Not Really Poker, But Definitely Gambling

[Bradley K. Sherman]

The cards in a standard, well-shuffled, 52-card deck will
be turned face-up, one at a time.

At some point during the deal you must say "red". If
the next card dealt is red you win $100. If black, you
lose $100.

Play, or don't play?

--bks

(If you say nothing, it will be assumed you said "red"
just before the last card is dealt.)

[Tim Norfolk]

It's really an even money proposition.

[Bradley K. Sherman]

Tim Norfolk <tims...@aol.com> wrote:
>On Sunday, April 19, 2015 at 12:37:20 PM UTC-4, Bradley K. Sherman wrote:
>> The cards in a standard, well-shuffled, 52-card deck will
>> be turned face-up, one at a time.
>>
>> At some point during the deal you must say "red". If
>> the next card dealt is red you win $100. If black, you
>> lose $100.
>>
>> Play, or don't play?
>

>It's really an even money proposition.

Answer/Spoiler below:


Correct, it is even money. Perhaps the easiest way to see this
is to start with a two-card deck consisting of one black and one
red card. It should be obvious that there is no way to gain an
advantage in that game. Then move to four cards, two red, two
black ...

--bks

[risky biz]

How is it even money if, for example, the first five cards dealt are black? Or, if the first five cards dealt are red? It isn't even money unless the count is even. I will almost always have the advantage because I choose when to bet. The one exception is if I haven't bet until the last card and the last card is black.

[da pickle]

I must take this opportunity to agree with your comment. Tim and bks,
when they grow up and visit drinking establishments, may fall prey to
many proposition bets that they believe they understand. As it is, they
will have to stick to the annual carnival cheats.

[Bradley K. Sherman]

risky biz <swing...@gmail.com> wrote:
>On Sunday, April 19, 2015 at 6:40:03 PM UTC-7, Tim Norfolk wrote:
>> On Sunday, April 19, 2015 at 12:37:20 PM UTC-4, Bradley K. Sherman wrote:
>> > The cards in a standard, well-shuffled, 52-card deck will
>> > be turned face-up, one at a time.
>> >
>> > At some point during the deal you must say "red". If
>> > the next card dealt is red you win $100. If black, you
>> > lose $100.
>> >
>> > Play, or don't play?
>> >

>> > (If you say nothing, it will be assumed you said "red"
>> > just before the last card is dealt.)
>>
>> It's really an even money proposition.
>
>How is it even money if, for example, the first five cards dealt are
>black?

Because *before you play* the chances of the first five cards
being black are the same as the chances of the first five cards
being red. The advantage gained in the first instance is the
same as the advantage lost in the second.

--bks

[Clave]

"da pickle" <jcpi...@nospam.hotmail.com> wrote in message
news:97Sdnaleg_eqQ6nI...@giganews.com...

> proposition bets that they believe they understand...

Oh, irony...

[Paul Popinjay]

On Monday, April 20, 2015 at 3:17:23 AM UTC-7, risky biz wrote:



> I will almost always have the advantage because I choose when to bet. The one exception is if I haven't bet until the last card and the last card is black.

Just curious, when would you wait until the last card?

If the first 25 cards were red, when would you make your bet?

[risky biz]

On Monday, April 20, 2015 at 6:06:48 AM UTC-7, Bradley K. Sherman wrote:

Granted- if most cards dealt at the front end of the deck are red and you've already bought into a commitment to bet on the last card then you probably bet a loser.

But I would play this all day because I have faith in the variation of the rate that red and black are dealt and the fact that I have the right of action at any time during the first 51 cards. I'm not that conversant with probablity calculations but I think the simplistic analysis you're using doesn't take enough account of those features of the game.

On the other hand, if there could only be ONE game- then maybe not.

[risky biz]

I would have bet it long before that just to get to the next deck. Assuming there was going to be more than one deck dealt.

[ramashiva]

On Monday, April 20, 2015 at 10:15:45 AM UTC-7, risky biz wrote:

> But I would play this all day

If you want to pay $11 to win $10, I will book your action all day.

> I'm not that conversant with probablity calculations

Thanks, Captain Obvious.

Do you agree you don't have an edge with a 2-card deck?

How about a four-card deck? Do you think you have an edge?

At what point do you imagine that adding a red card and a black card gives you an edge?


William Coleman (ramashiva)

[Paul Popinjay]

On Monday, April 20, 2015 at 10:16:52 AM UTC-7, risky biz wrote:



>
> I would have bet it long before that just to get to the next deck. Assuming there was going to be more than one deck dealt.

Well fucking humor me then. I asked two legitimate questions. Legitimate in my mind anyway.

[Paul Popinjay]

I had a machine game, not too long ago, where after winning a hand I would be offered a chance to double. One day I figured out that the card offered me to choose whether it was high or low was not from a freshly shuffled deck, but rather it was comprised of the cards leftover from my winning hand. In other words, say if it were blackjack, and I had KQ against the dealer's J9, I would double and bet "low", there actually being four less high cards offered. Sadly, this opportunity was lost, probably partly due to how much I abused them over the long run. Also, if it was Joker Poker, the Joker counted as a "low" card, so there was always an additional low card available if it had not been in my original hand. Pays to be observant. But I actually played it for a significant time without noticing.

[da pickle]

Twenty six red cards and twenty six black cards ... just like a roulette
wheel with 52 spots and no green. So the deck is just like a roulette
wheel ... a string of black does nothing as to the next spin because
there are still 26 and 26 ... right.

No, wrong ... the deck is not a roulette wheel ... the cards are placed
face up and not replaced in the deck and NOT shuffled again each time
another card is turned.

When 5 more blacks than reds have come up, those blacks are GONE ... the
chances of a red have increased.

On the 52 spot roulette wheel ... 26 blacks in a row says nothing about
the chance of a red or a black on the next spin.

In a deck of cards, 26 blacks in a row gives really good odds on the
next card being RED.

[ramashiva, the two card deck (one black and one red) has a 50/50 chance
of the first card being either red or black ... but if the first card is
turned over is BLACK ... I know the color of the second card.]

[Bradley K. Sherman]

da pickle <jcpi...@nospam.hotmail.com> wrote:
> ...

>[ramashiva, the two card deck (one black and one red) has a 50/50 chance
>of the first card being either red or black ... but if the first card is
>turned over is BLACK ... I know the color of the second card.]

