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#1
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Prop bets
Someone was tossing out these proposition bets at the home game last week. My math isnt up to snuff, how do these look number wise?
Bet 1: Pays 1:1 Take the bet and your betting any Ace, Seven or Jack will not appear on the flop. Bet 2: Pays 4:1? Betting that a flop will add up to 21 blackjack style. Have fun numbercruches, Im guessing these should be pretty simple to figure out but not my forte. |
#2
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Re: Prop bets
[ QUOTE ]
Bet 1: Pays 1:1 Take the bet and your betting any Ace, Seven or Jack will not appear on the flop. [/ QUOTE ] (40 choose 3)/(52 choose 3) = .4470 A fair bet would pay you 1.23:1 [ QUOTE ] Bet 2: Pays 4:1? Betting that a flop will add up to 21 blackjack style. [/ QUOTE ] EDIT: redoing this one -- original didn't take into account aces as 11's. |
#3
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Re: Prop bets
If bet 2 were a fair bet, it would pay about 10:1.
Java code below: <font class="small">Code:</font><hr /><pre> public class PropBet { public static void main(String[] args) { int[] deck = {1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,5,5,5,5,6,6,6,6,7 ,7,7,7,8,8,8,8,9,9,9,9,10,10,10,10,10,10,10,10,10, 10,10,10,10,10,10,10}; System.out.println(deck.length); int total = 0; int cnt21 = 0; int sum1, sum2, sum3, sum4; for (int i=0; i<52; i++){ for (int j=i+1; j<52; j++){ for (int k=j+1; k<52; k++){ sum1 = sum2 = sum3 = sum4 = 0; sum1 = deck[i] + deck[j] + deck[k]; if (deck[i] == 1) sum2 = 11 + deck[j] + deck[k]; if (deck[j] == 1) sum3 = deck[i] + 11 + deck[k]; if (deck[k] == 1) sum4 = deck[i] + deck[j] + 11; if (sum1 == 21 || sum2 == 21 || sum3 == 21 || sum4 == 21) cnt21++; total++; } } } System.out.println(total); double ans = (double)cnt21/(double)total; System.out.println(ans); System.out.println((1-ans)/ans); } } </pre><hr /> |
#4
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Re: Prop bets
Excellent work, thanks!
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#5
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Re: Prop bets
[ QUOTE ]
(40 choose 3)/(52 choose 3) = .4470 [/ QUOTE ] Assuming that you are offered this bet after looking at your cards, it's even worse if you have no A, J, or 7: (38 choose 3)/(50 choose 3) = 0.4304 However, if both your cards are either A, J, or 7 (ie AA, A7, J7, AJ, etc.) then you should take the bet: (40 choose 3)/(50 choose 3) = 0.5041 A fair payout here is only 98.38 cents on the dollar. EDIT: I assume hold'em here, not a crazy "house" game like pineapple. There the odds are even better if you have 3 As, Js, or 7s: (41 choose 3)/(49 choose 3) = 0.5786 ~ 0.7283:1 |
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