Math Is Fun Forum

This is a little challenging...
Prove

Hello elliot60,
           Welcome to the math forum!

Bobbym,
     The 5 empty boxes divide the 10 boxes into 6 segments.  So the average number of boxes with money until an empty box is reached is 10/6 = 5/3.  The empty box is next, so we have 8/3 boxes to reach an empty box.   I thought my answer was correct, but if you have wrote a simulation and done so correctly then I can't really argue with that.

BobbyM,
     I have a different answer to #1.

I am amazed that somebody got the exact answer!  Good job! I would really like to see the solution.  I wracked my brain trying to figure it out.

Well I got two equations...

Then it was just trial and error.
Yeah, I haven't posted in a while, I'm going to try to become more active on the forum.

Here's another one: problem #996

A sum of 35 integers is S. Two digits in one of the integers are interchanged and a new sum T is produced. Then the difference S - T is necessarily divisible by:

They think the answer is 11.

Take the sum of the first 35 counting numbers...(35*36)/2 = 630.
Now switch 31 into 13 and add again and you end up with 612.

630-612 = 18 which is divisible by 3 and not 11.

I've finally completed my analysis of Four Card Poker.  The difficult part was figuring out the probabilities for the dealer to wind up with various hands.
The player initially makes a bet on the Ante spot.  After the player receives his cards, he can either fold his hand and lose his ante, or he can bet from one to three times the ante bet.
If the player's hand ties or beats the dealer's hand the player wins and gets paid 1-to-1 on both his ante and raise bet.
If the dealer's hand beats the player's hand, the player loses both his ante and raise bet.
If the player gets Three of a kind, Straight Flush, or Four of a kind, the player receives a bonus based on the Ante bet as follows:

Three of a kind     pays 2-to-1
Straight Flush      pays 20-to-1
Four of a kind      pays 25-to-1
The bonus pays regardless of the dealer's hand.
The dealer receives six cards and the player receives five, giving the casino an edge.

Hands are ranked as follows:
Fouf of a kind
Straight flush
Three of a kind
Flush
Straight
Two pair
One pair
High card

The Ace can count both high and low in making the player's or dealer's hand.

Here are the various ways that the dealer can wind up with various hands:
(Just divide by

for the probability)

Four of a kind

Straight Flush

Three of a kind

Flush

Straight

         

Now subtract out the straights which are also flushes

                                                   

             

                 

Phew!

Two pair

One pair

Subtract out the straights

Subtract out the flushes

Add in the straights which are also flushes

High card

The number of ways for the player to wind up with the hands are as follows:

Four of a kind

Straight Flush

Three of a kind

Flush

Straight

Two pair

One pair

High card

The best strategy for the player is to fold if he only has high card, bet 1x the ante for a pair of 2's through 9's, bet 3x the ante for a pair of 10's or higher.
We are now in a position to estimate the player's mathematical expectation.
In order to simplify the calculation, I ignored the situations where both the player and the dealer have the same hand.
High card

pair of 2's through pair of 9's

pair of 10's through Aces

Two pair

Straight

Flush

Three of a kind

Straight Flush

Four of a kind

Now we need to add the bonus which is as follows:

So,

So the casino has an advantage of

on Four Card Poker.
According to James Grojean, the casino has an advantage of 3.396%.
So my analysis is very close.

Hmmm...
I'm not sure why that's the answer.  In step 3, we aren't counting A or B as being the umpire.  We are only counting the remaining 4 people, none of whom are A or B.
We can also count the remaining 7 people in turn as being the umpire.
We put A on one team and B on the other team so we have...

I get the same answer as before.

a) 5! = 120

b)

c)

    However, this is true only if the committees are distinguishable from one another, if the committees are not distinguishable from one another then I think the answer needs to be divided by 3! or 6 since we can have the same groups of 4 people assinged to different committees, this gives

d)

    If we pick the chairpersons of the committees as a group of their own we end up with
   

    but this doesn't include all of the possibilities because you could make the chairpersons head different groups and this would be a different arrangement but it wouldn't be counted as a new arrangement under this total, so to take care of this we need to multiply this number by 5! or 120 and then we get

Lo and behold we get the same answer!
However, this is true only if the committees are distinguishable from one another, if the committees are not distinguishable from one another then I think the answer needs to be divided by 3! or 6 since we can have the same groups of 4 people assinged to different committees this gives

Maybe I didn't make this clear enough, but the natural number 5832... is not the number 5832.  The three little dots means it continues forever.  So it is not listed in the 5832th spot.

Actually, that real number can be paired with a natural number it is
...111111111111111111111112

Every time you add a 1 to the end of your real number you create a new real number, so I add a 1 to the front of my number to match it.