And if the first card is red, you know that the last card is
black and you are a *guaranteed loser*.

BR win $100
RB lose $100

--bks

[risky biz]

On Monday, April 20, 2015 at 10:29:21 AM UTC-7, ramashiva wrote:

> On Monday, April 20, 2015 at 10:15:45 AM UTC-7, risky biz wrote:
>
> > But I would play this all day
>
> If you want to pay $11 to win $10, I will book your action all day.

You haven't established that any such odds exist but you apparently think that sounded good. Bradly already stated that it's even, based on HIS assumption of only one deck being played. Show your calculation that produces me giving anyone 11 to 10.

> > I'm not that conversant with probablity calculations
>
> Thanks, Captain Obvious.

If it's so obvious to you then you should be able to explain why blackjack can be beat even though the odds are "against" an accomplished player. I'm listening for this and your previous calculation, also.

> Do you agree you don't have an edge with a 2-card deck?
>
> How about a four-card deck? Do you think you have an edge?
>
> At what point do you imagine that adding a red card and a black card gives you an edge?
>
>
> William Coleman (ramashiva)

No cards have to be added. I just have to bet at the right time. If I'm able to make two bets during a single deck when I have the best of it it erases my theoretical disadvantage of the forced bet after the 51st card is desalt.

You're betting that 100,000 decks dealt will not have enough variation to allow me to do that often enough to take your money. I will take that bet all day.

These typical probability quizzes that produce so many "wrong" answers are highly dependent for those wrong answers on a presentation of the problem which leaves out key information. Left out of this one was how many times the games could be played. Bradley is assuming only once absent the presentation stating that. I'm assuming numerous decks played and my assumption is just as legitimate as his since assumptions have to be applied to an inadequately stated problem.

[Bradley K. Sherman]

risky biz <swing...@gmail.com> wrote:
>
>No cards have to be added. I just have to bet at the right time. If I'm able to make two bets during a single deck when I have the
>best of it it erases my theoretical disadvantage of the forced bet after the 51st card is desalt.
>

You only get to bet once per shuffle.

--bks

[risky biz]

On Monday, April 20, 2015 at 9:18:37 AM UTC-7, Paul Popinjay wrote:

I can humor you but it's the same answer. I would probably bet sometime between the first 8 - 11 consecutive blacks simply because the odds would be so stacked against me that the odds of a favorable variation wouldn't be worth the time invested to wait for it. At the 25th card there's zero positive variation for me so there would be no point in waiting longer. Bet.

[risky biz]

After the first card no one knows the what the second card will be or what the last card will be.

What I can know is when the count is in my favor and the odds of that happening several times before the 51st card are high enough for me to book consistent winnings. Because it isn't going to be just one deck dealt.

[risky biz]

On Monday, April 20, 2015 at 12:43:15 PM UTC-7, Bradley K. Sherman wrote:

That's correct, my error. But I still have 51 cards to produce favorable variation. My opponent has only the last card. Because if I'm required to bet the last card so is he. He has the worst of it.

[Bradley K. Sherman]

risky biz <swing...@gmail.com> wrote:
>On Monday, April 20, 2015 at 12:06:24 PM UTC-7, Bradley K. Sherman wrote:
>> da pickle <jcpi...@nospam.hotmail.com> wrote:
>> > ...
>> >[ramashiva, the two card deck (one black and one red) has a 50/50 chance
>> >of the first card being either red or black ... but if the first card is
>> >turned over is BLACK ... I know the color of the second card.]
>>
>> And if the first card is red, you know that the last card is
>> black and you are a *guaranteed loser*.
>>
>> BR win $100
>> RB lose $100
>

>After the first card no one knows the what the second card will be or
>what the last card will be.

I don't think you understand the game. In a two-card game,
*everyone* knows what the second card will be after the
first card is turned over.

--bks

[risky biz]

On Monday, April 20, 2015 at 12:53:36 PM UTC-7, Bradley K. Sherman wrote:

I wasn't referring to the two card hand of someone else. I was referring to the OP problem as stated.

[Bradley K. Sherman]

risky biz <swing...@gmail.com> wrote:
>On Monday, April 20, 2015 at 12:43:15 PM UTC-7, Bradley K. Sherman wrote:
>> risky biz wrote:
>> >
>> >No cards have to be added. I just have to bet at the right time. If
>I'm able to make two bets during a single deck when I have the
>> >best of it it erases my theoretical disadvantage of the forced bet
>after the 51st card is desalt.
>> >
>>
>> You only get to bet once per shuffle.
>

>That's correct, my error. But I still have 51 cards to produce favorable
>variation. My opponent has only the last card. Because if I'm required
>to bet the last card so is he. He has the worst of it.

But you will get "unfavorable variation" just as often as "favorable
variation". There is no guarantee that there will be more red than
black cards remaining in the deck at any time.

--bks

[ramashiva]

On Monday, April 20, 2015 at 12:34:04 PM UTC-7, risky biz wrote:

> On Monday, April 20, 2015 at 10:29:21 AM UTC-7, ramashiva wrote:

> > On Monday, April 20, 2015 at 10:15:45 AM UTC-7, risky biz wrote:

> > > But I would play this all day

> > If you want to pay $11 to win $10, I will book your action all day.

> You haven't established that any such odds exist but you apparently think that sounded good. Bradly already stated that it's even, based on HIS assumption of only one deck being played. Show your calculation that produces me giving anyone 11 to 10.

It's an even money gamble. No strategy exists which will give you an edge. You and Pickle are demonstrating that you are both innumerate.

I picked 11 to 10 because that gives me a healthy edge, but as long as you bet more than you can win, I have an edge.

> > > I'm not that conversant with probablity calculations

> > Thanks, Captain Obvious.

> If it's so obvious to you then you should be able to explain why blackjack can be beat even though the odds are "against" an accomplished player. I'm listening for this and your previous calculation, also.

The odds are NOT against an accomplished player with normal rules and a single deck. That is an established mathematical fact.

> > Do you agree you don't have an edge with a 2-card deck?

> > How about a four-card deck? Do you think you have an edge?

> > At what point do you imagine that adding a red card and a black card gives you an edge?

> No cards have to be added. I just have to bet at the right time.

There is no "right" time to bet which will give you an edge.

