Math probability dilemma

I'm sure there's some magic formula that lets you calculate all this in one brief stroke, but I'm not a math guy, so I simply write out all the possibilities. These are all the possibilities of drawing exactly two diamond cards in hand (D = diamond, N = not a diamond).

DDNNN
DNDNN
DNNDN
DNNND

NDDNN
NDNDN
NDNND

NNDDN
NNDND

NNNDD

You draw the cards one by one from the deck. The chances of the first card being a diamond is 6 in 40. The chances of the first card being a non-diamond is 34 in 40. Drawing a card, either a diamond or a non-diamond, alters the chances of the next card being a diamond or not, because there are now less cards of that type in the deck, and less cards overall in the deck. Therefore:

DDNNN = 6/40 * 5/39 * 34/38 * 33/37 * 32/36 = 1.36%
DNDNN = 6/40 * 34/39 * 5/38 * 33/37 * 32/36 = 1.36%
DNNDN = 6/40 * 34/39 * 33/38 * 5/37 * 32/36 = 1.36%
DNNND = 6/40 * 34/39 * 33/38 * 32/37 * 5/36 = 1.36%

NDDNN = 34/40 * 6/39 * 5/38 * 33/37 * 32/36 = 1.36%
NDNDN = 34/40 * 6/39 * 33/38 * 5/37 * 32/36 = 1.36%
NDNND = 34/40 * 6/39 * 33/38 * 32/37 * 5/36 = 1.36%

NNDDN = 34/40 * 33/39 * 6/38 * 5/37 * 32/36 = 1.36%
NNDND = 34/40 * 33/39 * 6/38 * 32/37 * 5/36 = 1.36%

NNNDD = 34/40 * 33/39 * 32/38 * 6/37 * 5/36 = 1.36%

Perhaps a bit surprisingly, the percentages don't change for each permutation (well, it surprised me!). Anyway, this adds up to a total chance of 13.6%, or about 1 in seven that a 5 card hand will contain exactly 2 diamond cards.

Now my question to you is: what are you going to do with this information? How is knowing the exact percentage of drawing 2 diamonds in a starting hand useful to you?

Johan, you are right, of course, and now it makes sense to me. I just didn't realize it when I spelled out all the possibilities!

You know, the nice thing about mathematics is that it always makes sense, but it is not always intuitive.

Infinity can be perplexing, thats why I had to be so careful in my Foundations of Analysis class. Keeping track of what is legal and what isn't when working with infinite sequences.

Here's a simple proof for your example though.
x = 0.999999...
10x = 9.999999... multiply by 10
10x = 9 + x subtract x from both sides
9x = 9 divide by 9
x = 1 QED

Before I diverge off topic a little, I agree with the sentiment... how is this useful?

Also, did you WANT the probability of drawing EXACTLY 2 Diamonds, or did you want the probability of drawing AT LEAST 2 Diamonds. It's easier to draw at least 2 than exactly 2...

Now, speaking of math proofs:

Assume that A=B
Multiply both sides by A
A^2 = AB
Now subtract B^2 from each side
A^2 - B^2 = AB - B^2
Factor both sides
(A+B)(A-B) = (A-B)B
(A-B) cancels out
(A+B) = B
Substitute A for B, since they are equal
2A = A
Finally, divide by A...
2 = 1

- Seth

[EDIT: Correct mistake... I had said that (A^2 + B^2) = (A+B)(A-B)... oops!]

Assume that A=B
Multiply both sides by A
A^2 = AB
Now add B^2 to each side
A^2 + B^2 = AB + B^2
Factor both sides
(A+B)(A-B) = (A+B)B
(A+B) cancels out
(A-B) = B
Add B to both sides
A = 2B
Substitute A for B, since they are equal
A = 2A
Finally, devide by A...
1 = 2

- Seth

Uh... A^2 + B^2 is not (A+B)(A-B)

As for the 0.99999... = 1, this is an identity. There are no tricks in proving it, the values are equal. Pt314's proof is perfectly valid. Another is to sum the infinte series 0.9 + 0.09 + 0.009....

Nice try but you can't say (A-B) cancels out as you are dividing by zero (A=B).

:)

Since this thread already is totally OT by now, my bad please forgive me, I feel that the classic non intuitive probability problem should be presented:

You are a contestant in a TV show, the host presents you with three doors and behind one door is a prize, the other two reveals a loss.
After you have "randomly" (you don’t know what’s behind the doors) chosen one door the host, to spice things up, without opening the door you first picked reveals one of the doors that does NOT contain a prize. (he knows what’s behind all doors)
You now have two doors to choose from, the one you first picked and the other non opened door.

What should you do to increase your chances of winning?
1) Stay with the door you first choose
2) Change to the door you have not chosen
3) Doesn’t matter, it’s fifty-fifty (two doors) anyways.

