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**OzMark**- Replies: 6

Hi people of mathsisfun

My name is mark, I am almost finished studying to be a draftsman for construction

I did some university to be a structural engineer, but i did not finish the course.

I did however complete all the mathematics subjects for the degree.

The math subjects I have interests in are ;

probability, mostly as applied to gambling though i know the house ALLWAYS wins so i dont gamble

engineering mathematics for design, this is what engineers do to get paid!

proofs of the engineering mathematics, seriously there are too many people who use equations in blind faith with little understanding of how they are made

using spreadsheets to produce neat and accurate engineering calculations quickly

graphical methods, which have fallen from use being replaced by software packages, but are so useful to gain understanding

Particularly i am a fan of the work of the spanish archirect Antoni Gaudi and Origami Math guy Eric Demaine

Anyway, I still have a lot to learn about these topics, and being here will help my motivation I think

I have already helped out with a few solutions so hopefully I can be a part of this forum bing a productive place

Later

Fifth card makes first card of straight, 4C1 = 4

4 x 40 x 12 x 8 x 4 x 4 = 2048 = 0.0788 %

52 51 50 49 48 2598960

Fifth card makes last card of straight, 4C1 = 4

4 x 40 x 12 x 8 x 4 x 4 = 2048 = 0.0788 %

52 51 50 49 48 2598960

Fifth card makes an inside card of straight, 4 x 3 = 12 (4 places lead card of straight, 3 places end card)

12 x 40 x 12 x 8 x 4 x 4 = 6144 = 0.2364 %

52 51 50 49 48 2598960

Sums to 0.3940%, so is correct

Remove the fifth card (last 4/48 term) to see the picture with 4 cards dealt

= Chance of getting the first 4 cards of a straight (fifth card inside or outside)

0.3940% *48/4 = 4.728 %

For outside straights only

(0.0788% + 0.0788%) *48/4 = 1.8912 %

ANSWER

Chance of getting the first 4 cards of a straight deal, where a fifth card can make a straight on the start or end = 1.8912 %

OK the redraw situation

If you get to throw out a card and get a fresh one, for the second chance straight probability

multiply the 0.3940 % by

(48/4 * 44/48 * 4/47) = 0.936

= 93.6%

almost same chance of getting a straight on the 6th card as on the 5th card

48/4 cancels the 5th card prob, because we no longer deal with a wanted card there

44/48 (= [48-4]/48) is the prob that the 5th card is unwanted for a straight

so you discard the incorrect 5th card and draw a 6th card, the chance the 6th card is what you want is 4/47

0.3940% (chance of straight on a 6 card deal, on the 5th card)

0.3940% x 0.936 = 0.3689 % (chance of straight on a 6 card deal, on the 6th card)

0.3940% x 1.936 = 0.7629 % (chance of straight on a 6 card deal, on the 5th or 6th card)

ANSWER:

Chance of a straight with a 1 card redraw = 0.7628 % (1 in 131.1)

(on a deal of either 5 or 6 cards)

Im happy to expand on the information in this post or the above post if anyone who wants to understand is still stuck

----------------------------------

Here is the probability of straight on a 6 card deal, on the 6th card in an expanded form

5 x 40 x 16 x 12 x 8 x 44 x 4 = 112640 x 4 = 0.3689 %

52 51 50 49 48 47 2598960 47

Say the first card must be the middle card in the straight, 3 4 5 6 7 8 9 10 J Q, these are the 10 types of straights

it can be any of 4 suits so 10 x 4 = 40, use 40/52

The second card must be from within cards 2 places lower or 2 places higher (2+2=4) side of the middle card,

eg. either slot A 2 4 or 5 for a 3 middle card

it can be any of 4 suits so 4 x 4 = 16, use 16/51

The third card must be from one of the 4 slots outside the middle card, but 1 has been used; 4-1 =3

it can be any of 4 suits so 3 x 4 = 12, use 12/50

The fourth card must be from one of the 4 slots outside the middle card, but 2 have been used; 4-2 =2

it can be any of 4 suits so 2 x 4 = 8, use 8/49

The fourth card must be from one of the 4 slots outside the middle card, but 3 have been used; 4-3 =1

it can be any of 4 suits so 1 x 4 = 4, use 4/48

But the middle card doesnt have to be the first, it can be dealt in any of the 5 positions

either 1st 2nd 3rd 4th or 5th, so multiply by 5

The order of the other cards does matter

5 x 40 x 16 x 12 x 8 x 4 = 10240 = 0.3940 %

52 51 50 49 48 2598960

ANSWER

Chance of a straight in the first deal of 5 cards = 0.3940 % (1 in 253.8)