> If I'm able to make two bets during a single deck when I have the best of it

You only get one bet per deck.

> it erases my theoretical disadvantage of the forced bet after the 51st card is desalt.

There is no theoretical disadvantage to the forced bet after the 51st card.

> You're betting that 100,000 decks dealt will not have enough variation to allow me to do that often enough to take your money. I will take that bet all day.

The number of times the game is played is irrelevant. It's an even gamble at the start of every deck.

> These typical probability quizzes that produce so many "wrong" answers are highly dependent for those wrong answers on a presentation of the problem which leaves out key information. Left out of this one was how many times the games could be played. Bradley is assuming only once absent the presentation stating that. I'm assuming numerous decks played and my assumption is just as legitimate as his since assumptions have to be applied to an inadequately stated problem.

Again, the number of games played is irrelevant. It's even money every time.


William Coleman (ramashiva)

[fffurken]

On Monday, April 20, 2015 at 8:53:36 PM UTC+1, risky biz wrote:
> On Monday, April 20, 2015 at 12:43:15 PM UTC-7, Bradley K. Sherman wrote:
> > risky biz wrote:
> > >
> > >No cards have to be added. I just have to bet at the right time. If I'm able to make two bets during a single deck when I have the
> > >best of it it erases my theoretical disadvantage of the forced bet after the 51st card is desalt.
> > >
> >
> > You only get to bet once per shuffle.
> >
> > --bks
>
> That's correct, my error. But I still have 51 cards to produce favorable variation.

lol

I told you.

Reza Risky:
http://tinyurl.com/meo3px2

[risky biz]

If twelve black cards have been dealt, only 3 red cards have been dealt, I can choose to bet now, and you've ALREADY faded my action in advance I don't have an edge?

Are you and ramashiva sure you want to say that?

I don't have to bet when the variation is unfavorable to me. You bet in advance when the variation is unfavorable.

It's OK for you and ramashiva to start getting it anytime now.

[Bradley K. Sherman]

risky biz <swing...@gmail.com> wrote:
> ...

>If twelve black cards have been dealt, only 3 red cards have been dealt,
>I can choose to bet now, and you've ALREADY faded my action in advance I
>don't have an edge?

> ...

No one is saying that you can't find a situation with a positive
expectation, just that you aren't guaranteed to find one but you
have to say "red" sometime even so. Have you accepted the fact
that in a two-card game it's an even-money proposition? Four
cards? Six?

--bks

[risky biz]

On Monday, April 20, 2015 at 1:10:47 PM UTC-7, ramashiva wrote:
> On Monday, April 20, 2015 at 12:34:04 PM UTC-7, risky biz wrote:
>
> > On Monday, April 20, 2015 at 10:29:21 AM UTC-7, ramashiva wrote:
>
> > > On Monday, April 20, 2015 at 10:15:45 AM UTC-7, risky biz wrote:
>
> > > > But I would play this all day
>
> > > If you want to pay $11 to win $10, I will book your action all day.
>
> > You haven't established that any such odds exist but you apparently think that sounded good. Bradly already stated that it's even, based on HIS assumption of only one deck being played. Show your calculation that produces me giving anyone 11 to 10.
>
> It's an even money gamble. No strategy exists which will give you an edge. You and Pickle are demonstrating that you are both innumerate.

No. You seem to have a mental block.

> I picked 11 to 10 because that gives me a healthy edge, but as long as you bet more than you can win, I have an edge.
>
> > > > I'm not that conversant with probablity calculations
>
> > > Thanks, Captain Obvious.
>
> > If it's so obvious to you then you should be able to explain why blackjack can be beat even though the odds are "against" an accomplished player. I'm listening for this and your previous calculation, also.
>
> The odds are NOT against an accomplished player with normal rules and a single deck. That is an established mathematical fact.

Do you think there is some reason I can't count cards in this game as easily as I do in blackjack? The wonderful advantage here over blackjack is that the house doesn't refuse my action. They book it in advance and swear they'll play me as long as I want. Assuming that you're the house, that is.

> > > Do you agree you don't have an edge with a 2-card deck?
>
> > > How about a four-card deck? Do you think you have an edge?
>
> > > At what point do you imagine that adding a red card and a black card gives you an edge?
>
> > No cards have to be added. I just have to bet at the right time.
>
> There is no "right" time to bet which will give you an edge.
>
> > If I'm able to make two bets during a single deck when I have the best of it
>
> You only get one bet per deck.

I already corrected that. But I still have the best of it.

> > it erases my theoretical disadvantage of the forced bet after the 51st card is desalt.
>
> There is no theoretical disadvantage to the forced bet after the 51st card.
>
> > You're betting that 100,000 decks dealt will not have enough variation to allow me to do that often enough to take your money. I will take that bet all day.
>
> The number of times the game is played is irrelevant. It's an even gamble at the start of every deck.

Only if you assume there will be no variation between red/black and that would be a terrible bet to make.

> > These typical probability quizzes that produce so many "wrong" answers are highly dependent for those wrong answers on a presentation of the problem which leaves out key information. Left out of this one was how many times the games could be played. Bradley is assuming only once absent the presentation stating that. I'm assuming numerous decks played and my assumption is just as legitimate as his since assumptions have to be applied to an inadequately stated problem.
>
> Again, the number of games played is irrelevant. It's even money every time.

Only if no cards are ever turned over. In other words, theoretically. But no one gambles theoretically. You said you would take my action all day. That isn't theoretical. It's real and reality is variation and I have the choice of when to bet.

How can you recognize that a counter in blackjack can have an edge because of variation but not in this game? That's completely contradictory reasoning.

[risky biz]

On Monday, April 20, 2015 at 2:59:22 PM UTC-7, Bradley K. Sherman wrote:

A two card game is self-completing in the problem as presented because I'm forced to bet on the second card.

If it's a six card game, you're my opponent, and the first two cards are black there are four cards left- one black and three red. I can choose to bet now rather than wait to bet on the last card but you have already bet on the last card. You think you have the advantage?

Like I said- all day.

[Clave]

"risky biz" <swing...@gmail.com> wrote in message
news:28972cd7-90f2-48a3...@googlegroups.com...

<...>

> If it's a six card game, you're my opponent, and the first two cards are
> black there are four cards left- one black and three red. I can choose to
> bet now rather than wait to bet on the last card but you have already bet
> on the last card. You think you have the advantage?

