[bushindo]

This problem is inspired by a recent trip to Las Vegas.

Mr. G is an avid gambler. After studying blackjack for a while, he realizes that the casino has a minor edge in winning, even if he plays with an optimal strategy. Consequently, he devises the following scheme.

For simplicity, assume that each blackjack hand cost 1 dollar to play, no double-down or splitting allowed. Winning will always pays double the bet. His chance of winning every game is .48, and his chance of losing is .52.

Everyday, he would take a bankroll of 20 dollars down to the casino. He would play as long as it takes until he either loses all his bankroll, or reaches a total bankroll of 21 dollars. As soon as he reaches either state, he would pack up his belongings and retire to his room. So for instance, suppose one day he loses 3 hands in a roll, and then wins 4 hands afterward, he would have 21 dollars total, at which point he would retire for the day.

1) Does this strategy allow him to beat the house advantage, i.e. get a positive expected winning per day?

3) Suppose that he uses this strategy for 100 days straight, what is his expected winnings, or losses, at the end of the 100 days period? Assume that he has a big enough bank account in the beginning so that he can always start with 20 dollars each day, even in the worst case scenario.

[Guest]

Do you think you can build a strategy to win money at a casino ? I don't think so...

Let's first suppose you don't make any successive lost / won (in this order) that cancel each other in terms of gain. The only possibilities to finish with 21 dollars is to lose n times and win n+1 times, with n more or equal to 0, and less or equal to 19. The probability of such an event to happen is 0.52^n * 0.48^(n+1). So the probability of leaving the casino with 21 dollars under this assumption is obtained by taking the sum over n : A = 0.48*(1+p+p^2+p^3+...+p^19) where p=0.48*0.52. I find A=0.64 approximatively.

However, you can also make any times of [one lost / one win] before and it doesn't change the outcome. So you have to multiply A by 1+p+p^2+p^3+...+p^infinity, which is around 1.33. So the overall probability of winning one dollar is W=0.85

The probability of losing the 20 dollars is of course 1-W. So each day, the expected loss is (1-W)*20$ + W*1$ = 2.10$. After 100 days, you should have lost 210$ in average. Also a lot of time...

Sorry my explanations are not very clear. Maybe the reasoning is not very clear in my head neither...

[bushindo]

Do you think you can build a strategy to win money at a casino ? I don't think so...

Let's first suppose you don't make any successive lost / won (in this order) that cancel each other in terms of gain. The only possibilities to finish with 21 dollars is to lose n times and win n+1 times, with n more or equal to 0, and less or equal to 19. The probability of such an event to happen is 0.52^n * 0.48^(n+1). So the probability of leaving the casino with 21 dollars under this assumption is obtained by taking the sum over n : A = 0.48*(1+p+p^2+p^3+...+p^19) where p=0.48*0.52. I find A=0.64 approximatively.

However, you can also make any times of [one lost / one win] before and it doesn't change the outcome. So you have to multiply A by 1+p+p^2+p^3+...+p^infinity, which is around 1.33. So the overall probability of winning one dollar is W=0.85

The probability of losing the 20 dollars is of course 1-W. So each day, the expected loss is (1-W)*20$ + W*1$ = 2.10$. After 100 days, you should have lost 210$ in average. Also a lot of time...

Sorry my explanations are not very clear. Maybe the reasoning is not very clear in my head neither...

0.52^n * 0.48^(n+1) is actually the probability of 1 particular configuration of n wins and (n+1) lost, where the order is important. For instance, the chance of 1 wins followed by 1 loss and then 1 win is .48 * .52 * .48 = .48^2 * .52. The chance of 2 wins and 1 loss, where order is not important, is equal to C( 2, 1) * .48^2 * .52, where C(2,1) means 2 `choose' 1. We need a combinatoric term in the bold part of this argument.

It is also possible for n in the bold part to exceed 19. For instance, we can win 100 games and lose 99 games, and still end up with 21 dollars.

The answer to question 1 is correct. The answer to part II is not, however. The true probability is slightly higher.

Edited May 7, 2009 by bushindo

[Guest]

0.52^n * 0.48^(n+1) is actually the probability of 1 particular configuration of n wins and (n+1) lost, where the order is important. For instance, the chance of 1 wins followed by 1 loss and then 1 win is .48 * .52 * .48 = .48^2 * .52. The chance of 2 wins and 1 loss, where order is not important, is equal to C( 2, 1) * .48^2 * .52, where C(2,1) means 2 `choose' 1. We need a combinatoric term in the bold part of this argument.

