Colored cards at Morty's

[bonanova]

It's been far too long, said Alex to no one in particular as
he tossed his hat skillfully onto a coak hook at Morty's
last night.

Ian and Davey knew that was their cue, and they joined
Alex at his favorite table. Not surprisingly, he pulled out a
deck of cards.

I'm feeling generous tonight laddies, he began. I'll play
a game with you that guarantees that you can double your
money. So to make it fair, I'll collect your initial stake
from you, and then you can keep what ever you win.
That way you can assure no worse than to break even.

Jamie had been listening, from the bar, and what he
heard drew him over to join the others. OK tell us,
he said, how does the game work?

I'll shuffle this deck here, and then deal them face up
one at a time. You start with $1. You can bet any fraction
of your current worth, before I deal the next card, on the
color of the next card. Even odds on each bet, no matter
what.

So how does the guarantee work then? asked Ian.

Drink yer ale, matey! Alex replied, If yer paying attention
at all, you'll know the color of the last card. You can just
wait, and then bet everything on the last card! I'll take
yer stake, you get your winnings, and you break even.

Now, who wants to play?

Jamie and Ian figured that if they were really lucky
they could win every bet and parlay their stake into
about $50 quadrillion. But that would be offset by the
fact that if they lost any of the 52 bets, they'd win
nothing, and Alex would get their stake. Seems like a
bunch of coin tosses to me, said Jamie. Not worth my
time just to break even, said Ian, And the two of them left.

Davey, trance-like for a moment as he stroked his beard,
finally said, I'll play. Deal the cards.

Did Davey have any expectation of winning money?
If so, how much?

[TwoaDay]

if you card count but with colors you increase the odds.. i don't want to do math now

[HoustonHokie]

There's going to be a fair amount of times (probably half the time?) that the last TWO or more cards are the same color. If he waits for those opportunities, and then bets it all, he should do quite well...

Edited July 31, 2008 by HoustonHokie

[Guest]

Davey would expect to win a minimum of 94 cents every deck.

If he bets 3 cents on each card being red until he knows what color the remaining card or cards are...the worst case senario will be that he will have 97 cents left, knowing what color the last card is and betting it all.

It's been far too long, said Alex to no one in particular as

he tossed his hat skillfully onto a coak hook at Morty's

last night.

Ian and Davey knew that was their cue, and they joined

Alex at his favorite table. Not surprisingly, he pulled out a

deck of cards.

I'm feeling generous tonight laddies, he began. I'll play

a game with you that guarantees that you can double your

money. So to make it fair, I'll collect your initial stake

from you, and then you can keep what ever you win.

That way you can assure no worse than to break even.

Jamie had been listening, from the bar, and what he

heard drew him over to join the others. OK tell us,

he said, how does the game work?

I'll shuffle this deck here, and then deal them face up

one at a time. You start with $1. You can bet any fraction

of your current worth, before I deal the next card, on the

color of the next card. Even odds on each bet, no matter

what.

So how does the guarantee work then? asked Ian.

Drink yer ale, matey! Alex replied, If yer paying attention

at all, you'll know the color of the last card. You can just

wait, and then bet everything on the last card! I'll take

yer stake, you get your winnings, and you break even.

Now, who wants to play?

Jamie and Ian figured that if they were really lucky

they could win every bet and parlay their stake into

about $50 quadrillion. But that would be offset by the

fact that if they lost any of the 52 bets, they'd win

nothing, and Alex would get their stake. Seems like a

bunch of coin tosses to me, said Jamie. Not worth my

time just to break even, said Ian, And the two of them left.

Davey, trance-like for a moment as he stroked his beard,

finally said, I'll play. Deal the cards.

Did Davey have any expectation of winning money?

If so, how much?

[HoustonHokie]

So Davey's counting cards, and when he sees that all of the red or black cards have been exhausted, he bets it all from there till the end. How much does he win on an average hand?