But you only have a 1:5 chance of that scenario happening.

[ramashiva]

On Monday, April 20, 2015 at 3:10:00 PM UTC-7, risky biz wrote:

> If it's a six card game, you're my opponent, and the first two cards are black there are four cards left- one black and three red.

What you don't seem to grasp is that every situation that is favorable to you is offset by another situation which is equally likely and equally unfavorable.

In this case, the probability that the first two cards are black is exactly the same as the probability that the first two cards are red.

> I can choose to bet now rather than wait to bet on the last card but you have already bet on the last card.

I am not betting on the last card. I am betting that the card you choose will be black.

> You think you have the advantage?

If you bet more than you win, like $11 to win $10, I KNOW I have the advantage.

> Like I said- all day.

Like I said, I will book your action all day as long as you bet more than you win.

Here is a basic fact about this game which may help you to understand why the game is even money --

If there are r red cards remaining and b black cards remaining, then the probability that the next card is red is --

r/(r+b)


William Coleman (ramashiva)

[Clave]

"Bradley K. Sherman" <b...@panix.com> wrote in message
news:mh0ljs$cj7$1...@panix2.panix.com...

> The cards in a standard, well-shuffled, 52-card deck will
> be turned face-up, one at a time.
>
> At some point during the deal you must say "red". If
> the next card dealt is red you win $100. If black, you
> lose $100.
>
> Play, or don't play?

A strange game...

[Paul Popinjay]

On Monday, April 20, 2015 at 11:53:17 AM UTC-7, da pickle wrote:



>
> Twenty six red cards and twenty six black cards ... just like a roulette
> wheel with 52 spots and no green. So the deck is just like a roulette
> wheel ... a string of black does nothing as to the next spin because
> there are still 26 and 26 ... right.
>

What are you babbling about? Does this have anything to do with what I asked? I can only assume this was posted by "Dumb Pickle", but I haven't checked the headers.

[Paul Popinjay]

On Monday, April 20, 2015 at 12:47:00 PM UTC-7, risky biz wrote:



>
> I can humor you but it's the same answer. I would probably bet sometime between the first 8 - 11 consecutive blacks simply because the odds would be so stacked against me that the odds of a favorable variation wouldn't be worth the time invested to wait for it. At the 25th card there's zero positive variation for me so there would be no point in waiting longer. Bet.

Uhh, I asked two questions. Two separate questions. Maybe I should have asked them in separate posts, but I'm trying to cut down on content-free ones.

[Paul Popinjay]

On Monday, April 20, 2015 at 3:28:56 PM UTC-7, ramashiva wrote:



>
> If you bet more than you win, like $11 to win $10, I KNOW I have the advantage.
>

I'm selling ten dollar bills on EBay for $9. Doing quite well.

[risky biz]

You have to bet regardless of what the odds are on any particular turn of a card if I CHOOSE to. That's because you agreed to do so before the first card was turned. We both agreed that there would be a bet on the last card if I chose to let it go that far. On that bet we're even odds BEFORE the cards are turned. But I can escape the last card bet at any time during the first 51 cards. You CAN'T. I have the best of it.

In short- you have committed yourself to an exploitable strategy. I can exploit you every time there are more red cards in the deck than black cards. Given your exploitable strategy there is no way I don't have the advantage. The worst I could do is witness card turns which are incredibly consistently red/black, red/black. Then I would have even money on the last card. You are deluded if you think there won't be enough variation toward a red rich deck for me to have the best of it. I will take this action all day long.

[risky biz]

On Monday, April 20, 2015 at 3:28:56 PM UTC-7, ramashiva wrote:

> On Monday, April 20, 2015 at 3:10:00 PM UTC-7, risky biz wrote:
>
> > If it's a six card game, you're my opponent, and the first two cards are black there are four cards left- one black and three red.
>
> What you don't seem to grasp is that every situation that is favorable to you is offset by another situation which is equally likely and equally unfavorable.

What you don't grasp is that I can bet the favorable situations and ignore the unfavorable situations. If I never get a favorable situation (likelihood: zero) the worst I have is even odds on the last card.

> In this case, the probability that the first two cards are black is exactly the same as the probability that the first two cards are red.

Then I decline to bet and get even money lasst card while I wait for a deck where the first two cards are red. You think you're even money with me like that? No cigar.

> > I can choose to bet now rather than wait to bet on the last card but you have already bet on the last card.
>
> I am not betting on the last card. I am betting that the card you choose will be black.

And what are the odds of that happening when I know most of the cards in the deck are red? Can you possibly be saying that more red cards in the deck when I choose to bet the next card is red is even money?

And, yes you are betting on the last card and also any other card when I CHOOSE to make you bet. You have no choice. You bet up front with an exploitable strategy.

> > You think you have the advantage?
>
> If you bet more than you win, like $11 to win $10, I KNOW I have the advantage.

I know if you bet more than YOU win, like $500 to win $10, I KNOW I have the advantage. Which has zero to do with what we're discussing. I don't know why you keep repeating that.

> > Like I said- all day.
>
> Like I said, I will book your action all day as long as you bet more than you win.
>
> Here is a basic fact about this game which may help you to understand why the game is even money --
>
> If there are r red cards remaining and b black cards remaining, then the probability that the next card is red is --
>
> r/(r+b)
>
>
> William Coleman (ramashiva)

Do your own math and figure it out. If there are more red cards than black cards I have the advantage and I can choose to bet that advantage. If I don't get the advantage in a particular deck I have even money on the last card. What else could I ask for?

All day.

[Clave]

"risky biz" <swing...@gmail.com> wrote in message

news:dc478f10-d13a-437f...@googlegroups.com...

> On Monday, April 20, 2015 at 3:19:16 PM UTC-7, Clave wrote:
>> "risky biz" <swing...@gmail.com> wrote in message
>> news:28972cd7-90f2-48a3...@googlegroups.com...
>>
>> <...>
>>
>> > If it's a six card game, you're my opponent, and the first two cards
>> > are
>> > black there are four cards left- one black and three red. I can choose
>> > to
>> > bet now rather than wait to bet on the last card but you have already
>> > bet
>> > on the last card. You think you have the advantage?
>>
>> But you only have a 1:5 chance of that scenario happening.
>
> You have to bet regardless of what the odds are on any particular turn of
> a card if I CHOOSE to. That's because you agreed to do so before the first
> card was turned. We both agreed that there would be a bet on the last card
> if I chose to let it go that far. On that bet we're even odds BEFORE the
> cards are turned. But I can escape the last card bet at any time during
> the first 51 cards. You CAN'T. I have the best of it.
>
> In short- you have committed yourself to an exploitable strategy. I can
> exploit you every time there are more red cards in the deck than black
> cards.

