Weird Math Bets

[sekrah]

I pocketed $120 from morons that I work with the other day..

First, the idiots didn't know the affect of multiple tickets had on the lottery. I tried to explain to them that if I had 20 tickets in the Big Four lottery (10,000:1), my odds were 500:1 to win.. I had 3 people INSIST that it doesn't work like that and my odds were much much worse.. Took $20 each from them.. The answer was common sense but these guys gave me the "I Know for a fact that you're wrong", so I bet them..

At same time, I gave them the Birthday paradox. Which states that in a group of 23 people, the odds that two of them share the same birthday is over 50%. They called me more names.. I bet them on this too...

We e-mailed PA Lottery website, got the answer on lottery tickets. Winner, Collect $60.

I printed off the birthday paradox off wikipedia and another source. Winner, Collect $60 again.


So I got to thinking.. Are there any other out of whack mathematical oddities that I can use to get someone on?

[cardcounter0]

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I tried to explain to them that if I had 20 tickets in the Big Four lottery (10,000:1), my odds were 500:1 to win..

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What is odd about this? Isn't this just standard math?

[ItalianFX]

One little trick that I always liked was:

Take two separate decks of cards. Shuffle them individually - have 2 people shuffle them so they don't get mixed and people can't accuse you of setting the deck.

Put both decks on a table and begin flipping the top card up into another pile. There is a highly likely chance that there will be the same card flipped up on both piles at the same time.

Experiment for yourself a few times just to prove to yourself that it works. I tried figuring out the odds/probabilities of this working with my Stats/Probabilities professor at my college and we couldn't figure it out.

So to set it up, you can bet someone (or people) that with two randomly shuffled decks of cards, when you flip over the cards, somewhere in the deck there will be the same card on both decks at the same time.

I think that is confusing, but I think you can understand it if you think a little harder.

[The Mayo]

[ QUOTE ]
One little trick that I always liked was:

Take two separate decks of cards. Shuffle them individually - have 2 people shuffle them so they don't get mixed and people can't accuse you of setting the deck.

Put both decks on a table and begin flipping the top card up into another pile. There is a highly likely chance that there will be the same card flipped up on both piles at the same time.

Experiment for yourself a few times just to prove to yourself that it works. I tried figuring out the odds/probabilities of this working with my Stats/Probabilities professor at my college and we couldn't figure it out.

So to set it up, you can bet someone (or people) that with two randomly shuffled decks of cards, when you flip over the cards, somewhere in the deck there will be the same card on both decks at the same time.

I think that is confusing, but I think you can understand it if you think a little harder.

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Two decks, A and B.
A_1 = first card in A, etc.

In general, P(A_i == B_i) = 1/52 for all i.
(That is, A_i is whatever, and there's a 1/52 chance that B_i is the same.)

So P(A_i != B_i for all i) = (1-1/52)^52 = 0.36
P(A_i == B_i for some i) is .74.

Also, either deck can be left unshuffled. Comparing two shuffled decks is not "more random" than comparing a shuffled deck to an unshuffled deck. But you'll get more bar bets made by shuffling both.

[pzhon]

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In general, P(A_i == B_i) = 1/52 for all i.
(That is, A_i is whatever, and there's a 1/52 chance that B_i is the same.)

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True.

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So P(A_i != B_i for all i) = (1-1/52)^52


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False. You are assuming independence, but this is not valid. If you know that there was a match on the first 51 cards, the probability of a match on the last card is 1, not 1/52.

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= 0.36

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This estimate is ok, though. The correct value is extremely close to 1/e ~ 0.37, closer than your estimate of (51/52)^52 is to 1/e. Permutations with no fixed points are known as derangments, and these are understood well by mathematicians.

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P(A_i == B_i for some i) is .74.


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P(A_i = B_i for at least 1 i) is 1-0.37 = 0.63.


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Comparing two shuffled decks is not "more random" than comparing a shuffled deck to an unshuffled deck. But you'll get more bar bets made by shuffling both.

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That is an interesting phenomenon showing how poorly people understand probability.

There are many common problems where people have faulty intuition in probability. One is that long streaks happen much more frequently than people imagine. I haven't done this, but try betting on whether there will be a streak of 6 matching coins in a row if you toss 100. The probability of a streak of 6 or longer (without specifying whether the streak must be heads or tails) is over 80%.

[GMan42]

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Also, either deck can be left unshuffled. Comparing two shuffled decks is not "more random" than comparing a shuffled deck to an unshuffled deck. But you'll get more bar bets made by shuffling both.

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Also, an easy way to do this, that only requires one deck, is to turn over each card as you call "ace, two, three, four, etc." out loud four times through. I used to do this as an extremely boring form of solitaire when I was a kid, mainly just to see how long it would finally take me to get through a whole deck without matching the card I called out...I think it was like 200x or so.

[SamIAm]

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I tried to explain to them that if I had 20 tickets in the Big Four lottery (10,000:1), my odds were 500:1 to win.. I had 3 people INSIST that it doesn't work like that and my odds were much much worse.

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Really? What are your chances if you buy 20,000 tickets? (Hint: It's not 20k * 1/10k.)

If your first ticket had a 1000:1 chance, you only have a 1019:20 chance to win with 20, not 500:1. In other words, I insist that it doesn't work like that and your odds were <u>MUCH MUCH</u> worse. [img]/images/graemlins/smile.gif[/img]

The way I figure it, you owe your office mates $120. Tell them they can send me my kickback via PokerStars.

[SamIAm]

Dammit, is the "Big Four" one of those lotteries where you can only buy 10k tickets? (I don't speak Yankee.)

[SheetWise]

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Dammit, is the "Big Four" one of those lotteries where you can only buy 10k tickets? (I don't speak Yankee.)

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I think it was the major source of funding for "The War of Northern Aggression". Damn Yanks.

[BruceZ]

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Really? What are your chances if you buy 20,000 tickets? (Hint: It's not 20k * 1/10k.)

If your first ticket had a 1000:1 chance, you only have a 1019:20 chance to win with 20, not 500:1. In other words, I insist that it doesn't work like that and your odds were <u>MUCH MUCH</u> worse. [img]/images/graemlins/smile.gif[/img]

The way I figure it, you owe your office mates $120. Tell them they can send me my kickback via PokerStars.

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Wanna bet? [img]/images/graemlins/laugh.gif[/img]

In any lottery, if you buy N tickets with no 2 tickets identical, the probability of matching all of the numbers is exactly N times higher than if you buy just 1 ticket.

Your calculation would apply to a raffle, in which the numbers on the tickets sold form the pool of possible winning numbers that can be drawn.