Gambling with Dice
Two gamblers decide upon a game of dice with the following rules:
Gambler Albert (A) wins if he rolls 6 before Gambler B rolls 7.
Gambler Edward (E) wins if he rolls 7 before Gambler A rolls 6.
A coin toss has determined that gambler Albert will roll first ...
Which gambler has the better chance of winning? ... and why?
Gambler Albert (A) wins if he rolls 6 before Gambler B rolls 7.
Gambler Edward (E) wins if he rolls 7 before Gambler A rolls 6.
A coin toss has determined that gambler Albert will roll first ...
Which gambler has the better chance of winning? ... and why?
posted by Zaux at
11:09 AM
17 Comments:
Is this another from a Christmas cracker? ;)
it is being brought to you in living ... uh ... "black and white" ... mmmmm... that sounds better when one can use the word "color".
It originates from Zaux's private collection of puzzles, the location of which is the bookshelf.
All refernces to puzzlers living and dead is purely intentional... mmm ... good pain meds.
ok i've never played dice, in dice are there a set number of dice for the game or is that irrelevent to the question.
rolling 2 die
I may be wrong, but I'm guessing Albert will win first? There are six possible ways for both Albert and Edward to roll 7 and 6, respectively. Since Albert goes first, he has the advantage over them both.
are you sure there's 6 ways for both to win?
There are more ways to roll a 7 than a six. So the there's a slight probability that a seven will be rolled before a six.
But since the first to roll might be gaining a slight probability. If Edward wins the flip, then he has two advantages. If Albert wins the flip, then the advantages will tend to even out.
Hi Zaux. I withdraw my first comment. I fell into your trap. I thought it was a trick question.
E has the advantage. This is because the probability of throwing a 6 is 5/36 and the probability of throwing a 7 is 6/36. So A will need 7.2 rolls on average to throw a 6, and E will need 6 rolls on average to throw a 7. But A rolled first, so he's stlll got an average of 6.2 rolls left to get a 6, whereas E only has 6 rolls left on average to get a 7.
Phew! I only thought of that while typing this; the calculations looked horrific when I started this response.
Fortunately the margin is safely in E's favour, so I'll stick with my answer above. I suspect that to calculate the probability of who wins could be quite tedious.
I'd definitely be inclined to find the probabilities by computer simulation.
the published solution including the probability does look tedious
Hi Zaux. You teaser. You've not said if I got it right (or not). So now I'm full of doubt.
Gambler E has a very slight probability advantage.
Phew! thanks. Out of curiosity, what are the probabilities? (Just the numbers will do). Thanks.
Chris ... there's 2 pages of mumbo-jumbo and probability formulation ... but if I understand it correctly:
A has a 30/61 chance of winning
and
E has a 31/61 chance of winning
as I previously mentioned, the odds are minutely in favor of E
Hi Zaux, thanks for that. You've provided exactly the info I wanted, and it is exactly what I guessed it to be. To wit, I suspected that for A, it'd be 6/(6+6.2) = 30/61 and for E it'd be 6.2/(6+6.2) = 31/61. But I was very far from confident that my guess was right.
I mean, wildly speculated, not guessed.
Really posting this to kick the last one through.
I was pretty sure that it was almost even stevens for who'd win. Then I noticed that 6.0 and 6.2 were close, so I made the completely non-rational guess that I posted above.
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