## Two Blackjacks

Before dealing from a fresh deck of cards, the dealer offers you two bets: You can bet any amount that you will be dealt a blackjack, and you can bet any amount that the dealer will be dealt a blackjack. If either hand is dealt a blackjack, you will be paid 17-to-1 for the respective bet. If BOTH hands get a blackjack, both bets will be paid 20-to-1 (as long as you originally placed both bets; placing a single bet will only be paid 17-to-1). So a few questions in probabilities arise. Find the probabilities as reduced fractions of:
1. One hand being dealt a blackjack from a freshly shuffled deck.
2. Two hands being BOTH dealt blackjacks from a fresh deck.
3. Neither of two hands being dealt a blackjack
• If you bet one dollar on both your own AND the dealer's hand for each fresh deck, what would be your expected rate of loss (the odds are ALWAYS in the house's favor.)
• If you only bet on your hand for each fresh deck, what would be your expected rate of loss?
• What odds would provide even gain/loss expectations?
A Blackjack is the combination of an Ace and one of [Ten, Jack, Queen, or King].

Solutions were received from: Jeremy Galvagni, Paul Botham, Jozef Hanenberg, Parthiv Shah, Denis Borris, Jimmy Chng Gim Hong, and Rich Polster. Jozef Hanenberg's solution is thorough and succinct:
```I know nothing of the art of playing Blackjack or betting, but I suppose that in this game each is being dealt two cards and that "17 to 1" means that in case you win, your net gain is 17 dollar, rather then 16 dollar for each dollar you bet.

1. Probability of getting a Blackjack = 16/52 * 4/51 * 2 = 32/663

2. Probability of two hands BOTH getting Blackjack =
16/52 * 4/51 * 2 * 15/50 * 3/49 * 2 = 11520/6497400 = 96/54145

3. This is easier calculated with the rule of the complement: we first
calculate the probability that there is exactly one Blackjack.

first second   first   second
Ace 10-King    Ace     rest     4/52 * 16/51 * 2 * 3/50 * 32/49 * 2
Ace 10-King    Ace     Ace      4/52 * 16/51 * 2 * 3/50 * 2/49
Ace 10-King    10-King rest     4/52 * 16/51 * 2 * 15/50 * 32/49 * 2
Ace 10-King    10-King 10-King  4/52 * 16/51 * 2 * 15/50 * 14/49
Ace 10-King    rest    rest     4/52 * 16/51 * 2 * 32/50 * 31/49

Adding up gives the probability that you get a Blackjack and the
dealer has no Blackjack :  302080/6497400 = 7552/162435
The probability that there is exactly ONE Blackjack is double this
number: 604160/6497400 = 15104/162435
So the probability that neither has a Blackjack is
1 - (11520+604160)/6497400 = 5881720/6497400 = 11311/12495.

Expected profit(loss) if you bet a dollar on each hand:
40 * 96/54145 + 16 * 15104/162435 - 2 * 11311/12495 = -0.251805 dollar

Expected profit(loss) if you bet a dollar on your own hand:
17 * 32/663 - 1 * 631/663 = -0.131222 dollar.

What odds would give a fair game?
Betting on your own hand with odds x-to-1 gives x * 32/663 - 1 * 631/663 = 0.
Solving gives odds 19.7-to-1 (or the be precise 19.71875-to-1)
Betting on both hands: suppose two Blackjacks pays y-to-1 and one Blackjack
pays x-to-1. We get the equation:

2y * 11520/6497400 + (x-1) * 604160/6497400 - 2 * 5881720/6497400 = 0

This can be reduced to the equation 288 * y + 7552 * x = 154595

There are several solutions (x,y) i.e.
(x,y) = (19.71875 , 19.71875) or    (x,y) = (18 , 64.79) or
(x,y) = (19 , 38.56)) or  (x,y) = (19.5 , 25.45).
```

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