HI Randy,

I feel compelled to comment here...

First, assuming your solution is correct, then 5 out of 9 is 55.55555... %. If you are rounding this to the nearest full number then it is 56%.

However, neither the problem nor the solution makes any comment on what happens when the die is rolled to a blank surface. Your solution is only correct if the rules allow for that die to be rolled again - however, the problem does not suggest that.

When David rolls 2 then he wins four times and loses twice

When David rolls 4, he will win five times, and lose once.

When David rolls 5, he will win five times and lose once.

When Brian rolls 1, he will lose three times and wins three times

When Brian rolls 3, he will win four times, and lose twice.

When Brain rolls 6, he will win six times.

So David wins 14 times, and Brian only 13.

yielding a result of 51.85% in favour of David.

Of course, that is an extremely simple approach, which gives only an approximate answer, and is not mathematically / statistically accurate.

So - as the problem is written, we have to allow for the blank sides to count as zero. If this is the case, then the ONLY truly correct answer provided was that of Federico. The end result is that David wins 52.52% of the time.

Hate to be pedantic - actually, - no I don't... 😉

My apologies to Michael, but your answer is not correct, as you go in to detail which does not allow for the blank sides.

Because David did not go into any incorrect detail, his is actually the first (approximately) correct answer...

Regards,

Trevor

HI Randy,

I feel compelled to comment here...

First, assuming your solution is correct, then 5 out of 9 is 55.55555... %. If you are rounding this to the nearest full number then it is 56%.

However, neither the problem nor the solution makes any comment on what happens when the die is rolled to a blank surface. Your solution is only correct if the rules allow for that die to be rolled again - however, the problem does not suggest that.

When David rolls 2 then he wins four times and loses twice

When David rolls 4, he will win five times, and lose once.

When David rolls 5, he will win five times and lose once.

When Brian rolls 1, he will lose three times and wins three times

When Brian rolls 3, he will win four times, and lose twice.

When Brain rolls 6, he will win six times.

So David wins 14 times, and Brian only 13.

yielding a result of 51.85% in favour of David.

Of course, that is an extremely simple approach, which gives only an approximate answer, and is not mathematically / statistically accurate.

So - as the problem is written, we have to allow for the blank sides to count as zero. If this is the case, then the ONLY truly correct answer provided was that of Federico. The end result is that David wins 52.52% of the time.

Hate to be pedantic - actually, - no I don't... 😉

My apologies to Michael, but your answer is not correct, as you go in to detail which does not allow for the blank sides.

Because David did not go into any incorrect detail, his is actually the first (approximately) correct answer...

Regards,

Trevor

If you want simplicity how about this.

David will win because rolling a number 1 will never win.

Ok, I'm definitely Monday morning quarterbacking here, but when I first saw the question, I just figured that David's average was higher than Brians', so of course he would win over time.

But then I figured that the zeros were some trick part of the question, like "You can't know unless you know all the rules", and I didn't bother.

But thanks for the game!

Hi Randy

That's clever, I jumped in too fast!!

Oh God!!! Is it all to start AGAIN???

P.S

My life long friend Bill Wynn whom I met on your forum STILL ste each other this sortt of thing now and again and last week I sent him this to which he hasn't YET replied!

What would you do with this

hijklmno?

Ok Trevor, you're killing me! This is a tough crowd. And Richard, obviously it's code for "Hi Dr. Jekyll, no I wouldn't fly KLM either!"

🙂

-RG

Actually you would DRINK IT!!!!

It's water h20

Revenge is SOOOO Sweet!

Love and Laughter

Richard

Here is the problem -

Rolling a 1 will win three times out of six, because the other die has three zeros.

Trevor

I win - because I used the same reasoning as Randy. As a man thinketh... and if I think like Randy then multimillion-dollar-deals here I come!

And exactly how does ijklmn equate to 2? I don't get this one.

-RG

Randy -

I would have thought you would get that one...

"h" to "o" progressively through the alphabet...

Well done Richard...

Regards,

Trevor