Maths heads and statisticians - what are the chances?
Discussion
I've just experienced what feels like the most improbable event!
For those statistically minded, what are the chances of the following happening?
My daughter and I were just playing a game that required the roll of a normal six sided dice. She rolled first, moved forward the appropriate number of squares and then I rolled and recorded the same score. She then rolled again, recorded a different score and I rolled and recorded the SAME score as her again..... this happened a total of EIGHT times in a row!
Ignoring the potential bias from insufficient randomising owing to insufficient shaking etc. (though after the first couple of times I made a point of making sure the dice was tumbling all over the place!) what are the odds or chances of rolling the same number as your opponent EIGHT times in a row! (16 rolls in total)
In the spirit of education, please show workings etc.
For those statistically minded, what are the chances of the following happening?
My daughter and I were just playing a game that required the roll of a normal six sided dice. She rolled first, moved forward the appropriate number of squares and then I rolled and recorded the same score. She then rolled again, recorded a different score and I rolled and recorded the SAME score as her again..... this happened a total of EIGHT times in a row!
Ignoring the potential bias from insufficient randomising owing to insufficient shaking etc. (though after the first couple of times I made a point of making sure the dice was tumbling all over the place!) what are the odds or chances of rolling the same number as your opponent EIGHT times in a row! (16 rolls in total)
In the spirit of education, please show workings etc.
Edited by larrylamb11 on Sunday 7th November 10:46
The chance of any specific number coming up On a roll is 1/6.
So if you roll the dicd twice, the chance of getting the same number twice is 1/6 x 1/6. A good way to think about this is say you have rolled a six already. You roll again. Your chance of getting another 6 (or indeed any number) is 1/6.
So you just extend the chain, 1/6 x 1/6 x 1/6. …
In this case the likelihood is 0.00006 %. ie about 1 in 1.65 million.
This has better not be your daughters homework.
So if you roll the dicd twice, the chance of getting the same number twice is 1/6 x 1/6. A good way to think about this is say you have rolled a six already. You roll again. Your chance of getting another 6 (or indeed any number) is 1/6.
So you just extend the chain, 1/6 x 1/6 x 1/6. …
In this case the likelihood is 0.00006 %. ie about 1 in 1.65 million.
This has better not be your daughters homework.
Edited by LostM135idriver on Sunday 7th November 10:38
larrylamb11 said:
16 throws in total - we each threw 8 times and each time threw the same number (each time different number to the previous go, so dice not loaded!).
So that is much more likely, and much more complicated to calculate.Edited by LostM135idriver on Sunday 7th November 10:56
LostM135idriver said:
Ah, well then it’s (1/6)^16. Blimey.
I don’t think that’s correct. I believe it is still (1/6)^8. This is because the first throw of each pair can be any of the 6 numbers, so effectively is just multiplying by 1. The equation would look like6/6 x 1/6 role 1
6/6 x 1/6 role 2 etc etc
(1x1/6)x(1x1/6)……. Giving us (1/6)^8 still
Yes, good point if you do not need a specific number. first two rolls odds are 1/6 that you get the same number.
The first roll of the second pair the chance is 5/6 that you get a different number. The chance that the second roll of that pair matched is 1/6.
And so:
1/6 x (5/6 x 1/6)^7
I think.
The first roll of the second pair the chance is 5/6 that you get a different number. The chance that the second roll of that pair matched is 1/6.
And so:
1/6 x (5/6 x 1/6)^7
I think.
Edited by LostM135idriver on Sunday 7th November 11:02
Edited by LostM135idriver on Sunday 7th November 11:02
The first dice roll of any pair won’t matter - they were not expecting a specific result. The only one that mattered was that the second dice match the first - this is 1/6.
It would be 1/36 if they wanted both dice to be a 6 for example. As at that point the first dice can only be 1 of its 6 possible outcomes
It would be 1/36 if they wanted both dice to be a 6 for example. As at that point the first dice can only be 1 of its 6 possible outcomes
I'm going for 1 in 1,679,616, or 6 to the power of 8.
You aren't predicting what the number will be in any pair of throws, just that the 2nd throw of the pair will match the first throw. So the chances of this happening, when they have a throw each is 1 in 6. It doesn't matter what his daughter throws, so no odds on that. He just has to match her number, so 1 in 6 each time. And they did it 8 times.
You aren't predicting what the number will be in any pair of throws, just that the 2nd throw of the pair will match the first throw. So the chances of this happening, when they have a throw each is 1 in 6. It doesn't matter what his daughter throws, so no odds on that. He just has to match her number, so 1 in 6 each time. And they did it 8 times.
LostM135idriver said:
It’s not, because the chance of it matching the previous pair has to be consideredZ.
I understand what you're saying - basically that the chance of my daughters NEXT roll is not 6/6 as it can't be the number already rolled, so for roll number two it would be 5/6 and mine 1/6, then roll three 4/6 (as it can't be either of the prior two rolls) and mine 1/6, roll four 3/6 and mine 1/6 etc.... yeah... that's gets more complicated! The net result is that the chance of it happening is at MOST 1 in 1.68 million, but probably significantly more.....
A few years back, I was on a plane coming back from Holiday on Malta, and got chatting to a woman next to me about thirty years old.
She explained she had married a Maltese guy who owned a sports shop in Malta, but was going back to England to visit her father, who was having a big 70th birthday bash in a marquee in his house near Clitheroe.
Thought no more about it.
Then, later on that year, on a plane going to Malta, I sat next to a lady about 70 years old, who explained she was from Clitheroe and was going visiting her daughter who had married a guy who owned a Sports Shop in Malta.
Bearing in mind there are around 250 different seats I could have sat on, what are the odds on that?
When I told her, we both couldn't believe it.
She explained she had married a Maltese guy who owned a sports shop in Malta, but was going back to England to visit her father, who was having a big 70th birthday bash in a marquee in his house near Clitheroe.
Thought no more about it.
Then, later on that year, on a plane going to Malta, I sat next to a lady about 70 years old, who explained she was from Clitheroe and was going visiting her daughter who had married a guy who owned a Sports Shop in Malta.
Bearing in mind there are around 250 different seats I could have sat on, what are the odds on that?
When I told her, we both couldn't believe it.
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