Well, the probability of a card being red or black is 0.5. The probability of the last two cards being the same color is 0.5^2, or 0.25. And so on. There are 26 possible solutions from one consecutive card at the end all the way to 26 consecutive cards, with decreasing likeliness as the number increases (p(n)=0.5^n). To make sure the probabilities sum to 1, p(26) = 1-p(1)-p(2)-...-p(25). However, his winnings increase as the number of consecutive cards increases. If n=1, he ends up with $1*2=$2. After giving 1$ to Alex and putting the original $1 back in his wallet, his profit is $0. Mathematically, profit = 2^n - 2. If n=2, wins 2^2-2=2. All that remains is to multiply the probability of a particular event by the winnings, sum them, and divide by the number of cases. I get that the sum of the probability by the winnings is 25. With the number of cases being 26, he wins, on average, 25/26, or about $0.96.

[HoustonHokie]

Davey would expect to win a minimum of 94 cents every deck.

If he bets 3 cents on each card being red until he knows what color the remaining card or cards are...the worst case senario will be that he will have 97 cents left, knowing what color the last card is and betting it all.

Doesn't Davey actually lose 6 cents there? He gives Alex $1 at the end (his original bet) and is left with $0.94, which is $0.06 less than the $1.00 he started with.

Edited August 1, 2008 by HoustonHokie

[Guest]

If he bets 3 cents on each card being red until he knows what color the remaining card or cards are...the worst case senario will be that he will have 97 cents left, knowing what color the last card is and betting it all. He then has $1.94, Alex takes a dollor leaving 94 cents for Davey.

Doesn't Davey actually lose 6 cents there? He gives Alex $1 at the end (his original bet) and is left with $0.94, which is $0.06 less than the $1.00 he started with.

[HoustonHokie]

If he bets 3 cents on each card being red until he knows what color the remaining card or cards are...the worst case senario will be that he will have 97 cents left, knowing what color the last card is and betting it all. He then has $1.94, Alex takes a dollar leaving 94 cents for Davey.

But he walked in with $1.00 and left with $0.94. Here's how the action unfolds:

Davey walks up to the table and puts $1.00 of his own money down, then bets $0.03 on red each time. 26 black cards and 25 red cards come out, and before the last card is dealt, there is $0.97 on the table. He bets it all on red, wins, and now there is $1.94 on the table. Alex takes $1.00, and Davey picks up the $0.94 left on the table - and walks out 6 cents poorer than he walked in.

It's much better to reserve betting until all the cards of one color have been dealt.

Edited August 1, 2008 by HoustonHokie

[Guest]

But he walked in with $1.00 and left with $0.94. Here's how the action unfolds:

Davey walks up to the table and puts $1.00 of his own money down, then bets $0.03 on red each time. 26 black cards and 25 red cards come out, and before the last card is dealt, there is $0.97 on the table. He bets it all on red, wins, and now there is $1.94 on the table. Alex takes $1.00, and Davey picks up the $0.94 left on the table - and walks out 6 cents poorer than he walked in.

It's much better to reserve betting until all the cards of one color have been dealt.

You are correct.

[Guest]

I may have missed something, but I read "you can bet any amount".

What if you bet zero and count cards, until the last card?

[HoustonHokie]

I may have missed something, but I read "you can bet any amount".

What if you bet zero and count cards, until the last card?

It's not that you can bet any amount - you can bet "any fraction of your current worth". So if Davey doesn't make any bets until the last card, he'll double his money to $2, but Alex will take $1 and leave Davey with $1, which is exactly what he started with - no gain, no loss.

[HoustonHokie]

So Davey's counting cards, and when he sees that all of the red or black cards have been exhausted, he bets it all from there till the end. How much does he win on an average hand?

Well, the probability of a card being red or black is 0.5. The probability of the last two cards being the same color is 0.5^2, or 0.25. And so on. There are 26 possible solutions from one consecutive card at the end all the way to 26 consecutive cards, with decreasing likeliness as the number increases (p(n)=0.5^n). To make sure the probabilities sum to 1, p(26) = 1-p(1)-p(2)-...-p(25). However, his winnings increase as the number of consecutive cards increases. If n=1, he ends up with $1*2=$2. After giving 1$ to Alex and putting the original $1 back in his wallet, his profit is $0. Mathematically, profit = 2^n - 2. If n=2, wins 2^2-2=2. All that remains is to multiply the probability of a particular event by the winnings, sum them, and divide by the number of cases. I get that the sum of the probability by the winnings is 25. With the number of cases being 26, he wins, on average, 25/26, or about $0.96.

