Can you guarantee a win on a true 50/50

If you have a deck of cards with betting whether the next card will be red or black. What is the best way to maximize profit on a $5 minimum $500 max with a $500 bankroll?

count cards

Quote: odiousgambit

count cards

... which will not guarantee a profit, but you have a better than even chance of a profit and can limit a loss to $10 maximum, guaranteed. Playing to maximize profit and playing to guarantee a win are two separate strategies.

How would you go about this?

Quote: stevenm0519

How would you go about this?

Flat bet $5 for 51 cards, maximum loss $255. With counting cards the last one is known, bet the remaining bankroll which at worst will be $245 and result in a total of $10 lost, guaranteed worst case.

Quote: stevenm0519

If you have a deck of cards with betting whether the next card will be red or black. What is the best way to maximize profit on a $5 minimum $500 max with a $500 bankroll?

Dunno if it's optimum, but here's my starter......

Bet min that the first card will be red. keep track of count of red and count of black and continue to bet min that the next card will be whatever there are most left of.
When you know that all reds are gone, bet max on black until the end of the deck, alternatively when you know that all blacks are gone, bet max on red till the end of the deck.

Anyone care to estimate the ev on that or come up with a better system?

I am guessing using Kelly criterion betting strategy would optimize profit, but I can't quantify that.

Cheat. Have you learned nothing from this forum?

Quote: Joeshlabotnik

Cheat. Have you learned nothing from this forum?

That's funny. I'm trying to use this to test a theory

Quote: Joeshlabotnik

Cheat. Have you learned nothing from this forum?

Card counting has already been mentioned. :-)

Two questions:

First, how many cards are dealt from the deck before shuffling?

Second, what do you mean by "maximize profits"?

Without any additional information, my gut answer is, if you can't count the number of black and red cards dealt so far (because the deck is reshuffled after each play), then bet the whole $500 at once.

1) 0

2) a way to get the absolute most money you can on that kind of game. For instance if I told you I had the fifty two cards that have been shuffled and with 100% penetration how would you go about making money on it

Quote: stevenm0519

What is the best way to maximize profit on a $5 minimum $500 max with a $500 bankroll?

Hang on. One thread and two very different questions.
There's the question in the subject "Can you guarantee a win" and there is the question in the first post."What is the way to maximise profit"

Let's do the first one.
Let's say "I'll bet $5 Red" for the first 51 cards and count what has passed up to that point. The very worst you can do is encounter 26 black cards and 25 red cards. So you will be down by $5 leaving you $495 to wager on that last card which you know will be black. Even if the first 25 cards to come out are black you would only ever be 25x5 = $125 down and of course you would then bet black with some amount of enthusiasm. So if the cards are not shuffled, then YES you can guarantee a win. BUT, of course there was never more than one 50/50 wager and lots of wagers that were not 50/50... So in conclusion, you never did guarantee a win with a 50/50 proposition.

Then question 2.
Maximise profit? Is that maximise guaranteed profit, or maximise expected profit? For the latter, Kelly has to be the way to go, but you would need to re-evaluate your edge after each card. A right royal PITA.

Quote: stevenm0519

How would you go about this?

Keep track of red vs black... Just a simply running tally of what you've seen would suffice. If there are more of one color than the other, bet max on that color. This is assuming as you said "fair" 50/50 and you're not being short paid or any house commission for a win.

The closer you got to the end of the deck the more valuable this would become, just as a true count conversion in blackjack.

Ex. 26 and 26 to start... remove one red card... now 26 black, 25 red... odds of black = 26/51 = 50.98%... so almost a 1% edge.

Ex. only 16 cards left... 8 and 8... remove one red card... now 8 black, 7 red... odds of black = 8/15 - 53.33%... so now a 3.33% edge.

As someone else pointed out this would be a fantastic time to employ "end game" if they deal down to the last card (which no one in their right mind would/should when playing for money). You had better know the color of the last card and have a max bet on it is all I need to say here.

Quote: stevenm0519

1) 0...

You must be misinterpreting his question. When playing the game from the beginning... Freshly shuffled deck... how many times can you bet on the next card before the next time the deck is shuffled? If you say 0 that means you're never actually playing a round and no one is playing the game, ever.

