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08-05-2002, 10:38 AM
1. How many times must you throw two dice before you are more likely than not to roll a seven?

2. Assume the chances of getting a Royal Flush on a poker machine is one in 40,000. You walk up to that machine just as someone is being paid for a Royal on it. You start playing from that point and refuse to quit until you hit. If you were to predict the EXACT number of plays you need to make to hit it, the prediction that would have the best chance of being right is what? (I happen to know the precise answer. Can you come within a hundred?)

08-05-2002, 11:01 AM
a) Infinity, since you're NEVER more likely to throw a 7 than the rest of the numbers....

... or only once, if you're looking to throw a 7 before any other SPECIFIC number.

b) 30K... I can't WAIT to see the precise answer.

08-05-2002, 11:21 AM
1. 4

2. 27726

08-05-2002, 11:53 AM
... just a wild guess.

2: 20.000

Assuming the machine has no memory.

08-05-2002, 12:36 PM
1) An infinite number of times. There will always be 36 combinations with only 6 of these combinations equalling 7. You will therefore always be more likely to throw a number other than 7. (five times more likely)

2) The fact that it just hit a Royal should be irrelevant. My answer is 8000 attempts.

Jimbo

08-05-2002, 01:32 PM
1) Undefined, you're always more likeley to roll a seven.

2) 1. Since every play is independent of each other no given play is more likeley to give you a royal. The first play is the best guess then because you're at least not guessing to high, you know for certain that at that time the event has not yet occured.

I think...

Sincerely, Andreas

08-05-2002, 03:13 PM
1. 3.80

2. 27726 by using

log(.500)/(log(39999)-log(40000))

08-05-2002, 04:23 PM
1. Four rolls. 1-(30/36)^4 = 0.5177469

2. 25,285 hands. ( 1-(1/e))*40000 = 25284.812

08-05-2002, 04:32 PM
Would you mind walking though the though process to getting to those equations.

08-05-2002, 04:40 PM
Maybe somebody has gotten it already, but i haven't read the others.

1. four times

2. 27726

08-05-2002, 05:06 PM
My Guesses:

1) Depending on how you read the question: 4 or

Never.

2) 1.

(The number of plays needed to have a greater than 50% to hit would be 27726 .)

08-05-2002, 05:34 PM
1. The probability of a 7 is 1/6 since out of 36 combinations the 6 combinations that give 7 are 1-6, 2-5, 3-4, 4-3, 5-2, 6-1. The probability of not getting a 7 on a roll is 5/6. The probability of not getting 7 for n rolls is (5/6)^n. This becomes less than .5 for n = 4 rolls.

2. This is intended to be a trick. The probability of being right if you predict it will hit on the nth play is the probability that it will not hit on the first n-1 plays and then hit on the nth play. This is

(1/40,000)(39999/40000)^n. This is maximum for n=1, so you should predict it will happen on the first play for the prediction to have the highest probability of being right.

This is not the same thing as asking how long it will take to have a greater than 50% chance of hitting as we did in the first problem, nor is it the same as asking the average time it will take. It also doesn't matter that the machine just hit as long as the machine has no memory.

08-05-2002, 10:23 PM
1: Never

2: 1

08-05-2002, 11:22 PM
many years ago before almost any poker player could figure out these things i won large winnings from two other players who were thought to be more knowlegedable than me with each of these props. funny thing is on one of them we even used david as the deciding person to determine who was right. do you remember david the 25,000 bet i had with mickey.

08-06-2002, 02:29 AM
Minor correction:

The probability of being right if you predict it will hit on the nth play is the probability that it will not hit on the first n-1 plays and then hit on the nth play. This is

(1/40,000)(39999/40000)^n

Make that (1/40,000)(39999/40000)^(n-1) with the same conclusion. The idea is that the first play has the highest probability (1/40000) because all the other plays require n-1 consecutive failures immediately prior to the first success.

08-06-2002, 02:38 AM
Are you going to give us the answers?

appears like you keep giving questions, which has been great, but never the answers.

08-06-2002, 04:04 AM
4 1

08-06-2002, 04:25 AM
1) After four rolls you are more likely to have rolled at least one seven.

2) Because each trial is equally likely to have a royal flush, and we will play until we get a royal flush, the most likely time to get the Royal flush is on the first trial, because the chance that the royal occurs then is 1 out of 40,000. The chance that it occurs on the second trial is slightly less, 1 minus the chance that it occured on the first trial times 1 out of 40,000. So if I have to get within 100 of the correct answer, I will choose the 100th trial.

Bob T.

08-06-2002, 05:26 AM
I suspect David's answer would have been longer if we had all fallen for his trap. Perhaps something like this:

I told you to think and not plug numbers mindlessly into your formulas and calculators. I asked for the prediction most likely to be correct, not the time it would take to be more likely than not to win. I only required you to be within 100 and many of you were off by thousands. The correct prediction is the first play since its probability of 1/40000 is maximum. If you don't know how I got that you should stop reading this forum at once. You don't know enough about probability, and you deserve to be taken out back and shot.

How'd I do?

08-06-2002, 09:40 AM
David,

This is the same fallacy that allows all of those gambling system-hawkers to tout their crap.

Each roll ALWAYS gives you 6/36 chances to roll a 7. If you were going to phrase the question as "Over how many rolls, all considered together, would it take until rolling a 7..." then you might be able to sell the 5/6 *5/6 *5/6 *5/6 = 48.22% chance of NOT rolling a 7 anytime during those 4 rolls....

But since I know you know that already, I'm wondering what your point was... unless it was to set up the next question (good trick, by the way)

08-08-2002, 09:21 PM
Just a) for now.

Let P(n)= the probability of throwing a pair of dice n times w/o throwing a seven.

P(n)={5/6}^n

Now P(3)> 1/2 while P(4)

08-09-2002, 05:33 AM
question a is misleading. it can be interpreted as: 'how many times must you throw 2 dice that the next throw is more likely than not to be a 7?' or as 'how many times must you throw 2 dice so that you will be more likely than not to have thrown a 7 in that sequence?'

08-14-2002, 01:42 AM
1. Never 2. 20,000 If the chances are 1 out of 40,000 then half the time i will hit it before 20,000 and half the time I will hit it after 20,000.