| How Odds are Calculated
In How to Compute House Advantage, the odds for the outcome of
the dealer's hand when he must hit to 17, or for a player who follows the same
rules in playing his hands, were given as:
| 17 | 18 | 19 | 20 | 21 | BUST |
|---|
| 14.61% | 13.87% | 13.27% | 18.12% | 6.99% | 33.15% |
The material that follows illustrates the calucaltions by
which those odds were derived.
Odds for initial hands
The standard deck contains cards of thirteen possible values, so the odds of receiving
a single card of any given value the same: there is a 1 in 13 chance (7.69%) a card will
be of any value, and this is equal for every value. The net outcome for the
value of the initial hand is as follows:
| 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | J | Q | K | A |
|---|
| 2 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 12 | 12 | 12 | 13 |
|---|
| 3 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 13 | 13 | 13 | 14 |
|---|
| 4 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 14 | 14 | 14 | 15 |
|---|
| 5 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 15 | 15 | 15 | 16 |
|---|
| 6 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 16 | 16 | 16 | 17 |
|---|
| 7 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 17 | 17 | 17 | 18 |
|---|
| 8 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 18 | 18 | 18 | 19 |
|---|
| 9 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 19 | 19 | 19 | 20 |
|---|
| 10 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 20 | 20 | 20 | 21 |
|---|
| J | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 20 | 20 | 20 | 21 |
|---|
| Q | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 20 | 20 | 20 | 21 |
|---|
| K | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 20 | 20 | 20 | 21 |
|---|
| A | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 21 | 21 | 21 | 12 |
|---|
However, the odds are not evenly dispersed because there are four ten-value cards
(10, jack, queen, king). Since the chances of drawing a ten-value card is 4 in 13
(30.7%) rahter than 1 in 13 (7.69%), which increases the probability for hands with
a value of 12 to 20. The values of the total hands, above, are combined to form
the table below.
| VAL | # | % | DISTRIBUTION |
|---|
| 4 | 1 | 0.59% |  |
|---|
| 5 | 2 | 1.18% | |
|---|
| 6 | 3 | 1.78% | |
|---|
| 7 | 4 | 2.37% | |
|---|
| 8 | 5 | 2.96% | |
|---|
| 9 | 6 | 3.55% | |
|---|
| 10 | 7 | 4.14% | |
|---|
| 11 | 8 | 4.73% | |
|---|
| 12 | 16 | 9.47% | |
|---|
| 13 | 16 | 9.47% | |
|---|
| 14 | 15 | 8.88% | |
|---|
| 15 | 14 | 8.28% | |
|---|
| 16 | 13 | 7.69% | |
|---|
| 17 | 12 | 7.10% | |
|---|
| 18 | 11 | 6.51% | |
|---|
| 19 | 10 | 5.92% | |
|---|
| 20 | 18 | 10.65% | |
|---|
| 21 | 8 | 4.73% | |
|---|
Granted, once the first card is dealt, the odds are minutely less that a card of
identical value will be brawn as the second card in the hand, because there is one
less of that value in the remaining deck. This difference would decrease the liklihood
of drawing any even-numbered hand - but only by a minute percentage that becomes even more
minute in multiple-deck games. (see Footnote A).
Odds of outcome for each hand
The odds of outcome, after all hits are taken, for each of these hands may likewise
be calculated by applying the same likelihood (1 in 13 of each possiblevalue) to any
hit that may added, up to the point that hits are no longer taken according to the rules
by which a hand is played (in this case, upon reaching a total of 17).
As an example, an initial hand of fifteen (ten-five, nine-four, etc., omitting the
four-ace combination that can fluctuate in value for the present time) would be hit
once. Since the hand has not yet reached 17, there is a 100% chance a hit will be taken.
