Math and Probability Problem...

[b fold the truth]

This is due tomorrow morning. Help...please.

There are 2 identical urns and ten identical red balls and ten identical blue balls.

The twenty balls, which are identical except for color, will all be placed in the two identical urns.

You will leave the room and the urns will then be moved around and the contents shaken up. You will return and try to draw one ball in the following manner.

a) first you will choose an urn (you cannot approach the urns until you choose one),
b) then you will draw a ball from the chosen urn (without looking inside).

If you draw a blue ball, then you win. Otherwise you lose.

Suppose that you could choose the original placement of the balls in the two urns. You may arrange them in any way you would like except that all 20 of the balls must go into the two urns.

1) How should you place the balls to give yourself the best chance of winning?


2) What is the probability of winning with your placement?

 

[ConstantChange]

I love and hate these problems at the same time. I wish I could help, but it's midnight, I work at 8:00am, and I was on campus from 8:30am to 6:00pm tonight. I'm mentally exhausted.

It would take me awhile to figure this one out.

There has to be some way to improve the 50/50 chance.

If you put 19 balls in one and one blue ball in another.......ahhhh.....sorry B....I'm to tired for this.

When you find out the answer.....please post it.

Sorry I can't help.

 

[b fold the truth]

I am thinking that if I put it like this...

Urn I: 9 blue, 10 red
Urn II: 1 blue

I have a 50/50 chance of picking each urn. There are 9 to 10 odds (9/19) that I will get a blue ball if I choose Urn I.

I have a 50/50 chance of picking Urn II. There are 1 to 1 odds (1/1) that I will get a blue ball in I choose Urn II.

Would the probability of getting a blue ball be 0.473 in Urn I?
Would the probability of getting a blue ball be 1.000 in Urn II?

So...with a 50/50 chance of getting each Urn...does that leave me with a probability of getting a blue ball 0.73?

B True

 

[KAYNE]

THE ANSWER IS C.



KAYNE

 

[Steroid_Virgin]

I like your idea.. 1 blue ball in one urn... that has to yield the best probability...

 

[WODIN]

It's a basic 50/50 proposition because you are using the two urns up front. The 20 balls are irrelivant. This is a descrete probability because you are to approach one of two urns without knowing the content of either.

 

[Steroid_Virgin]

yup you are right bfold... 0.7365 odds.. is the max. The worst thing you could do is put a red by ball itself, or more than one blue ball alone in an urn with no red balls..

nice work, you dont need help bro!

 

[Steroid_Virgin]

wodin.. no... Bfold was right, I had originally thought what you did, but worked through the problem and found that his first guess was right.

 

[WODIN]

I think you can say that if he draws a blue ball then there is a 73~ % probability that the ball came from the urn with the fewest balls. I don't think you can say that he has a greater chance up front though because the initial pick is one of two urns.

Oh well...

Bfold post up the answer when you get it. This helps pass the time.

 

[Cyclemein]

2Thick, of all of the threads, answer this one.

 

[Fonz]

This is due tomorrow morning. Help...please.

There are 2 identical urns and ten identical red balls and ten identical blue balls.

The twenty balls, which are identical except for color, will all be placed in the two identical urns.

You will leave the room and the urns will then be moved around and the contents shaken up. You will return and try to draw one ball in the following manner.

a) first you will choose an urn (you cannot approach the urns until you choose one),
b) then you will draw a ball from the chosen urn (without looking inside).

If you draw a blue ball, then you win. Otherwise you lose.

Suppose that you could choose the original placement of the balls in the two urns. You may arrange them in any way you would like except that all 20 of the balls must go into the two urns.

1) How should you place the balls to give yourself the best chance of winning?


2) What is the probability of winning with your placement?

I know the answer.

I'm still going to wait until Warik wakes up.

It is fairly simple.

Want to see if he's getting smarter.......

Fonz

 

[KHMER ROGUE]

2Thick, of all of the threads, answer this one.

I'll bump it up for ya!

 

[Steroid_Virgin]

ok fonz... you say 73% is not the best odds?

 

[Intenceman]

I got to this thread to late to help, sorry.

