A Much Simpler Version Of The \"Blackjack Paradox\"
 

It wasn't meant to be about black jack. So here is an extremely trivial example of the concept I was trying to illustrate.

For some reason a casino decides to offer a dead even simplistic table game. They shuffle a single deck and deal out one card. Players can bet whether it is red or black. That's all there is to it. Reshuffle after every round.

Two players at the table have found an edge and unbeknownst to each other are betting the maximum with their edge. One of them is catching a glimpse of the bottom card. And he of course bets the opposite color. The other player is also catching aqglimpse of one card but not the bottom one. So they are often betting the opposite way. Meanwhile logic tells us that they will both win 26 out of 51 bets on average. Does that make sense? If so, can someone NOT an expert in probability make the numbers work out?

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

This and the blackjack thread are stupidly simple. The edge comes from the extra information. Someone else knowing extra information does not change anything. The actual information changes the specific example to neutral EV in this case, but both players will think they have an edge. Repeated indefinitely, their edge will be the same 26/51 no matter how many people know what (as long as no additional info can be gained from others' knowledge).

The "error" in both threads is confusing the edge/odds in the single instance, where the 3rd party (us) has more information than the betting parties, with the edge when the example is repeated infinitely and cards remain random.

I can't believe any intelligent person would think this is a question worth posing.

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

Half the time they bet the same way, half the time they bet opposite ways.
When they bet opposite colors, one person has seen a red card, one has seen a black, so on average they will break even, ie. win 26 out of 52 bets on average.
When they bet the same way however, they have both seen the same color. Here they will win 26 out of 50 bets on average.
So altogether they win on average 26/51 bets.

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

[ QUOTE ]
Half the time they bet the same way, half the time they bet opposite ways.

[/ QUOTE ]

no

51 players
 

Suppose the bank plays against 51 players dealing everyone a card and let them bet independnently. The bank will lose 1 bet out of 51 everytime the game is played won by the bettors. Since the conditions are symetric every player has an edge of 1/51. this edge doesn’t disappear if 1 or 50 or 49 stop playing and isn’t dependent on whether you now or not. Kind regards Artur

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

If one player has seen a red card the other is less likely to see a red one as well. So on average they will bet oposite colors less often. This would mean that their edge decreases.

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

[ QUOTE ]
If one player has seen a red card the other is less likely to see a red one as well. So on average they will bet oposite colors more often. This would mean that their edge decreases.

[/ QUOTE ]
FYP

Although their advantage does decrease (to zero) when they bet "opposite," their advantage increases proportionally when they bet "same."



The numbers work out as follows:

The odds of winning when playing alone are 26/51 which yields an expectation of
.0196 or 1/51 (((26/51)*2)-1).

When playing together they will bet "opposite" 26 out of 51 times. When they bet "opposite" they each have a 25 out of 50 chance of winning. Therefore, they have an expectation of 0 (((25/50)*2)-1) on the "opposite" bets.

When playing together they will bet "same" 25 out of 51 times. When they bet "same" then they have a 26 out of 50 chance of winning. Therefore, they each have an expectation of .04 (((26/50)*2)-1) on "same" bets.

Therefore, when playing together each has the same expectation as the other.

Their expectation when playing together is the same as when playing alone because: They have a 26 in 51 chance of an expectation of 0 and a 25 in 51 chance of an expectation of .04. (26/51*0)+(25/51*.04) = .0196 or 1/51

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

I don't really see what your point is?

So what if they're often betting opposite ways? If we combine their information, then on hands where they're betting opposite ways, they both have an exactly even money bet. On hands where they bet the same way, they each win slightly more than double what they won on their own (the "slightly more" part is counterbalanced by the fact that they are more likely to have seen opposite color cards).

There's nothing paradoxical about different pieces of incomplete information recommending different plays. That's why it's incomplete.

Of more interest to me (and I think more "paradoxical") is something I realised when typing the above. If the player who can see the bottom card started being able to see the two bottom cards instead, then even though his information is doubled, his earn per hand would only increase very slightly. This is because when he sees one red and one black, his information is useless and he has no edge at all. The other 50% his earn more than doubles, but he can only bet half his hands, so his earn per hand doesn't benefit much. Interesting that the value of the information decreases so drastically.

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

[ QUOTE ]
Two players at the table have found an edge and unbeknownst to each other are betting the maximum with their edge. One of them is catching a glimpse of the bottom card. And he of course bets the opposite color. The other player is also catching a glimpse of one card but not the bottom one. So they are often betting the opposite way.

[/ QUOTE ] If the other player glimpses one card which is not always the bottom card, does this mean that it can also be the top card?

Mickey Brausch

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

....

i dont think this illustrates your point too well? it seems too simple so i think ive missed it

both players will average 26/51, that doesnt mean they will get exactly 26/51. they do both have the same ev yet will go through different variances.

