Some basic probabilities help

[pyrohornIII 4817]

My son is taking an online math class at Sam and some of the probabilities problems are giving him issues.   This is not my forte and it's due tonight.  He's done about 16 so far, but still has two left.  Any help would be much appreciated.  

I am going to post screenshots of the problems, so you can see exactly how the problems are worded.  

The first one on the lottery I asked him if he could work it backwards using the formula he used to get the -$4.99, but the question is what are they after for fair game?  break even?  Set the value to $0 or $5.00?   

 

Here's the second one.  Maybe we have looked at it too long.  I mean it's possible that no hands in any one deal would have two hearts and two clubs.   And actually the question should read "contain" exactly, not "consist".    No 7 card hand can consist solely of 4 cards.   It can contain specific types of cards and still be 7 cards.  Probably reading too much into it.   Thanks folks.   

 

[mchookem 16346]

obligatory

[Huckleberry 33053]

1 hour ago, pyrohornIII said:

The first one on the lottery I asked him if he could work it backwards using the formula he used to get the -$4.99, but the question is what are they after for fair game?  break even?  Set the value to $0 or $5.00?   

 

 

Lotteries are, generally speaking, combination problems versus permutations because in combinations order does not matter. So you're looking for the number of combinations of 7 items selected from 65. 

So there are 696,190,560 possible lottery results. Expected value is [(696,190,559*-5) + (1*9,699,995)] / 696,190,560 = -$4.986 which rounds to -$4.99

I used $9,699,995 instead of $9,700,000 for the winning possibility because you don't get your $5 back with most lotteries.

To determine the fair game payout value you set the EV equal to zero. So we have

 [(696,190,559*-5) + (1*(x-5))] / 696,190,560 = 0

(696,190,559*-5) + (1*(x-5)) = 0

1*(x-5) = 696,190,559*5

x-5 = 3,480,952,795

x = 3,480,952,800

The real problem here is if the teacher/website are going to use x or x-5 for their answer. To me the correct answer is as I show above. They might have the value without the 5 added as the answer though if they're not accounting for losing the 5 dollars when you win the payout.

1 hour ago, pyrohornIII said:

Here's the second one.  Maybe we have looked at it too long.  I mean it's possible that no hands in any one deal would have two hearts and two clubs.   And actually the question should read "contain" exactly, not "consist".    No 7 card hand can consist solely of 4 cards.   It can contain specific types of cards and still be 7 cards.  Probably reading too much into it.   Thanks folks.   

 

Yeah you're right about how it's phrased but you know what they're looking for. Shorthand notation is C(13,2) x C(13,2) x C(26,3) equals 78 x 78 x 2600 = 15,818,400 hands that will have 2 hearts and 2 clubs. Basically the number of ways you can get 2 of the 13 hearts times the number of ways you can get 2 of the 13 clubs times the number of ways you can get 3 of the 26 spades and diamonds. All three of those things have to happen so you need to multiply the number of possible events of each of them to get your total number of possibilities. This comes out to roughly 11.8% of all 7-card hands which passes the smell test. 

 

Edited December 2, 2021 by Huckleberry

[pyrohornIII 4817]

Thanks Huck!

[Steel Shank 9430]

edit: nevermind, Huck got it first.

[Upgrayedd 5817]

2 hours ago, Huckleberry said:

 

Lotteries are, generally speaking, combination problems versus permutations because in combinations order does not matter. So you're looking for the number of combinations of 7 items selected from 65. 

So there are 696,190,560 possible lottery results. Expected value is [(696,190,559*-5) + (1*9,699,995)] / 696,190,560 = -$4.986 which rounds to -$4.99

I used $9,699,995 instead of $9,700,000 for the winning possibility because you don't get your $5 back with most lotteries.