IF I am reading the question right, then you have two trials or chances- one for each urn. The total probability when you have two trials since they are discrete trials- is the first plus the second-therefore if you put one blue ball alone in the first- the probability for the first is,as you stated, 1.0. The total probability would then be 1.0 plus the probability of the second, which is then is 9/19 =.47., so 1.0+.47= 1.47 divided by two =.73. If you put 5 of each in each urn, it becomes 5/10=.50 + 5/10= .50 or .50+.50= 1.00 divided by two which equals .50. Your solution is better.

 

[HappyScrappy]

This is a common problem - the one blue ball in one, and then all the rest in the other is the given solution - yes, it is near 75% prob, which is the best you can do over the straight 50/50.

It is easy when it is basic like this.

For discrete calculations they usually have some changing variable over iterations - like take a ball out, whatever color it is, another ball appears that is also that color, put them both back in, then take a ball out, etc etc - and then you calculate at what point they are all the same color - at what iteration point.

 

[Fonz]

ok fonz... you say 73% is not the best odds?

Intencemen beat me to it.....growl

This question is geared towards what is refreerd to an ati-probability. (1-P) Simple concept that some people just don't get.

I just wanted to see if Warik could get it right or if he's
wasting his talents in menial math courses.

Fonz

 

[Steroid_Virgin]

so wodin was wrong.... and I was right.. the exact probability is 0.7365 as I stipulated earlier... sit down Fonz, sit down Wodin

BOOOYAAA

 

[Fonz]

This is a common problem - the one blue ball in one, and then all the rest in the other is the given solution - yes, it is near 75% prob, which is the best you can do over the straight 50/50.

It is easy when it is basic like this.

For discrete calculations they usually have some changing variable over iterations - like take a ball out, whatever color it is, another ball appears that is also that color, put them both back in, then take a ball out, etc etc - and then you calculate at what point they are all the same color - at what iteration point.

I got a problem for you Scrappy.

Came up in my cousins math exam.

Whats the probability of winning the lottery?

Assume you have 48 choices and 6 number picks.

You have 2 hours.

Fonz

 

[Warik]

Re: Re: Math and Probability Problem...

I know the answer.

I'm still going to wait until Warik wakes up.

It is fairly simple.

Want to see if he's getting smarter.......

Fonz

Thanks, chief.

So, we are choosing the urn randomly and thus we do not know which urn contains which set of balls. However, we DO know that our chances of picking either urn are identical, so 50/50 is part of the equation.

Now, your arrangement of 19 in 1 and 1 blue in the other would be awesome if you always chose that particular urn, but you can't guarantee that.

So, what are the chances of pulling a blue ball?

(P(Choosing Urn 1) * P(Choosing Blue Ball)) + (P(Choosing Urn 2) * P(Choosing Blue Ball))

We'll assume Urn 1 is the 9/19 urn and urn 2 is the one with 1 blue ball.

So, P(Choosing Urn 1) = .50
P(Choosing Blue Ball) = 9/19 = .474

P(Choosing Urn 2) = .50
P(Choosing Blue Ball) = 1

(.50)(.474) + (.50)(1) = .237 + .50 = .737

So, fonzy, I believe he is correct. The probabilities of choosing the two urns are identical, so that is irrelevant. The best thing to then do is to maximize the probability in each urn of getting a blue ball.

P in urn 1 is .474
P in urn 2 is 1.00

If you put any blue balls from urn 1 into urn 2, that will reduce urn 1's probability without increasing urn 2s, and if you take any out of urn 2, that will raise urn 1s to 50/50 and drop urn 2s to zero.

Plus, I ran a computer sim to test it and this seems like the best solution.

So, he's right. I'd love to see your solution.

-Warik

 

[HappyScrappy]

I got a problem for you Scrappy.

Came up in my cousins math exam.

Whats the probability of winning the lottery?

Assume you have 48 choices and 6 number picks.

You have 2 hours.

Fonz

48 choices and 6 number picks? please clarify.

meaning that I can choose from the pool of numbers 1-48, and I can pick 6 numbers out of that?

