# Thread: This is advanced for me but hopefully basic for you guys!

1. ## This is advanced for me but hopefully basic for you guys!

I have been posed a question. I'm an engineer and my logical mind has come up with an answer. However, I'm aware that it is possible to calculate the answer, and that it may not match what my logical brain came up with! I just don't know how to do it!! I hope you can help me?

2 scenarios, which is more likely to succeed?

Pick an number from 1 to 6. Roll a die. The chance of being correct is 1/6. this much I know!

First scenario. You have 4 throws, you can keep the same number or change it for each throw. What are the chances of your number coming up? Does it increase with the number of throws you have?

Logical brain says that the fact that your number didn't come up in the first 3 throws has no effect on the probability that it will come up in the 4th throw. Is this relevant?

Second scenario. You only have one throw, but can pick 4 numbers. this gives you a 2/3 chance of success, 4 times the probability if you only chose 1 number, right?

2. Hi would would say they are both the same,

1st: you get 4 chances, Each throw gives you 1/6 chance to hit your number but you have 4 of these chances 1/6 + 1/6 +1/6 +1/6 = 2/3 (doesnt matter if you keep the same number or change it its still the same as each throw is independant)

2nd is the same. 2/3

Hope this helps,

Niall101

3. Originally Posted by Niall101
Hi would would say they are both the same,

1st: you get 4 chances, Each chance gives you 1/6 chance to hit your number but you have 4 of these chances 1/6 + 1/6 +1/6 +1/6 = 2/3 (doesnt matter if you keep the same number or change it its still the same as each throw is independant)

2nd is the same. 2/3

Hope this helps,

Niall101
Thanks Niall, I guess this was my question, can you simply add the individual tries to get the likelyhood over 4 tries.

How about this though, if you extend it to 6 tries, the probability of getting your number with 6 tries would be 1. IS this correct?

It seems to me that you'd be more likely to succeed if you took the other option and chose all 6 numbers for 1 throw!

Paul.

4. Yes you are right, We need the #times the event occurs over the sample space.

No of possible outcomes for 4 throws is 6*6*6*6=1296

No of these times which your number appears is ( 1 for when your number appears 5 when it doesnt appear)
1 *5*5*5 = 125
1*1*5*5 = 25
5*1*1*5 = 25
1*1*1*5 = 5
1*1*1*1 = 1
1*5*1*1 = 5
1*1*5*1 = 5
5*1*1*1 = 5
5*1*1*5 = 25
5*5*1*5 = 125
5*5*1*1= 25
5*5*5*1 = 125
5*1*5*5 = 125

So 621/1296 = 0.4791

Maybe my arithmetic is off somewhere but I think this method is right??

Maybe this is better

Cheers Paul

5. Originally Posted by Niall101
Yes you are right, We need the #times the event occurs over the sample space.

No of possible outcomes for 4 throws is 6*6*6*6=1296

No of these times which your number appears is ( 1 for when your number appears 5 when it doesnt appear)
1 *5*5*5 = 125
1*1*5*5 = 25
5*1*1*5 = 25
1*1*1*5 = 5
1*1*1*1 = 1
1*5*1*1 = 5
1*1*5*1 = 5
5*1*1*1 = 5
5*1*1*5 = 25
5*5*1*5 = 125
5*5*1*1= 25
5*5*5*1 = 125
5*1*5*5 = 125

So 621/1296 = 0.4791

Maybe my arithmetic is off somewhere but I think this method is right??

Maybe this is better

Cheers Paul
Your basic method is correct but I think you have made a mistake somewhere.

The easy way to find the probability your number comes up at least once in 4 throws is to consider the probability that it does not come up at all, which is $(5/6)^4$.

So the probability that your number comes up at least once is
$1 -(5/6)^4$, which is 0.5177, approximately.

6. Nice. Much simpler lol. I will be taking probability this year so hopefuly I will learn all these short methods. Thanks!