To determine the fair game payout value you set the EV equal to zero. So we have

 [(696,190,559*-5) + (1*(x-5))] / 696,190,560 = 0

(696,190,559*-5) + (1*(x-5)) = 0

1*(x-5) = 696,190,559*5

x-5 = 3,480,952,795

x = 3,480,952,800

The real problem here is if the teacher/website are going to use x or x-5 for their answer. To me the correct answer is as I show above. They might have the value without the 5 added as the answer though if they're not accounting for losing the 5 dollars when you win the payout.

Yeah you're right about how it's phrased but you know what they're looking for. Shorthand notation is C(13,2) x C(13,2) x C(26,3) equals 78 x 78 x 2600 = 15,818,400 hands that will have 2 hearts and 2 clubs. Basically the number of ways you can get 2 of the 13 hearts times the number of ways you can get 2 of the 13 clubs times the number of ways you can get 3 of the 26 spades and diamonds. All three of those things have to happen so you need to multiply the number of possible events of each of them to get your total number of possibilities. This comes out to roughly 11.8% of all 7-card hands which passes the smell test. 

 

[Redneck Mutha 8007]

"You miss 100% of the shots you don't take"

- Wilt Chamberlain

[Amos Moses 795]

@jeevsie there is a probability in physics that things that fall down will come back up.


However, it is very small.

Sincerely,
Prof. Hans Koch

[NeverMarryAStripper 21913]

17 hours ago, Huckleberry said:

The real problem here is if the teacher/website are going to use x or x-5 for their answer. To me the correct answer is as I show above. They might have the value without the 5 added as the answer though if they're not accounting for losing the 5 dollars when you win the payout.

It's easier if you think of it as expected return against cost.  The expected return is the jackpot times the probability of winning, or $9,700,000 * 1/696,190,560 = 0.01393297.  Your cost is $5.  Expected return - Cost = -$4.98606703.  A fair game would have the expected return equal the cost.   Jackpot * 1/696,190,560 = $5.  Jackpot must equal $3,480,952,800 for the expected return to equal the cost.

[South Austin 52526]

[jeevsie 3737]

5 hours ago, Amos Moses said:

@jeevsie there is a probability in physics that things that fall down will come back up.


However, it is very small.

Sincerely,
Prof. Hans Koch

Truer words have never been spoken.

Sincerely,

Cough Boy

[deadshank 24653]

On 12/1/2021 at 6:27 PM, Huckleberry said:

 

Lotteries are, generally speaking, combination problems versus permutations because in combinations order does not matter. So you're looking for the number of combinations of 7 items selected from 65. 

So there are 696,190,560 possible lottery results. Expected value is [(696,190,559*-5) + (1*9,699,995)] / 696,190,560 = -$4.986 which rounds to -$4.99

I used $9,699,995 instead of $9,700,000 for the winning possibility because you don't get your $5 back with most lotteries.

To determine the fair game payout value you set the EV equal to zero. So we have

 [(696,190,559*-5) + (1*(x-5))] / 696,190,560 = 0

(696,190,559*-5) + (1*(x-5)) = 0

1*(x-5) = 696,190,559*5

x-5 = 3,480,952,795

x = 3,480,952,800

The real problem here is if the teacher/website are going to use x or x-5 for their answer. To me the correct answer is as I show above. They might have the value without the 5 added as the answer though if they're not accounting for losing the 5 dollars when you win the payout.

Yeah you're right about how it's phrased but you know what they're looking for. Shorthand notation is C(13,2) x C(13,2) x C(26,3) equals 78 x 78 x 2600 = 15,818,400 hands that will have 2 hearts and 2 clubs. Basically the number of ways you can get 2 of the 13 hearts times the number of ways you can get 2 of the 13 clubs times the number of ways you can get 3 of the 26 spades and diamonds. All three of those things have to happen so you need to multiply the number of possible events of each of them to get your total number of possibilities. This comes out to roughly 11.8% of all 7-card hands which passes the smell test. 

 

 

[Dr Fear 1268]

LOL, total throwback to my actuarial science days at UT.  All I remember is the formula involves factorials !!!!!!!