A Much Simpler Version Of The "Blackjack Paradox"

[Gabe]

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.

[diddle]

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?

[felson]

[ 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.

[Buccaneer]

[ 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!

[Mickey Brausch]

[ 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]

[Mickey Brausch]

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

[PairTheBoard]

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

[goofball]

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.

[Guy McSucker]

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.

[Gabe]

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.