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TTChamp 12-31-2005 12:32 PM

Can you beat this game?
 
By "beat" I mean play it for a long term profit.

I call the game "Kings".

The face cards are separated out of several standard decks of cards. For the purpose of this post assume that the cards are dealt from and infinite shoe and that the shoe is fair. Also, the dealer turns up all the cards at the end of each hand so that the players can make sure they wern't cheated.

The player places their first bet (any amount they want) and the dealer deals 3 cards face down and looks at them. If there are no Kings, then the player wins three times their first bet and the hand is over.

If there is a king then the dealer shows the player a King and the player loses their bet. Then the player can place a second bet (again any amount they want). The player then choses one of the two remaining cards. If the card they choose is a King then they win 4 times their second bet. If the card they choose is not a King then they lose their second bet.

Can you develope a betting scheme that makes this game +EV?

MegumiAmano 12-31-2005 12:50 PM

Re: Can you beat this game?
 
If I'm reading your rules correctly, then all you have to do to beat the game is place your original bet.

There are 3 face cards in 27 different combinations. 8 times in 27, you'll get zero kings and win. The other 19 times you'll lose.

(3)(8/27) - (1)(19/27) = +5/27

BruceZ 12-31-2005 01:17 PM

Re: Can you beat this game?
 
[ QUOTE ]
If I'm reading your rules correctly, then all you have to do to beat the game is place your original bet.

There are 3 face cards in 27 different combinations. 8 times in 27, you'll get zero kings and win. The other 19 times you'll lose.

(3)(8/27) - (1)(19/27) = +5/27

[/ QUOTE ]

I agree, it's a no-brainer if the only cards are J,Q,K in equal numbers, and the first bet pays 3-to-1 while the second bet pays 4-to-1. Then both bets are +EV. If he meant that your own bet is included in the payoff, so that the first bet pays 2-to-1 while the second bet pays 3-to-1, then both bets are -EV no matter how you bet. The way to make this problem more interesting is to make the first bet pay only 2-to-1 so that it is -EV, while the second bet pays 4-to-1, so that it is +EV. Then you can only make the game +EV by betting enough on the second bet relative to the first bet.

Here are the relevant probabilities:

no King: 2/3 * 2/3 * 2/3 = 8/27 = 2.375-to-1

1 King: 2/3 * 2/3 * 1/3 * 3 = 12/27

2 Kings: 1/3 * 1/3 * 2/3 * 3 = 6/27

3 Kings: 1/3 * 1/3 * 1/3 = 1/27

If 1 king is exposed, the second bet wins all the time with 3 kings, half the time with 2 kings, and none of the time with 1 king. So it wins 4 times out of 19, and the odds to make the second bet even-money are 15-to-4 or 3.75-to-1.

Note that we can gain an advantage if the dealer does not choose the exposed card randomly when there are 2-kings. For example, if the dealer always exposes the leftmost card, then we will bet that card 3 is the king when he exposes card 2, and we won't bet at all when he exposes card 3. Then we will win 4 out of 6 with 2 kings and only bet on 8 of the 12 with 1 king, so that all together we will win 5 out of 15 on the second bet with odds of exactly 2-to-1. We can do even better if his choice of card always telegraphs where the other king is. In that case we win the second bet 7/19 with odds of 1.7-to-1.

TTChamp 12-31-2005 03:17 PM

Re: Can you beat this game?
 
[ QUOTE ]
If I'm reading your rules correctly, then all you have to do to beat the game is place your original bet.

There are 3 face cards in 27 different combinations. 8 times in 27, you'll get zero kings and win. The other 19 times you'll lose.

(3)(8/27) - (1)(19/27) = +5/27

[/ QUOTE ]

Sorry, bad terminology.

I meant that if you bet $5 the first time, the dealer will pay you $15 total (including the $5 you put on the table). So it is actually 2:1 on the first bet. Same idea for the second bet.


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