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

each player has an advantage of 1/51 over the house. the house expected loss is 2/51 with 2 players. with 51 players each seeing a different card they'd each have a 1/51 expected profit per hand and the house would be guaranteed a 1 bet loss.

in practice the more players there are seeing different cards, the sooner the house catches on, so i guess the players ev would go down with more players.

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

Sometimes they glimpse cards of different colors so they have no edge in this hand. Sometimes they glimpse cards of the same color so they have DOUBLE the edge in this hand.

What is the paradox?

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

[ QUOTE ]
[ QUOTE ]
Two players at the table have found an edge and unbeknownst to each other are betting the maximum with their edge. One of them is catching a glimpse of the bottom card. And he of course bets the opposite color. The other player is also catching a glimpse of one card but not the bottom one. So they are often betting the opposite way.

[/ QUOTE ] If the other player glimpses one card which is not always the bottom card, does this mean that it can also be the top card?

[/ QUOTE ]

no, because DS said that each player wins 26 out 51 bets.

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

[ QUOTE ]
It wasn't meant to be about black jack. So here is an extremely trivial example of the concept I was trying to illustrate.


[/ QUOTE ]If it was not about BJ then why did you use such a screwed up game to illistrate a simple game.

When is the last time you got some sleep? I can get this type of mental anguish from my wife thank you very much!

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

[ QUOTE ]
[ QUOTE ]
If the other player glimpses one card which is not always the bottom card, does this mean that it can also be the top card?

[/ QUOTE ]

no, because DS said that each player wins 26 out 51 bets.

[/ QUOTE ]

[img]/images/graemlins/wink.gif[/img]

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

OK, diddle, I'll be very brief. [img]/images/graemlins/smile.gif[/img]

[ QUOTE ]
Does that make sense?

[/ QUOTE ]Yes, absolutely.

[ QUOTE ]
If so, can someone ... make the numbers work out?

[/ QUOTE ] The situation described by Sklansky is the same as this: Me and Joe are playing that game at two different tables, against the same deck of cards (both decks have the same order of cards).

Joe glimpses the bottom card. He proceeds to bet accordingly.

The dealer at my table shows me the bottom card (its the same card as Joe's) and then takes it from the bottom and places it somewhere else in the deck. (Not at the top.) I proceed to bet accordingly. Guess what? I'm betting exactly the same way as Joe! Same edge for both of us.

It's like I glimpsed that card after the dealer placed it somewhere in the deck.

In the long run, me and Joe are glimpsing the same cards and get the same information.

Mickey Brausch

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

A six sided die is rolled and the house pays off 5-1 for bets on a number. Both Joe and Ted know the die is loaded and Never Rolls a Six but rolls the other numbers with equal probabilty. Joe always bets on 1. Ted always bets on 5. Joe and Ted both win a dollar for every $5 they bet. YET THEY ARE BETTING ON DIFFERENT NUMBERS!!! HOW IS THAT POSSIBLE?

PairTheBoard

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

The knowledge of the other player can't possibly change the edge of the first player, obviously the opposite will be true.
It seems like sure they'll be betting opposite colors plenty (and will have no edge the times they bet opposite colors) but the times they bet the same color their edge will be doubled. It will even out.

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

Like many others, I fail to see the problem here.

They each win 26/51 of the time.

Sometimes they bet the same way and both win. Sometimes they bet the same way and both lose. Sometimes they bet opposite ways and one wins while the other loses.

Are we supposed to think that they can't both win 26/51 because 26 + 26 > 51? Well, sometimes they both win. That is all.

To "make the numbers work out":

- chances player B sees the same colour as player A are 25/51. So 25/51 they bet the same way. Then they both win 26/50 of the time.

- chances player B sees the opposite colour are 26/51. Then they bet opposite ways and each wins 50% of the time.

So the probability of winning for either player is (25/51 * 26/51) + (1/2 * 26/51) = 26/51.

Still don't see what the problem is.

Guy.

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

if there were 5 guys instead of 2 they each bet on a different number 1-5. 4 guys lose 1 each roll. one guy wins 5 bets. house loses a bet each roll. they can go up to their room and split the money. each guy made 1/5 bet profit per roll.

take the original example but make it a four card deck and have three players instead of 2. one player sees the top, one guy sees the 2nd card, and the other guy sees the 3rd card. one guy will be wrong every time. two guys right. the house will lose a bet every time. when the guys go up to their room they split up the money and made 1/3 bet on each hand.

with a 52 card deck, same thing, but house gets suspicious when 51 guys try to squeeze into a room.

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

[ QUOTE ]

I can't believe any intelligent person would think this is a question worth posing.

