Side Bet

[Dov]

I was in a game last night where someone made a proposition bet which was obviously skewed in his favor. Someone else at the table actually took the bet. I tried to calculate the odds for it, but wanted to double check with you guys.

The bet is 1-1 that no J, 7, or 2 will appear on the flop. It was later modified to include the entire board.

The way I did it was to take 12/52+(12/51*39/51)+(12/50*38/50). This was to calculate the flop without regard to any exposed cards. (Like my own hand) Is this the right way to do this problem?

BTW - I got an answer of around %59 chance that none of the cards will show up on the flop.

[bigpooch]

It's best to consider the chances that NONE of the cards
appear on the flop: the odds of that happening are
40/52 x 39/51 x 38/50 = 38/85 so the odds of at least one
of three different ranks appearing on the flop is 47/85 or
about 55.29412%.

For the entire board, the odds of no card appearing is then
38/85 x 37/49 x 36/48 = 2109/8330.

[mosta]

one note. these answers (55.3% that there is at least one card of these three different ranks on the flop and 2109/8330 = 74.7% that there is at least one by the river) assume the bet is made before any cards are dealt. if he has a hand and the hand has no J or 7 or 2 then there is a 57% chance of getting at least one on the flop and a 76.3% chance of getting one by the river.

[Dov]

Thanks guys.

I appreciate it.

[BruceZ]

You can also do it the way you tried to do it, but it goes like this:

12/52 +
(12/51*40/52) +
(12/50*40/52*39/51)
= 55.2941%

So each term multiplies the probability of not making it on the previous cards by the probability of making it on that card. It looks like your products were shifted off by a term.

For the whole board, this method gets longer, but it follows the same pattern:

12/52 +
(12/51*40/52) +
(12/50*40/52*39/51) +
(12/49*40/52*39/51*38/50) +
(12/48*40/52*39/51*38/50*37/49)
=74.6819%