License plate method example:

Find the number of possible combinations for a license plate in the format: [0-9][0-9][A-Z][A-Z][A-Z] Answer: 10*10*26*26*26

I know that using C(13,2)[...]/C(52,5) is the appropriate way of determining the probability of a full house. I found a strange outcome using the license plate method though, so please humor me.

The license plate method I used, which assumes the probability of a full house is equal to the probability of a pair * the probability of three of a kind.

[1 * (3/51)] * [1 * (3/50) * (2/49)]

In English:

Deal one card. To create a pair, the next card must be one of the 3 remaining with the same denomination.

Deal another card. To create three of a kind, the next card must be one of the remaining 3 with the same denomination, and the final card must be one of the remaining 2 with the same denomination.

That produces: 0.00014405762304922

The correct answer: (3744/2598960) = 0.0014405762304922

My incorrect license plate method result is exactly one order of magnitude less than the correct answer. It's actually 10x more likely to get a full house.

My questions:

  • Is it possible to calculate the probability of a full house using the license plate method, and if so, how?
  • Is the results being off by exactly an order of magnitude a coincidence, and if not what is the correlation?

Assumptions: 52 card deck, 13 denominations, 4 suits

  • $\begingroup$ I realize now why my license plate method was incorrect. The two events (getting a pair, getting 3 of a kind) are not independent. The denominations have to be different. Even with those adjustments, I don't get the correct answer. I still would like to know what a correct license plate method would look like in this case. $\endgroup$ – Animism Mar 22 '13 at 15:25
  • $\begingroup$ I also found this question: math.stackexchange.com/questions/117462/… Not a duplicate, but very related. $\endgroup$ – Animism Mar 22 '13 at 15:40

The "license plate" method that you used found, correctly, the probability of a full house that occurs in the order "First three of one kind, next two of another."

But the $3$ of a kind could occur in $\dbinom{5}{3}$ places, so we need to multiply the "license plate" answer that you got by $\dbinom{5}{3}$, which happens to be $10$.

Because of the special role $10$ plays in the decimal system, this produces a visually striking "coincidence."

The license plate idea is absolutely fine, if appropriate care is taken to make sure we really do account for all possible orders.

  • $\begingroup$ Thank you, this explains it very well. The remaining question I have though is: why did my license plate method result in the correct answer for a fixed-order hand? Shouldn't the [1 * (3/50) * (2/49)] calculation take into account that the pair has already been dealt? The 3 of a kind couldn't use the same denomination, so there would be less cards to choose from. [(48/50) * (3/49) * (2/48)]? $\endgroup$ – Animism Mar 22 '13 at 15:39
  • $\begingroup$ Wow. So [1 * (3/51)] * [1 * (3/50) * (2/49)] produces exactly the same answer as [1 * (3/51)] * [(48/50) * (3/49) * (2/48)]. I believe my original method (the former) was completely incorrect, and that the latter is the correct "license plate" method for finding the fixed-order hand probability. As for why they produce the same answer, that I can't explain. $\endgroup$ – Animism Mar 22 '13 at 15:47
  • $\begingroup$ What I thought was your license plate method was $(1)(3/51)(48/50)(3/49)(2/48)$, or a similar thing with the $3$ of a kind first. That is correct, apart from a missing $\binom{52}{2}$ term, or equivalently a missing $\binom{5}{3}$. $\endgroup$ – André Nicolas Mar 22 '13 at 16:21

I don't think a direct license plate method will work. Let me describe the process I'd use if I wanted to do something similar to that:

  1. What is the chance that the first card will be suited to a full house? 1, of course.

  2. What is the chance that the second card will be suited to a full house? 1 again - chance of match is 3/51.

  3. What is the chance that the third card will be suited to a full house? If the first and second matched, then 1 (chance of match - 2/50). If the first and second didn't match, then 6/50.

  4. What is the chance that the fourth card will be suited to a full house? If all three matched, then 48/49 (1+3). If two matched, then 5/49 (2/49 for 1+3, 3/49 for 2+2).

  5. Fifth card? If 1+3, then 3/48 = 1/16. If 2+2, then 4/48 = 1/12.

Now, here are the possible sequences:

Card, Match, Card, 2+2, FH -> 1(3/51)(48/50)(3/49)(1/12) = 6/20825

Card, Match, Card, 1+3, FH -> 1(3/51)(48/50)(2/49)(1/16) = 3/20825

Card, Match, 3, Card, FH -> 1(3/51)(2/50)(48/49)(1/16) = 3/20825

Card, Card, Match, 1+3, FH -> 1(48/51)(6/50)(2/49)(1/16) = 6/20825

Card, Card, Match, 2+2, FH -> 1(48/51)(6/50)(3/49)(1/12) = 12/20825

Summing them up, you get 30/20825 = 6/4165 = 0.0014405762304922


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