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Transference
12-08-2004, 03:55 AM
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.

gaming_mouse
12-08-2004, 05:23 AM
[ 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.

gaming_mouse
12-08-2004, 05:41 AM
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&lt;52; i++){
for (int j=i+1; j&lt;52; j++){
for (int k=j+1; k&lt;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 />

Transference
12-08-2004, 08:06 AM
Excellent work, thanks!

IsaacW
12-08-2004, 11:24 AM
[ 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