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schmidtdancer
Back to the cards! In poker, a flush is when all five cards are the same suit. Find the probability of being dealt a flush (when being dealt five cards). Start by just considering clubs.
a) What is the probability that the first card dealt is a club? b) What is the probability that the second card dealt is a club given that the first one was a club? c) What is the probability that the third card dealt is a club given that the first two were clubs? d) What is the probability that the fourth card dealt is a club given that the first three were clubs? e) What is the probability that the fifth card dealt is a club given that the first four were clubs? f) The probability of being dealt all five clubs is the product of the above probabilities. Why is this true and what is this probability? g) You have now found the probability of being dealt a flush in clubs. This is the same as the probability of being dealt a flush in diamonds, hearts, or spades. Then, what is the proability of being dealt a flush?
Umm... im not sure
There are 52 cards in a standard deck. Divide by 4 to get the number of cards of each suit.
That would mean there are 13 club cards out of 52 cards. Wouldn't that mean the probability of drawing a club is 1 out of 4?
Yes, drawing the first club out of the full deck of 52 cards has a probability of 1 out of 4, or 0.25. What about the probabiliity of the second card: - how many clubs are left (assuming first one is a club) ? - how many cards are left in the deck?
Now on the 2nd draw there would be 13 club cards out of 51 3rd draw 12 out of 50 4th draw 11 out of 49 5th draw 10 out of 48
Wouldn't that be the chances for each draw assuming he draws into a 5 card flush. Now take the product of all those chances to get the probability. 1/4 * 13/51 * 6/25 * 11/49 * 5/24
I would guesstimate a little less than 1400 to 1
so a i got 1/4. what is b?
Chance according to Google is 0.003940. or about 254 to 1 so it looks like I probably erred.
I am afraid I have been away from those kind of problems too long. Hopfully, @mathmate will provide further assistance.
Do u know anyone online right now that can give me assistance now?
The Google solution involved a flush in any suit not just clubs.
The probability of a flush can occur in hearts, diamonds, spades or clubs, just as long as all cards are the same suit!
So what would my answers be?
You requested the probability of a flush in clubs only.
Yea. thats what the question said
how would I start answering B?
I answered that, if the first one was a club, you now have a deck of 51 cards of which 12 are clubs; 12/51 or 0.235294
Ok so b would then be 12/51?
Here is mathmate. hopefully shed some light on this.
I have (a) 1/4 (b) 12/51 now c?
Ask yourself how many cards are now in the deck, how many are clubs and figure out the probability using the method you have been taught.
Sorry for being away. @radar sorry, I was just questioning in case there was a typo. For (b) After the first card, there are 13-1=12 clubs left out of 51. So the probability is 12/51. Or, using conditional probabilities: P(1)=13/52 P(1&2)=13/52*12/51 P(2|1)=P(2&1)/P(1)=(13/52*12/51) / (13/52) = 12/51, same as before.
Okay I got that for (b) too!:) Im not sure how to find c now
Sorry but I have to now run, you are in good hands.
Would (c) be 11/51 then?
@schmidtdancer The questions are made in such a way to guide you to the final answer. I suggest that after a and b have been explained and answered, it would be advantageous for you to continue the logic and post your suggested responses for verification. What do you think?
? I just need clarification on how to find C. can u guide me through the steps....and ill figure it out by myself then u can check?
Thanks! how do i begin c?
c) What is the probability that the third card dealt is a club given that the first two were clubs?
For the third club in a row, how many clubs are left? and how many cards are left in the deck?
since there are 14 clubs in a set, then we would have 11 left right?
because b is 12/51? and were losing another so would c be 11/51?
I'll make it clear: After the first two clubs are drawn and before we draw the third card, how many clubs remain in the deck, and how many cards total remain?
there are 14 clubs total in a deck... and 52 cards total in a deck... so, after two are drawn, then we have 12 clubs and 50 cards?
is that for b or c?
Each deck has 52 cards, divided by 4 suits gives 13 cards per suit to start with (not 14). (a) before drawing any card, we have 13 clubs and 52 cards. (b) given the first card drawn was a club, before drawing the second card, we have 12 clubs and 51 cards. (c) given the first 2 cards drawn were clubs, before drawing the third card, we have how many clubs and how many cards in the deck?
We have only drawn 2 clubs out of 13, how many left?
Sorry, the OS seems to be very selective in response. I cannot get to your question unless I go by your profile. When I go by "mathematics", it never responds for the past two days.
Its ok. but is 11 right?
(c) Also, drawn two cards (clubs) out of 52, how many left?
so 11/52 would be c?
Yes, that is correct for (c). Again, your response was not updated. I had to check through you profile every time I suspect a response. You're comfortable continuing?
How would I find (d) now @mathmate
Each deck has 52 cards, divided by 4 suits gives 13 cards per suit to start with (not 14). (a) before drawing any card, we have 13 clubs and 52 cards. (b) given the first card drawn was a club, before drawing the second card, we have 12 clubs and 51 cards. (c) given the first 2 cards drawn were clubs, before drawing the third card, we have how many clubs and how many cards in the deck? (d) given the first 3 cards drawn were clubs, before drawing the fourth card, we have how many clubs and how many cards in the deck?
so (d) is: 10/49?
This is what I have.... is it correct @mathmate ? (a) 1/4 (b) 12/51 (c) 11/50 (d) 10/49 (e) 9/48
Looks good so far. Keep it up, you're almost there!
Im a little confused on f and g!
Have you done conditional probability before?
When it says product, do I just multiply a-e?
Yes, the numerical part is just the product of the 5 probabilities. Give me a minute for the explanation part of (f). Once you have the numerical probability of clubs obtained in (f), how would you propose to find (g)?
so f is then: 11880/23990400??
Then would g be the decimal value of f?
It's almost correct, but you need to simplify it to the simplest form. Can you do that?
I was gonna ask u lol, I'm not sure how to simplify it... im trying
No. Think of (g) is for the case where there are 4 suits instead of just clubs. Imagine buying raffle tickets. What are the changes of winning if you bought 4 instead of one?
I have 495/999600 so far
Is this f?? 33/66640
Factors that you can cancel are like 36...
is this right @mathmate 33/66640
Ok thank you! now, g?
g) You have now found the probability of being dealt a flush in clubs. This is the same as the probability of being dealt a flush in diamonds, hearts, or spades. Then, what is the proability of being dealt a flush?
im not sure what its asking for an answer....@mathmate
Means a flush of any of the 4 suits. Imagine buying raffle tickets. What are the changes of winning if you bought 4 instead of one?
4 times as much?@mathmate
Right! That's. I hope you are better prepared for the next question.
But what would g be??
Is the answer: 4 times as much? @mathmate
4 times what you got for (f).
132/266560 in fraction form <-- is this the answer @mathmate
I would rather put it as 4*(33/66640) =33/16660. In probabilities, small numbers like 0.002 is better represented by fractions.
thx! so g @mathmate is 33/16660?
but thats lower than f.
Yes. I am sorry it is probably painful for you as much as for me because the system does not respond (does not update). So I don't really know when you put in a response.
its ok but @mathmate how is g 33/16660? when f is higher than that value?
When the denominator is 4 times smaller, it means that the fraction is 4 times bigger. For example, 1/4 is smaller than 1/1.
Oh oak!! thanks for all your help
You're welcome! Good luck with your homework/exam! :)