Ask The Wizard #345
When drawing 13 cards from a thoroughly shuffled deck, how many distinct card ranks should we anticipate seeing?
This scenario is a classic example of a Markov Chain problem.
The table below outlines the anticipated quantity of card ranks, ranging from 0 to 4 cards, for any total number of cards drawn, from 1 through 52.
Expected Ranks by Cards Dealt
Cards | 0 Ranks | 1 Rank | 2 Ranks | 3 Ranks | 4 Ranks | Expected Ranks |
---|---|---|---|---|---|---|
1 | 12.000000 | 1.000000 | 0.000000 | 0.000000 | 0.000000 | 1.000000 |
2 | 11.058824 | 1.882353 | 0.058824 | 0.000000 | 0.000000 | 1.941176 |
3 | 10.174118 | 2.654118 | 0.169412 | 0.002353 | 0.000000 | 2.825882 |
4 | 9.343577 | 3.322161 | 0.324994 | 0.009220 | 0.000048 | 3.656423 |
5 | 8.564946 | 3.893157 | 0.519088 | 0.022569 | 0.000240 | 4.435054 |
6 | 7.836014 | 4.373589 | 0.745498 | 0.044178 | 0.000720 | 5.163986 |
7 | 7.154622 | 4.769748 | 0.998319 | 0.075630 | 0.001681 | 5.845378 |
8 | 6.518655 | 5.087731 | 1.271933 | 0.118319 | 0.003361 | 6.481345 |
9 | 5.926050 | 5.333445 | 1.561008 | 0.173445 | 0.006050 | 7.073950 |
10 | 5.374790 | 5.512605 | 1.860504 | 0.242017 | 0.010084 | 7.625210 |
11 | 4.862905 | 5.630732 | 2.165666 | 0.324850 | 0.015846 | 8.137095 |
12 | 4.388475 | 5.693157 | 2.472029 | 0.422569 | 0.023770 | 8.611525 |
13 | 3.949628 | 5.705018 | 2.775414 | 0.535606 | 0.034334 | 9.050372 |
14 | 3.544538 | 5.671261 | 3.071933 | 0.664202 | 0.048067 | 9.455462 |
15 | 3.171429 | 5.596639 | 3.357983 | 0.808403 | 0.065546 | 9.828571 |
16 | 2.828571 | 5.485714 | 3.630252 | 0.968067 | 0.087395 | 10.171429 |
17 | 2.514286 | 5.342857 | 3.885714 | 1.142857 | 0.114286 | 10.485714 |
18 | 2.226939 | 5.172245 | 4.121633 | 1.332245 | 0.146939 | 10.773061 |
19 | 1.964946 | 4.977863 | 4.335558 | 1.535510 | 0.186122 | 11.035054 |
20 | 1.726771 | 4.763505 | 4.525330 | 1.751741 | 0.232653 | 11.273229 |
21 | 1.510924 | 4.532773 | 4.689076 | 1.979832 | 0.287395 | 11.489076 |
22 | 1.315966 | 4.289076 | 4.825210 | 2.218487 | 0.351261 | 11.684034 |
23 | 1.140504 | 4.035630 | 4.932437 | 2.466218 | 0.425210 | 11.859496 |
24 | 0.983193 | 3.775462 | 5.009748 | 2.721345 | 0.510252 | 12.016807 |
25 | 0.842737 | 3.511405 | 5.056423 | 2.981993 | 0.607443 | 12.157263 |
26 | 0.717887 | 3.246098 | 5.072029 | 3.246098 | 0.717887 | 12.282113 |
27 | 0.607443 | 2.981993 | 5.056423 | 3.511405 | 0.842737 | 12.392557 |
28 | 0.510252 | 2.721345 | 5.009748 | 3.775462 | 0.983193 | 12.489748 |
29 | 0.425210 | 2.466218 | 4.932437 | 4.035630 | 1.140504 | 12.574790 |
30 | 0.351261 | 2.218487 | 4.825210 | 4.289076 | 1.315966 | 12.648739 |
31 | 0.287395 | 1.979832 | 4.689076 | 4.532773 | 1.510924 | 12.712605 |
32 | 0.232653 | 1.751741 | 4.525330 | 4.763505 | 1.726771 | 12.767347 |
33 | 0.186122 | 1.535510 | 4.335558 | 4.977863 | 1.964946 | 12.813878 |
34 | 0.146939 | 1.332245 | 4.121633 | 5.172245 | 2.226939 | 12.853061 |
35 | 0.114286 | 1.142857 | 3.885714 | 5.342857 | 2.514286 | 12.885714 |
36 | 0.087395 | 0.968067 | 3.630252 | 5.485714 | 2.828571 | 12.912605 |
37 | 0.065546 | 0.808403 | 3.357983 | 5.596639 | 3.171429 | 12.934454 |
38 | 0.048067 | 0.664202 | 3.071933 | 5.671261 | 3.544538 | 12.951933 |
39 | 0.034334 | 0.535606 | 2.775414 | 5.705018 | 3.949628 | 12.965666 |
40 | 0.023770 | 0.422569 | 2.472029 | 5.693157 | 4.388475 | 12.976230 |
41 | 0.015846 | 0.324850 | 2.165666 | 5.630732 | 4.862905 | 12.984154 |
42 | 0.010084 | 0.242017 | 1.860504 | 5.512605 | 5.374790 | 12.989916 |
43 | 0.006050 | 0.173445 | 1.561008 | 5.333445 | 5.926050 | 12.993950 |
44 | 0.003361 | 0.118319 | 1.271933 | 5.087731 | 6.518655 | 12.996639 |
