Harpo1224
Harpo1224
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February 3rd, 2010 at 3:33:53 PM permalink
About once every six months for the past 4 years, my friend and I have the following discussion, with the latest iteration involving an office NFL squares pool.

Assuming a standard game, (10x10 grid, random digit assignment, ignoring number preferences), the odds of winning should be 1%, just like a zero house edge lottery. I assert that the purchase of a second square, even though your chances of winning increase, reduces the expected return slightly because the chance of both squares paying out simultaneously is zero. Essentially, the second bet is less efficient than the first.

My questions are:
1. Is this sound reasoning?
2. Does it hold true for a multiple payout system (major payouts for the end of each quarter plus a grand prize for the final score)?
3. What is the optimum/worst number of squares to purchase?

The answer will be much appreciated in settling some long-time disputes of the nature of lotteries, roulette, blackjack and statistics in general. Thanks.
Wizard
Administrator
Wizard 
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February 3rd, 2010 at 4:37:53 PM permalink
1. No. If the first square loses, the odds increase to 1/99 that the second square will win.
2. No.
3. Assuming nobody skims the pool, it doesn't make any difference.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
dwheatley
dwheatley
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February 3rd, 2010 at 8:23:33 PM permalink
I agree with the Wizard. Take a simple example: You bet $1 on a coin flip. You pick heads, winning $1 for heads, losing $1 for tails. Pretending coins are a 50-50 flip, your expectation is 0. Now... pick a 2nd result so you have a better chance to win. You pick tails as well. Your expectation on this bet is... less than $0? I don't think so...
Wisdom is the quality that keeps you out of situations where you would otherwise need it
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