It is also possible for n in the bold part to exceed 19. For instance, we can win 100 games and lose 99 games, and still end up with 21 dollars.

The answer to question 1 is correct. The answer to part II is not, however. The true probability is slightly higher.

Thank you for your comment. I'm familiar with combinatoric probabilities, but I thought I had find a trick to include all possibilities of ending up with 21 dollars, by grouping all lose/win (L/W) pairs, that's the reason of the 1.33 coefficient. Let's see if your examples are included in my counting. The example of W-L-W is a bad example because if you win the first time, you just quit the game.

For 2 win and 1 loss, the only possibility is L-W-W, and it is included in the first part of my count (with n=1).

For 3 win and 2 loss, the possibilities are L-L-W-W-W and L-W-L-W-W only. The first case is included in the first part of my count (for n=2). The second case is included in the 1.33 coefficient because it correspond to one L-W pair followed by the L-W-W series.

For 100 win and 99 loss, similarly, you can group all pairs of L-W and the following is one of the cases with n=1,2,...,19

However, I'm not sure of what I'm saying.

[Guest]

Thank you for your comment. I'm familiar with combinatoric probabilities, but I thought I had find a trick to include all possibilities of ending up with 21 dollars, by grouping all lose/win (L/W) pairs, that's the reason of the 1.33 coefficient. Let's see if your examples are included in my counting. The example of W-L-W is a bad example because if you win the first time, you just quit the game.

For 2 win and 1 loss, the only possibility is L-W-W, and it is included in the first part of my count (with n=1).

For 3 win and 2 loss, the possibilities are L-L-W-W-W and L-W-L-W-W only. The first case is included in the first part of my count (for n=2). The second case is included in the 1.33 coefficient because it correspond to one L-W pair followed by the L-W-W series.

For 100 win and 99 loss, similarly, you can group all pairs of L-W and the following is one of the cases with n=1,2,...,19

However, I'm not sure of what I'm saying.

Now I see my mistake. Sequences like L-L-W-W-L-W-W are not included in my count. You were right, it seems you need combinatorics after all. I can not find an easy way to evaluate the formula without a computer program.

[Guest]

Wow... you guys are very skilled!!! O.O

I am going to go try to understand what you guys posted...

[Guest]

Maybe I'm wrong but this is making no sense to me.

He either loses $20 or wins $1.

I don't see how the odds match up here. He wins 48 out of 100 hands regardless... Assuming the probability doesn't change. Over the course of an extended time he would lose.

edit: this is also removing the number of cards as a variable. assuming each hand of blackjack is a unique event.

Edited May 7, 2009 by palmerc7

[Guest]

He will win 2$.... the house pays double...

For simplicity, assume that each blackjack hand cost 1 dollar to play, no double-down or splitting allowed. Winning will always pays double the bet. His chance of winning every game is .48, and his chance of losing is .52.

[Guest]

Unless we're stretching the rules of blackjack here, if you bet $1, you win $1, you take back $2, but your winning is only $1.

[Guest]

True, very true.... So no matter what you bet you will always get double....

but what if you bet 10$ or even 20$? The chances of you winning and loosing are changed.

[Guest]

I think the probability will be the same regardless of the bet, but the question posed is if he bets $1 every time.

[bushindo]

Maybe I'm wrong but this is making no sense to me.

He either loses $20 or wins $1.

I don't see how the odds match up here. He wins 48 out of 100 hands regardless... Assuming the probability doesn't change. Over the course of an extended time he would lose.

edit: this is also removing the number of cards as a variable. assuming each hand of blackjack is a unique event.

It is true that he either loses $20 dollars or win $1 every day. Let p be the probability that he wins 21 dollars in a single day. The chance that he'll lose all 20 dollars is (1- p), since we are working with the assumption that he continues to play until one of those events occur.

Expected winnings per day = - $ 20 * (1- p) + $1 * p

The key is finding the probability that he'll win 21 dollars for any given day, p. If p is close enough to 1, then the expected earnings might be positive, giving this scheme an advantage over the house.

[bushindo]

Maybe I'm wrong but this is making no sense to me.

He either loses $20 or wins $1.

I don't see how the odds match up here. He wins 48 out of 100 hands regardless... Assuming the probability doesn't change. Over the course of an extended time he would lose.

edit: this is also removing the number of cards as a variable. assuming each hand of blackjack is a unique event.

It is true that he either loses $20 dollars or win $1 every day. Let p be the probability that he wins 21 dollars in a single day. The chance that he'll lose all 20 dollars is (1- p), since we are working with the assumption that he continues to play until one of those events occur.

