Warrior General Games

1 in a billion chance to call the same play 3 times in a row?

I have 15 outside run plays and this play has a 3 block.

https://glb2.warriorgeneral.com/game/replay/772103/1768292

That's what happens when you hire Roy "Tin Cup" McAvoy as your coach. "Give me another ball..."

(maybe that's too old a reference )

Originally posted by Ace of Spades 7
That's what happens when you hire Roy "Tin Cup" McAvoy as your coach. "Give me another ball..."

(maybe that's too old a reference )

Obviously you are a golfer

For someone who claims to know programming, it amazes me you keep getting stumped over simple things.

Computers don’t generate random numbers. They use a table. In small quantities, the table can be streaky. In large, its not.

Math is a thing.

Originally posted by Xars
For someone who claims to know programming, it amazes me you keep getting stumped over simple things.

Computers don’t generate random numbers. They use a table. In small quantities, the table can be streaky. In large, its not.

Math is a thing.



Yeah...I was just being stupid with that post. Obviously it isn't a 1 in a billion chance...but you gotta admit, the odds are pretty low for that to happen

If you flip a coin 10 times and the first 9 times it lands on heads, what are the odds that the last flip will land on tails?

Originally posted by Cybertron
If you flip a coin 10 times and the first 9 times it lands on heads, what are the odds that the last flip will land on tails?

Uh, 50/50. Like the man said, math is a thing.

Originally posted by atlbruce
Uh, 50/50. Like the man said, math is a thing.

I agree. I have argued with people who include the 1st nine flips into the equation and say it is near impossible that the 10th flip will land on heads.

Originally posted by Cybertron
1 in a billion chance to call the same play 3 times in a row?

I have 15 outside run plays and this play has a 3 block.

https://glb2.warriorgeneral.com/game/replay/772103/1768292

Given it ran against 3 different defenses and didn't pick up any yardage - might be better as not being in there

Originally posted by william78
Given it ran against 3 different defenses and didn't pick up any yardage - might be better as not being in there

Just wait until the Empire's S* All American HB plays next season

On a slightly different note, isn't there some logic/calculation where plays that "work well" early in the game are more likely to be called later in the game?

Originally posted by Cybertron
I agree. I have argued with people who include the 1st nine flips into the equation and say it is near impossible that the 10th flip will land on heads.

However the next 10 flips could go the other way

Here's a fun math quiz:

On an old show, a host would have contestants choose one of 3 doors. After that, he would open one of the two unchosen doors to show it was a loser. He'd then give the contestant an option to switch to the remaining door before revealing which door was the winner.


What's the optimal choice? To switch, or not to switch?

example:

door 1,2,3 - winner is 2 contestant picks 2.

Host shows 3 is loser, gives contestant option to switch to 1, they decline, host reveals they won.

Originally posted by Raid
Here's a fun math quiz:

On an old show, a host would have contestants choose one of 3 doors. After that, he would open one of the two unchosen doors to show it was a loser. He'd then give the contestant an option to switch to the remaining door before revealing which door was the winner.


What's the optimal choice? To switch, or not to switch?

Let's Make a Deal?

I would stick with my gut....original choice. If there is only 1 "winning door", then that means there are 2 "losing" doors. There will always be at least 1 losing door that he didn't pick, so they will show him that door. Then it just comes down to a coin flip.

Originally posted by Cybertron
Let's Make a Deal?

I would stick with my gut....original choice. If there is only 1 "winning door", then that means there are 2 "losing" doors. There will always be at least 1 losing door that he didn't pick, so they will show him that door. Then it just comes down to a coin flip.

Actually you switch mathematically, most players stick to their first choice.

The host shows - you have a 1 in 3 chance (33%) of being correct. Whatever option picked is totally irrelevant , whether you picked right or wrong it doesn't matter the host is opening one of the doors.

When the host offers to open the 2nd door you now have a 50/50 chance of winning. The odds are better.

Originally posted by william78
When the host offers to open the 2nd door you now have a 50/50 chance of winning. The odds are better.

Nice job contradicting your own point. If the odds are 50-50, there is no "better" choice. Either staying or switching has exactly the same chance of being correct. Why is probability such a difficult topic for so many?