We took our AP Calc test last wednesday, so for the rest of the year we are making a movie on a math topic. Our topic is combinations and we are using poker to explain it.

 

My question is about odds ( and there are a few)

 

Let's say Chris has an Ace, a King, a Queen, a 3 and a 4 (A,K,Q,4,3) What are his odds of picking up the straight?

 

Is it 4/47 * 4/46, or 4/47 * 4*47?

 

Let's say Tom has two fives , a 6, a 7, and an 8 (5,5,6,7,8) What are his odds of picking up the three of a kind?

 

Is it 2/47 + 2/47 + 2/47?

 

Much help appreciated.

for the second case i think the odds of 3 of a kind would be 2/47 + (45/47)*2/46+ (45/47)*(45/46)*(2/45)

Edited May 8, 200619 yr by AssHatSoxFan

Go to the library and pick up Doyle Brunsons Book. Super System. The last 50 pages or so are all odds.

There are 8 cards out of 47 that will work for his 4th card (J or 10) and 4 cards out of 46 that would then work for his 5th card. (whichever is missing)

 

First card 1:5.875

Second card odds 1:11.5

Total to fill 1:17

 

edit: after figuring this, I checked in Advanced Concepts of Poker by Frank Wallance and he shows 1:24 (3 straight double open).

Three of a kind problem. How exact do you want to be?

 

2 fives remaining / 47 cards.

 

Odds of first draw being a five is 2:47

Second draw (if first not a five) is 2:46

Third Draw (if first two not a five) is 2:45

 

The more you draw and don't get a five, the more the odds are you will get a five on the next one, until you reach 2:2 and you will have to get a five.

 

So on to the problem. Your chances of pulling trips is about 1 in 9.

You guys aren't factoring in that you can get 3 of a kind by picking up running 6-7or8's, which I suppose would give you a full house, but it's still 3 of a kind.

After my AP Calc test I shunned math forever.

 

I shall never go back.

 

On a side note, though, in none of my math classes, ever, have we discussed probability. Not until I took Stat this spring have I done anything at all with it.

QUOTE(maggsmaggs @ May 8, 2006 -> 09:43 AM) <{POST_SNAPBACK}>

We took our AP Calc test last wednesday, so for the rest of the year we are making a movie on a math topic. Our topic is combinations and we are using poker to explain it.

 

My question is about odds ( and there are a few)

 

Let's say Chris has an Ace, a King, a Queen, a 3 and a 4 (A,K,Q,4,3) What are his odds of picking up the straight?

 

Is it 4/47 * 4/46, or 4/47 * 4*47?

 

Let's say Tom has two fives , a 6, a 7, and an 8 (5,5,6,7,8) What are his odds of picking up the three of a kind?

 

Is it 2/47 + 2/47 + 2/47?

 

Much help appreciated.

I assume the rules are draw poker and he's playing against himself? (up to 3 new cards)

First one: Holds onto A,K,Q= 4/47 *4/46 (4 J's, and 4 10's left in the deck)

Holds onto A, 3, 4=4/47 *4/46 (4 2's, and 4 5's left in the deck)

 

2nd one: Holds onto two 5's. (3/47*2/46*1/45)*10 (3 of Aces, 2, 3, 4, 9. 10, J, Q, K,) + 2/47 (two 5's left in the deck)<---Now that I think about it, that's not considered a three of a kind with a full house

Holds onto 6 and 7 = (3/47*2/46)*2 (catching 2 7's or 2 6's)+ (3/47*2/46*1/45)*10 (3 of Aces, 2, 3, 4, 9. 10, J, Q, K,)

Same odds for holding only 7 and 8, or 6 and 8.

Holds onto two 5's and one of the other cards: 2/47 (5 on the first new card draw) + 2/46 (5 on the second new card draw if the first card was not a 5)

Holds on to two 5's (avoid full house): 2/47 (5 on the first new card draw) + 2/46 (5 on the second new card draw if the first card was not a 5) +2/45 (5 on the third card if niether the first or second card is a 5) [Texsox did this one]

 

For the first one my TI-89 tells me it's .74%

Second hand first method is 4.31%

Second hand second mehtod is .62%

Second hand third method is 8.60%

Second hand fourth method is 13%

Edited May 17, 200619 yr by santo=dorf

Are wrap around straights allowed? How many cards are coming? You need to be more specific to get better answers.

 

If you are looking for a 2 (wrap around straight) then the odds that the first card is a 2 is 4/47 which is 8.5%. If there are 2 cards coming its 4/47 + 4/46 which is 17%. I would assume you have those 5 cards playing Texas Hold 'Em after the flop? If that's the case then it is 17% odds of catching it on the turn or the river, but only 8.5% of catching it on the turn.

 

Also, was it the AB or BC test? I'm going to be taking the BC test next year and would like to hear some input.

Thanks, we just turned in the movie. I took BC, you got to pay attention in class (big,) ask your teacher for help a lot (big,) read the text book even in if you think you understand it right. The hardest stuff is the complex derivatives and integrals and the related rates (damn them.) Hope this helps, the BC test is not as bad as everyone makes it out to be.

I assume the rules are draw poker and he's playing against himself? (up to 3 new cards)

First one: Holds onto A,K,Q= 4/47 *4/46 (4 J's, and 4 10's left in the deck)

Holds onto A, 3, 4=4/47 *4/46 (4 2's, and 4 5's left in the deck)

 

2nd one: Holds onto two 5's. (3/47*2/46*1/45)*10 (3 of Aces, 2, 3, 4, 9. 10, J, Q, K,) + 2/47 (two 5's left in the deck)

Holds onto 6 and 7 = (3/47*2/46)*2 (catching 2 7's or 2 6's)+ (3/47*2/46*1/45)*10 (3 of Aces, 2, 3, 4, 9. 10, J, Q, K,)

Same odds for holding only 7 and 8, or 6 and 8.

Holds onto two 5's and one of the other cards: 2/47 (5 on the first new card draw) + 2/46 (5 on the second new card draw if the first card was not a 5)

Holds on to two 5's (avoid full house): 2/47 (5 on the first new card draw) + 2/46 (5 on the second new card draw if the first card was not a 5) +2/45 (5 on the third card if niether the first or second card is a 5) [Texsox did this one]

 

For the first one my TI-89 tells me it's .74%

Second hand first method is 4.31%

Second hand second mehtod is .62%

Second hand third method is 8.60%

Second hand fourth method is 13%

 

Santo, on the straight, you don't have to catch it in order, so the first card drawn could be a J or a 10 so you are looking at 8/47 and the next card has to be the other or 4/46. Doesn't that make sense?

QUOTE(Texsox @ May 17, 2006 -> 10:32 PM) <{POST_SNAPBACK}>

Santo, on the straight, you don't have to catch it in order, so the first card drawn could be a J or a 10 so you are looking at 8/47 and the next card has to be the other or 4/46. Doesn't that make sense?

Good catch.

Good catch.

 

Checking the only poker book I have here that lists odds, and it's a 40 year old book, doesn't exactly match. I was beginning to think I was wrong.