What are the odds of tossing a pair of threes, twice, with two normal dice?
1/6^4 = 1/36^2 = 1/1296
Thanks, I told a couple of people that the dice were weighted (they weren't) and always landed on three and got them to role them (to prove me wrong) and to my utter surprise...this is what happened!
The thing is here a three is easier to make than "snake-eyes," or a two. Why? because we might have a duce and a pip or a pip and a duce. Thus they are 2 ways out of 36 to throw a three, and to do it twice is 1/18^2 = 1/324.
Yes, but the OP wanted to roll two threes twice, not a combined three twice.
O.K., yes, I see, 1/6^4.
This was sort of an experiment. I was so surprised by the result from the first person (rolling a pair of threes on their first role) that I repeated the same experiment on a different person a couple of days later, different dice (still non-weighted) but the same experiment with the same results! Does this greater the odds? In both cases there were only two dice and they only rolled once to get the double threes.
With this kind probability it's important to remember that the previous experiment has no bearing on the outcome of future experiments. The next time you hand someone a pair of dice then probabilty will still be 1 in 36, even though taken as a group three consecetive pairs of three would only havea chance of 1 in 46656.
Remember, every possible outcome is very unlikely, it's just that most of them are so similar to the other ones that they are more likely (for example, the dice summing to 7 is more likely than summing to 2). But the roll of 3-4 is just as likely as 1-1. Why aren't you amazed when someone rolls 3-4, 2-5, 4-6? There was only a 1/6^6 chance of that happening! AMAZING
I had this discussion with my brother a long time ago, but about the lottery.
I couldn't convince him that he might as well pick the sequence 1,2,3,4,5,6,7, since that was as likely as any other sequence. He argued that 1,2,3,4,5,6,7 was never going to win, that it was MUCH more likely that other sequences (such as random ones, like 1,13,17,21,24,35,41) would win. I couldn't convince him (in a short time) that his sequence had one chance in the same zillion as my sequence did.
Just because it *looks* random to us, doesn't mean it's more likely.
Actually, 3-4 is more likely than 1-1 to turn up: twice as likely - because you can also roll 4-3. There is only one way to roll 1-1.
On the lottery, where the winners shares the prize, it seems likely that many people will play numbers in sequence like 1,2,3,4..., so you might have to share the prize should it win.
I was refering specifically to 3 first, then 4 second, not the other way around.
As for the lottery: the only strategy I've ever heard is to pick numbers which are out of the range of birthdays so you don't have to share.
I was refereing to 2 dies rolled simultaneous. If only one of them landed on 3, would the odds then be 1 in 12?
No, the odds of exactly one of the dice landing on 3 would be 5/18 and the odds of at least one dice landing on 3 would be 11/36
5/18 = 1/6 * 5/6 + 5/6 * 1/6
11/36 = 1/6 * 5/6 + 5/6 * 1/6 + 1/6 * 1/6 (or 1 - 5/6 * 5/6 of you prefer).
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