanks for clarification of your question. I don't know the odds/fair odds of your bets you want to take, but I assume that they are even and fair.
You then are looking at a onedimensional random walk. I will cite from Wikipedia's random walk page:
If a and b are positive integers, then the expected number of steps until a onedimensional simple random walk starting at 0 first hits b or −a is ab. The probability that this walk will hit b before a steps is a / (a + b), which can be derived from the fact that simple random walk is a martingale.
In your scenario a and b are equal (either double or bust), hence probability of busting is a / (a + b) = 50%. So probability of double up is also 50%. This is given a fair bet and even oods, and infinite WR.
For uneven odds / unfair bets things are different. However if your game is EV, there is no chance that probability of double up will be > 50%. (Otherwise you could play for eternety with a fixed amount either doubling up or bust, then proceed with the next amount and make profit from more doublings than busts  giving you +EV in the long run.... forbidden by the strong law of large numbers).
If you carefully read the Wiki page they show you how to calculate those probabilities with binomial coefficients.

Originally Posted by MangoJ
Thanks for clarification of your question. I don't know the odds/fair odds of your bets you want to take, but I assume that they are even and fair.
You then are looking at a onedimensional random walk. I will cite from Wikipedia's random walk page:
In your scenario a and b are equal (either double or bust), hence probability of busting is a / (a + b) = 50%. So probability of double up is also 50%. This is given a fair bet and even oods, and infinite WR.
For uneven odds / unfair bets things are different. However if your game is EV, there is no chance that probability of double up will be > 50%. (Otherwise you could play for eternety with a fixed amount either doubling up or bust, then proceed with the next amount and make profit from more doublings than busts  giving you +EV in the long run.... forbidden by the strong law of large numbers).
If you carefully read the Wiki page they show you how to calculate those probabilities with binomial coefficients.
the bets i will take are not even and fair or 50% or equal as it is blackjack. thus meaning a house edge of around .5% (complicated slightly if dealt a split or double on the last chip)

Premium Member
Originally Posted by tronikHOUSE
the bets i will take are not even and fair or 50% or equal as it is blackjack. thus meaning a house edge of around .5% (complicated slightly if dealt a split or double on the last chip)
This is something I am looking at at the moment  Mango's post about the random walk theory is spot on. The onlly trouble is, it gets a bit more complex if the bets are anyting other than a fair, even chance with an EV of zero
Blackjack is generally around 0.46%, since you can't double down or splits because of your bet size limitations, at an estimate the House Edge would probably extend to around 0.55%.

the option is always there to reload your account when dealt a double or split although its not always wanted!
in theory, if you are stuck playing between your 2 targets (bust and double or thereabouts) you are losing .46% or .55% per hand. but i think in this case, the parameters are so small maybe that factor could be ignored?

Premium Member
Originally Posted by tronikHOUSE
the option is always there to reload your account when dealt a double or split although its not always wanted!
in theory, if you are stuck playing between your 2 targets (bust and double or thereabouts) you are losing .46% or .55% per hand. but i think in this case, the parameters are so small maybe that factor could be ignored?
I think so yeah  at the moment i'm not sure how much of a difference it makes to the random walk theory not having EV neutral bets. Though it would be nice to be able to work this EV out exactly by use of a formula, i'm not even sure it's possible, I think statisticians would probably opt for running a simulation on a computer programme  repeating the situation thousands if not millions of times.
But for the purposes of making a judgement on your potential EV either on a casino game or a sports bet, if the actual edge of the bet is that fine then yes, my gut is agreeing with you that it could be ignored.
So I guess that takes you back to your original question;
although i can't put the whole lot on one hand i assume the chances are the same as if i could put the lot on one hand
Yep, I think that's a fair comment as long as we realise that it is somewhat of an approximation.
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