# Thread: 98%: Next Card 21

1. ## 98%: Next Card 21

You may place this bet after your first two cards have been dealt and your bet may not exceed your regular blackjack bet. You bet on what your next card will be. You may bet on any number of cards, A,2,3,...,K, and, if your card comes, you receive 10 to 1 on your bet. If your card is not the next to come, you lose. For example, say you bet \$5 each on J, Q and K and the next card turns out to be a K. Your J and Q bets will lose, but your K bet will pay \$50, for a net gain of \$40.

So, in order for this bet to show a profit, the card you bet to come next must compose more than 1/10 of the remainder of the deck. If you were playing single deck, this could probably occur quite frequently.

So the existence of this sidebet begs the question as to whether it can be beaten using a standard count, without resorting to some special count or complicated side count. On sufficiently negative counts, you could wager that a low card will show up and, on sufficiently positive counts, you could wager that a high card will show up. It seems that what constitutes a sufficiently positive or negative count will be highly dependent on the number of decks in use. Anyone out there care to surmise what TCs (just using hi-lo I guess) would indicate a positive expectation on a Next Card 21 bet for 1, 2 and 6 deck games?

nt

3. ## ET Fan: Sorry. Correct answer is ...

+/-80/11 ~= 7.3, any number of decks. I got the 10.4 based on your premise: "in order for this bet to show a profit, the card you bet to come next must compose more than 1/10 of the remainder of the deck." But this caused a paradox, and then I realized the correct ratio is 1/11.

If you have Griffin, you can use his bivariate normal assumption to solve this.

ETF

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