Consecutive Pairs vs 4 of a kind (Poker Variant)

[RocketManJames]

Hi, there is a card game that is played amongst my friends and I on occasion. I believe it is of Vietnamese descent. Anyway, in the game there are two things called bombs. The "lesser" bomb is a 4 of a kind. The "greater" bomb is a set of 6 cards consisting of 3 consecutive pairs.

The game is played with exactly 13 cards. Which is actually more probable? The 6 consecutive pairs or the 4 of a kind?

My intuition tells me that 3 consecutive pairs would be more common. If this is true, I'm surprised that it is the more powerful bomb.

Anyone want to tackle this for me?

Thanks in advance.

-RMJ

[BruceZ]

The "greater" bomb is a set of 6 cards consisting of 3 consecutive pairs....
The game is played with exactly 13 cards. Which is actually more probable? The 6 consecutive pairs or the 4 of a kind?


Assuming you mean 3 consecutive pairs as you said previously, that is much more probable than quads. It will occur 20% of the time or 4-to-1. Quads will occur 3.4% of the time or 28.2-to-1.

P(quads in 13 cards) =
[ 13*C(48,9) -
C(13,2)*C(44,5) +
C(13,3)*40 ] / C(52,13)
= 3.4% or 28.2-to-1

P(3 consecutive pairs in 13 cards) =
[ 11*6^3*C(46,7) -
(11 + 10*6^2 + 9*6^4 + 36*6^6)*40 ] / C(52,13)
= 20% or 4-to-1

The second term subtracts the cases where we have two of these. There are 11 which overlap in 3 places (consecutive quads), 10 which overlap in 2 places, 9 which overlap in 1 place, and 8+7+6+5+4+3+2+1 = 36 which overlap in 0 places with different spacings from 0 to 7. This whole term is negligible.

[BruceZ]

The probability for 3 consecutive pairs is less than 20% because I am over counting sets and quads. I may correct the calculation later; however, since the question was whether this is more or less probable than quads, we can answer that without calculating this probability exactly. Note that:

P(3 consecutive pairs in 13 cards) > [ 11*6^3*C(40,7) - 10*6^4*C(44,5) ] / C(52,13) = 4.8%

The actual probability will be greater than this since this only considers cases where we get exactly 3 consecutive pairs (no set or quads). The subtracted term is for 4 or more in a row, and this is an upper bound, so subtracting it gives a lower bound. This is more probable than quads even just considering this subset, since the probability of quads is only 3.4%.

[RocketManJames]

Thanks, BruceZ. I don't care too much about the actual probability, so long as I know that the 3 consecutive pairs occur a lot more often than the quads.

-RMJ