[/ QUOTE ]

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

They're both playing independently against the house, despite being at the table and seeing the same cards. They both have the same edge against the house. Player A's action has no effect on player B's action, yet some people wrongly think that A has an effect on B.

So, to answer the actual questions posed....yes, it makes sense, since they break even the times they bet opposite colors and have an edge when they bet the same way, and they bet the same way a non-zero number of times, and yes, a non-expert in statistics can show that, as has been demonstrated in this thread.

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

4 Cases:

They both see red- 1/2*25/51 = 24.5%
They both see black 1/2*25/51 = 24.5%
Bottom is black, other is red 1/2*26/51 = 25.5%
Bottom is red, other is black 1/2*26/51 = 25.5%

Using cummulative knowlege, knowing what both of those cards are, we can assess the probability that they will win-

51% of the time, there is no advantage. 49% of the time, there is an advantage, but it is even MORE than the player figured.

So 49% of the time, the player bets more with a chance of winning 52% of the time. The other 51% of the time, it wins 50% of the time. Overall, the player wins 51% of the time, which is identical to 26/51.

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

"To "make the numbers work out":

- chances player B sees the same colour as player A are 25/51. So 25/51 they bet the same way. Then they both win 26/50 of the time.

- chances player B sees the opposite colour are 26/51. Then they bet opposite ways and each wins 50% of the time.

So the probability of winning for either player is (25/51 * 26/51) + (1/2 * 26/51) = 26/51.

Still don't see what the problem is."

Guy.

Cool. Most people can't answers problems they see. You answered one you couldn't see.

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

DS,

Perhaps you will find this similarly complex paradox equally fascinating!

Three men go to a hotel. The desk clerk charges them $30 for the room. Each man gives the clerk $10. Shortly afterwards the desk clerk realizes that he has overcharged the three men by $5. He calls the Bell Boy over, gives him the $5 and explains to him that he has overcharged the three men and asks him to go up and give them the $5. On his way up, the Bell Boy thinks they will be happy to get a refund so why don't I give them each $1. They will be happy and I will have picked up $2. But when he does that and gives each man $1, that means they only paid $9 each for the room, which is a total of $27; plus the $2 the Bell Boy pocketed is a total of $29. But they gave the desk clerk $30! Where did the extra $1.00 go??!!?!

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

El D,

Rake.

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

[ QUOTE ]
DS,

Perhaps you will find this similarly complex paradox equally fascinating!

Three men go to a hotel. The desk clerk charges them $30 for the room. Each man gives the clerk $10. Shortly afterwards the desk clerk realizes that he has overcharged the three men by $5. He calls the Bell Boy over, gives him the $5 and explains to him that he has overcharged the three men and asks him to go up and give them the $5. On his way up, the Bell Boy thinks they will be happy to get a refund so why don't I give them each $1. They will be happy and I will have picked up $2. But when he does that and gives each man $1, that means they only paid $9 each for the room, which is a total of $27; plus the $2 the Bell Boy pocketed is a total of $29. But they gave the desk clerk $30! Where did the extra $1.00 go??!!?!

[/ QUOTE ]

Oldie, but goodie...

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

[ QUOTE ]
Three men go to a hotel. The desk clerk charges them $30 for the room. Each man gives the clerk $10. Shortly afterwards the desk clerk realizes that he has overcharged the three men by $5. He calls the Bell Boy over, gives him the $5 and explains to him that he has overcharged the three men and asks him to go up and give them the $5. On his way up, the Bell Boy thinks they will be happy to get a refund so why don't I give them each $1. They will be happy and I will have picked up $2. But when he does that and gives each man $1, that means they only paid $9 each for the room, which is a total of $27; plus the $2 the Bell Boy pocketed is a total of $29. But they gave the desk clerk $30! Where did the extra $1.00 go??!!?!

[/ QUOTE ]

there is no extra dollar.

they paid $27 for a $25 room, with the bell boy taking the extra $2.

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

[ QUOTE ]
The other player is also catching aqglimpse of one card but not the bottom one.

[/ QUOTE ]

If the card he catches a glimpse of is sometimes the card that is being dealt, but he doesn't know that, and bets the opposite way, this could be a disadvantage.

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

Would it have anything to do with the fact that one gets to see the bottom card, meaning his information is useful for the first 51 decisions, whereas the second person has seen card X, where X < 52, and thus this information is only valuable up to card X, and not afterwards? This might be idiotic...

Re: A Much Simpler Version Of The \"Blackjack Paradox\"
 

[ QUOTE ]
Would it have anything to do with the fact that one gets to see the bottom card, meaning his information is useful for the first 51 decisions, whereas the second person has seen card X, where X < 52, and thus this information is only valuable up to card X, and not afterwards?

[/ QUOTE ]
No, the casino reshuffles after every deal.