But that's only going to happen half the time.

[risky biz]

How can they be separated? What I do depends on what's been dealt.

I answered both questions. Twice.

[Tim Norfolk]

I believe that the problem is you have the impression that you can choose to play or not during the course of play. That is not the case, in the problem expressed.

[risky biz]

I wish I could find one.

[Tim Norfolk]

Read the original post. You have a choice on 'when', not 'what'.

[risky biz]

I do not have that impression. I have the best of it by the rules as stated in the OP.

"At some point during the deal you must say "red". If
the next card dealt is red you win $100. If black, you
lose $100."

If twenty blacks have been turned and five reds have been turned and I then say "red"- are the odds against me or against you?

[risky biz]

"What" is "when".

[Paul Popinjay]

On Monday, April 20, 2015 at 7:19:50 PM UTC-7, Clave wrote:



>
> But that's only going to happen half the time.

Y'know, for a guy who considers himself a semi-savvy troll, I'm astonished at how you allow Risky to hook the ever-loving fuck out of you.

[Pepe Papon]

On Mon, 20 Apr 2015 10:29:17 -0700 (PDT), ramashiva
<ramas...@gmail.com> wrote:

>On Monday, April 20, 2015 at 10:15:45 AM UTC-7, risky biz wrote:
>
>> But I would play this all day
>

>If you want to pay $11 to win $10, I will book your action all day.

>
>> I'm not that conversant with probablity calculations
>
>Thanks, Captain Obvious.
>

>Do you agree you don't have an edge with a 2-card deck?
>
>How about a four-card deck? Do you think you have an edge?
>
>At what point do you imagine that adding a red card and a black card gives you an edge?

With a 2-card or 4-card deck, if the first card is black, then the
majority of the remaining cards will always be red and if the first
card is red, then the majority of the remaining cards will always be
black. You can never have an edge.

With a 6-card deck, the majority of the remaining cards can vary as
more cards are turned over. If the first card is black, there are 3
red and 2 black remaining, so you have an edge and should bet at this
moment. If the first card is red, there are 3 black and 2 red, so
you wouldn't want to bet at this point.

But at that point, the next card will be black 3/5 of the time, and of
those times, the next card will be black half the time. At that
point, there will be 2 red and one black remaining, and you would have
an edge.

So, half the time, the first card will be black, and you will have an
edge. The other half of the time, you won't have an edge after the
first card, but you will have an edge later on 30% of the time.

You'll have an edge at some point during the deal 65% of the time.
Since you get to decide when to bet, you'll have an edge in the
majority of the deals.
--

Pepe Papon

[mo ron]

On Monday, April 20, 2015 at 8:18:44 PM UTC-6, risky biz wrote:

> What you don't grasp is that I can bet the favorable situations and ignore the unfavorable situations. If I never get a favorable situation (likelihood: zero) the worst I have is even odds on the last card.

this is where you're not understanding. if you haven't found a favorable situation to bet in this particular deck, you're far worse than even money on the end.

for every deck you get a favorable count to bet, there's another deck with the same unfavorable count. the odds you get one or the other are the same.

[mo ron]

pepe, why don't you keep your idiot mouth closed when we're discussing things that are way out of your depth?

[fffurken]

And break the habit of a lifetime? Hardly.

[mo ron]

On Tuesday, April 21, 2015 at 6:43:59 AM UTC-6, fffurken wrote:

> > pepe, why don't you keep your idiot mouth closed when we're discussing things that are way out of your depth?
>
> And break the habit of a lifetime? Hardly.

i bet he bit his tongue three times writing that bizarrely stupid post. maybe rama will ride to the rescue and tell us how smart seth is again ... pure comedy GOLD.

[Tim Norfolk]

That is not the issue. The problem is that, once you decide to play the game, then you only have the choice of when to say 'red'.

[risky biz]

On Tuesday, April 21, 2015 at 6:54:35 AM UTC-7, Tim Norfolk wrote:
> On Monday, April 20, 2015 at 10:34:49 PM UTC-4, risky biz wrote:
> > On Monday, April 20, 2015 at 7:22:28 PM UTC-7, Tim Norfolk wrote:
> > > On Monday, April 20, 2015 at 10:06:40 PM UTC-4, risky biz wrote:
> > > > On Monday, April 20, 2015 at 3:19:16 PM UTC-7, Clave wrote:

> > > > > "risky biz" wrote in message

> > > > > news:28972cd7-90f2-48a3...@googlegroups.com...
> > > > >
> > > > > <...>
> > > > >
> > > > > > If it's a six card game, you're my opponent, and the first two cards are
> > > > > > black there are four cards left- one black and three red. I can choose to
> > > > > > bet now rather than wait to bet on the last card but you have already bet
> > > > > > on the last card. You think you have the advantage?
> > > > >
> > > > > But you only have a 1:5 chance of that scenario happening.
> > > >
> > > > You have to bet regardless of what the odds are on any particular turn of a card if I CHOOSE to. That's because you agreed to do so before the first card was turned. We both agreed that there would be a bet on the last card if I chose to let it go that far. On that bet we're even odds BEFORE the cards are turned. But I can escape the last card bet at any time during the first 51 cards. You CAN'T. I have the best of it.
> > > >
> > > > In short- you have committed yourself to an exploitable strategy. I can exploit you every time there are more red cards in the deck than black cards. Given your exploitable strategy there is no way I don't have the advantage. The worst I could do is witness card turns which are incredibly consistently red/black, red/black. Then I would have even money on the last card. You are deluded if you think there won't be enough variation toward a red rich deck for me to have the best of it. I will take this action all day long.
> > >
> > > I believe that the problem is you have the impression that you can choose to play or not during the course of play. That is not the case, in the problem expressed.
> >
> > I do not have that impression. I have the best of it by the rules as stated in the OP.
> >
> > "At some point during the deal you must say "red". If
> > the next card dealt is red you win $100. If black, you
> > lose $100."
> >
> > If twenty blacks have been turned and five reds have been turned and I then say "red"- are the odds against me or against you?
>
> That is not the issue. The problem is that, once you decide to play the game, then you only have the choice of when to say 'red'.