 

[Fonz]

Re: Re: Re: Math and Probability Problem...

Thanks, chief.

So, we are choosing the urn randomly and thus we do not know which urn contains which set of balls. However, we DO know that our chances of picking either urn are identical, so 50/50 is part of the equation.

Now, your arrangement of 19 in 1 and 1 blue in the other would be awesome if you always chose that particular urn, but you can't guarantee that.

So, what are the chances of pulling a blue ball?

(P(Choosing Urn 1) * P(Choosing Blue Ball)) + (P(Choosing Urn 2) * P(Choosing Blue Ball))

We'll assume Urn 1 is the 9/19 urn and urn 2 is the one with 1 blue ball.

So, P(Choosing Urn 1) = .50
P(Choosing Blue Ball) = 9/19 = .474

P(Choosing Urn 2) = .50
P(Choosing Blue Ball) = 1

(.50)(.474) + (.50)(1) = .237 + .50 = .737

So, fonzy, I believe he is correct. The probabilities of choosing the two urns are identical, so that is irrelevant. The best thing to then do is to maximize the probability in each urn of getting a blue ball.

P in urn 1 is .474
P in urn 2 is 1.00

If you put any blue balls from urn 1 into urn 2, that will reduce urn 1's probability without increasing urn 2s, and if you take any out of urn 2, that will raise urn 1s to 50/50 and drop urn 2s to zero.

Plus, I ran a computer sim to test it and this seems like the best solution.

So, he's right. I'd love to see your solution.

-Warik

OK. Sehr Gut.

Now do the lottery question.

The last question was easy. The lottery question is a much better challenge for you.

Fonz

 

[Warik]

I got a problem for you Scrappy.

Came up in my cousins math exam.

Whats the probability of winning the lottery?

Assume you have 48 choices and 6 number picks.

You have 2 hours.

Fonz

The urn question was harder.

Assuming order matters here, there are about 8,835,488,640 different permutations of numbers you can get.

Only one of them will win.

1/8835488640 = 0.000000011% chance of winning the lottery.

If order doesn't matter, then your chances of winning are much better: 1/1712304 = 0.000058%.

If you played the lottery every day of your life, you would need to live 4,700 years to win.

If you played the lottery once per hour, you would need to live 195 years to win.

i.e.

Thank God people aren't smart enough to realize that the lottery is a futile effort, else I might have to actually pay for my own college.

-Warik

 

[Fonz]

48 choices and 6 number picks? please clarify.

meaning that I can choose from the pool of numbers 1-48, and I can pick 6 numbers out of that?

Ok, I'll clarify.

This is based on the spanish lotto system.

There are 6 boxes of 48 numbers each.

You have to tick one number in each box.

They have to be all different too. You can't have 48 in Box 1
and 3, or 23 in Box 3 or 5.

I think this one is a bit beyond Warik.

Fonz

 

[Warik]

Re: Re: Re: Re: Math and Probability Problem...

OK. Sehr Gut.

Now do the lottery question.

The last question was easy. The lottery question is a much better challenge for you.

Fonz

nigga please. That urn question was hard as fuck. Took me almost 5 minutes. I wipe my ass with your lottery question.

I think I'll keep you busy for a while.

Assume a poker game with Deck = {1,1,1,1,2,2,2,2,3,3,3,3} with 2 players. You are player one and are dealt a 3. Player 2 is dealt a 2. You are to receive 2 more cards. What are the chances of your final hand being 3 high?
-Warik

 

[HappyScrappy]

yeah, unless I'm missing something, I don't get how the lottery one is hard - I'm assuming I'm misreading it?

it is just (1/48)(1/47)(1/46)(1/45)(1/44)(1/43)

 

[Fonz]

yeah, unless I'm missing something, I don't get how the lottery one is hard - I'm assuming I'm misreading it?

it is just (1/48)(1/47)(1/46)(1/45)(1/44)(1/43)

Thats the mistake most people made.(Me too) It has to
be done doing Hypergeometric distribution.

The odds are actually approx: 14 million to one.

