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View Full Version : Odds of Winning and EV in a Basketball Pool Promotion

ykcirT
01-17-2003, 09:07 PM
I was recently playing some Hold ?em at a casino where they were offering a basketball pool to the seated poker players. I?m sure most of you are familiar with these, but this is how it works: In the first half of a college hoops game, each player is given a card with 2 numbers on it... one representing the home team?s score and one representing the visitor?s score. If, at the end of the first half, the last digit of the score of each team matches what you have on the card, you win (payoffs for both correct and reverse scores). In the second half, they give you another 2 tickets. One is just like the first half (forward and reverse wins) and the other has to be exactly correct. The first half card is also good in the second half. Payouts are as follows: Correct on the ?any way? card = \$500, reverse on the ?any way? card = \$250, and correct on the ?correct only? card = \$300.

A young man sitting at our table starting spouting off statistics, and I just know he was wrong. He said you are a 5:1 dog against winning something in the game, and your EV per hour is \$50. Here are my calculations? what do you think?

- First of all, assume that all two digit scores are equally probable of coming up (I think this is probably true in hoops, but definitely not in football)
- There are 100 possible unique combinations on your card
- In the first half, you have a 1/100 chance of winning \$500 and a 1/100 chance of winning \$250. That means your overall odds of winning something are 1/50, and your EV for the first half is \$7.50 (1% x \$500 = \$5 and 1% x \$250 = \$2.50)
- In the second half, you now have 5 possible combinations of numbers, assuming you have no duplicate cards (2 each on the two ?any way? cards and 1 on the ?correct only? card).
- Thus, you have a 2/100 chance of winning \$500, 2/100 of winning \$250, and 1/100 of winning \$300. So, your overall odds of winning something are 1/20. Your EV is \$18 for that half.
- So, bringing everything together, your overall odds of winning something in the course of the game is 7/100, or 7% chance, or about 13:1 against. Your EV for the entire game is \$25.50. Assuming that a game lasts 3 hours, your EV per hour is \$8.50

So, how did I do? I am sure that this is not correct, so please correct me.

Thanks,
Rich

BruceZ
01-18-2003, 04:56 AM
The only thing your answer doesn't account for are the times that you get a number with both digits the same like 33, in which case if you match the forward, you will also match the reverse. Your chance of getting a number like this is 1/10, and assuming you get paid for both forward and reverse, those numbers have an ev twice as high. Here is how you would compute the total ev:

(.9*.01 + .1*.02)(3*250 + 3*500) + .01(300) = \$27.75

Now suppose you only got paid for the forward when you match both. In that case, the ev of these 10 numbers will be reduced. In that case the ev is:

(.9*.01)(3*250 + 3*500) + (.1*.01)*500 + .01(300) = \$23.75

In both cases I've assumed you get paid double if you have duplicate cards. If this isn't the case, it won't matter much since this is rare.

BruceZ
01-18-2003, 05:38 AM
In both cases I've assumed you get paid double if you have duplicate cards. If this isn't the case, it won't matter much since this is rare.

I should make a more accurate statement. I've actually assumed that you don't get any duplicate cards. If you get duplicate cards and you can win multiple prizes when they win, then the ev will be a little higher. If you can't win multiple prizes, then it will be a little lower. You will get duplicate cards about 3% of the time, and I estimate this should only affect the ev by about 1% from the values I've calculated (about 25 cents).

Fraubump
01-22-2003, 06:46 AM
This is not your situation and is a lot more straightforward, but at ACR you get a max of one square with specific numbers associated with sepcific teams i.e., 9 celtics 7 lakers. Those must be the ending numbers of the teams scores at the end of a quarter for you to win. If you accept that any number is equally likely as any other number (this is probably not quite true due to the vagaries of basketball scoring, but it's probavbly very close) you have a pretty clear 1% chance per quarter to win the pool, so your ev is 1%x4xprize.