- 06 Jun '08 15:40 / 1 editin Blackjack, let's assume you don't play perfectly (and aren't counting), and your chances of winning each hand is 0.48

If you start with $2048 and the minimum bet is $1, what are your odds for doubling your bankroll using a strategy of:\

1) doubling your bet ( b[n] = 2 * b[n-1] )

2) doubling your bet plus 1 ( b[n] = 2 * b[n-1] + 1 )

*edit* Just to clarify if you aren't familiar with these strategies, you start with a $1 bet. Each time you win, you reset your bet to $1. Each time you lose, you follow one of the strategies above. - 07 Jun '08 09:05

Option 2*Originally posted by forkedknight***in Blackjack, let's assume you don't play perfectly (and aren't counting), and your chances of winning each hand is 0.48**

If you start with $2048 and the minimum bet is $1, what are your odds for doubling your bankroll using a strategy of:\

1) doubling your bet ( b[n] = 2 * b[n-1] )

2) doubling your bet plus 1 ( b[n] = 2 * b[n-1] + 1 )

*edit* Jus ...[text shortened]... win, you reset your bet to $1. Each time you lose, you follow one of the strategies above.

When the odds are against you it is best to minimise the number of bets. The classic situation of having $1,000 to play at roulette is to put the whole lot on red and walk out afterwards (win or lose). - 08 Jun '08 06:59

with method 1 ther is a 99.855444894050942976% chance of winning the first dollar before you have to quit.*Originally posted by forkedknight***in Blackjack, let's assume you don't play perfectly (and aren't counting), and your chances of winning each hand is 0.48**

If you start with $2048 and the minimum bet is $1, what are your odds for doubling your bankroll using a strategy of:\

1) doubling your bet ( b[n] = 2 * b[n-1] )

2) doubling your bet plus 1 ( b[n] = 2 * b[n-1] + 1 )

*edit* Jus ...[text shortened]... win, you reset your bet to $1. Each time you lose, you follow one of the strategies above.

if this succceeds on the first or second attempt

a 99.999791038213440568985886757252% chance

then all you have to do is win 2048 mor times than you lose

which gives appx. 2050 cycles to double your money. - 08 Jun '08 08:50

As a math teacher I would not give full points for this answer. Why?*Originally posted by preachingforjesus***with method 1 ther is a 99.855444894050942976% chance of winning the first dollar before you have to quit.**

if this succceeds on the first or second attempt

a 99.999791038213440568985886757252% chance

then all you have to do is win 2048 mor times than you lose

which gives appx. 2050 cycles to double your money. - 08 Jun '08 17:03

well It might be because I didn't say:*Originally posted by FabianFnas***As a math teacher I would not give full points for this answer. Why?**

the only way to not get the first dollar is to lose 10 times in succession which has a prbability (.52)^10 ~ 0.15%

now ~0.15% losing cycles makes for 1.0015*2048 total cycles

this gives ~2050 betting cycles with a ~99.85 success rate

you get 72.96% success in two hands soyou will average 1.x hands per cycle so it should come in at less than 4000 hands. - 13 Jun '08 18:10

With method (1), you have enough money for 11 bets, as you will for the entire game.*Originally posted by preachingforjesus***with method 1 ther is a 99.855444894050942976% chance of winning the first dollar before you have to quit.**

if this succceeds on the first or second attempt

a 99.999791038213440568985886757252% chance

then all you have to do is win 2048 mor times than you lose

which gives appx. 2050 cycles to double your money.

Your chances of losing 11 in a row are:

0.52 ^ 11 = .00075168

which means your chances of winning a dollar before losing all your money is:

1 - .00075168 = 0.99924831 (I don't know where you got .9985444...)

In order to win $2048, you will need to win exactly 2048 hands, which means you will have 2048 opportunities to lose 11 times in a row.

The oddes of _not_ losing 11 hands in a row on 2048 different cycles is:

.99924831 ^ 2048 = 0.214374

Therefore, there is a 21.4374% chance you will double your money before losing it. - 13 Jun '08 18:21 / 1 editIt's really case (2) that I'm more interested in (and comparing it with the first).

With method (2), unlike with method (1), you win more money per cycle by losing more hands. Also, you only have enough money for 10 bets. Your odds of winning x dollars in a cycle are:

x => odds

1 => .048

2 => .52 * .48 = .2304

3 => .52^2 * .48 = .129792

4 => .0674918

5 => .0350958

6 => .0182498

7 => .0094899

8 => .0049347

9 => .0025661

10 => .001334

11 => you lose

Therefore, there is not a set # of cycles you need to play (not a set # of wins).

How, then, do you find your odds of winning $2048?