Care to try again Warik? (J/k)

What the hell is 3 high(I don't play poker. I'm a BlackJack fan)

Fonz

 

[Warik]

Thats the mistake most people made.(Me too) It has to
be done doing Hypergeometric distribution.

The odds are actually approx: 14 million to one.

Care to try again Warik? (J/k)

What the hell is 3 high(I don't play poker. I'm a BlackJack fan)

Fonz

Doh.

Shouldn't have used the combination formula for the order doesn't matter part. If order doesn't matter then it's (6 * 5 * 4 * 3 * 2 * 1) / (48 * 47 * 46 * 45 * 44 * 43). Should thus take longer than 195 years to win.

Oh, and 3 high means "I have no hand whatsoever, and the highest card I have is a 3."

Also assume that straights, flushes, etc... do not exist.

-Warik

 

[Steroid_Virgin]

Re: Re: Re: Re: Re: Math and Probability Problem...

nigga please. That urn question was hard as fuck. Took me almost 5 minutes. I wipe my ass with your lottery question.

I think I'll keep you busy for a while.

Assume a poker game with Deck = {1,1,1,1,2,2,2,2,3,3,3,3} with 2 players. You are player one and are dealt a 3. Player 2 is dealt a 2. You are to receive 2 more cards. What are the chances of your final hand being 3 high?
-Warik

Do I assume each of the two players both get two more cards, or just me (player 1)?

 

[Steroid_Virgin]

Also, does drawing a 3 high mean he pulls a straight 1,2,3 or can a 3 and a pair of 1's or three and pair of 2's also satisfy "3 high".

 

[b fold the truth]

Thanks everyone...I'm going to fill out my test now and take it to class...I'll let you know what he says is the correct answer today...

B True

 

[Fonz]

Re: Re: Re: Re: Re: Math and Probability Problem...

nigga please. That urn question was hard as fuck. Took me almost 5 minutes. I wipe my ass with your lottery question.

I think I'll keep you busy for a while.

Assume a poker game with Deck = {1,1,1,1,2,2,2,2,3,3,3,3} with 2 players. You are player one and are dealt a 3. Player 2 is dealt a 2. You are to receive 2 more cards. What are the chances of your final hand being 3 high?
-Warik

OK.

First Sequence:

12 cards.

Player One: 3
Player Two: 2

Odds cut down to 10.

Second Sequence:

Prob of player one getting dealt a 3: 3/10(Or not: 7/10)
Prob of player one getting dealt a 2: 3/10(Or not: 7/10)
Prob of Player on getting dealt a 1: 4/10(Or not: 6/10)

Prob of player 2 getting dealt a 3: 3/10(Or not: 7/10)
Prob of player 2 getting dealt a 2: 3/10(Or not: 7/10)
Prob of Player 2 geting dealt a 1: 4/10(Or not: 6/10)

Deck, cut down to 8.

Third Sequence:

Prob of player one getting a three: 3/8(Or not: 5/8)

Probability = P(OneHand2)(Not a 3) * P(OneHand3)(Not a 3)

P = (7/10) * (5/8) = 35/80 = 43.75%

Fonz

 

[Warik]

Re: Re: Re: Re: Re: Re: Math and Probability Problem...

Probability = P(OneHand2)(Not a 3) * P(OneHand3)(Not a 3)

P = (7/10) * (5/8) = 35/80 = 43.75%

Fonz

That's all fine and dandy, but what if player 1 is dealt 2 1s or 2 2s? That would give him either a pair of 1s or a pair of 2s, which is no longer 3 high.

muhaha.

-Warik

 

[Warik]

Also, does drawing a 3 high mean he pulls a straight 1,2,3 or can a 3 and a pair of 1's or three and pair of 2's also satisfy "3 high".

I said no straights or flushes exist... so the only possible hands are 1 pair, 3 of a kind, or <X> high. Pulling a pair of 1s or a pair of 2s kill a 3 high, so no that won't work either.

Be very afraid Fonz. This is just a dip in the water. My next problem will keep you here for an hour if you persist in using that silly P = 1 - Q strategy.

-Warik

 

[Fonz]

Re: Re: Re: Re: Re: Re: Re: Math and Probability Problem...

