Tien Len/ Thirteen Probability Question

In summary, the conversation revolves around the idea of using the card game 13/Tien Len to teach students about probability. The teacher is confident in their theoretical probability calculation for a 4 of a kind, but unsure about the calculation for a double sequence. They ask for help in correcting their work, explaining it in a way that is easy for high school students to understand. The conversation also includes a reference to a website for calculating Tien Len probabilities.
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Hello,

I am a first year Algebra II teacher, and Probability is an Algebra II standard. I thought it would be fun to teach my students 13/Tien Len and then have them learn about probability after they are familiar with the game, for example the difference between experimental and theoretical probability.

If you are unfamiliar with 13, you have to get rid of your 13 cards as quickly as possible by beating combinations played be earlier cards. You get to lay down whatever combination you want whenever someone passes.

The highest ranking card is a 2. Whenever someone lays down a 2 though, they run the risk of it being beat by a "2 killer" which comes in 2 forms. A 4 of a kind, or a double sequence.

I'm confident of my 4 of a kind theoretical probability calculation, but the double sequence one I'm very unsure on. In fact it prompted me to find this great website.

A double sequence is like 33-44-55 all the way up though KK-AA-22. So I know there are 11 different double sequences you can get.

My calculation (I'm trying to do them w/o any formulas other than nCr or nPr as well since the kids I'm teaching it to, are not that strong in math, I'm teaching in an inner city school)

So can you please correct my work if possible.

I figured the chance of getting any particular card is 13/52 since everyone is dealt 13 cards. Then the chance of getting another particular card is 12/51. There are 6 combinations of any pair though so I used that in my calculation. (4c2 = 6)

So I multiplied (13/52 x 12/51 x 6) (11/50 x 10/49 x 6) (9/48 x 8/47 x 6) and then I multiplied that times 11 since there are 11 different double sequences. When I multiplied it all out I came out with a probability of around 20% which is close to the experimental probability my students are getting during our games.

I know the people that frequent this forum are my intellectual superiors, however can you please keep that in mind with your answer and dumb it down to the level that I could explain it to a hormone filled HS Sophmore or Junior? Thanks in advance, Mark
 
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  • #2

What is Tien Len/ Thirteen Probability Question?

Tien Len/ Thirteen Probability Question is a mathematical puzzle that involves determining the probability of a certain hand combination in the card game Tien Len.

How do you solve the Tien Len/ Thirteen Probability Question?

The Tien Len/ Thirteen Probability Question can be solved by using the concept of combinatorics, specifically the combination formula. The formula is nCr = n! / (r!(n-r)!), where n is the total number of cards in the deck and r is the number of cards in the hand combination being considered.

What is the significance of Tien Len/ Thirteen Probability Question?

Tien Len/ Thirteen Probability Question is a popular puzzle among mathematicians as it showcases the application of combinatorics in a real-life scenario. It also helps in developing critical thinking and problem-solving skills.

Can Tien Len/ Thirteen Probability Question be applied to other card games?

Yes, the concept of Tien Len/ Thirteen Probability Question can be applied to other card games that involve forming specific hand combinations, such as poker or rummy.

Is there a shortcut to solving Tien Len/ Thirteen Probability Question?

No, there is no shortcut to solving Tien Len/ Thirteen Probability Question. It requires careful analysis and application of the combination formula to determine the probability of a specific hand combination.

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