Expected winnings per day = - $ 20 * (1- p) + $1 * p

The key is finding the probability that he'll win 21 dollars for any given day, p. If p is close enough to 1, then the expected earnings might be positive, giving this scheme an advantage over the house.

I think the probability will be the same regardless of the bet, but the question posed is if he bets $1 every time.

Let's assume that the bet amount is always $1. Winning always give him a net gain of $1, and losing a net gain of -1.

Edited May 7, 2009 by bushindo

[Guest]

Wouldn't the probability of winning 21 dollars be the same as winning one hand. (please note that I am not a statistician, I'm just working on reason here)

($20) First Hand - 48% chance of winning. (If he wins, walks away, loses, keeps playing)

($19) Second Hand - 48% chance of winning, therefore statistically speaking, he will lose the $20.

I think the game of quitting at $21 has zero effect on the overall probabilities over the long term.

Could be WAY off here, but I don't see how the $21 game has any effect on his winnings, when you treat each hand as unique event.

[Guest]

but I'd love to be proven wrong here. I think my favorite puzzles/BT on this website is the "3 door" probability rule where your chances go up when a door is eliminated and given the chance to switch, took me forever to wrap my brain around that one.

[plasmid]

I'll call the probability of winning a hand "x" and the probability of losing a hand "y".

Define PW(n) as the probability of winning for the day (reaching $21) if you currently have $n.

Then we know that PW(20) = x + yPW(19)

Likewise PW(19) = xPW(20) + yPW(18)

And in general PW(n) = xPW(n+1) + yPW(n-1) unless n=20 or n=1.

For cases other than n=20 and n=1, we can rearrange this to

0 = -xPW(n+1) + PW(n) - yPW(n-1)

For the case n=20, we can take PW(20) = x + yPW(19) and rearrange it to

x = PW(20) - yPW(19)

And for the case n=1, we can take PW(1) = xPW(2) and write it as

0 = -xPW(2) + PW(1)

Set it up as a system of linear equations with 20 equations of 20 unknowns.

x =  PW(20) - yPW(19)

0 =-xPW(20) +  PW(19) - yPW(18)

0 =		  -xPW(19) +  PW(18) - yPW(17)

0 =					-xPW(18) +  PW(17) - yPW(16)

0 =							  -xPW(17) +  PW(16) - yPW(15)

etc.

0 =-xPW(4) +  PW(3) - yPW(2)

0 =		 -xPW(3) +  PW(2) - yPW(1)

0 =				  -xPW(2) +  PW(1)

I know that you can use linear algebra to take all of the equations of the form 0 = ... (all of the equations except the first one) and multiply them by constants and add them together to get an equation of the form

0 = aPW(20) + bPW(19)

Spoiler for Proof of that:

You would take the next-to-last equation and multiply it by 1/y and add it to the last one to get rid of PW(1), to get

0 = cPW(3) + dPW(2) where c=-x/y and d=(1/y)-x

Then take the third-to-last equation and multiply it by d/y and add it to the equation above to get

0 = ePW(4) + fPW(3) (the math gets nasty)

Keep doing that until you've added the second equation to eliminate PW(18) and be left with something of the form

0 = aPW(20) + bPW(19)

So PW(19) = -a/b PW(20)

Substituting in the first equation would give

x = PW(20) - y( -a/b PW(20) ) = PW(20) + yaPW(20)/b = (ya+b)PW(20)/b

PW(20) = bx / (ya+b)

The math is too nasty for me to actually work it out by hand, but that should prove that if you have something like Matlab to plug the numbers into, you will get a solution for PW(20). As long as ya+b is not zero. Which it isn't, because we know that a solution must exist. (The only thing I was worried about disproving was non-independence of the equations.)

Then it's just a question of whether $1 x probability of winning > $20 x probability of losing, or 1 x PW(20) > 20 x (1-PW(20))

[Guest]

Q: Why did I choose Chemistry over Math??

A: See spoiler above.

[Guest]

The only way I ever thought of to win is this (but i've never tried to prove it out):

Bet $1 (if you win, you continue to bet $1)

Once you lose, you double the bet ($2). If you win you go back to betting $1.

If you lose again, you double the bet again ($4). This can increase fairly quickly but only if you lose many times in a row (which the odds against that are fairly small). Once you win, you go back to betting $1.

Edited May 7, 2009 by tbrophy

[Guest]

The only way I ever thought of to win is this (but i've never tried to prove it out):

Bet $1 (if you win, you continue to bet $1)

Once you lose, you double the bet ($2). If you win you go back to betting $1.