That's all fine and dandy, but what if player 1 is dealt 2 1s or 2 2s? That would give him either a pair of 1s or a pair of 2s, which is no longer 3 high.

muhaha.

-Warik

See...LOL

That is why I hate Probability questions.

If the wording isn't 100% right on ,the whole problem changes dramatically.

OK.....Give me a sec. I have to answer some mail first and make a few calls then i'll get on you problem. Its not difficult, just time consuming.

(Still a bit dazed from yesterday.....)

Fonz

 

[Warik]

I said no straights or flushes exist... so the only possible hands are 1 pair, 3 of a kind, or <X> high. Pulling a pair of 1s or a pair of 2s kill a 3 high, so no that won't work either.

Be very afraid Fonz. This is just a dip in the water. My next problem will keep you here for an hour if you persist in using that silly P = 1 - Q strategy.

-Warik

Actually, I'll give it to you now just to be evil.

Deck = {1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,5,5,5,5}

4 players

Player 1 is dealt a 1
Player 2 is dealt a 2
Player 3 is dealt a 1
Player 4 is dealt a 4.

You, and all other players, are to receive 3 more cards.

Find the probability of getting 5 high.

Possible hands = <X> high, one pair, three of a kind, four of a kind, two pair.

You have until after my lunch break. I'll take it around 2pm ET today.
-Warik

 

[Fonz]

Actually, I'll give it to you now just to be evil.

Deck = {1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,5,5,5,5}

4 players

Player 1 is dealt a 1
Player 2 is dealt a 2
Player 3 is dealt a 1
Player 4 is dealt a 4.

You, and all other players, are to receive 3 more cards.

Find the probability of getting 5 high.

Possible hands = <X> high, one pair, three of a kind, four of a kind, two pair.

You have until after my lunch break. I'll take it around 2pm ET today.
-Warik

LOL.....Stop being evil.

Halloween is like one of the only 5 days of the year when i drink.

Hope i don't get that Q wrong.........

Fonz

 

[Warik]

LOL.....Stop being evil.

I don't know how to be otherwise.

Halloween is like one of the only 5 days of the year when i drink.

Oh shit... the preemptive excuses.

Hope i don't get that Q wrong.........

I'm sure you'll do just fine. I've given you an hour and twenty minutes. If I can do it in 10, you can do it in 1:20. Might be cutting it a bit close. Tell ya what, you have until I get BACK from my lunch break. So you have an hour and 25 minutes. Gogogoo!!!

-Warik

 

[Warik]

Heeereeee fonzy fonzy fonzy?

 

[Fonz]

Heeereeee fonzy fonzy fonzy?

Don't start.....

I can't even do 2 + 2 now........

God I hate hangovers......

Did you stay home?

Fonz

 

[Fonz]

Actually, I'll give it to you now just to be evil.

Deck = {1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,5,5,5,5}

4 players

Player 1 is dealt a 1
Player 2 is dealt a 2
Player 3 is dealt a 1
Player 4 is dealt a 4.

You, and all other players, are to receive 3 more cards.

Find the probability of getting 5 high.

Possible hands = <X> high, one pair, three of a kind, four of a kind, two pair.

You have until after my lunch break. I'll take it around 2pm ET today.
-Warik

OK.....just popped 2g Enteric coated Aspirin.

Deck is 1-5 with choice =4

20 Cards total(1st hand dealt):

Hand 1: {1,2,1,4}

Prob 5: 1-1/5 (1-P) = 4/5

16 cards total2nd Hand dealt)

Prob 1: 4-2/16 = 1/8
Prob 2: 4-1/16 = 3/16
Prob 3: 4/16 = 1/4
Prob 4: 4/16 = 1/4
Prob 5: 4/16 = 1/4(No 5.........3/4)

12 Cards Total3rd Hand dealt)

Prob 5:

8 Cards Total4th Hand dealt)

Prob 5:

This is what is referred to as a decreasing numerical monotic
series....LOL

Might be beyond the scope of your course.(J/k)

Using (1-P)

4/5.........3/4........X............Y

X obviously has a denominator of 3 and Y of 2.