45 | 0.001681 | 0.075630 | 0.998319 | 4.769748 | 7.154622 | 12.998319 |
46 | 0.000720 | 0.044178 | 0.745498 | 4.373589 | 7.836014 | 12.999280 |
47 | 0.000240 | 0.022569 | 0.519088 | 3.893157 | 8.564946 | 12.999760 |
48 | 0.000048 | 0.009220 | 0.324994 | 3.322161 | 9.343577 | 12.999952 |
49 | 0.000000 | 0.002353 | 0.169412 | 2.654118 | 10.174118 | 13.000000 |
50 | 0.000000 | 0.000000 | 0.058824 | 1.882353 | 11.058824 | 13.000000 |
51 | 0.000000 | 0.000000 | 0.000000 | 1.000000 | 12.000000 | 13.000000 |
52 | 0.000000 | 0.000000 | 0.000000 | 0.000000 | 13.000000 | 13.000000 |
This inquiry is being posed and explored on my forum at Wizard of Vegas .
In your video poker programming tips , where you clarify that despite there being 2,598,960 potential starting hands in video poker using a 52-card deck, only 134,459 unique hand classifications are crucial for analysis.
I would like to know how many unique hand classifications exist when using two to six decks of cards?
For this topic, I consulted my knowledgeable colleague, Gary Koehler, who specializes in video poker mathematics. Here are the insights he provided based on the number of decks:
Video Poker Classes of Hands
Decks | Combinations | Classes |
---|---|---|
1 | 2,598,960 | 134,459 |
2 | 91,962,520 | 202,735 |
3 | 721,656,936 | 208,143 |
4 | 3,091,033,296 | 208,468 |
5 | 9,525,431,552 | 208,481 |
6 | 23,856,384,552 | 208,481 |
If we roll five red dice and five blue dice, what is the likelihood that both sets yield the same outcome, irrespective of their order? For instance, achieving a result of 1-2-3-3-6 for both sets.
The table below illustrates the various types of rolls possible:
- The count of distinct combinations that can result in this roll. For a full house, for instance, there are six ways to select three dice showing the same number and five possibilities for the pair, which results in a total of 30 unique full house combinations.
- The arrangements possible. For a full house, you can choose three of the five dice for the triplet in combin(5,3)=10 different ways, leaving the other two for the pair.
- The total means of achieving a specific hand as a roll. This product results from the previous two columns. For example, there are 30 outcomes to form a full house multiplied by 10 arrangements, equating to 300 distinct ways to roll a full house.
- The likelihood of attaining that hand. For the full house scenario, this translates to a probability of 300 out of 6.5= 0.038580.
- The odds that both rolls yield the same result as well as match the specified hand. This is calculated by squaring the probability from the fourth column and dividing it by the second column. For example, the chance that two rolls are both a full house stands at 0.038580.2However, the chance that they match the same full house configuration is 1/30. Therefore, the probability of both rolls resulting in the same full house comes to 0.038580.2/30 = 0.00004961.
In the bottom right cell, the cumulative probability that both sets of rolls are identical is 0.00635324.
Matching Roll
Type of Roll |
Different Types |
Orders | Total Combinations |
Probability One Roll |
Probability Two Rolls |
|
---|---|---|---|---|---|---|
Five of a kind | 6 | 1 | 6 | 0.00077160 | 0.00000010 | |
Four of a kind | 30 | 5 | 150 | 0.01929012 | 0.00001240 | |
Full house | 30 | 10 | 300 | 0.03858025 | 0.00004961 | |
Three of a kind | 60 | 20 | 1,200 | 0.15432099 | 0.00039692 | |
Two pair | 60 | 30 | 1,800 | 0.23148148 | 0.00089306 | |
Pair | 60 | 60 | 3,600 | 0.46296296 | 0.00357225 | |
Five singletons | 6 | 120 | 720 | 0.09259259 | 0.00142890 | |
Total | 7,776 | 1.00000000 | 0.00635324 |