Sorry I guess it was two different questions. The first one can you guarantee a profit then how would you play that game to get the most money. Say I was offering this game and whoever won the most got an extra bonus what would be your strategy. Hopefully I didn't just add another question lol

Quote: stevenm0519

Sorry I guess it was two different questions. The first one can you guarantee a profit then how would you play that game to get the most money. Say I was offering this game and whoever won the most got an extra bonus what would be your strategy. Hopefully I didn't just add another question lol

Hrm..... Why would you offer this game even without a bonus? Unless your players are too stupid to count, you are guaranteed to end up dealing a game that you would lose.

My strategy...
Try to negotiate stakes 1000x higher and/or take 52 seats in your game :o)

Please describe absolutely and explicitly the rules of the game. Absolute and explicit and complete or else I for one won't show much further interest. :o)

Because the idea is for having a strategy to a game where you know the end result will be 50/50 instead of casinos which may have...let's say ...27 losses in a row :)

Quote: Romes

If there are more of one color than the other, bet max on that color.

Romes, Do you meant that?

On this game where we we have a game plan that cannot conceivably lose, I don't think it would be wise to place any wager of $500 that would risk ruin.
I'd be inclined to maintain a calculation of my rolling edge and bet that percentage of my rolling bankroll. When the edge becomes 100%, then of course max bet until the dealer storms off in a fury.

Quote: stevenm0519

Because the idea is for having a strategy to a game where you know the end result will be 50/50 instead of casinos which may have...let's say ...27 losses in a row :)

I don't understand. What part of the game has an end result that's 50/50 if the player is able to count and vary his bets?

Simplify. have 2 cards, one red and one black.
Bet between 5 and 500 on guessing the outcome.

There is only one strategy. bet $5 on round one and depending on the outcome of that, bet either 495 or 500 on round two.
ONLY round one is a 50/50 proposition. Round two is a 100/0 sure fire, cannot lose, proposition unless you can't count to 1.

Quote: stevenm0519

Sorry I guess it was two different questions. The first one can you guarantee a profit then how would you play that game to get the most money. Say I was offering this game and whoever won the most got an extra bonus what would be your strategy. Hopefully I didn't just add another question lol

You can guarantee profit OVER TIME. Not on JUST the next bet. The only way you could do that is wait until only 1 color remains, then bet max on that color. If you're going to play every hand pending the color for max EV, then you will have variance, and only overcome that variance and realize your EV once you've hit "the long run." Also referred to in blackjack as N0.

...You still didn't answer the question about when the deck is shuffled. This is part of the game and we need to know EXACTLY how the game is played in order to tell you the specifics you're looking for.

Quote: stevenm0519

Because the idea is for having a strategy to a game where you know the end result will be 50/50 instead of casinos which may have...let's say ...27 losses in a row :)

You could still lose a bunch in a row... losing 10 in a row in this game would be .5^10 = .000976 = .01% of the time, or about 1 in 1100. If you play this game with any kind of frequency you'll for sure see that streak. That's not even accounting for the smaller streaks too.

edit: apparently double posted

Quote: stevenm0519

1) 0

2) a way to get the absolute most money you can on that kind of game. For instance if I told you I had the fifty two cards that have been shuffled and with 100% penetration how would you go about making money on it

So, every bet is 50/50 (as if you're calling a coin toss rather than drawing a card from a deck)?

And the answer to "a way to get the absolute most money you can" is, "you can't." As long as there is a low point you can reach (-500) but not a high point, "eventually" (as in "eventually six monkeys put in front of six computer keyboards will write Hamlet - backwards") you will lose everything in a 50-50 game. If you have infinite time and infinite bankroll, that's another matter, but you don't - even if you're immortal, eventually the Sun will have burned through its hydrogen and expand to the point where Earth's orbit is also expanded (because of the decrease in the Sun's mass) to the point where it is outside of the habitable temperature range.

The reason I said a deck is because there is 52 total bets where as a coin can be flipped 52 times in a row just heads or tails and with the cards if you bet one color you are guaranteed to win 50% of your bets. but with this game you can only lose a fixed number of times and because of max bets you obviously can't do a Martingale but I wanted to see what options people would use

Quote: stevenm0519

The reason I said a deck is because there is 52 total bets where as a coin can be flipped 52 times in a row just heads or tails and with the cards if you bet one color you are guaranteed to win 50% of your bets...

This is HUGE apples and oranges. The coin "resets" before each flip. However the deck "remembers" what it's missing. So if you take 10 black cards out of the deck, that would be like weighting one side of the coin.