That's all I need.

[risky biz]

First question: 52 card deck, 4 cards have been dealt, 3 black, 1 red. That leaves 23 black, 25 red. What's my percent edge if I say "red"?

Second question: How many times during a 52 card deck could I reasonably expect red to be +2?

Answer the question rather than just disappear from the thread.

[risky biz]

This is unbelievable, isn't it? It's a gambling newsgroup.

[mo ron]

On Tuesday, April 21, 2015 at 8:42:59 AM UTC-6, risky biz wrote:

> First question: 52 card deck, 4 cards have been dealt, 3 black, 1 red. That leaves 23 black, 25 red. What's my percent edge if I say "red"?

25 / (23 + 25)

> Second question: How many times during a 52 card deck could I reasonably expect red to be +2?

the same number as you could reasonably expect black to be +2.

> Answer the question rather than just disappear from the thread.

no.

[Tim Norfolk]

The edge is 25/(23+25)-23/(23+25)

The question about when you can expect this advantage is much more interesting, and I'm playing with it.

[Tim Norfolk]

On Sunday, April 19, 2015 at 12:37:20 PM UTC-4, Bradley K. Sherman wrote:
> The cards in a standard, well-shuffled, 52-card deck will
> be turned face-up, one at a time.
>

> At some point during the deal you must say "red". If
> the next card dealt is red you win $100. If black, you
> lose $100.
>

> Play, or don't play?
>

> --bks
>
> (If you say nothing, it will be assumed you said "red"
> just before the last card is dealt.)

I have to revise my answer, but I don't have the numbers quite yet.

Consider that the player wins if the last card is red, and they haven't yet decided. The probability of this is 1/2. If the player does nothing, the edge is 0. So, does there exist a strategy for which we can have a positive edge?

[Bradley K. Sherman]

You will have an edge of +1 at some point in about 96% of the deals.
You will have an edge of +2 at some point in about 85% of the deals.
You will have an edge of +3 at some point in about 71% of the deals.

(Simulated runs of 1,000,000 deals. Caveat: No one has checked my code.)

--bks

[Bradley K. Sherman]

Tim Norfolk <tims...@aol.com> wrote:
>On Sunday, April 19, 2015 at 12:37:20 PM UTC-4, Bradley K. Sherman wrote:
>> The cards in a standard, well-shuffled, 52-card deck will
>> be turned face-up, one at a time.
>>
>> At some point during the deal you must say "red". If
>> the next card dealt is red you win $100. If black, you
>> lose $100.
>>
>> Play, or don't play?
>>

>> (If you say nothing, it will be assumed you said "red"
>> just before the last card is dealt.)
>
>I have to revise my answer, but I don't have the numbers quite yet.
>
>Consider that the player wins if the last card is red, and they haven't
>yet decided. The probability of this is 1/2. If the player does nothing,
>the edge is 0. So, does there exist a strategy for which we can have a
>positive edge?

No, but if you think you have such a strategy, I can simulate it.

--bks

[mo ron]

On Tuesday, April 21, 2015 at 10:08:47 AM UTC-6, Tim Norfolk wrote:

> The edge is 25/(23+25)-23/(23+25)

the first time in my life i've been wrong.

> The question about when you can expect this advantage is much more interesting, and I'm playing with it.

you can never expect the advantage. as i said earlier, you'd expect the same disadvantage to be equally likely.

[Tim Norfolk]

I prefer doing it combinatorially, but here is a start for risky:

Take 4 cards. There are 6 equally likely cases. Suppose that we say 'red' when there is an advantage. Mark this with a '|':

RRBB lose
RBRB lose
RBB|R win
B|RRB win
B|RBR win
B|BRR lose

Expectation 0.

[Tim Norfolk]

As I said, the more interesting question is when you do have an edge, as in mid-shoe entry in BJ.

[mo ron]

On Tuesday, April 21, 2015 at 10:29:30 AM UTC-6, Tim Norfolk wrote:

> As I said, the more interesting question is when you do have an edge, as in mid-shoe entry in BJ.

how is that a more interesting question?

[mo ron]

On Tuesday, April 21, 2015 at 10:29:41 AM UTC-6, T.Bagger wrote:

> This thread is too long and boring. Anytime a puzzle or a prop bet appears to be simple and seems to be X. It's not X.

it seem interesting because on precursory reading it's difficult to understand the real problem. once you realize you're forced to play each deck, and you're forced to play only one way (red), it becomes trivially dumb.

[da pickle]

It may be a 50/50 game but the unevenness will vary as the game
proceeds. The problem as stated allows you to quit while you are ahead.
Just like a coin flip, the ahead and behind will vary.

Or is the position of the group that you will NEVER be ahead at any time?

[mo ron]

On Tuesday, April 21, 2015 at 11:09:53 AM UTC-6, da pickle wrote:

> > it seem interesting because on precursory reading it's difficult to understand the real problem. once you realize you're forced to play each deck, and you're forced to play only one way (red), it becomes trivially dumb.
>
> It may be a 50/50 game but the unevenness will vary as the game
> proceeds. The problem as stated allows you to quit while you are ahead.
> Just like a coin flip, the ahead and behind will vary.
>
> Or is the position of the group that you will NEVER be ahead at any time?

the position of the universe i believe is that you'll be ahead about as often as you'll be behind.

[da pickle]

That means I can quit while I am ahead ... doesn't it?

[da smart pickle]

And just like a coin flip there is no advantage. Or do you think it's possible to gain an edge coin flipping by choosing when to quit? Poor dumb pickle. Quit embarrassing me.

[T.Bagger]

On Tuesday, April 21, 2015 at 10:24:08 AM UTC-7, da pickle wrote:

So?

I'm re-posting the links along with the explanation that should make it easier to see why there is no advantage available in this proposition.

-----
So I have a well-shuffled deck of 52 cards. I turn them over one by one, starting at the top. You can raise your hand whenever you like. After you raise your hand, you win a prize if the next card I turn over is red.

