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Old 12-31-2005, 01:17 PM
BruceZ BruceZ is offline
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Join Date: Sep 2002
Posts: 1,636
Default 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

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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.
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