Would you take this sidebet? $10 to win $1000

Greetings,


Was at a very loose, agressive $80-120 HE game last night and an interesting sidebet broke out. One player was the banker for this sidebet. He collected a $10 chip from anyone who was interested in the proposition and that player had to guess the RANK ONLY (regardless of suit) of his next hand. (i.e.: AK, 25, 88, etc.) For $10 he was offering a 100-1 payout of $1000 if you were correct. Obviously, if you won, you would forfeit your hand and collect the guaranteed $1000 rather than play. I think it made sense that the only guy who won chose 2-5 so he could take the bet and not reveal too much about his hand. Anyway, would you bank this proposition, or try and guess your cards and why?

I calculated about a .6 % chance of guessing right (assuming you always guess a non-pair), thus banking offers an advantage, but I haven't gone to the trouble of figuring out what the bank's expectation is.


-MD

There's no differentiation between suits, so assuming you picked a hand at random:


If you pick a non-pair: There are 16 combinations of correct cards, and 2652 - 16 combinations of incorrect cards. Probability: .00606


If you pick a pair: There are 12 combinations of correct cards, and 2652 - 12 combinations of incorrect cards. Probability: 0.00454


Of course, this is assuming you don't look at your own cards. That adds to the cases involved.


But anyway,


EV on non-pair = 0.00606 * 1000 - 0.99394 * 10 = $6.06 - $9.94 = -$3.88. Even money is about 164:1.


EV on pair = 0.00404 * 1000 - 0.99596 * 10 = $4.04 - $9.96 = -$5.92. Even money is about 247:1.


That's a sucker bet if ever I've seen one. The latter case makes the lottery attractive.

He stands to lose about $2 on each bet (I assume that nobody would pick a pair here). I think there's no way that he can recover that loss by the gained information. I would take this bet anytime (picking 72) except when I'm in the big blind (where I might get a free play and cannot safely dump the hand.)


cu


Ignatius

Your calculation is only correct if the order how you get the cards matters, which should not be the case with poker hands.


cu


Ignatius

The guy who was taking this bet probably worked the numbers using faulty logic like that cited above. An easy way to figure the probablity:


Say you choose 72. There are eight ways for the first card to come off the deck good, four 7s and four 2s, out of 52. If your first card is good, there are only four good cards that can come for your second card, out of the 51 left (if you catch a 7, only a 2 helps you, and vice versa). Therefore, the probability that you will catch the hand you predict is


(8/52) * (4/51) = .012 or about 82:1


Take the bet every hand the next time you see this guy. Maybe skip the bet on your big blind.


Where do they spread $80/120? /images/smile.gif

I get if you dont' pick a pair you will pick the right combo aobut 1/83. So you have the best of this bet.


What Mr. Kincy says is incorrect.


P(picking a non pair)= 16/ 52C2


16/ 52(51)/2= 16/1326= 0.01206.


I was wondering why he was offering this bet though i guess if he wins or loses he gets some info but i suspect knowing someone doesn't have 25

doesn 't really help you that much.

Crap, I forgot to divide by two didn't I?

You of course would pick a non-pair.


You can assume the draws are with replacement, since no replacement actually helps you (it removes a card of the 1st rank when you're hoping for a card of the 2nd rank).


If you draw twice with replacement, there are 13 squared combinations of rank with order important. You have 2 outcomes that are good for you since order is actually unimportant.


2/169 is clearly better than 100:1.

You don't need to divide by two. Well, I guess it depends on precisely how you do your calculation, but I did it like this:


I pick 2 and 7 as my cards. The probability of getting a 2 or 7 on the first card is:


8/52


times the probability of getting the other one as my second card:


4/51


= .012066 = 1.2%


I'm relatively new to probability and poker math and such, but can those that initially corrected his math explain to me why they would take this bet? You will only win slightly more than 1 time out of 100, but you are only getting a return of 9 to 1.


-Ryan!

rromanik@umich.edu

Obviously you would not choose a pocket pair (only 6 variations of, say, 77) but a suited or offsuit hand (16 variations of, say, 52). Now there are only 1326 hands, and 16-to-1326 is clearly better than 10-to-1000, so, yes, go ahead and take it if you´ve got enough money (and if the guy wont run away when you make your hand).