I just realized that I messed up the probability math here. This is better:

So Davey's counting cards, and when he sees that all of the red or black cards have been exhausted, he bets it all from there till the end. How much does he win on an average hand?

The probability of the last n cards being the same color is actually the same as the probability of dealing n straight cards of the same color from the front of the deck, which is a whole lot easier to get my mind around. The probability that the first card is red is 26/52. If the first card is red, the probability that the second card is a black is 26/51. So the probability of a run of one red card is 26/52*26/51 = 0.255. The probability of dealing only n red cards in a row is the probability of dealing the first red card times the probability of dealing the second red card times the probability of dealing the third, fourth, ... nth red card times the probability of dealing the n+1th as a black card. Mathematically, the probability of dealing n cards in a row of a particular color is then (26/52)*(25/51)*(24/50)*...*[(27-n)/(53-n)]*[26/(52-n)]. Because there are both black and red cards, and it doesn't matter which color the run is in, that probability is multiplied by 2 to get the final probability of a run of n cards in the same color. The sum of probabilities of n=1 to n=26 equals 1 on its own without forcing it the way I did last time. The rest of the problem remains unchanged, except that his average winnings decreases significantly, from $0.96 to $0.27.

[Guest]

I just realized that I messed up the probability math here. This is better:

So Davey's counting cards, and when he sees that all of the red or black cards have been exhausted, he bets it all from there till the end. How much does he win on an average hand?

The probability of the last n cards being the same color is actually the same as the probability of dealing n straight cards of the same color from the front of the deck, which is a whole lot easier to get my mind around. The probability that the first card is red is 26/52. If the first card is red, the probability that the second card is a black is 26/51. So the probability of a run of one red card is 26/52*26/51 = 0.255. The probability of dealing only n red cards in a row is the probability of dealing the first red card times the probability of dealing the second red card times the probability of dealing the third, fourth, ... nth red card times the probability of dealing the n+1th as a black card. Mathematically, the probability of dealing n cards in a row of a particular color is then (26/52)*(25/51)*(24/50)*...*[(27-n)/(53-n)]*[26/(52-n)]. Because there are both black and red cards, and it doesn't matter which color the run is in, that probability is multiplied by 2 to get the final probability of a run of n cards in the same color. The sum of probabilities of n=1 to n=26 equals 1 on its own without forcing it the way I did last time. The rest of the problem remains unchanged, except that his average winnings decreases significantly, from $0.96 to $0.27.

I didn't quite follow that last bit, but I believe the probability of the last two cards being of the same color is roughly 0.5.

Anyway, I haven't worked it all out yet, but I think there may be a way to increase your winnings, if you bet a portion of your stack proportional to the odds of the next card turning up a particular color. For instance, if you don't bet on the first card and it turns up black, the odds of the next card being red is 26/51. So you bet $0.50 on red. If I understand the rules correctly, you keep the $0.50 when you win, only paying the initial $1 stake at the end of the deck. You therefore expect to win about $0.01 on this bet ((26-25)/51*0.5). If you lose this bet, your stack is cut in half, but your odds of winning the next bet increases to 26/50, so you now bet $0.25. Your expectation is now exactly $0.01 ((26-24)/50*0.25). If you had won your first bet, you pass on the second bet, and your odds of winning on the third card is 25/49 (the largest proportion of your stack smaller than your odds of winning). Either way, as you progress through the deck, your odds of winning each bet increases whenever there is not an equal number of cards of each color remaining. Since you are only betting when the odds favor you, you expect to have more than $1 to bet on the last card, or on the last cards that are known to be of a particular color.

[edited: grammar]

Edited August 1, 2008 by d3k3

[Guest]

I didn't quite follow that last bit, but I believe the probability of the last two cards being of the same color is roughly 0.5.