This is the figure given in wiki for a straight

en.wikipedia.org/wiki/Poker_probability

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Also notice that poker tables list Royal Flushs and Straight Flushs seperate to the other flushes

4 Royal Flush + 36 Straight Flush + 10200 Other Straights = 10240 All Straights

The guy at the first post made a similar mistake for For Flushes, its 1 in 504.8 for any flush in the first hand,

not 1 in 508 as he says, which is for other flushes but not the royal or straight flushes

Hi Bob Bundy,

Thanks, your right, I only had the case : 4 females 1 male,

I should have had the cases: 4 females 1 male and 5 females 0 males

which is 45/143 + 9/143 = 54/143

This is great, Im learning here

So what is the etiquitte now

Should I redo the spreadsheet, or make a second one, or

leave it (which might be the best reflection of this dialogue and give other readers a bit to think about)

?

Again Thanks,

Mark

Hi

I get 45/143 (bobbym you get 54/143, maybe just a transcription transpose error)

Which equals 0.3147 (bobbym=0.3776)

Ive calculated the entire set of outcomes and summed them to 1.0000, so im pretty sure I'm right.

Sorry I dont use the nPr or nCr functions, I think its too easy to make mistakes with them.

I do probability from first principals, so my terminology is not conventional

I hope my spreadsheet makes sense, here it is with all the outcomes

imgur.com/DKhMcUu

Oh and here is how I obtained the any order multipliers (dont know what they are really called)

imgur.com/6CtcOsU

Hello,

Bob Bundy, I think your Venn diagram is slightly wrong,

it should be 70-60-70 and 50 play no sport (your diagram says 60-60-70 and 60 play no sport)

BobbyM,I think your Venn diagram is correct, you could add that the play no sports guys are 50

Do not play either sport 50/250 = 0.200

Play exactly one sport 140/250 = 0.560

Play soccer not baseball 70/250 = 0.280

Play soccer given that he play baseball 60/130 = 0.461 (actually 0.46154)

Play baseball given that he did not soccer 70/120 = 0.583

did not play baseball (50), given that he did not play soccer(120) = 50/120 = 0.461 (actually 0.41666)

There are

13 black spades (including the ace of spades),

13 black clubs (including the ace of clubs),

the ace of diamonds,

the ace of hearts,

= 13 + 13 + 1 + 1 =28

28/52 = 7/13

Thanks

Q1, refer also also

http://www.mathsisfun.com/puzzles/a-que … -time.html

The question asks for the distance from the bottom of the clock face at 6 o'clock to the tip of the clock hands, what they mean is when do both the hands make an angle to a vertical line on the clock face through 6 o'clock.

The answers for when the hands are at either side of the clock face are ;

12:55 and 23sec, 1:50 46s, 2:46 9s, 3:41 32s, 4:36 55s, 5:32 18s, 6:27 42s, 7:23 5s, 8:18 28s, 9:13 51 s, 10:09 14s & 11:04 37s

There are 11 more solutions when the minute and hour hand are directly on top of each other ;

12:00 00s, 1:05 27s, 2:10 55s, 3:16 22s, 4:21 49s, 5:27 16s, 6:32 44s, 7:38 11s, 8:43 38s, 9:49 5s & 10:54 33s

Also we need to assume the length of the minute and hour hands is the same, or else we cant solve the question at all.

The solution given only cites one of these solutions 8:18 28s, I suppose that is made with reference to the picture.

I can make an explanation of the equations I used if I get a reply to this post, I wasnt going to be bothered if noone reads these.

Have Fun !

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