rndaryam
rndaryam
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November 15th, 2016 at 8:26:17 AM permalink
In a game, a fair dice is thrown and afterwards a fair coin. If you cast a 1, you lost the game. If you cast one of the numbers 2 to 5, you throw the coin once. If you throw a 6, you throw the coin 2 times.
You win when you got the "head" in one of your coin.
1. How big is the probability of winning the game?
2. The game has been won. What was the probability of getting a 6?
CrystalMath
CrystalMath
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November 15th, 2016 at 8:41:07 AM permalink
Homework? Show us your work so far.
I heart Crystal Math.
MathExtremist
MathExtremist
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November 15th, 2016 at 9:41:35 AM permalink
And don't forget Bayes' Theorem. Anytime you see a question worded with past tense probabilities, you're most often looking at conditional probabilities. I'm assuming that's what this lesson is about anyway...

P(B|A) = P(B)*P(A|B)/P(A).

In other words, the answer to the first question goes right into the second.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
RS
RS
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November 15th, 2016 at 9:44:34 AM permalink
Inb4 55 pages of "2 dice problem" posts.
billryan
billryan
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November 15th, 2016 at 2:35:03 PM permalink
You have a five out of six chance to advance to round two.
In round two ,you have a 50-50 shot if you rolled a 2-5, or two 50-50 shots if you rolled a six.
So in round one, one in six, you lose, and you'll lose six out of twelve in the second round.
Or so it seems to me.
The older I get, the better I recall things that never happened
charliepatrick
charliepatrick
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November 15th, 2016 at 3:06:31 PM permalink
Quote: rndaryam

..."head" in one of your coin...

(i) When analysing games it's worth checking what the exact rules are. I'm guessing you win if both of your coins are "Heads".
(ii) Assume you always toss two coins but only take the first toss for a 2-5 and ignore them for a 1. Thus there six outcomes of the first round and each of these can have HH HT TH TT. So 24 equally likely outcomes. 11 of these win. Similarly 3 ways from the 11 come from throwing a 6.

A more normal Bayes problem features things like people being tested for a disease where you know (a) how good the test is (e.g. the chances of false positives and false negatives) (b) how many people tend to have the disease and (c) what the test result was. Typically this means comparing a freak result for a normal person against a correct result for something that is highly unlikely.
Joeman
Joeman
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November 15th, 2016 at 3:24:57 PM permalink
Quote: CrystalMath

Homework?

I was about to answer, but you may be right, CM. One could argue he wants to know the answer to the first question because he has encountered this game and wants to know what kind of edge he's up against. But the second question is purely academic (unless there's some type of wack-a-doodle side bet going on).
"Dealer has 'rock'... Pay 'paper!'"
kimma2366
kimma2366
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November 25th, 2016 at 11:30:48 AM permalink
Hi Joeman ,
you seem to be a smart person on probability .
Can you please help me .
What is the probability ( % ) of 2 heads come out in a row in every 9 times you flip a coin ?
What is the probability ( % ) of 2 heads or 2 tails ( % ) come out in a row in every 9 times you flip a coin ?
Thank you for your help .
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