For that hit, there is a 1 in 13 (7.69%) chance for each value:
| HIT CARD | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | J | Q | K | A |
|---|
| TOTAL HAND | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 25 | 25 | 25 | 16 |
|---|
| 16 | 17 | 18 | 19 | 20 | 21 | BUST |
|---|
| 1 | 1 | 1 | 1 | 1 | 1 | 7 |
| 7.69% | 7.69% | 7.69% | 7.69% | 7.69% | 7.69% | 53.84% |
Since one of the outcomes is a sixteen, and since the rules dictate the hand must be
hit until it reaches seventeen there's a 7.69% chance the hand will need to take a
second hit, with these outcomes:
| HIT CARD | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | J | Q | K | A |
|---|
| TOTAL HAND | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 26 | 26 | 26 | 17 |
|---|
| 17 | 18 | 19 | 20 | 21 | BUST |
|---|
| 1 | 1 | 1 | 1 | 1 | 8 |
| 7.69% | 7.69% | 7.69% | 7.69% | 7.69% | 61.52% |
Since there is a 0% chance that there will be any result under 17 after the second hit,
all possibilities have been exhausted, and the total results can be calculated by adding
the probabilities. The outcomes for the second hit are mutliplied by the chance (7.69%)
that the second hit would need to be taken, to produce the following porbabilties for the hand:
| 17 | 18 | 19 | 20 | 21 | BUST |
|---|
| 1ST | 7.69% | 7.69% | 7.69% | 7.69% | 7.69% | 46.14% |
|---|
| 2ND | 0.59 | 0.59 | 0.59 | 0.59 | 0.59 | 4.73% |
|---|
| TTL | 8.28% | 8.28% | 8.28% | 8.28% | 8.28% | 58.58% |
|---|
This completes the calculations necessary to determine the total odds for a "hard"
fifteen. The "soft" fifteen-an ace-four combination-is different because the
original ace would revert to a value of one when the hand exceeds twenty-one. Instead
of a 22 resulting in a busted hand, its value would revert to 12 and the player would
continue to take hits, and it would require several more iterations to calculate the
possible outcomes.
Odds of outcome for all hands
To determine the odds of outcome for the outcome of all hands, when the same rules
are applied consistently, would be determined by calculating the possible outcome of
each possible hand, multiplying it by the likelihood of drawing that hand, and adding
them together.
This would require several pages of tables to illustrate completely, all derived
from the basic tables shown above. To tabulate the odds for busting, it would be
necessary to multiply the chances of drawing each hand by the chances of busting
that hand, then add those results together for all hands.
For example, there is a 7.10% chance of drawing a hard 15 to begin, and a 58.58%
chance you will bust before reaching any acceptable total (17+)-to arrive at a
net chance of 4.1% that you will draw and bust a hard 15 from a freshly-shuffled deck.
That net chance would need to be added to the net chances of drawing and busting
every other possible hand. These calculations have been done, and the results are ...
| 17 | 18 | 19 | 20 | 21 | BUST |
|---|
| 14.61% | 13.87% | 13.27% | 18.12% | 6.99% | 33.15% |
________________________________________________
Footnote A: Absolute Precision
In deference to those who rejoice at any opportunity to niggle, it would require some
minor adjustments to compute these odds with absolute precision. For example, it is less
likely to draw a paired hand than any other because, once the first card has been removed,
there is one fewer card of the exact same value left in a single deck when compared to
cards of every other value. It is also less likely that the next hit on a six-nine hand
(15) would be a six or a nine, so the calculations would have to be done for every unique
combination rather than every unique value to arrive at absuolutely precise odds.
However, these differences are minute. The odds of drawing a paired hand is 0.591%
if only the values of the cards are considered, which would be 1.8% less likely to occur
(0.580%) in a single-deck game, 1.2% less likely (0.589%) in a double-deck game, and
so on, with the difference becoming more minute as decks are added. The net result is
a difference of a few thousandths of a percent per pair.
While it would, indeed, derive a more precise estimation of odds and strategy to adjust
the calculation for cards that have already been dealt, te result would affect only 1 in
100,000 hands in most games. A difference this minute does not lead to any significant
differences, hence it is hardly worth considering in practical application of the results.
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