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# Math and Odds and Blackjack

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posted by IAmBroom at 7:24 PM on October 28, 2008

In Mathematica:

deck = Flatten[Table[{11, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 10, 10}, {4}]];

Table[

{dealer, you1, you2} = RandomSample[deck, 3];

Boole[dealer == 10 && you1 + you2 == 15],

{10^6}

] // N // Mean

>> 0.026119

To use cvoixjames' 4-deck shoe, change the number of suits from {4} to {16}.

posted by hAndrew at 8:44 PM on October 28, 2008 [1 favorite]

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# Math and Odds and Blackjack

October 28, 2008 2:50 PM Subscribe

What are the chances, in blackjack, that I will start with 15, and the dealer will be showing a 10 (or face card worth 10)?

Despite the fact that my buddy thinks the odds of this are near 100%, I've tried to calculate the actual odds (assuming a single deck).

I've come up with 2.62% -ish. Not really sure if I'm right.

I started with all the possible hands I could have.

2652 (52 * 51)

Then I figured out how many hands would give me 15.

64 hands with a 10 (or equivalent) and a 5.

64 hands with a 5 and a 10.

16 hands with a 9 and a 6.

16 hands with a 6 and a 9.

16 hands with an 8 and a 7.

16 hands with a 7 and an 8.

16 hands with an ace and a 4.

16 hands with a 4 and an ace.

These fall into two groups, those that have a 10, and therefore deprive the dealer of one, and those that don't.

128 hands have a 10.

96 hands are 15 some other way.

If I have a 10, she has 15 cards left, out of 50, to get a 10.

If I don't she has 16 out of 50.

So my odds of making a 15 using a 10 (or face card) are 4.8%.

Her odds of then having a 10 showing are 30%.

So we have a 1.44% chance of that happening.

-Plus-

Me making 15 some other way: 3.7%.

And her having a 10 showing: 32%.

Gives us 1.18% for that second scenario.

So 2.62% of my having 15 and the dealer showing a 10 or face card.

Two questions.

Assuming a single deck, is this correct?

Assuming multiple decks, what changes (if anything)?

If I'm wrong, where did I go wrong? (It would help my brain to have a combination of english and math to explain where I went wrong, rather than just something like "you should have used a factorial for possible hands 52!-4!", etc...)

Thanks.

Despite the fact that my buddy thinks the odds of this are near 100%, I've tried to calculate the actual odds (assuming a single deck).

I've come up with 2.62% -ish. Not really sure if I'm right.

I started with all the possible hands I could have.

2652 (52 * 51)

Then I figured out how many hands would give me 15.

64 hands with a 10 (or equivalent) and a 5.

64 hands with a 5 and a 10.

16 hands with a 9 and a 6.

16 hands with a 6 and a 9.

16 hands with an 8 and a 7.

16 hands with a 7 and an 8.

16 hands with an ace and a 4.

16 hands with a 4 and an ace.

These fall into two groups, those that have a 10, and therefore deprive the dealer of one, and those that don't.

128 hands have a 10.

96 hands are 15 some other way.

If I have a 10, she has 15 cards left, out of 50, to get a 10.

If I don't she has 16 out of 50.

So my odds of making a 15 using a 10 (or face card) are 4.8%.

Her odds of then having a 10 showing are 30%.

So we have a 1.44% chance of that happening.

-Plus-

Me making 15 some other way: 3.7%.

And her having a 10 showing: 32%.

Gives us 1.18% for that second scenario.

So 2.62% of my having 15 and the dealer showing a 10 or face card.

Two questions.

Assuming a single deck, is this correct?

Assuming multiple decks, what changes (if anything)?

If I'm wrong, where did I go wrong? (It would help my brain to have a combination of english and math to explain where I went wrong, rather than just something like "you should have used a factorial for possible hands 52!-4!", etc...)

Thanks.

How do you do a quick monte carlo simulation? Is there a software specific for that? I often want to run one but writing the program usually discourages me.

posted by Brennus at 3:22 PM on October 28, 2008

posted by Brennus at 3:22 PM on October 28, 2008

Incidentally, your original calculation shows the danger of rounding intermediate values:

posted by smackfu at 3:32 PM on October 28, 2008

*So we have a 1.44% chance of that happening.***1.44796***Gives us 1.18% for that second scenario.***1.15837***So 2.62% of my having 15 and the dealer showing a 10 or face card.***2.60633**posted by smackfu at 3:32 PM on October 28, 2008

*How do you do a quick monte carlo simulation? Is there a software specific for that? I often want to run one but writing the program usually discourages me.*

**Brennus**, I can't speak for what

**0xFCAF**did, but I could put one together in 15 min or so in Excel. A bit longer in Matlab or Mathematica.

posted by IAmBroom at 7:24 PM on October 28, 2008

Vegas: it's a 4 deck shoe.

Blackjack rules are basic: split 8's and aces, double down if you have an eleven, stick on anything if the dealer up card is 5 or below,

stick on 17 or greater.

that will give you 45/55 against the house

posted by cvoixjames at 7:31 PM on October 28, 2008 [3 favorites]

Blackjack rules are basic: split 8's and aces, double down if you have an eleven, stick on anything if the dealer up card is 5 or below,

stick on 17 or greater.

that will give you 45/55 against the house

posted by cvoixjames at 7:31 PM on October 28, 2008 [3 favorites]

This isn't really a direct answer to your question, but I think that linking to the Wizard of Odds page on Blackjack is still appropriate. It has tons and tons of information on Blackjack math/probabilities/etc.

posted by Perplexity at 8:37 PM on October 28, 2008

posted by Perplexity at 8:37 PM on October 28, 2008

*How do you do a quick monte carlo simulation? Is there a software specific for that? I often want to run one but writing the program usually discourages me.*

In Mathematica:

deck = Flatten[Table[{11, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 10, 10}, {4}]];

Table[

{dealer, you1, you2} = RandomSample[deck, 3];

Boole[dealer == 10 && you1 + you2 == 15],

{10^6}

] // N // Mean

>> 0.026119

To use cvoixjames' 4-deck shoe, change the number of suits from {4} to {16}.

posted by hAndrew at 8:44 PM on October 28, 2008 [1 favorite]

This thread is closed to new comments.

960 possibilities if your first card is a five (4 x 16 x 15)

960 possibilities if your first card is worth ten (16 x 4 x 15)

6 x 256 possibilities if your first card is A, 4, 6-9 (4 x 4 x 16)

= 3456 out of a total of (52 x 51 x 50) or 132,600 possibilities

= 2.61%

I think an infinite # of decks would change that to 2.54%, since the dealers third card would not be affected by your ten, and the number of possibilities would be 52 x 52 x 52 since cards aren't used up.

posted by smackfu at 3:11 PM on October 28, 2008