You really don't like having to tax and insure your car properly do you.It actually started with SORN on your vehicle.
One in 380,204,032?Here's one for you.
5 people each have a standard deck of 52 cards. They each shuffle their pack randomly. Each person draws one card, places it face down, and after everyone has a card, they all turn them over.
What is the probability that everyone has the same card?
(eg 5 x 6 of clubs; 5x jack of spades, etc)
Maybe the first guy doesnt have to match with anyone so his draw could be any card? Its the only reason i could think of to not x the last 52Why is that Wij? I got the same as Scouse, 52^5 is that wrong?
Without putting too much thought into it, I'd say 52 ^ <number of packs> as well. If there were only 2 packs there wouldn't be 52 possibilities, there'd be 52^2.Why is that Wij? I got the same as Scouse, 52^5 is that wrong?
Yeah, that's it isn't it. The first guy picks any card, so the second guy has a 1 in 52 chance to match it. The third guy 1 in 52^2, and so the 5th 1 in 52^4.Maybe the first guy doesnt have to match with anyone so his draw could be any card? Its the only reason i could think of to not x the last 52
Thats where probability blows my skull.Here's one for you.
5 people each have a standard deck of 52 cards. They each shuffle their pack randomly. Each person draws one card, places it face down, and after everyone has a card, they all turn them over.
What is the probability that everyone has the same card?
(eg 5 x 6 of clubs; 5x jack of spades, etc)
That's horrid.Pakistan passenger plane crashes in Karachi
Carrying ~90 passengers (the Indy article says 107 on board), it crashed into a "residential area" in Karachi.
Why is that Wij? I got the same as Scouse, 52^5 is that wrong?
I don't follow that. Pls explain.Because while 52^5 is the odds of one particular identical combination (say all Ace of Spades), since there are 52 possible combinations where all cards match the odds are then 52^5 / 52 = 52^4.
That's unrelated to @Zarjazz' explanation which is about cards, not packs?Because if you have 2 packs of cards and you pick one card from the first pack, there's a 1 in 52 chance that you pick the same card from the second pack, so it's 52 ^ <number of packs - 1>.
If A and B are independent events, the probability of this event happening can be calculated as shown below:
I do get your point about packs. But yoir thinking seems to imply conditionality.Because if you have 2 packs of cards and you pick one card from the first pack, there's a 1 in 52 chance that you pick the same card from the second pack, so it's 52 ^ <number of packs - 1>.
Celebrity deaths.That's horrid.
Not to belittle in any way - it's interesting to make the point that we were at over 8 times that (at 854) on daily deaths from coronavirus.
I don't follow that. Pls explain.
If one card combo is 52^5 the existence of other combos is independent and the chances of one of those other combos is the same, independent, probability. No?
My favourite type of agreeingWe are furiously agreeing.
Nope. It's 1 in 52.The chance of drawing the 'right' card from the first deck is 1. It's then multiplied by 52 for each of the subsequent decks.