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1800GAMBLER
09-11-2003, 04:06 PM
Need a few good math sites to explain how to work out the following:

The number of combinations (not really combinations) that you could set up pool balls in, 7 red, 7 yellow and 1 black.

In a higher/lower card guessing game, the probability of someone getting to the end of X amount of cards. X on to X means a lose to. Example, first card 6, person goes higher, next card 9, person carries on. Assuming the person always goes with the correct probability.

Last question, is a game i made up and had a hard time making the formula but think i've got it. You get offer a game in which you pay \$1 to play, you have a deck of 52 cards and you have to remove a card randomly (without replacing) untill only the 4 aces are left, if you hit an ace you have to stop. If you win you win \$1 000 000. If you hit an ace on your first card you remove you get another free go. Same again if it happens on your free go. Boring game; but profitable?

Thanks.

Wake up CALL
09-11-2003, 05:15 PM
Last question, is a game i made up and had a hard time making the formula but think i've got it. You get offer a game in which you pay \$1 to play, you have a deck of 52 cards and you have to remove a card randomly (without replacing) untill only the 4 aces are left, if you hit an ace you have to stop. If you win you win \$1 000 000. If you hit an ace on your first card you remove you get another free go. Same again if it happens on your free go. Boring game; but profitable?

Unless you have more money than Bill Gates I think you will need a part-time job to supplement your losses on this game. /images/graemlins/smile.gif

crockpot
09-11-2003, 10:38 PM
do the math here, and you will find that even gates should be long broke before he wins the million.

chaos
09-12-2003, 01:22 PM
I got an answer of 51,480 for the pool ball problem.

To simplify things lets say there are numbers on the 15 ball. There are 15! was to arrange the 15 balls. (You have fifteen choices for the first postion, 14 choices for the second postion, ...).

The seven red balls can be put in order in 7! ways as can the 7 yellow balls. Since we counted each ordering of the the red and yellow balls distintly, we need to divide our total by 7! twice.

15!/ (7! * 7!) = 51,480

Cyrus
09-12-2003, 03:25 PM
"Need a few good math sites to explain how to work out the following: The number of combinations (not really combinations) that you could set up pool balls in, 7 red, 7 yellow, and 1 black."

Yes, you obviously mean arrangements.

We have n kinds of objects (e.g. 3 categories of balls), for each of which there are an equal m number of copies (like 7 reds, 7 yellows, 7 blacks, etc). Then the number of all possible arrangements of those objects is

(n*m)! / (m!)^n

To answer your question, leave the one black ball out for a moment. Plugging in n=2 kinds of balls, and m=7 of each kind, i.e. 7 reds and 7 yellows, we get 3432 possible arrangements of the 14 red &amp; yellow balls.

There's still the one black ball left. The black ball can be placed anywhere in an arrangement of reds &amp; yellows. Each such arrangement contains 14 balls in all, so the black ball can be placed in any of the 15 available positions (there's one extra position because the black ball can go at the front as well, i.e. in position zero). So 15*3432=51,480 is the answer.

For more complicated calculations, e.g. unlimited repetition of objects of n kinds and no restriction on the number of times any object can appear, one needs enumerating generating functions. Try www.mathworld.com (http://www.mathworld.com)

I don't know of websites when you can plug in your numbers for such problems and get the answer. Sorry.

RocketManJames
09-12-2003, 04:46 PM
Maybe I'm missing something...

If you have 52 cards, and you keep choosing cards randomly until you either a) hit an A or b) go through 48 w/o, here's what I get.

Choosing 48 cards from 52 such that none are Aces is the same as choosing 4 cards from 52 such that all 4 are Aces. So, 52 c 4 = 270,725.

Doesn't this mean that you should be able to do it once every 270,725 times on average? And, note that this doesn't even include the free games you win when you hit an Ace right off the bat (1/13 chance).

So I think if you run this game, you basically go broke, since people will win a ton from you. Where is my math wrong?

-RMJ

chaos
09-15-2003, 01:57 PM

Copernicus
09-15-2003, 11:43 PM
not even close. What you have is just the number of ways to pick 4 cards out of 52. The correct answer is astronomical:

(44/48)*(47/51)*(46/50)*(45/49).....(6/10)*(5/9)*(4/8)

At first you might think that the first term should be 52/52, since you get a free roll with an A. However, you dont get to take your money back...you are committed to play again with the free roll, so the game doesnt really start until youve picked one of the 48 non Aces first...ie there "are no Aces in the deck" for the first draw that really starts the game.

If you dont buy the formula, reduce it to a simpler game...4 Aces, and you win if you miss the spade A on your first three draws but you get a free ride if you hit the Heart ace first. The winning non-heart-first combinations on the first 3 cards are
CHD
CDH
DCH
DHC

So you win on 4 combinations. If you do draw a heart first, the game hasnt really started yet...you may as well have picked a joker off the top of a new "4 card deck". so you start again till there is a non-heart, and win on the 4 above, with as many Hs as you want up front. But what is the total number of hands? Its not 4*3*2*1=24, because you "never" draw a heart first. Its only 3*3*2=18 possible hands, so youve won on 4/18.

In terms of a formula your chances of winning are:

(2/3)(2/3)(1/2)=4/18, which is equivalent to the first formula. (total-losers-draws)/(totals-draws)*(remaining-losers)/(remaining) and so on..

BruceZ
09-16-2003, 01:45 AM
This is crazy. I'll just pick the top 48 cards, and I'll win if all 4 aces are at the bottom of the deck. What are the odds that the bottom 4 cards are all aces? They are just 1 in C(52,4) = 1 in 270,725 like RocketMan said. I win all of these without a free draw. Obviously the free draw gives me more chances to win, not less. If you think it gives me less, I'll just forfeit those cases and be back to 270,725. /images/graemlins/grin.gif

The chance of winning your ace game is 1/4, and of course I get another play if I draw the Ah, why wouldn't I? /images/graemlins/confused.gif

Bozeman
09-16-2003, 02:40 AM
This may be astronomical, but it is not astronomically small.

First, you forgot the 3/7, 2/6, and 1/5 terms at the end.

If you do the game without the special first rule, you have
48/52 .... 1/5, and all the terms cancel except
1*2*3*4
----------------
52*51*50*49

which just happens to equal C(52,4).

It is not clear to me how the rules treat when you first draw an ace, but if it means you reshuffle and try again, then the first term is 48/52*(1+4/52+(4/52)^2+(4/52)^3)...), which just happens to equal 1. If there were no aces in the deck, it would be 48/48=1.

Now the remaining terms are 24/(51*50*49*48), so the answer is 1 in 249900 (not surprisingly, C(51,4)).

Craig