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View Full Version : Please help settle a bitter debate at Canterbury


11-27-2005, 06:58 AM
buddy calls river card, says, "what's the odds of that?" I say 1/46, he says no way, it turns into a side bet.

Even money bet. I get 52 chances to call either a turn or a river card exactly. I cannot continue to call the same exact card.

I am using my hole cards, any exposed cards, the board cards, and general hand reading ability to make the best possible guesses.

The debate is who has the best of it? The one calling the cards with 52 opportunities or the one holding the bet?

There is now a $1000 side bet on who has the best of it.

Thanks!

Bryan

11-27-2005, 06:59 AM
btw: mathematical proofs will help in this discussion. Both sides have tried all matters of debating.

BruceZ
11-27-2005, 07:16 AM
[ QUOTE ]
buddy calls river card, says, "what's the odds of that?" I say 1/46, he says no way, it turns into a side bet.

Even money bet. I get 52 chances to call either a turn or a river card exactly. I cannot continue to call the same exact card.

I am using my hole cards, any exposed cards, the board cards, and general hand reading ability to make the best possible guesses.

The debate is who has the best of it? The one calling the cards with 52 opportunities or the one holding the bet?

There is now a $1000 side bet on who has the best of it.

Thanks!

Bryan

[/ QUOTE ]

Even if you use no hand reading and simply guess a random river card out of the 46 unseen cards, you have a 45/46 chance of missing each guess, so your chance of missing 52 times in a row is (45/46)^52 =~ 31.9%, so you are better than a 2-1 favorite to get at least one right (2.1-to-1).

AaronBrown
11-27-2005, 01:09 PM
In addition to BruceZ's excellent post, I would add that I doubt there is much additional to be gained by hand reading. You might put one of the players on, say, AJ; but you won't be sure, and the expected numbers of Aces and Jacks among 9 unseen starting hands is over 3 anyway, so being pretty sure there are 2 in one hand doesn't do much for you.

It doesn't matter that you are not allowed to guess the same card every time, that's a silly restriction.

The big problem with the bet is that if you have 1 chance in 52 of guessing a card, the odds are much better than 50% that you will do it right 1 time in 52. For large N, if you have 1 chance in N per turn and N independent turns, your chance of losing tends toward 1/e = 37%.

You expect to win N*(1/N) = 1 time. If you had a 50% chance of zero wins, you would have to have a large chance of 2 or more wins to get up to expectation.

A fair bet is to give you approximately ln(2)*N chances. In this case, that's 32. Then you'll have a 50% chance of getting 1 or more right with 45/46 chance per turn.

daryn
12-25-2005, 04:04 AM
[ QUOTE ]
I cannot continue to call the same exact card.

[/ QUOTE ]

damn.. that's where your edge was!

how are you supposed to guess just randomly??

ohnonotthat
12-25-2005, 04:33 AM
If someone offers to bet you $1,000 at even money that you cannot guess the exact card ONCE they either hate money or love you.

If you are given 52 chances and need only succeed once your chance of winning this bet is excellent.

The principle is the same as that used to calculate the odds of AK flopping a pair (or better), or of a pocket pair flopping a set (or better).

Do not add the possibilities of succeeding; multiply the chances of failing - and subtract the answer from "1".

1- (51/52 to the 52nd power) will do it; it's called "multiplication of negative expectation".

*

I assume you were joking when you asked whether it mattered how you chose the card you used for your guess ?

If you were not [joking], here's a hint.

- It does not matter.

*

Oops, I just noticed that Aaron already responded. Expect his reply to be clearer and likely more concise than mine but in this one case it's doubful he got it any righter than I did - this is stats 101 stuff.

*

Pascal is alleged to have gotten rich betting people they could not roll a 6 on one die if given 4 tries then to have gone broke betting they could not roll 6-6 using two dice if given 24 tries.

5 / 6 to the 4th is .48225 (chance of getting at least one 6)

35/36 to the 24th is .50896 (chance of getting at least one 6-6

*

He too tried adding the chances of success rather than multiplying the chances of failure.

Then again, he was French. /images/graemlins/smirk.gif

ohnonotthat
12-25-2005, 04:46 AM
Sounds like a hammer and a bucket of ice water might work better. /images/graemlins/shocked.gif /images/graemlins/grin.gif