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Suppose you are playing 2 hands of blackjack at the same time and the dealer guarantees if you lose hand A they will pay whatever you bet for hand B but if you win hand A you automatically lose hand B. Can you find a way to mathematically prove a way to win instead of continuing even money? Let's say you only brought $2000 to the casino this night!

Hand A     Hand B 
  W          L
  W          L
  L          W
  W          L
  W          L
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  • $\begingroup$ What are the pay-out rates? And which simplified set of rules do you use? And can you still win with hand B in case you lose with hand A? $\endgroup$
    – Hetebrij
    Commented Jul 26, 2016 at 20:35
  • $\begingroup$ Betting $A$ dollars on the first hand and $B$ dollars on the second hand is the same as betting $A-B$ dollars only on the first hand. Clearly this cannot be used to guarantee a money win. $\endgroup$
    – Asinomás
    Commented Jul 26, 2016 at 20:41
  • $\begingroup$ Carry on smiling what it ensures is that you will not have 2 continuous losing strings! $\endgroup$
    – Triumph
    Commented Jul 26, 2016 at 20:49
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    $\begingroup$ So, bet \$1 on hand A, play to lose, then bet \$1999 on hand B? Although I'm not sure what "they will pay whatever you bet for hand B" means, exactly. Also, I squared up your table but I don't understand what it signifies. $\endgroup$
    – Joffan
    Commented Jul 26, 2016 at 21:24
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    $\begingroup$ @Triumph I'm not the dealer - I can always take another card :-) $\endgroup$
    – Joffan
    Commented Jul 26, 2016 at 23:06

1 Answer 1

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A complete answer depends on the exact rules for the game, but here is one possibility.

One (old) set of rules for blackjack stipulates an automatic 3:2 payout on a blackjack -- e.g., a \$1 bet wins \$1.50. In that case, betting \$800 on hand A and \$1200 on hand B produces a profit of \$400 by intentionally losing hand A if it isn't a blackjack (assuming you can continue to "hit" even if your card count reaches 21, which, of course, one would ordinarily never do) and neither profit nor loss if hand A is a blackjack (i.e., you win \$1200 for the blackjack but automatically lose your hand B bet). In general, betting $40\%$ of your money on hand A and $60\%$ on hand B will never lose money and usually make money.

The one caveat here is that the OP doesn't specify what happens in the result of a tie, which in particular occurs when both the player (specifically hand A) and the dealer have blackjack. In normal play, hand A would neither win nor lose, while hand B cannot win and can only avoid losing by also starting at blackjack. But if we assume a dealer-hand-A blackjack tie is an overall tie, then the strategy described above guarantees the player will never lose (and will eventually get banned from the casino).

Remark: Since blackjacks are relatively rare, one can probably accelerate the average rate of return by reserve a small amount of money and betting heavily with the rest on hand B, for hand A to lose -- e.g., putting \$2 on hand A and \$1993 on hand B for the first round, holding onto \$5 in case hand A "wins" \$3 with blackjack, which would leave you with \$10 for the next round (after which you could play the conservative guaranteed strategy until you've recouped enough to bet heavily on hand B again.) It might be worth looking into what the optimal no-risk betting strategy is, if the OP clarifies the exact rules, especially regarding ties, but also regarding blackjack payouts and minimum/maximum bets.

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