high G |

The comment is, If you go last it's more likely the short (bad) straw has already been chosen. Or, the opposite, say you’ve got N peeps. The first person to draw a straw is the least likely to draw the short one. While the later people are more likely to draw the short straw.

(Instead of straws, you can use cards with all being black but one.)

Steve Geddes |

The odds are identical for all players.

Perhaps one way to understand it is to play without looking at your draw. Everyone draws a straw in any old order, then simultaneously look at what they've picked - there's no advantage there whether you happened to be first, last or anything in between.

The only difference the way it's actually played is that the player drawing the short straw knows immediately that they've lost (instead of waiting til the end to find out). The probability of losing doesn't change.

Jail House Rock |

The odds are identical for all players.

Perhaps one way to understand it is to play without looking at your draw. Everyone draws a straw in any old order, then simultaneously look at what they've picked - there's no advantage there whether you happened to be first, last or anything in between.

Not sure this sheds new light as the drawing sequence is the same as before. The guy picking last still appears to have the advantage of the situation, even if everyone waits till the end before looking.

Steve Geddes |

6 people marked this as a favorite. |

Steve Geddes wrote:The odds are identical for all players.

Perhaps one way to understand it is to play without looking at your draw. Everyone draws a straw in any old order, then simultaneously look at what they've picked - there's no advantage there whether you happened to be first, last or anything in between.

Not sure this sheds new light as the drawing sequence is the same as before. The guy picking last still appears to have the advantage of the situation, even if everyone waits till the end before looking.

Nope. Imagine there are four players:

Player one gets the short straw 1/4 of the time.

Player two only gets the short straw if P1 didn't AND if P2 chooses the short straw out of the remaining three. The probability of two events occurring in this situation is the probability of the first multiplied by the probability of the second.

This chance is 3/4 (chance of the first player drawing a long straw) x 1/3 (chance of the second player drawing the short straw given that P1 didn't)= 3/12 = 1/4

Player three gets the short straw if P1 didn't take it (out of the initial four straws), P2 didn't take it (out of the remaining three straws) and then P3 chooses it out of the remaining two. This chance is 3/4 x 2/3 x 1/2 = 6/24 = 1/4

Player four gets the short straw if P1, P2 and P3 all avoided it. This chance is 3/4 x2/3 x 1/2 = 6/24 = 1/4

The same calculations hold true for any number of players - it just takes longer. I suspect you're thinking of it as "the player at the end of the queue has more and more chances for someone else to get it" this is balanced out perfectly by the fact that the players at the end of the queue are choosing from far fewer straws. Thus, if nobody has picked it they are at ever increasing risk of being the loser.

Steve Geddes |

As another way to see it - try and allocate what probability you think each player has. They have to add up to one, since SOMEONE has to get the short straw.

If the last player in a game with n players has less than 1/n chance of winning - whose chance is the worst? Presumably you'll allocate the extra losing probability to the first player, but how can someone choosing from n straws have anything other than a 1/n chance of picking the short one?

Jail House Rock |

Jail House Rock wrote:Steve Geddes wrote:The odds are identical for all players.

Not sure this sheds new light as the drawing sequence is the same as before. The guy picking last still appears to have the advantage of the situation, even if everyone waits till the end before looking.

Nope. Imagine there are four players:

Player one gets the short straw 1/4 of the time.

Player two only gets the short straw if P1 didn't AND if P2 chooses the short straw out of the remaining three. The probability of two events occurring in this situation is the probability of the first multiplied by the probability of the second.

This chance is 3/4 (chance of the first player drawing a long straw) x 1/3 (chance of the second player drawing the short straw given that P1 didn't)= 3/12 = 1/4Player three gets the short straw if P1 didn't take it (out of the initial four straws), P2 didn't take it (out of the remaining three straws) and then P3 chooses it out of the remaining two. This chance is 3/4 x 2/3 x 1/2 = 6/24 = 1/4

Player four gets the short straw if P1, P2 and P3 all avoided it. This chance is 3/4 x2/3 x 1/2 = 6/24 = 1/4

The same calculations hold true for any number of players - it just takes longer. I suspect you're thinking of it as "the player at the end of the queue has more and more chances for someone else to get it" this is balanced out perfectly by the fact that the players at the end of the queue are choosing from far fewer straws. Thus, if nobody has picked it they are at ever increasing risk of being the loser.

