Mathematical Induction Example:

If 10/2 =5 and 10/1 = 10 then 10/0 must equal 15, right?

Any other crazy math/prob. statements that might appear correct on first look?

If 1! = 1 and 2! = 2, then 3! = 3?

If a coin is tossed 5 times and they are all heads, then the 6th toss will most likely be a tail because "a tail is due."

If the square root of 1 is 1, then the square root of 3.4 must be 3.4.

If the square root of 4 is 2, then the square root of 2 must be 1

I’m doing research on some numbers other than 3.4 and 2 so stay tuned.

Meanwhile, please study and understand these solutions before reading my next square root post, which is a bit math-harder but perhaps more interesting.

I warned you!

We know that the square root of 1 is 1. (Euclid; 321BC)

If the square root of 4 is 2 then the square root of 2 must be 1 (Statmanhal; 2022)

If the square root of 100 is 10 then the square root of 10 must be 1.

We also know that 10 = 5*2 (Euclid; ibid)

Therefore,

SQRT(10) = SQRT(5*2) = SQRT(5)*SQRT(2) = SQRT(5) * 1 = SQRT(1) = 1

We have now proved the following:

SQRT(10) = SQRT(5) = SQRT(2) = SQRT(1) = 1

Voila! So Be It! Q.E.D.

We are getting a little more action on this forum thanks mostly to heehaww. I’m no heehaww by any means and also no Euclid but I have fun doing this even if no one else does. So, my next Math Crazy gem is the following:

Consider a heads up game where each player Is dealt two cards from a deck consisting of 2,2, 3,3, 4,4; thus, there are 3 possible pairs; twos, threes and fours.

If player A is dealt a pair, then there are only two pairs left so the probability you get a pair is reduced. Correct?

Think of a quick answer without calculation then see the solution??? below.










Answer

There are 6 cards and C(6,2) = 15 ways for two cards to be dealt to a player. And 3 of those ways are pairs. Therefore, the probability a player is dealt a pair is 3/15.

A simpler way -- given a first card is dealt, then to have it pair, only 1 of the remaining 5 cards can give a pair, for a probability of 1/5

Whoa, why two different answers???

Let a=b. Then a+a=a+b. This implies 2a=a+b. Subtract 2b from each side and you get
2a-2b = a-b, or equivalently 2(a-b) = a-b. But this implies 2=1!

In the spirit of the thread I will let the bold question go. However you never did provide a solution for the probability that player 2 gets a pair given that player 1 has one. There are 4 cards so C(4,2) = 6 possible combos, of which 2 are pairs, so the probability of player 2 having a pair if player 1 does is 2/6.

Alternatively player 2 gets a first card. There are now 3 remaining possible second cards, one of which pairs the first. Thus we get 1/3. Wow! Two different answers again!

1+2+3 = 1x2x3.
Conclusion: multiplication is redundant.

You didn't make it clear if it is 2 = 1 or 2 = 1 factorial

We can't have ambiguous statements.

If a!=b!, then a = b.

If 1! = 0! = 1, then 0 = 1.

If we had 7 fingers and 2 thumbs, we would use an 18 digit math system. Our multiplication tables would be (18^2)/(10^2) = 3.24 times as complicated. Science would be 3.24 times as advanced. 2022*3.24 = it would already be year 6651.

Also:

This statement is false.

This statement is true.

Who's on first?

This can't be true, we already have 8 fingers and 2 thumbs.

I think Bob148 meant for each hand.

I discovered a serious error in the ranking of Hold’em hands.

A flush is rated higher than a straight. BUT - If you get a four flush, you have 9 outs and with an open-end straight, you only have 8 outs. Therefore, a straight is harder to get than a flush.

I wonder if Mason figured this out but has remained silent so that no one knows of this error in all the books he sold .

I think you're onto something here.
A straight flush draw has two outs and a set only one for quads too. Obv quads are twice as hard to make.

Also, the easiest flush to get is Ace high, yet it beats a 7 high flush which is the most rare.

My friend told me that the size of pienapple is about 22/7 that of an apple

If 0!=1 then 1!=2 and 2!=3. By Lofi’s Law of Factorial Induction, 3!=4

Now 3! = (1 + 2)! = ( 1! + 2!) = 2 + 3 = 5. So, 3! = 4 and 3! = 5. By Floa’s Factorial Law of Averaging, 3! = (4 + 5)/2 = 4.5. In general, you can show the following: n! = (2n +3)/2

This is a poker forum so let’s see how we can use this result. We will calculate the probability you are dealt a pair in Hold’em, say a pair of aces..

There are about 4 aces in a deck of 52 cards, so there are C(4,2) ways to get an ace pair and C(52,2) ways to deal two cards. Therefore

Pr(Ace Pair) = C(4,2)/C(52,2).

For the numerator, C(4,2) = 4!/(2! * 2!) and by the Lofi factorial result, that equals 5.5/(3.5*3.5). By Lorpa’s Law of Rounded Probabilistic Averaging, that equals 5/10 = 0.5. Similarly, the denominator = 52!/(2!*50!) = 53.5/(3.5*51.5) and Lorpa tells us this is equal to 1/3

So Pr(Ace Pair) = 0.5/(1/3) = 3*0.5 = 1.5 = 150%

Now I don’t know about you, but in games I have played you rarely see ace pairs occurring that frequently. Therefore, card deck manufacturers are producing defective stacks and/or gambling sites are manipulating the decks to receive more rake.

If anybody knows of a good card deck scamming lawyer please relay these results to him/her. Oh, better yet, relay them to Rudy Giuliani in care of The (3!-0.5) Seasons Landscaping Company

In complex numbers we define i^2=-1. Therefore i=sqrt(-1). i*i = sqrt(-1) * sqrt(-1) = sqrt(-1*-1)= sqrt(1) = 1. Therefore i^2 =1, and i^=-1, so 1=-1. For any positive x x=-x. Therefore negative numbers don’t exist!

Just realize I can go further. Since x =-x for all x, add x to both sides, and 2x=0. Hence for all x, x=0. Numbers other than zero don’t exist! Kind of makes this whole thread pointless doesn’t it?

I'm pretty sure most lookers see this thread as pointless without this particular crazy math, which I think may be crazy!

From Numen's Thread:

Do you think you have ever held a 52-card deck that is unique in the way cards are ordered (please only consider the shuffled decks)?

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I think 52! is a pretty big number. I tried multiplying it out but got stuck at 44, e.g., 52 * 51 * …* 45 * 44 * ??

From this failed analysis I have concluded that the probability that probability is probably correct is uncertain, probably.

For example, getting a pair has low probability of about 6%. I once had a pair of 3’s. On the very next hand I got 4 3, almost another same 3-pair. I’m still trying to work out how often that would happen.

If you always bet 50% of your bankroll, you'll never go broke.

Not true if your bankroll is 0.

On second thought, it might be true if you bet 40%