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.