The only way those 2 are equivalent is if you SHUFFLE the deck after EVERY single round. i.e... Place your bets! 52 cards left. The card is black. Resolve the bets... Shuffle the deck and start again. That's the only way drawing red/black from a deck is "equivalent" to heads/tails on a coin.

I think we are saying the same thing...

A coin flip has been looked at and everyone comes up with betting systems and then people say they don't work because the coin doesn't have a memory...but cards would in the game I described. So you bet one side, your guaranteed to win 26 times so how would you bet the game to make the most money using the $5 minimum $500 max

Depends on your edge and bankroll.

Quote: stevenm0519

I think we are saying the same thing...

A coin flip has been looked at and everyone comes up with betting systems and then people say they don't work because the coin doesn't have a memory...but cards would in the game I described. So you bet one side, your guaranteed to win 26 times so how would you bet the game to make the most money using the $5 minimum $500 max

This is getting to be like pulling teeth, lol... Okay, so you're saying they ARE dealing more than one card before Shuffling... Now the same question for the 3rd time... HOW MANY cards are they dealing? All the way to the bottom, or do they shuffle after 10 cards? 20 cards?

If they deal the WHOLE way through including the last card then you would ONLY have a guaranteed win when only one suit remains. However, you would have an edge and positive EV on EVERY SINGLE BET except for when the same number of colors remain.

i.e. X = black cards, Y = red cards.

X = Y -> No Edge
X > Y -> Edge (size of which depends on X and Y and the number of TOTAL cards remaining)
Y > X -> Edge (size of which depends on X and Y and the number of TOTAL cards remaining)
X = 0 -> 100% Edge (this is your guaranteed win... bet max here)
Y = 0 -> 100% Edge (this is your guaranteed win... bet max here)

Quote: stevenm0519

That's funny. I'm trying to use this to test a theory

What's your theory?

Quote: Romes

This is getting to be like pulling teeth, lol... Okay, so you're saying they ARE dealing more than one card before Shuffling... Now the same question for the 3rd time... HOW MANY cards are they dealing? All the way to the bottom, or do they shuffle after 10 cards? 20 cards?

I'm pretty sure he inferred that ALL 52 cards are dealt, with each 'deal' being one round of the game and no shuffling after any card dealt.

Quote: stevenm0519

I think we are saying the same thing...

A coin flip has been looked at and everyone comes up with betting systems and then people say they don't work because the coin doesn't have a memory...but cards would in the game I described.

Right, but that's not a 50/50 proposition. If you want to guarantee a win, as has been said, just wait for the first 51 cards to come out and stake all your money on the value of the last card. The probability of a win is 100%, not 50%. That's the only way you guarantee anything.

Quote: OnceDear

I'm pretty sure he inferred that ALL 52 cards are dealt, with each 'deal' being one round of the game and no shuffling after any card dealt.

He implied. You inferred. I was asking for a concrete answer past implications =).

I'm sorry someone asked me how many times the cards are shuffled and I said 0. So it's 52 cards once all 52 are dealt that's the end of the game and its one card per one bet. Hopefully that clears it up sorry.

Quote: stevenm0519

I'm sorry someone asked me how many times the cards are shuffled and I said 0. So it's 52 cards once all 52 are dealt that's the end of the game and its one card per one bet. Hopefully that clears it up sorry.

All good! See my last post with X and Y in it... That should pretty much be your answer.

Quote: stevenm0519

I'm sorry someone asked me how many times the cards are shuffled and I said 0. So it's 52 cards once all 52 are dealt that's the end of the game and its one card per one bet. Hopefully that clears it up sorry.

In that case, the "count the number of red and black cards as they come out, and bet the minimum until 26 cards of either color have come out, at which point all of the remaining cards will be the other color, so bet the maximum from that point forward" strategy previously mentioned applies. You can "guarantee" a profit if your starting bankroll is more than 102 times the minimum bet, as you can lose 51 times the minimum bet with the first 51 cards, but the 52nd card is a guaranteed win.

This problem would be greatly simplified, without changing its applicability to whatever "theory" we're talking about, by reducing the deck to two cards, one red and one black.

Quote: BleedingChipsSlowly

I am guessing using Kelly criterion betting strategy would optimize profit, but I can't quantify that.

Allow me to " The Kelly Criterion takes into account both the size of your advantage (I.e the value available) and the size of your bankroll, so as to minimise risk and maximise your advantage.

Zero Advantage = Zero Bet

The Kelly Criterion should never be used if betting on a safety in any Superbowl.