So, X = 2/3 and Y =2/2 =1

Makes perfect sense because as you play longer the probability
of getting a 5 increases. As you can see (1-P) = 1 at term Y.
Y being the third hand dealt. Also, there are 4 5's and 4 players.

So, P = (4/5(Irrelevant)) * 3/4 *2/3 *1/1 = 1/2 or 50%

Fonz

 

[Warik]

You're not even in the ballpark.

Damn... not even in the galaxy.

Since player 1 got a 1, the only way for him to get 5 high is if he gets a 2,3,4 and then a 5. Any other combination of cards would result in a pair, 3 of a kind, or two pair.

A 1,2,3,4,5 is also known as a straight - a hand which is far less likely to occur than a pair of 3 of a kind. You're given 1 card and think there's a 50% chance you'll get the other 4? muahaha.... perhaps my problem is "beyond the scope" of your statistical comprehension.

-Warik

 

[Fonz]

You're not even in the ballpark.

Damn... not even in the galaxy.

Since player 1 got a 1, the only way for him to get 5 high is if he gets a 2,3,4 and then a 5. Any other combination of cards would result in a pair, 3 of a kind, or two pair.

A 1,2,3,4,5 is also known as a straight - a hand which is far less likely to occur than a pair of 3 of a kind. You're given 1 card and think there's a 50% chance you'll get the other 4? muahaha.... perhaps my problem is "beyond the scope" of your statistical comprehension.

-Warik

Anyways...jokes aside.

I don't think I understood the ghist of your problem.

If you want to include straights, pairs etc... then its much more
complicated(obviously).

I certainly didn't.

No way you can do it in 10-15min.

Well...maybe. Just not me and not today.

I'll have an answer for you tommorrow. I'm actually wearing
ear mufflers to not hear anyhting... I look like the easter Bunny.
According to my sister(She's laughing her head off)

Alcohol and Fonz doesn't mix very well.

You win, for now......(Stress now)

Fonz

 

[Warik]

Anyways...jokes aside.

I don't think I understood the ghist of your problem.

If you want to include straights, pairs etc... then its much more
complicated(obviously).

Ignore that part about the straight. I thought I said 4 cards... it's really 3.

Here's the problem:

Deck = {all that shit I said}

Player 1,2,3, and 4 get those cards I mentioned.

So, now with the remaining deck = {1,1,2,2,2,3,3,3,3,4,4,4,5,5,5,5}, each player is to be given 3 cards.

So, what are the chances of player 1 getting 5 high?

That would mean that you would have to a) get a 5 and b) NOT get any kind of pair. No 1,1, no 2,2,2, no 1,1,3,3 or any shit like that. You would HAVE to get either:

1,2,3,5
or
1,2,4,5
or 2,3,4,5

etc..

in some arrangement.

No way you can do it in 10-15min.

I'll take that as a compliment, because I did. =)

You win, for now......(Stress now)

*flex* I Beat Fonz.

-Warik

 

[Fonz]

Ignore that part about the straight. I thought I said 4 cards... it's really 3.

Here's the problem:

Deck = {all that shit I said}

Player 1,2,3, and 4 get those cards I mentioned.

So, now with the remaining deck = {1,1,2,2,2,3,3,3,3,4,4,4,5,5,5,5}, each player is to be given 3 cards.

So, what are the chances of player 1 getting 5 high?

That would mean that you would have to a) get a 5 and b) NOT get any kind of pair. No 1,1, no 2,2,2, no 1,1,3,3 or any shit like that. You would HAVE to get either:

1,2,3,5
or
1,2,4,5
or 2,3,4,5

etc..

in some arrangement.



I'll take that as a compliment, because I did. =)



*flex* I Beat Fonz.

-Warik

LOL

You have one day to enjoy it........

My hangover(probably not the cold) will be gone tommorrow.

Thats what i get for wearing a Lycra Superman costume
in 40F weather up here in DC.

Damn it was cold.

Fonz

 

[Warik]

LOL

You have one day to enjoy it........

My hangover(probably not the cold) will be gone tommorrow.