If you lose again, you double the bet again ($4). This can increase fairly quickly but only if you lose many times in a row (which the odds against that are fairly small). Once you win, you go back to betting $1.

This technique actually has a name for it. Casinos have implemented a rule to cut off this trick at the knees: betting caps. Tables usually have a range, too (i.e., minimum $5, maximum $200). This means a billionaire like Bill Gates can't hit up a roulette table and bet $1 mill on black, then 2, then 4, and so forth. If he went buck wild, this would give the casino a 99.9% chance to lose $1 million by the time Gates is willing to bet up to $1.024 billion 1 / 2^10.

[bushindo]

I'll call the probability of winning a hand "x" and the probability of losing a hand "y".

Define PW(n) as the probability of winning for the day (reaching $21) if you currently have $n.

Then we know that PW(20) = x + yPW(19)

Likewise PW(19) = xPW(20) + yPW(18)

And in general PW(n) = xPW(n+1) + yPW(n-1) unless n=20 or n=1.

For cases other than n=20 and n=1, we can rearrange this to

0 = -xPW(n+1) + PW(n) - yPW(n-1)

For the case n=20, we can take PW(20) = x + yPW(19) and rearrange it to

x = PW(20) - yPW(19)

And for the case n=1, we can take PW(1) = xPW(2) and write it as

0 = -xPW(2) + PW(1)

Set it up as a system of linear equations with 20 equations of 20 unknowns.

x =  PW(20) - yPW(19)

0 =-xPW(20) +  PW(19) - yPW(18)

0 =		  -xPW(19) +  PW(18) - yPW(17)

0 =					-xPW(18) +  PW(17) - yPW(16)

0 =							  -xPW(17) +  PW(16) - yPW(15)

etc.

0 =-xPW(4) +  PW(3) - yPW(2)

0 =		 -xPW(3) +  PW(2) - yPW(1)

0 =				  -xPW(2) +  PW(1)

I know that you can use linear algebra to take all of the equations of the form 0 = ... (all of the equations except the first one) and multiply them by constants and add them together to get an equation of the form 0 = aPW(20) + bPW(19)

Spoiler for Proof of that:

You would take the next-to-last equation and multiply it by 1/y and add it to the last one to get rid of PW(1), to get 0 = cPW(3) + dPW(2) where c=-x/y and d=(1/y)-x Then take the third-to-last equation and multiply it by d/y and add it to the equation above to get 0 = ePW(4) + fPW(3) (the math gets nasty) Keep doing that until you've added the second equation to eliminate PW(18) and be left with something of the form 0 = aPW(20) + bPW(19)

So PW(19) = -a/b PW(20) Substituting in the first equation would give x = PW(20) - y( -a/b PW(20) ) = PW(20) + yaPW(20)/b = (ya+b)PW(20)/b PW(20) = bx / (ya+b) The math is too nasty for me to actually work it out by hand, but that should prove that if you have something like Matlab to plug the numbers into, you will get a solution for PW(20). As long as ya+b is not zero. Which it isn't, because we know that a solution must exist. (The only thing I was worried about disproving was non-independence of the equations.) Then it's just a question of whether $1 x probability of winning > $20 x probability of losing, or 1 x PW(20) > 20 x (1-PW(20))

This is correct. There are several ways to solve the equations plasmid set up.

The following system of equations are what we need to solve.

x =  PW(20) - yPW(19)

0 =-xPW(20) +  PW(19) - yPW(18)

0 =		  -xPW(19) +  PW(18) - yPW(17)

0 =					-xPW(18) +  PW(17) - yPW(16)

0 =							  -xPW(17) +  PW(16) - yPW(15)

etc.

0 =-xPW(4) +  PW(3) - yPW(2)

0 =		 -xPW(3) +  PW(2) - yPW(1)

0 =				  -xPW(2) +  PW(1)

We can break the above system of equation into the following

A t = b

where t is a 20x1 matrix of unknown variables PW(n).

t = [ PW(20), PW(19), ... , PW(1) ]

b is also a 20 x 1 matrix of coefficients on the left side

b = [ x, 0, 0, 0, ... , 0 ]

And A is is matrix. With ones on the diagonal, -y to the right, and -x to the left of every diagonals. All other entries are 0.

A is a band matrix, and it can easily proven that A is invertible, allowing for a unique solution. consequently,

t = A^-1 b

Another alternative solution is to iteratively solve for the following system

PW( 21 ) = 1

PW(0) = 0

PW( n ) = x * PW( n + 1 ) + y * PW( n - 1 )

For n between 1 and 20.