Quote: ThatDonGuy

In that case, the "count the number of red and black cards as they come out, and bet the minimum until 26 cards of either color have come out, at which point all of the remaining cards will be the other color, so bet the maximum from that point forward" strategy previously mentioned applies. You can "guarantee" a profit if your starting bankroll is more than 102 times the minimum bet, as you can lose 51 times the minimum bet with the first 51 cards, but the 52nd card is a guaranteed win.

Not if you always bet black until you get to the point where all red cards (or all black cards) have been removed.

Bet $5 on black every round, first 26 cards are red, you lose $5*26 = $130. Now you bet $370 on black, guarantee the win, now you're up $240. You bet $500 on black rest of game, for a profit of $240 + $500*25 = $12,700.

Worst case scenario, is you get 1 red 1 black repeating. You bet $5 black every round. After the 51'st round, you're down 26 red bets and UP 25 black bets, for a net loss of $5. Then you wager $495 (your remaining BR), guarantee the win, and profit $490 overall.

That isn't taking into account card-counting.

But with card-counting and varying your bet, you are introducing some level of variance. I'm not gonna spend the time to figure out how bad it can all get and all that. But say you're going to bet full kelly (Edge*BR / Variance), while betting $5 when you have no advantage.

Worst case scenario I can immediately think of:

26 black then 26 red.

$5 on red : -5, $95 remaining

now 26 to win, 25 to lose, EV 1/50 ~ 2% edge
2% of $495 = $10 : now down $15, $480 remaining

26 reds / 50 total ~ EV 2/50 ~ 4% edge
4% of $480 = $20 : now down $35, $465 remaining

26 reds / 49 total ~ EV of 6.1%
6.1% of $465 = $28 (call it $25) -- now down $60, remaining $440

26 reds / 48 total ~ 8.3% edge
8.3% of $440 = $36 (call it $35) -- now down $95, remaining $405

26 reds / 47 total ~ 10.6% advantage
10.6% of 405 is $43 (call it $40) -- now down $135, remaining $365

26 reds / 46 total ~ 13% advantage
13% of $365 is $47 ($45) -- now down $180, remaining $320

26 reds / 45 total ~ 15.5% edge
15.5% of $320 is $49.6 ($45) -- now down $225, remaining $275

26 reds / 44 total ~ 18.18% edge
18.18% of $275 is $50 exacly -- now down $275, remaining $225

26 reds / 43 total ~ 20.9% advantage
20.9% of $225 is $47 ($45) -- now down $320, with $180 remaining

26 reds / 42 total ~ 23.8% advantage
23.8% of $180 is $42 ($40) -- now down $360, with $140 remaining

26 red / 41 total ~ 26.82% edge
26.82% of $140 is $37 ($35) -- now down $395, with $105 remaining

26 red / 40 total ~ 30% edge
30% of $105 is $31 ($30) -- now down $425, with $75 remaining

26 red / 39 total ~ 33.333% edge
33.333% of $75 is $25 -- now down $450, with $50 remaining [at this point, guaranteed ruin]

26 red / 38 total ~ 36.8% edge
36.8% of $50 is $18 ($15) -- now down $465 with $35 remaining

26 red / 37 total ~ 40.5% edge
40.5% of $35 is $14 ($10) -- now down $475, with $25 remaining.

26 red / 36 total ~ 44.44% edge
44.44% of $25 is $11.11 ($10) -- now down $485, with $15 remaining

26 red / 35 total ~ 48.5% edge
48.5% of $15 is $7 ($5) -- now down $490

bet $5 - now down $495

bet $5 -- now down $500


EDIT: Actually, you'd be betting MORE than what I wrote above, since the variance decreases.





RE: "The bonus" -- it really depends. How many people are you playing against? What's the value of the bonus relative to the stakes you're playing? If it's a $1M bonus and you're playing with a $500 BR, that's much different than getting a $5 bonus. Mid-game, can you change your strategy based on how others at the table are playing or winning/losing? If everyone else is going to bet $5 every round on either side at random (without information), then the best they can do is $52*$5 is $260. Using information and betting $5 until the last round, you can guarantee a profit of at least $490.

Or maybe you're in a situation like this: You're 20 cards in, 9 reds and 11 blacks have come out (ie: 17 reds remain, 15 blacks remain). Everyone has $400. Everyone else bets $400 on red (the favorite). What's the optimal bet? (Hint: $5 on black.)