Thats what i get for wearing a Lycra Superman costume
in 40F weather up here in DC.

Damn it was cold.

Fonz

Ah... the old hangover excuse. take a seat kiddo =)

-Warik

 

[Fonz]

Ah... the old hangover excuse. take a seat kiddo =)

-Warik

LOL

B-Day, Halloween, New Years, GF's B'Day

4 days/365...

Fonz

 

[Warik]

You drink on Halloween but not St. Patrick's or Christmas? nigga please. It's OK to admit that there is something you don't know.

There are things I don't know either... like why you persist in your silly hangover excuse. I don't know that.

-Warik

 

[Fonz]

You drink on Halloween but not St. Patrick's or Christmas? nigga please. It's OK to admit that there is something you don't know.

There are things I don't know either... like why you persist in your silly hangover excuse. I don't know that.

-Warik

LOL

St. Pat's? Err....why? I'm not Irish.

Christmas? Again...no.

New Years? Yes. Champagne usually does me in quite well.

Warik......Halloween is just well...Halloween.

Can't really explain it. What other day can you dress up
in a costume and do whatever you want to do?(Well...close)

Fonz

 

[Warik]

Doesn't matter if you aren't Irish. It is St. Patrick's day; therefore, you must drink a whole lot and be cool and stuff.

It's OK, though... you won't get it even when sober. muhaha

-Warik

 

[KING]

1/8835488640 = 0.000000011% chance of winning the lottery.


-Warik

And since in Statistics one usually rounds up to the 4 decimal than the chances of winning the lotto are .0000

The odds are also found on the backside of a lotto ticket. The lotto one is pretty simple, just multiplying and dividing. The original one requires you to actually do P-1 and I really didn't even feel like trying. Although they are fun.

 

[Lestat]

You almost have as much of a chance of winning the lottery if you don't play...

 

[Fonz]

Actually, I'll give it to you now just to be evil.

Deck = {1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,5,5,5,5}

4 players

Player 1 is dealt a 1
Player 2 is dealt a 2
Player 3 is dealt a 1
Player 4 is dealt a 4.

You, and all other players, are to receive 3 more cards.

Find the probability of getting 5 high.

Possible hands = <X> high, one pair, three of a kind, four of a kind, two pair.

You have until after my lunch break. I'll take it around 2pm ET today.
-Warik

Ok.....feel much better now.

Hand 1:

P1: 1
P2: 2
P3: 1
P4: 4

Remaining: 2(1) 3(2) 4(3) 3(4) 4(5) Total: 16

Hand 2:

P1 or P2 pr P3 or P4

Probability of a 1: 1/8
Probability of a 2: 3/16
Probability of a 3: 1/4
Probability of a 4: 3/16
Probability of a 5: 1/4 1-P = 3/4

P(T) = 1.0

So, in order to get 5 high....

P = Prob of getting a 5*Prob of not getting 5(Hand1)*(Hand2)

P(No5)(Hand 1) = (1-(4-1)/(16-4)) = 1-3/12 = 3/4

P(No5)(Hand 2) = 1-((3-1)(12-4)) = 1-2/8 =3/4

P = (1/4) * {1-1/4} *{1-1/4}

P = 1/4 * 3/4 * 3/4 = 9/64

QED.......Warik.

Fonz

 

[Warik]

QED.......Warik.

Fonz

I don't know what quantum electrodynamics has to do with the fact that you got the answer wrong again.

Keep trying. You will lose partial ego rights unless you answer this correctly.

-Warik

 

[Warik]

bump for Fonzy. $

 

[Fonz]

I don't know what quantum electrodynamics has to do with the fact that you got the answer wrong again.

Keep trying. You will lose partial ego rights unless you answer this correctly.

-Warik

QED is latin Warik.

Ask your math professor one day, he'll know.

Fonz

 

[Warik]

QED is latin Warik.

Ask your math professor one day, he'll know.

Fonz

I've seen it before. I was making a joke about the fact that you got the problem wrong.

You have been beaten. Do you concede defeat?

-Warik

 

[Warik]

SURRENDER FONZ.

IT IS USELESS TO RESIST.