Solving the system from bottom up by substituting is messy, like plasmid indicated. However, my attempts at it shows that approach will end up with some formed of continued fractions.

Solving the above, we see that the chance of leaving the table everyday with 21 dollars is .905. Consequently, the expected value of winnings everyday is -2.625 dollars. After 100 days of applying this strategy, we expect to see a loss of 262.5 dollars.

[bonanova]

A similar problem described a hypothetical village of 100 child-bearing couples.

We must assume equal gender probability for each birth.

If each couple produced children until a boy child is born, and then they have no more

children, what is the expected gender distribution of the children in these 100 families?

In the present puzzle, the man bets all day until one of two stopping conditions obtains.

And the question is asked whether this strategy can overcome a negative expectation

for each individual bet.

For the first puzzle it seems there will be more boys than girls, because the birth of at least one son is ensured.

But if you find the probabilities of B, GB, GGB, GGGB, ... and take the weighted average, you get equal numbers of girls and boys.

But the better solution is simply to note that the probability of [say] a boy at each birth is 50%.

This does not depend at all on the stopping condition.

The expected percentage of boys is 50% - that's a given.

So the expected number of male children is one half of the total children.

There is no strategy the child-bearing couples can employ to avoid this result.

-----------------------

The same analysis applies to the second puzzle.

The expectation for each bet is negative - say it's -$0.02 - that's a given.

That fact does not depend on a stopping strategy.

That means however many bets the man makes in a day before he's either $1 up or $20 down,

[say it's n bets] his expected outcome for the day is -$0.02n.

Thus it is inescapable that for every day he finishes $20 behind, the expected number of days

that he finishes $1 ahead is less than 20.

At the end of the day, so to speak, no matter when or why he quits for the day,

he will expect to lose 2 cents for each hand that he plays.

There is no way to escape this - it's a given.

After 100 days, his expected losses are $2 times the average number of hands played per day.

One can calculate that average number, I'm not that interested, but since it will be a positive number, the house will unavoidably win.

------------------

In the related strategy of doubling the bet after each loss until a hand is won,

a string of losses eventually occurs that wipes out any finite initial stake.

[Guest]

A similar problem described a hypothetical village of 100 child-bearing couples.

We must assume equal gender probability for each birth.

If each couple produced children until a boy child is born, and then they have no more

children, what is the expected gender distribution of the children in these 100 families?

In the present puzzle, the man bets all day until one of two stopping conditions obtains.

And the question is asked whether this strategy can overcome a negative expectation

for each individual bet.

For the first puzzle it seems there will be more boys than girls, because the birth of at least one son is ensured.

But if you find the probabilities of B, GB, GGB, GGGB, ... and take the weighted average, you get equal numbers of girls and boys.

But the better solution is simply to note that the probability of [say] a boy at each birth is 50%.

This does not depend at all on the stopping condition.

The expected percentage of boys is 50% - that's a given.

So the expected number of male children is one half of the total children.

There is no strategy the child-bearing couples can employ to avoid this result.

-----------------------

The same analysis applies to the second puzzle.

The expectation for each bet is negative - say it's -$0.02 - that's a given.

That fact does not depend on a stopping strategy.

That means however many bets the man makes in a day before he's either $1 up or $20 down,

[say it's n bets] his expected outcome for the day is -$0.02n.

Thus it is inescapable that for every day he finishes $20 behind, the expected number of days

that he finishes $1 ahead is less than 20.

At the end of the day, so to speak, no matter when or why he quits for the day,

he will expect to lose 2 cents for each hand that he plays.

There is no way to escape this - it's a given.

After 100 days, his expected losses are $2 times the average number of hands played per day.

One can calculate that average number, I'm not that interested, but since it will be a positive number, the house will unavoidably win.

------------------

In the related strategy of doubling the bet after each loss until a hand is won,

a string of losses eventually occurs that wipes out any finite initial stake.

[Guest]

Any casino owner with any red blood in his veins would allow Bill Gates to use a double up system on

roulette with a much greater spread than $1- $200, if Bill started his sequence at $!; perhaps even allowing a million dollar bet or higher. Assuming one bet per minute and eight hours of play each day Bill will win $480 per day, which he will do every day until he has a really bad day {which will happen much sooner than you'd expect}, He'd drop upwards of 2 million on that day,

[Guest]

this is gonna be pretty hard to write but the probabilty of leaving with $21 i think is

[summation from p=1 to infinity]((.48^p)(.52^(p-1))((2p-1)!)/(p!)(q-1!))

which can be simplified but frankly who gives a ****

[Guest]

sorry the q's suppose to be a p