Okay, you've just raised your hand. At this point I say to you: "Hey. Would you mind if I turn over the bottom card instead of the top one?". It clearly makes no difference whether you say yes or no. If, for example, I have 17 cards left, 12 red and 5 black, then you've got a 12/17 chance of winning, regardless of whether I turn over the top card, the bottom card, the middle card, or any other randomly chosen card.

In other words, we might as well change the rules of the game so that I deal from the top until you raise your hand, and then I deal one card from the bottom. What's your chance of winning that game? Regardless of your strategy, you win when the bottom card is red and lose when it's black -- so regardless of your strategy, you win with probability 50%.

In summary:

If I deal the last card from the bottom, your strategy doesn't matter.
If I deal the last card from the top, your chances of winning -- under any strategy -- are exactly the same as if I deal the last card from the bottom.
Therefore, when I deal the last card from the top (i.e. in the original game) your strategy doesn't matter.
-----

This puzzle along with 143 comments:
http://www.thebigquestions.com/2013/12/17/tuesday-puzzle-4/

The answer along with some more comments:
http://www.thebigquestions.com/2013/12/19/thursday-solution-2/

[Paul Popinjay]

On Tuesday, April 21, 2015 at 10:24:08 AM UTC-7, da pickle wrote:



>

> That means I can quit while I am ahead ... doesn't it?

Hey Dumb Pickle, please stop.

[Tim Norfolk]

I'm a Mathematician. I like to count things.

[da pickle]

I did not say that it was worth it, but do I ever get ahead in any coin
flipping contest or NOT? Just answer the question. A dollar a flip ...
do I EVER, ever, ever, ever get ahead a dollar?

[da pickle]

Does not answer the question ... we agree that it is a 50/50 game but we
play it many times ... apparently we will be ahead for a while and then
behind for a while and then ahead and then behind and the ahead and then
behind ... but we are ahead sometimes ... we can QUIT while we are ahead
... even when flipping coins.

[da pickle]

Even flipping coins, for a dollar a flip ... don't you get ahead some
times and then behind sometimes and then ahead some times during that
march to infinity? You stop!

[da pickle]

On Tuesday, April 21, 2015 at 11:57:01 AM UTC-7, da DUMB pickle wrote:

". . . do I ever get ahead in any coin flipping contest or NOT?"
YES!

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

. . . do I ever get ahead at blackjack or NOT?
YES!

. . . do I ever get ahead at roulette or NOT?
YES!

. . . do I ever get ahead craps or NOT?
YES!

. . . do I ever get ahead at baccarat or NOT?
YES!

. . . do I ever get ahead on slot machines or NOT?
YES!


OMFG, if dumb pickle's strategy gets out Las Vegas will be ruined in a matter of months. Why hasn't someone figured this out before now? Just get ahead then quit. So simple.

On Tuesday, April 21, 2015 at 11:57:01 AM UTC-7, da DUMB pickle wrote:

". . . do I ever get ahead in any coin flipping contest or NOT?"
YES!

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

. . . do I ever get ahead at blackjack or NOT?
YES!

. . . do I ever get ahead at roulette or NOT?
YES!

. . . do I ever get ahead craps or NOT?
YES!

. . . do I ever get ahead at baccarat or NOT?
YES!

. . . do I ever get ahead on slot machines or NOT?
YES!


OMFG, if dumb pickle's strategy gets out Las Vegas will be ruined in a matter of months. Why hasn't someone figured this out before now? Just get ahead then quit. So simple.

[da pickle]

Sorry for the repeat. Computer error.

John, don't let this become like your stubborn refusal to admit Harry Reid didn't lie. Just quit now.

[da pickle]

On 4/21/2015 2:20 PM, da FAUX pickle

is one sick puppy

[risky biz]

I don't have to say "red" when black is positive.

[risky biz]

You're assuming that the player will say "red" when black is positive. I have no idea why you're assuming that.

[risky biz]

That isn't the same problem. When a coin turns up heads it doesn't reduce the number of heads by one.

[Pepe Papon]

On Tue, 21 Apr 2015 05:20:07 -0700 (PDT), mo ron
<amontillad...@gmail.com> wrote:

>On Tuesday, April 21, 2015 at 12:27:36 AM UTC-6, Pepe Papon wrote:
>> On Mon, 20 Apr 2015 10:29:17 -0700 (PDT), ramashiva
>> <ramas...@gmail.com> wrote:
>>
>> >On Monday, April 20, 2015 at 10:15:45 AM UTC-7, risky biz wrote:
>> >
>> >> But I would play this all day
>> >
>> >If you want to pay $11 to win $10, I will book your action all day.
>> >
>> >> I'm not that conversant with probablity calculations
>> >
>> >Thanks, Captain Obvious.
>> >
>> >Do you agree you don't have an edge with a 2-card deck?
>> >
>> >How about a four-card deck? Do you think you have an edge?
>> >
>> >At what point do you imagine that adding a red card and a black card gives you an edge?
>>
>> With a 2-card or 4-card deck, if the first card is black, then the
>> majority of the remaining cards will always be red and if the first
>> card is red, then the majority of the remaining cards will always be
>> black. You can never have an edge.
>>
>> With a 6-card deck, the majority of the remaining cards can vary as
>> more cards are turned over. If the first card is black, there are 3
>> red and 2 black remaining, so you have an edge and should bet at this
>> moment. If the first card is red, there are 3 black and 2 red, so
>> you wouldn't want to bet at this point.
>>
>> But at that point, the next card will be black 3/5 of the time, and of
>> those times, the next card will be black half the time. At that
>> point, there will be 2 red and one black remaining, and you would have
>> an edge.
>>
>> So, half the time, the first card will be black, and you will have an
>> edge. The other half of the time, you won't have an edge after the
>> first card, but you will have an edge later on 30% of the time.
>>
>> You'll have an edge at some point during the deal 65% of the time.
>> Since you get to decide when to bet, you'll have an edge in the
>> majority of the deals.
>
>pepe, why don't you keep your idiot mouth closed when we're discussing things that are way out of your depth?

Speaking of depth this is as deep as your insight ever gets. Now,
do you care to post some actual substance and show where my reasoning
is wrong?
--

Pepe Papon

[Pepe Papon]

On Tue, 21 Apr 2015 09:25:47 -0700 (PDT), Tim Norfolk

Now, try it with 6 cards.
--

Pepe Papon

[risky biz]

If more black cards have been dealt it's less likely that the bottom card is red just as it's less likely that the top card is red. Try 'The Big Picture' next time.

