math help: probability
i need some help with my math homework.....yeah yeah its probably really easy for most of you but im kinda stuck though.
example of what im working on would be:
drawing a 6 from a deck of cards.
or
drawing 2 cards that are all 2 from a deck of cards.
an explination on how to do these would be very helpfull, thanks.
example of what im working on would be:
drawing a 6 from a deck of cards.
or
drawing 2 cards that are all 2 from a deck of cards.
an explination on how to do these would be very helpfull, thanks.
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A normal deck contains 52 cards, divided into four suits, each numbered from one (ace) to 13 (king).
1) Drawing a 6: there are four sixes in the deck, one per suit. So the chance of drawing any one of them is just 4 in 52 (= 1/13).
2) Drawing a pair of twos: For the first card, the probability is the same as in (1). For the second card, you now have 51 cards, of which 3 are two's, so the chance of drawing any of them is just 3/51. When you want the chances of a "compound event" (like this), you multiply the probabilities of each event separately. So the chances of drawing a pair of two's is 4/52 * 3/51. I leave the arithmetic to you.
Now, here's a question to see if you've understood. Suppose you're playing poker, solitaire (so you are the only person getting cards). What are the chances of being dealt of royal flush (10, J, Q, K, A, all the same suit)?
Please, let the original author answer before anyone else does!
1) Drawing a 6: there are four sixes in the deck, one per suit. So the chance of drawing any one of them is just 4 in 52 (= 1/13).
2) Drawing a pair of twos: For the first card, the probability is the same as in (1). For the second card, you now have 51 cards, of which 3 are two's, so the chance of drawing any of them is just 3/51. When you want the chances of a "compound event" (like this), you multiply the probabilities of each event separately. So the chances of drawing a pair of two's is 4/52 * 3/51. I leave the arithmetic to you.
Now, here's a question to see if you've understood. Suppose you're playing poker, solitaire (so you are the only person getting cards). What are the chances of being dealt of royal flush (10, J, Q, K, A, all the same suit)?
Please, let the original author answer before anyone else does!
well i was thinking that but we were getting these strange answers and i didnt know really if this was the right way or if there was more than one way to do this.
heres an example: drawing 3 cards that are all queen from a deck of cards
the "answer" was something like 24/132600....if you can explain that would be great but ima have to do it the way you did and i'll get back to you on your question haha
heres an example: drawing 3 cards that are all queen from a deck of cards
the "answer" was something like 24/132600....if you can explain that would be great but ima have to do it the way you did and i'll get back to you on your question haha
Drawing three queens: 4/52 * 3/51 * 2/50 = 1/13 * 1/17 * 1/25 = 1/5525 which is exactly what you got. It's the same logic as drawing a pair of twos.
Notice that these odds are much smaller (by a factor of 13) than what you might read in a poker book. The reason is that the odds for poker (except for the special case of a royal flush) do not specify the first card.
For example, let's take your example and generalize: what are the odds of drawing three of a kind with three cards? The first card is "free" -- it doesn't matter what card you get. The second card has to match the first, so that is 3/51 (=1/17), and the third card has to match the second, so that is 2/50 (=1/25). The total probability is 1*1/17*1/25 = 1/425.
Now, let's take the case of a real 5-card poker hand. What are the odds of being dealt three-of-a-kind? Here it gets more complicated, because you're asking for three cards out of five to match, but you don't care which three (first, second, third; third, fourth, fifth; or first, second, and fourth; doesn't matter).
So what you do is you say, let's take the odds for the first three cards, which 1/425 (see above). The last two cards are free (they contribute factors of 1). Then you can work out the different hand arrangements, and that multiplies the odds for a single deal (approximately!).
How many ways can you pull three cards of out of a set of five? The answer is C(5,3), often spoken as "five choose three", and equal (in this case) to 5! / [3! (5-3)!] = 10. So there are ten different arrangements for any given three-of-a-kind hand, and hence the overall odds are 10/425 = 1/42.5 (or 2:95).
But this is only an approximation! If you think about it, as you deal the cards, the denominators used for the three-of-a-kind calculation above are different (e.g., 3/49 instead of 3/51). It also gets more interesting if multiple hands are being dealt. Working through the exact math is more complicated, but not impossible. There's a very nice Wikipedia article about it.
Notice that these odds are much smaller (by a factor of 13) than what you might read in a poker book. The reason is that the odds for poker (except for the special case of a royal flush) do not specify the first card.
For example, let's take your example and generalize: what are the odds of drawing three of a kind with three cards? The first card is "free" -- it doesn't matter what card you get. The second card has to match the first, so that is 3/51 (=1/17), and the third card has to match the second, so that is 2/50 (=1/25). The total probability is 1*1/17*1/25 = 1/425.
Now, let's take the case of a real 5-card poker hand. What are the odds of being dealt three-of-a-kind? Here it gets more complicated, because you're asking for three cards out of five to match, but you don't care which three (first, second, third; third, fourth, fifth; or first, second, and fourth; doesn't matter).
So what you do is you say, let's take the odds for the first three cards, which 1/425 (see above). The last two cards are free (they contribute factors of 1). Then you can work out the different hand arrangements, and that multiplies the odds for a single deal (approximately!).
How many ways can you pull three cards of out of a set of five? The answer is C(5,3), often spoken as "five choose three", and equal (in this case) to 5! / [3! (5-3)!] = 10. So there are ten different arrangements for any given three-of-a-kind hand, and hence the overall odds are 10/425 = 1/42.5 (or 2:95).
But this is only an approximation! If you think about it, as you deal the cards, the denominators used for the three-of-a-kind calculation above are different (e.g., 3/49 instead of 3/51). It also gets more interesting if multiple hands are being dealt. Working through the exact math is more complicated, but not impossible. There's a very nice Wikipedia article about it.
Feb 7, 2011. 11:52 AMArano
says:
do you know bernoulli? those treelike structures are wonderfull for more complex probability problems
Feb 7, 2011. 3:02 PMArano
says:
hell i wasn't aware of that number of bernoullis, but i was reffering to jacob bernoulli , after who the bernoulli process/trial/distribution is named... the tree like structure i meant seems to be connected to someone else i think now(was something else than pascals triangle)... i will look that up again in the math book i had at school
On a side note (for other people reading this) -- thank you very much for writing your questions as you did! You made it clear that this was homework, but you also made it clear that you wanted to understand how to get the answer, not just "do my work for me." Great approach!
Probability is very counter-intuitive (our brains have not evolved to do this correclty). I hope that I've been able to explain the way to solve these sorts of problems.
Probability is very counter-intuitive (our brains have not evolved to do this correclty). I hope that I've been able to explain the way to solve these sorts of problems.
thanks, some stuff just catches me off guard and i'd rather learn how to do it then have someone else do it for me.
Feb 7, 2011. 12:06 PMorksecurity
says:
Well-written questions draw well-written responses. Good going, everyone!
Well, there are 52 cards in a pack.
A six appears in a pack 4 times.
So the probability would be 4/25 or 6.25%
Hope this helps,
NM
P.S. Also, you can cut down the sums like this; if ther probability was 6/10, you can cut it down by halving both numbers so it would be 3/5. This can be done so you divide both numbers by 2, 3, 4 etc.
A six appears in a pack 4 times.
So the probability would be 4/25 or 6.25%
Hope this helps,
NM
P.S. Also, you can cut down the sums like this; if ther probability was 6/10, you can cut it down by halving both numbers so it would be 3/5. This can be done so you divide both numbers by 2, 3, 4 etc.
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