Blackjack Problem (Count Vs. Insurance) - Gambling and Probability

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Blackjack Problem (Count Vs. Insurance)

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12-05-2013 , 03:37 AM

Jimulacrum

Pooh-Bah

Join Date: May 2010 Posts: 3,670

Andy is seated at a blackjack table. The running count is +15, and exactly half of the initial 6-deck shoe is remaining. Andy and three other players place their bets. Andy is dealt 7-2; the other players are dealt 8-9, 10-2, and A-K. The dealer receives a face-up ace and offers the table an insurance bet, which pays the bettor 2:1 if the dealer turns out to have blackjack. What is the expected value of this insurance bet for a wager of x dollars, to the nearest hundredth?

Note: The running count increases by one for each low card dealt (2, 3, 4, 5, 6), decreases by one for each high card (10, J, Q, K, A), and is not affected by middle cards (7, 8, 9).

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12-05-2013 , 03:37 AM

Jimulacrum

Pooh-Bah

Join Date: May 2010 Posts: 3,670

Forgive the newbie phrasing. This was originally written for a different audience.

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12-05-2013 , 04:23 AM

Tropics

grinder

Join Date: Feb 2009 Posts: 526

Quote:

Originally Posted by Jimulacrum

A-K is the clear equity favorite. 10-2 can only be profitably played by Doyle Brunson. 8-9 can call if it's suited. 7-2 is the worst hand so we don't want to put any more money in the pot with it. Finally, the dealer should be taken out of the hand for revealing one of his cards. Use a card protector for Christ's sake!

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12-05-2013 , 04:35 AM

Jimulacrum

Pooh-Bah

Join Date: May 2010 Posts: 3,670

Quote:

Originally Posted by Tropics

I think you may be playing a different game than Andy, sir.

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12-06-2013 , 11:11 PM

Canceler

grinder

Join Date: May 2009 Posts: 481

My guess:

Spoiler:

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12-06-2013 , 11:53 PM

Jimulacrum

Pooh-Bah

Join Date: May 2010 Posts: 3,670

Quote:

Originally Posted by Canceler

My guess:

Spoiler:

You have work to show with that, Canceler, or is it just a pure guess?

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12-07-2013 , 12:54 AM

Canceler

grinder

Join Date: May 2009 Posts: 481

Here’s the way I did it. Of course, we’re interested in whether the dealer has a T down.

Before the deal, there are 156 unseen cards. We have an excess of 15 high cards, and on average 12 of them will be Ts. We also have all the Ts that would be in the three remaining decks. 12 + 48 = 60.

After the deal, there are 147 unseen cards, and two of the Ts were used up, so 58 left. 58/147 = 0.39, the chance of the dealer having a T down.

EV = (0.39 * 2x) - 0.61x
EV = 0.78x - 0.61x
EV = 0.17x

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