[risky biz]

Isn't he referring to being ahead in odds? i.e., the deck stops when you call "red"? I don't think he's referring to be ahead in winnings. The coin example was just to illustrate that there is variation in coin flips just as there is in the red/black game.

[Tim Norfolk]

Because that is the only time when you have a positive edge?

[Will in New Haven]

This is not untrue but the ability to quit when ahead does not give an advantage unless your opponent cannot quit when _he_ is ahead and there is no chance that you will go broke before you are ever ahead.

The conditions of this game, or any game at a public casino, indicate that your opponent won't quit on you while you are behind but the chance of a long losing streak and not being able to continue exists in all games at non-trivial stakes.

There is no analysis of this game that makes it worth playing, let alone giving eleven to ten.

--
Will in New Haven

[da pickle]

It is not only not untrue, it is true. In the "game" proposed, only
"you" have the power to quit. There is no "bankroll" limit as a
condition.

> The conditions of this game, or any game at a public casino, indicate that your opponent won't quit on you while you are behind but the chance of a long losing streak and not being able to continue exists in all games at non-trivial stakes.

But it is NOT a condition in the game under discussion.

> There is no analysis of this game that makes it worth playing, let alone giving eleven to ten.

Oh this, we agree again. As I said, it is not worth playing ... but if
you have the power to quit when you are ahead, you can win.

[da pickle]

On 4/19/2015 11:37 AM, Bradley K. Sherman wrote:
> The cards in a standard, well-shuffled, 52-card deck will
> be turned face-up, one at a time.
>
> At some point during the deal you must say "red". If
> the next card dealt is red you win $100. If black, you
> lose $100.
>
> Play, or don't play?
>

> --bks

>
> (If you say nothing, it will be assumed you said "red"
> just before the last card is dealt.)

Before you deal the first card, I say red ... if it is red, I take my
money and go home.

If the card is black, we reshuffle and I say red for the first card.
When I finally get ahead $100, I quit. End of game.

Stupid game and I need a lot of money to win your $100, but I will
before we age too much. [Note: it will always be an "odd" number of
trials. We may get even at some point, but we do not quit. You may get
ahead a LOT of money along the way, but you cannot quit. And it does
not matter whether we use all 52 cards ... two will do.]

[Bradley K. Sherman]

da pickle <jcpi...@nospam.hotmail.com> wrote:
>On 4/19/2015 11:37 AM, Bradley K. Sherman wrote:
>> The cards in a standard, well-shuffled, 52-card deck will
>> be turned face-up, one at a time.
>>
>> At some point during the deal you must say "red". If
>> the next card dealt is red you win $100. If black, you
>> lose $100.
>>
>> Play, or don't play?
>>

>> (If you say nothing, it will be assumed you said "red"
>> just before the last card is dealt.)
>
>Before you deal the first card, I say red ... if it is red, I take my
>money and go home.

If it is black. I take your money and go home. You lose.

--bks

[Paul Popinjay]

On Wednesday, April 22, 2015 at 10:02:29 AM UTC-7, da pickle wrote:



>
> If the card is black, we reshuffle and I say red for the first card.
> When I finally get ahead $100, I quit. End of game.
>

What do you mean "when you finally get ahead"? That's bullshit. That's not the way I read it.

[Tim Norfolk]

And you read it correctly. There was nothing about repeat trials. Nonetheless, it appears that the expectation is exactly 0.

[Paul Popinjay]

Even if there is repeat trials, what guarantees that he will EVER be ahead?

[da pickle]

Ah, changing the rules again, I see. Why not stick with what you
started with?

[da pickle]

The player will eventually "get ahead" ... it will be an odd trial ...
the first, the third or on of the many odd trials ahead.

black, red, black, red, black, red, "even so far" ... one red will do it
on this sequence. Others might take many more trials, all seeking to
make the blacks and the reds come out even. Either the player or the
opponent will be "ahead" after the first trial. If the player is ahead
on number one, he quits. If the player is ahead on number three, he
quits. If it takes 51 trials (or it may be a Fibonacci number
sequence), eventually (before infinity), the player will get ahead, he
quits.

Is it worth it, no ... but it is so.

That is why bks has added that the opponent can quit and that you have a
limited bankroll.

[T.Bagger]

On Wednesday, April 22, 2015 at 10:02:29 AM UTC-7, da pickle wrote:

I don't think it's all that unusual for someone to think like you, but it does seem odd that someone in a gambling newsgroup would be so uninformed. Of course you could win but you could also lose. On some occasions you would get so far behind it would be clear you would never ever recoup. When you do win it's only $100. And on those occasions when things go bad you will lose thousands before giving up.

There is no way to gain an edge in any gambling game by "quit timing". Your whole argument is incredibly silly.

[da pickle]

Well, he is correct ... if you assume only one trial (which bks has not
yet adopted as what he meant, but he will in a minute) there is no
positive expectation for that one trial.

One wonders why bks started with that limited bankroll talk if he was
only talking about a one trial case? Humm

What do you, Paul, think are the chances of getting on an infinite line
of trials? There need only be one of those. Ups and downs are likely
to be much more likely "in the real world" even quite short of infinity.

[da pickle]

But you have an unlimited amount of money (and apparently time) ... so
you just have to wait it out.

I have always said it was not worth it, but that is not part of the game
either.

If you quit behind, you lose. That is not part of the answer.

[Paul Popinjay]

On Wednesday, April 22, 2015 at 1:15:48 PM UTC-7, da pickle wrote:



>
> What do you, Paul, think are the chances of getting on an infinite line
> of trials? There need only be one of those. Ups and downs are likely
> to be much more likely "in the real world" even quite short of infinity.

Pickle, old pal, do I have the book for YOU! Sklansky on Martingale Theory.

I have it in hardback, first edition, very rare, I'll let you have it for $100. But pay me before you play Bradley's card game.

[Bradley K. Sherman]

da pickle <jcpi...@nospam.hotmail.com> wrote:
> ...

>One wonders why bks started with that limited bankroll talk if he was
>only talking about a one trial case? Humm

> ...

Because you went off on a tangent and I was trying to educate you.
Peals before swine.

--bks