"Do you take this gamble or not?"
Is not a valid mathematical question given your data, you must specify what you want to know so that you can mathematically use the information provided to make an odds calculation:

Better questions given your data would be:
Should you gamble to have the highest chance to go home with 100.000?
Or
Should you gamble to have the highest chance to go home with at least 20.000?
Or
What would have the highest odds of leaving you with most money after 1000 tries be; gamble or not gamble.
And so on…

ha! We (5 or 6 software engineers) spent an entire evening once at a party passionatly arguring this question. I will let others have that fun and not reveal the logic.

- Dwight

I disagree with your thoughts on this one. You shouldn't always take the gamble, the reason why is it only works out if you can repeat the risk 3 times or more. In that way, over time you will make more money. Granted, there is a chance that you will never win the 100K regardless of number of chances you have, but if you play the odds enough, it's in your favor to go for it. If you don't have the chance to take the same risk 3 times then the odds are against you.

In a board game I don't know if I like those odds - especially if it causes you to loose the game, because having 5 times as much in one game isn't the same as saying you have 100K in the span of 3 games, you simply have it once and kicks butt, otherwise, you are playing the game from a loosing position in the other 2 games, and we all know - 2 out of 3 wins, not the sum of the 3 games ... unless it yahtzee ...

My point was that you CANT say whats "mathematically clear" if it is not a mathematical question.

It's like saying:
There are 3 strawberries in a jar, for every strawberry you eat you get 1.5% more chance of getting stomach ache, do you eat the strawberries?
Mathematically the choice is clear: You should not eat the strawberries because they increase the chance of getting stomach ache!

... this is incorrect, you cant apply math to a psychological question. Just because you have some odds and numbers in a question doesnt make it possible to "mathematically" say whats "clear" or not. If the question was; Should you eat the berries to decrease the chance of stomach ache? Then you'd have a valid Mathematical question.

Clearly defined the Monty hall dilemma is not so much "deep" as it is one of understanding. Just like the "0.9999..forever = 1" and the fact that the more precise you messure a subatomic particles speed the less it has an exact position in space and vice versa.

Most "deep" problems arise from the same define principle, let's take the "If a tree falls" example; If you clearly define what each word means, in the "If a tree" example "sound" is most important, then it's not that difficult to answer.

But if youre into "deep" answer this:
What choices you make depends on who you are.
Who you are is defined by the past.
You have no control over the past therefor no control over who you are and thus no control over any choices you make.
Where does "free will" come in?

With a lottery ticket that costs 100$ with a 1 in a billion chance to win 101$ you would still be maximizing the chance to win as much money as possible by gambling.
But is normally something I assume is not what you want to do.

1% x $2.000.001 is also > 100% x $20.000
You’d have no problem having 99% chance of loosing $20.000 to have a go at 2 million one dollars right?
You shouldn’t, IF you are allowed try a hundred times that is, because statistically you would have gained 1 dollar for every 100th time you’d gamble.
To bad that most TV Shows only gives you one chance.

Both are perfectly mathematically valid, the second premise being identical to"maximize the chance to win as much money as possible." since $100.000 is as much money you could possibly win in your scenario.

Jeff, I wouldn't call this "flawed playing". It's just a typical example of risk avoiding behaviour, which may or may not be jusitified. Also, the point of playing a game may not always be trying to win. Different people may have different objectives, and they may not always be 100% logical or rational. Some people just want to try and not come in last place, some people want to try that odd ball strategy just to see what happens, etc. If your design is robust, it will be able to cope with all these kinds of "irrational" play.

However, I do think that a design does not have to be able to cope with intentional destructive behaviour. Sometimes players will just take certain actions to wreck the game to prove that it is broken. For example, a group of Puerto Rico players who never pick Builder, Mayor or Captain, just because that way they can say it is broken because the game never ends. That would just be silly.

And then there's the big gray area between "irrational" and "intentional destructive" which we could debate forever, but I think the point is clear ;)

If it hasn't been covered before, you should always switch. The easiest way to describe this is that your original pick is not really a pick, but the host is letting you divide the three doors into two groups, one with a single door (your pick) and one with two doors (the two doors you did not pick).

The single door group gives you a 1 in 3 chance on winning. The two door group has a 2 in 3 chance of winning. Since there is always at least one ZONK! door in the 2 door group (and the host knowing which door it is), having the host show it to the player or not is irrelevant to the math.

...

What should you do to increase your chances of winning?
1) Stay with the door you first choose
2) Change to the door you have not chosen
3) Doesn’t matter, it’s fifty-fifty (two doors) anyways.

I just love trying to explain that you have a 2/3 chance of winning if you flip. If they don't get it, I increase the number of doors.

Suppose there are 100 doors, and the host knows which one has the prize. You pick one, and the host opens 98 other doors and shows that they are empty. Do you stick with the door you first picked? Or pick the door left over from the other 99? Of course you would be better off switching. You can only lose if you picked the right door at first, which has a 1% chance of happening.

Hmm, it seems that a lot of the time a combinatorics/probability topic is started here it morphs into game theory, psychology, then into philosophy. Well, at least when we go off topic in this forum it doesn't end up in politics & religion, and thus a flame war.