- #1
markh26
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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
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
