Percentage question (probability , game)

[reenmachine]

Suppose you throw a dice and on the first throw you win if you score 1 , you lose if you score 2 & 3 and have to throw the dice again if you hit 4 , 5 or 6.If you have to throw the dice again the rules changes and you win if you hit 1 but lose if you hit 2.If you hit any other number , you continue to throw it until you win or lose (with the same rules as the second throw).How do I find out what are my odds to the nearest percentage of winning in this game before the first throw?

I tried this , please help me understand what I did wrong or what I did right:

First throw:

1/6 = win (16,67%)
2/6 = lose (33,33%)
3/6 = throw again (50%)

Second throw and beyond:

1/6 = win (16.67%)
1/6 = lose (16.67%)
4/6 = throw again (66.67%)

Since the odds are the same for winning or losing beyond the first throw , once I get passed it my odds becomes 50% to win overall.

Calculations(?):

50(0,50) + 16,67(1) + 33,33(0)
25 + 16,67 + 0 = 41,67 = 42%

any thoughts would be appreciated , I'm pretty insecure everytime I encounter a probability problem so go easy on me please :)

 

[mathman]

Calculation looks OK. Style question - why did you use comma for decimal points?

 

[reenmachine]

Calculation looks OK. Style question - why did you use comma for decimal points?

Thanks a lot , I had no idea how to attack the problem , I have almost no background in probability so I just went with my intuition.

As for the style question , I guess I always used commas since my youth.Is this considered wrong or is it just a computer problem? I rarely use computers to do or communicate mathematics.Do the vast majority of programs don't count the commas as decimal numbers? If that's the case I guess the more math I'll do on computers the more I'll have to change that habit ;)

 

[reenmachine]

(Out of context , but if on a random first throw (ignoring the previous problem and rules)):

If I have a 25% chance to have a 25% chance to win on my next throw , a 60% chance to have a 60% chance on my next throw , and a 15% chance to have a 15% chance on my next throw , how do I calculate my odds of winning on my next throw?

I tried:

25(0.25) + 60(0.60) + 15(0.15)
6.25 + 36 + 2.25 = 44.5%
So I would have a 44.5% chance to win on my next throw.

does that make any sense to you guys?

thanks

 

[SteamKing]

N.B.: In most English speaking countries, the decimal point is a dot (.), whereas on the Continent it is a comma (,). The separator for thousands is the opposite.

A similar situation occurs with dates: in the US, it is month-day-year; elsewhere day-month-year.

 

[reenmachine]

Ok I'll ask it here also even though I asked it in the homework subforum , but if I have a 5.7% chance to win and a 8.33% to lose a game (with the additionnal 85.97% meaning I have to roll the dice (or whatever) again with the same odds until I win or I lose , how do I calculate my odds of winning the game overall before the game even begins?

In the other thread they talked about conditionnal probabilities , a concept I'm not knowledgeable about.

I thought my odds of winning were 40.6% because of 5.7+8.33=14.03. 5.7/14.03*100 = 40.6 , but it seems I was wrong.

If I have to add up the probability of winning round 1 + P of winning round 2 etc... how do I do it?

My try:

round 1: 0.057
round 2: 0.8597 (0.057)
round 3: 0.8597 (0.8597)(0.057)
etc...

I have a very strong feeling that I'm wrong , and even if I'm right how do I know when to stop adding up? (can't add up forever)

thanks!

 

[pbuk]

I thought my odds of winning were 40.6% because of 5.7+8.33=14.03. 5.7/14.03*100 = 40.6 , but it seems I was wrong.

No, that's correct.

round 1: 0.057
round 2: 0.8597 (0.057)
round 3: 0.8597 (0.8597)(0.057)
etc...

I have a very strong feeling that I'm wrong , and even if I'm right how do I know when to stop adding up? (can't add up forever)

You are right again! The trick is that you can add up forever, this is called the sum of a geometric series
whose properties are well known. In this case the sum is $\frac{0.057}{1-0.8597} = \frac{0.057}{0.1403}$, just the same as your intuitive result.

 

[reenmachine]

wow I'm actually impressed that I was right :D

Last question , very basic , but 40,6% is the odds of winning right? If I used the word probability instead of odds , would the result be the same?

Thank you for your input :X

cheers

 

[mathman]

wow I'm actually impressed that I was right :D

Last question , very basic , but 40,6% is the odds of winning right? If I used the word probability instead of odds , would the result be the same?

Thank you for your input :X

cheers

Don't get caught up in language quibble. Odds of winning and probability of winning are the same.