Anyway, I haven't worked it all out yet, but I think there may be a way to increase your winnings, if you bet a portion of your stack proportional to the odds of the next card turning up a particular color. For instance, if you don't bet on the first card and it turns up black, the odds of the next card being red is 26/51. So you bet $0.50 on red. If I understand the rules correctly, you keep the $0.50 when you win, only paying the initial $1 stake at the end of the deck. You therefore expect to win about $0.01 on this bet ((26-25)/51*0.5). If you lose this bet, your stack is cut in half, but your odds of winning the next bet increases to 26/50, so you now bet $0.25. Your expectation is now exactly $0.01 ((26-24)/50*0.25). If you had won your first bet, you pass on the second bet, and your odds of winning on the third card is 25/49 (the largest proportion of your stack smaller than your odds of winning). Either way, as you progress through the deck, your odds of winning each bet increases whenever there is not an equal number of cards of each color remaining. Since you are only betting when the odds favor you, you expect to have more than $1 to bet on the last card, or on the last cards that are known to be of a particular color.

[edited: grammar]

Er, oops. "(the largest...odds of winning)" should come after "... so you now bet $0.25."

[bonanova]

Overheard at Morty's this afternoon:

Did you hear?

Davey took Alex to the cleaners last night!

[Guest]

Overheard at Morty's this afternoon:

Did you hear?

Davey took Alex to the cleaners last night!

dont give any clues yet, im working on this one

[Guest]

dont give any clues yet, im working on this one

$17.61

$277247380825.62

Edited August 1, 2008 by bonanova
Bonanova added spoilers for taliesin's results

[Guest]

If you use this strategy, you are guaranteed to come out even or ahead. Your expected winning can be calculated as follows:

The probability of a sequence of length N>1 of all one color at the end is 1*25/51*24/50*...*(27-N)/(53-N)*26/(52-N). This is counting up from the end to make it easier. The first number is a 1 because it could be either a black or red sequence. The 26/(52-N) comes from the fact that the sequence must be preceeded by the other color. This avoids double counting.

Your winnings for a sequence of length N is 2^(N-1)-1, with the -1 accounting for paying back the $1 stake. Multiplying all these terms out (I used Excel) gives an expected value of about $3.54

You can get a higher expected value by betting money earlier in the deck, but I'm not sure what the optimum strategy would be in that case. That also would allow the possibility of a loss of money (but you only started with $1, so I guess it's not that big a deal).

[Guest]

(assuming p is the money in your pot at go number i)

I'm thinking about it with two phases to the game:

Phase A: (which lasts, say, n goes) you can bet with the odds, but you're not guaranteed to win.

Phase B: (which then lasts (52-n) goes) you know every card left is the same colour, so are guaranteed to win them all.

Phase B is simple, because whatever you enter Phase B with, gets doubled until the end of the game: 2^(52-n) x p[n]

But Phase A has three scenarios at each go, assuming you always folow the best odds:

the odds are even - you probably don't bet. p[n+1] = p[n]

the odds are in your favour and you win. p[n+1] = p[n] x (1+b) where b is the proportion of your pot that you stake.

the odds are in your favour and you lose. p[n+1] = p[n] x (1-b).

The bit that makes it interesting is that every time you lose, it benefits you because you improve the odds of minimising n and spending longer in Phase B, where the mega profits are made.

So the bit I can't work out is the best strategy for deciding on 'b' which:

earns you enough if Phase B is short (ie you have won most of your Phase A bets), but

ensures you have enough money to take into Phase B if it's long (ie you have lost most of your Phase A bets).

[bonanova]

$17.61

$277247380825.62

No clues for a while. Several are still working.

Care to put your worst case strategy into a spoiler?

[Guest]

No clues for a while. Several are still working.

Care to put your worst case strategy into a spoiler?

Thanks for the spoilers on my other post, my apoligies.

I think I have kind of messed up by a few dollors. ( sorry about the spoilers)

I think the wost would become statically 4.82 and best 30047290338.2

The best case senerio is when you have the most cards where you are certain of the order. So you can be the maximun a maximun number of times in a row. This ocurs when there is either 26 red / or 26 black cards delt first. Then it is ovious you put all your money on the last 26 cards..

