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Would you make this bet?

kranky

kranky

Elite Member
The bet is that you get 22 chances to roll snake-eyes (double ones) using two dice. If you roll snake-eyes in 22 or fewer throws of the dice, you win.

EDIT:
Reply #40 has the background story and spoilers.
 
you have a 1 in 36 chances of winning. You have 22 rolls. 1 times 22 equals 22. 22 over 36 =11 over 18.

I have a weird way of doing math in my head, but it usually works out 😀
 
I think DeadByDawn is correct in his calculation of 11:18. I would take this bet as the odds of you winning are better than 50:50, which is better than the best bet in Vegas.
 
Originally posted by: Beattie
the probability of rolling "snakeyes" is 1/36. This bet has a pretty large expected loss.
Click to expand...

It's a little more complicated than that, though. If you have 22 rolls, then the probability of rolling snake eyes (only one possible outcome) equals 1-P(22 rolls without rolling snake eyes), which is an easy probability to calculate:

P(22 rolls w/o snake eyes) = (35/36)^22 = .538
So it follows that the probability of rolling at least one snake eyes is 1-.538 = .462.

It's a little less than 50/50 that it'll happen, so if you get better than a 1 to 1 payout, you probably could take the wager.
 
Probability of throwing a 1 is 1/6.
Pr of throwing snake eyes is 1/6 times 1/6 = 1/36.

Pr of throwing snake eyes at least once in 22 rolls is 1/36 + 1/36 + .... 22 times
=22/36 =11/18

or did i miss something?
 
Probabilty has nothing to do with this. Each throw is independant of the other.... it's a 1 in 36 chance EVERY time.
 
Originally posted by: DeadByDawn
you have a 1 in 36 chances of winning. You have 22 rolls. 1 times 22 equals 22. 22 over 36 =11 over 18.

I have a weird way of doing math in my head, but it usually works out 😀
Click to expand...

I'm not a statistician, but I took a couple prob&stats classes when I was in school, and I think I know why my answer isn't the same as yours.

If you extend your logic further, then if I had 36 rolls, 1 times 36 equals 36, so 36 over 36 = 1.

That would mean that if I had 36 rolls, I could guarantee that I would roll a snake eyes. The problem with your calculation is that you haven't factored in the possiblity of any given outcome occurring more than once. If every time I rolled, I could eliminate that outcome from occurring again (I roll a 1 & a 3, and then never that combination again), your calculation would be correct.

Sorry to be so nitpicky, I just felt like I had to pre-emptively defend my calculations 😛
 
The probability of winning this bet is 0.46193, so if it's an even money bet, you shouldn't make it.

To calculate the probability of winning the bet, first calculate the probability of rolling 22 non-snake eye rolls in a row. The probability of rolling non-snake eyes on one roll is 35/36. The probability of rolling 2 non-snake eyes in a row is (35/36) * (35/36). By extension, the probability of rolling 22 non-snake eyes in a row is (35/36) to the 22nd power, which is 0.53807. This is the probability of losing the bet. The probability of winning is 1 minus this number: 0.46193. A bad bet.
 
Originally posted by: Modeps
Probabilty has nothing to do with this. Each throw is independant of the other.... it's a 1 in 36 chance EVERY time.
Click to expand...



Ok, i get it. By my way it would figure that if you had 36 rolls then you would have a 100% probability of hitting snake eyes. That is not correct.
 
no.

but then again, I don't gamble at all. except when I drive, and in that situation, I'm only gambling with my life 😉
 
Originally posted by: DingDingDao
Originally posted by: DeadByDawn
you have a 1 in 36 chances of winning. You have 22 rolls. 1 times 22 equals 22. 22 over 36 =11 over 18.

I have a weird way of doing math in my head, but it usually works out 😀
Click to expand...

I'm not a statistician, but I took a couple prob&stats classes when I was in school, and I think I know why my answer isn't the same as yours.

If you extend your logic further, then if I had 36 rolls, 1 times 36 equals 36, so 36 over 36 = 1.

That would mean that if I had 36 rolls, I could guarantee that I would roll a snake eyes. The problem with your calculation is that you haven't factored in the possiblity of any given outcome occurring more than once. If every time I rolled, I could eliminate that outcome from occurring again (I roll a 1 & a 3, and then never that combination again), your calculation would be correct.

Sorry to be so nitpicky, I just felt like I had to pre-emptively defend my calculations 😛
Click to expand...

ding is right.
 
Originally posted by: DingDingDao
Originally posted by: Modeps
Probabilty has nothing to do with this. Each throw is independant of the other.... it's a 1 in 36 chance EVERY time.
Click to expand...

Huh?
Click to expand...

Unless for whatever reason the previous die rolls have an effect on each other, there's no real way of calculating the chance you'll come up with snake eyes once. Each roll is a 1 in 36 chance of getting snake eyes.
 
Originally posted by: DingDingDao
Originally posted by: Beattie
the probability of rolling "snakeyes" is 1/36. This bet has a pretty large expected loss.
Click to expand...

It's a little more complicated than that, though. If you have 22 rolls, then the probability of rolling snake eyes (only one possible outcome) equals 1-P(22 rolls without rolling snake eyes), which is an easy probability to calculate:

P(22 rolls w/o snake eyes) = (35/36)^22 = .538
So it follows that the probability of rolling at least one snake eyes is 1-.538 = .462.

It's a little less than 50/50 that it'll happen, so if you get better than a 1 to 1 payout, you probably could take the wager.
Click to expand...

Good stuff! It's been over 3 years since I had my stats class where we covered this. I use to have a VB program that would figure this out for me (did it for another class) but who knows where that is.
 
Originally posted by: Modeps
Originally posted by: DingDingDao
Originally posted by: Modeps
Probabilty has nothing to do with this. Each throw is independant of the other.... it's a 1 in 36 chance EVERY time.
Click to expand...

Huh?
Click to expand...

Unless for whatever reason the previous die rolls have an effect on each other, there's no real way of calculating the chance you'll come up with snake eyes once. Each roll is a 1 in 36 chance of getting snake eyes.
Click to expand...

What you're saying is correct, but my calculation is for 22 independent events.
 
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