It might seem like an easy question or like they are all the same: "What's the best way to guarantee a profit?" or "What's the best way to play this game to make money?" or "What if the winner gets a bonus?" as clearly...they are all VERY different questions.

Quote: MathExtremist

Right, but that's not a 50/50 proposition. If you want to guarantee a win, as has been said, just wait for the first 51 cards to come out and stake all your money on the value of the last card. The probability of a win is 100%, not 50%. That's the only way you guarantee anything.

Edge-sort. Demand that you do it twice, and have the dealer turn Red Cards one way and Black Cards the other. Use an automatic shuffler. Win 100% of the time.

I feel like this is being made too hard... Why wouldn't you just count wins?

bet minimums one color until you won 26 bets, then bet the bank on the other color after you've won 26. OR

if there is one card left and you are at 25 wins, bet the bank.

Worst case is small win, best case is multiple doubling of your bankroll.

I was thinking of a +4,-3 progression with different rules for consecutive wins and losses but I just realized testing the different first 51 outcomes is not as easy as I thought :(

Quote: stevenm0519

I was thinking of a +4,-3 progression with different rules for consecutive wins and losses but I just realized testing the different first 51 outcomes is not as easy as I thought :(

This is a scary statement... You DON'T BEAT CASINO GAMES WITH BETTING SYSTEMS. You must tract the house edge in a game that swings from side to side (such as your red/black deck example) and you must bet on the ADVANTAGE. Betting systems WILL NOT win unless your betting system is "bet more on color X when there's more of those left int he deck!"

Please take notice since you asked the question on page 1 how many people have chimed in with a "betting system" of some type where your bets have nothing to do with the actual cards removed?

Quote: stevenm0519

I was thinking of a +4,-3 progression with different rules for consecutive wins and losses but I just realized testing the different first 51 outcomes is not as easy as I thought :(

It's typically not super easy to figure this kind of stuff out...because it can branch out many different ways (ie: RBRBRBRB, RRBBRRBB, RBRRBRBBRBR....etc).


And I really hope you don't try/practice with this type of an example of a betting system based on wins & losses. Unfortunately, under these circumstances (the way the game is played and cards are removed), I think it'd be possible for someone to try to beat it using a betting system -- where they increase their winnings when on a losing streak -- which would (likely) end up being profitable for the player due to the removal of the cards and shifting advantage.

IOW: If you played this game and used a betting system, you'd probably win but not because of the betting system, but because the game is flawed.

Quote: RS

IOW: If you played this game and used a betting system, you'd probably win but not because of the betting system, but because the game is flawed.

Actually, a Martingale on black can guarantee a win in a game like this one without independent trials. All you'd have to do is bet a Martingale on black until you've won 26 units, then stop betting because there are no more black cards to win with. Regardless of the permutation of the cards in the deck, you'll always win 26 units.

But the reason you win is because the game doesn't use independent trials, not because the odds (off the top) are 50/50. If you wrote "red" and "black" on two sides of a coin and flipped it 52 times, the same Martingale on black couldn't guarantee a win. Selection without replacement is required.

Quote: stevenm0519

If you have a deck of cards with betting whether the next card will be red or black. What is the best way to maximize profit on a $5 minimum $500 max with a $500 bankroll?

There is no way to maximize winnings on a fair bet. You are equally likely to win as lose. Your EV is 0, so you would expect to break-even over the long term

Anyone know how to calculate the different wlwlw combinations for 52 bets?

Yes.
Define your calling strategy.

Quote: stevenm0519

Anyone know how to calculate the different wlwlw combinations for 52 bets?

There are 52! / (26! x 26!) = 495,918,532,948,104 different ways.
If you mean, how do you list them all out, I can't think of any way other than brute force - for example, let the black cards be in positions:
1, 2, 3, 4, ..., 24, 25, 26
1, 2, 3, 4, ..., 24, 25, 27
1, 2, 3, 4, ..., 24, 25, 28
...
1, 2, 3, 4, ..., 24, 25, 51
1, 2, 3, 4, ..., 24, 25, 52
1, 2, 3, 4, ..., 24, 26, 27
1, 2, 3, 4, ..., 24, 26, 28
...
1, 2, 3, 4, ..., 24, 26, 52
and so on, until you finally get to
27, 28, 29, ..., 51, 52
Actually, you only need to look at the ones that begin with 1; each one has a "partner" consisting of the other 26 numbers. (Every ordering has a "partner" ordering where the black and red cards are reversed; exactly one ordering in each pair has the desired color as card 1.)