The worst case seniro is when centered between the two best cases (all red first, and all back first) and therefore it is when the pack is delt red then black through all there cards this makes the change of getting a red/black guess correct 50% (which is the minimum possible)

So working on this I decided I was going to bet ever card, and hope for the best.

The first card I would vote red with 50c. (because there is a 50% chance to be correct, with a return of 2 statically you should get a return of 50c leaving you with a dollor.

The second card you know that there is once red or black card gone, so you vote against the it. (because there is more than on than the other). There is 51% chance you will be right if you vote against the first card so you vote 51% of your money against it. So vote 51c to will with a return of 2, and 51% chance of been correct statically you will win 52c which brings you total up to 1.01

Now there are two cards gone (taking worst case senerio again) one will be red and one black keeping the chance of been correct 50% again. now bet have you money on your lucky money and statically you will win no money and lose none.

Forth card same as the second 51% chance of been correct, with 1.01 to play with, and return 2:1 still. you bet 51% of your money which is 52c and you will get get a stastic return of 0.53c earning you another cent

The chance of becoming correct climbs slowly through the pack until you get to the end where it skyrockets to high chances and last card is 100%. You just keep betting the odds, and because you always have a change 50% or greater to win you should never lose more than your initial input.

let r be the number of red cards already player

let b be the number of black cards player

Chance of winning on red ( Cr ) is

Cr( r, b ) = ( 26 - r ) / ( 52 - ( r + b )

Chance of winning on black ( Cb ) is

Cb( r, b ) = ( 26 - b ) / ( 52 - ( r + b )

your bet will be the greater of the two ( you dont want to bet on losing odds do you), and this chance will be called

C( ) asuming you bet on the card with the bigger cahnce

let p be the cash in your pocket

let t be the cash you put on the table

let w be the winnings (statically, which means the expected average over infinity bets)

let n new amount in your pocket after a round ( one card )

t = p * C()

w = t * 2 * C()

n = p - t + w

n = p - p * C() + p * C() * 2 * C()

n = p ( 1 - C() + 2 C()^2 )

after each round p = n

expanding this out

i have a pdf showing best and worst card by card

then minus the $1 "put in" of the total and the end results are

$4.82 for worst and

$30047290338.2 for best

so you winnings using this system should be somewhere in between these two

Edited August 1, 2008 by taliesin

[Guest]

Every player loses all they put down with no chance of winning anything: the dealing never starts...

[Guest]

Every player loses all they put down with no chance of winning anything: the dealing never starts...

can you explain

[Guest]

So to make it fair, I'll collect your initial stake

from you, and then you can keep what ever you win.

After collecting the initial stake, Alex takes off to never return again. Or well, not in a long while at least...

[Guest]

I've been thinking about this recently, and I am now leaning towards a counter-intuitive answer:

The optimum strategy (from an expectation point of view, at least) seems to be to bet it all on every card. Sure, it's absurdly unlikely that you'll hit 52 cards in a row (prob = approx. 2x10^-15), but you would earn such an absurdly huge amount in that case ($4.5x10¹⁵) that your expected profit is about $8.08.

My reasoning here is that, at any point I've tried, your optimum strategy is to be it all. Say you have $1 left, and there are two cards of idfferent colors. If you bet it all both times, your expected profit is $1.50. If you bet $.50 on the first card, then everything on the last card (which seems intuitive), your expected profit is $1.

Am I on the right track here?

[Guest]

Am I on the right track here?

I've been thinking about this recently, and I am now leaning towards a counter-intuitive answer:

The optimum strategy (from an expectation point of view, at least) seems to be to bet it all on every card. Sure, it's absurdly unlikely that you'll hit 52 cards in a row (prob = approx. 2x10^-15), but you would earn such an absurdly huge amount in that case ($4.5x10¹⁵) that your expected profit is about $8.08.

My reasoning here is that, at any point I've tried, your optimum strategy is to be it all. Say you have $1 left, and there are two cards of idfferent colors. If you bet it all both times, your expected profit is $1.50. If you bet $.50 on the first card, then everything on the last card (which seems intuitive), your expected profit is $1.

The only problem with this if you lose one card you lose ALL your money. Therefore you need to get all 52 cards correct to win ( very slim chance)