i was playin @ a home game yesterday, and these two guys had a prop bet. each time a Jack Seven or Deuce would fall on the flop, one guy would pay the other. if it didnt, the other guy paid. just out of curiosity, i want to know that math behind it. does anybody know how to figure out who has the advantage/disadvantage and how you figured it out? if anyone could let me know the math behind this, i’d appreciate it, as its really bugging me. i came up with something on my own but i wanna see what you guys think as well. thanks.

http://www.mathhelpforum.com/math-help/las-vegas-math/ Hey bud, post this question in this forum. These guys are real sharp. I used this alot for GMAT help and fished around in the vegas section for fun.

Easy, right? Call a J, 7, or 2 a Pay card and all others not pay cards. P(pay card in flop) = 1 - P(no pay cards in flop) = 1 - 40 choose 3/52 choose 3 = 1 - 40!*49!/52!*37! = 1 - 40*39*38/52*51*50 = whatever.

4.5 to 1 against

Joey I am having a bit of trouble following you. Maybe I’m drunk but wouldnt the odds just be [12-(number of pay cards dealt)]/[52-(number of cards already dealt)]? I’ve only taken level one so maybe I have not learned enough about the n choose r applications?

remember J, Deuce or 7 could come up in any suit, and then they would win. so 40 cards are bad, 12 cards are good. so you would lose if 3 cards were in your 40 set. and your total probability for that is 40C3 / 52C3 so you would win with a prob of 1 - 40C3/52C3 I am using the 40C3 to indicate 40 Choose 3.

ok im still kind of not following. I understand that 40 are good 12 are bad. I lose you when you say “you would lose if 3 cards are in your 40 set” Please explain.

When do you lose? – if the cards drawn are not J, 7 & 2 When can the drawn cards not be J, 7 & 2? - when they are from the 52 - 4(J-card) - 4(7-card) - 4(2-card) = 52 - 12 = 40, So you lose when the 3 chosen cards are from the lot of 40 cards you don’t need. The probability of which is 40C3/52C3 Probability of losing = 40C3/52C3 Probability of winning = 1 - Probability of losing Probability of winning = 1 - 40C3/52C3 Probability of winning = 1 - 9880/22100 Probability of winning = 0.447059 Probability of winning = 44.7%