Hey, this makes sense. I think you have done it.

Steve Geddes |

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Yeah. Probability is a minefield - our intuitions usually lend us astray.

A maths degree, a fascination with the game of bridge and a short lived career playing online poker helps. :p

Steve Geddes |

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=Steve Geddes winsby being the first with a complete answer.

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Only for four players. The general case was left as an exercise for the reader..

Vod Canockers |

It depends upon whether or not the straws are shown as drawn or all at once.

For example if there are 5 straws and all revealed together, then the odds are 1 in 5.

If each person reveals their draw at once, then the first person has 1 in 5, the second has 1 in 4 or 0 in 4, the third 1 in 3 or 0 in 3, the fourth 1 in 2 or 0 in 2, the fifth has 1 in 1 or 0 in 1.

GreyWolfLord |

You could also view it as the statistical anamoly of the three doors question.

If you were asked which door had a car behind it...and there were three doors, which one gives you the best ability to get it.

If you choose a door, and then you are asked if you want to switch your answer to one of the other two doors before it's chosen, what are the odds?

Are your odds better if you stick with your original choice, or better if you make a new choice?

Odds ironically show that you should always disregard your first choice to make a second choice...as your odds are better.

In thought we would say, hey, it's still just 1/3 chance no matter what...

but the math for some strange reason says differently.

if I recall the conundrum correctly.

Kirth Gersen |

If each person reveals their draw at once, then the first person has 1 in 5, the second has 1 in 4 or 0 in 4...

The second person has a chance of (0.8)(0.25) + (0.2)(0) = 0.2 = 1 in 5. (That's the chance the 1st person didn't get it x the chance the second does, plus the chance the first person did get it x the chance the second does (0 if already accounted for). It works the same for the third person, and so on.

Everyone has a 1 in 5 chance.

BigDTBone |

1 person marked this as a favorite. |

You could also view it as the statistical anamoly of the three doors question.

If you were asked which door had a car behind it...and there were three doors, which one gives you the best ability to get it.

If you choose a door, and then you are asked if you want to switch your answer to one of the other two doors before it's chosen, what are the odds?

Are your odds better if you stick with your original choice, or better if you make a new choice?

Odds ironically show that you should always disregard your first choice to make a second choice...as your odds are better.

In thought we would say, hey, it's still just 1/3 chance no matter what...

but the math for some strange reason says differently.

if I recall the conundrum correctly.

This is only applicable because Monty Hall knows which door has the big prize and which ones don't. The omniscient choice giver changes the math.

You have a 1/3 choice, 1 good, 2 bad. No matter your choice there is at least one bad remaining. Monty Hall knows which one of the remaining choices is bad (perhaps both) and he shows you the bad choice. So, your original choice represents a 1/3 possibility of winning, where as if you now switch you get the benefit of Monty's knowledge and the likelihood of winning increases to 2/3rds.

In drawing straws you neither have the option to switch straws later (before or after the reveal) nor is their an omniscient host removing options based on his knowledge. It's a 1/x chance, no matter when you draw.

Misroi |

Yeah, the Monty Hall problem never made much sense at first blush. The first time someone tried to explain it to me, the tack they took was to expand the problem to 100 doors. "Let's say you pick door 5. Monty then opens all the doors except door 5 and door 37. What are the chances that you picked correctly on your initial guess, and what are the odds that switching to door 37 is a better choice?" It didn't really ring true, as while that's one way of viewing the three door problem, the other way is that Monty isn't opening all but one doors, but just one door. I didn't see the explanation as compatible to the solution.

In the end, simple probability ended up solving it for me. At the outset, you have three doors, and you have no knowledge of what's behind any of them. Your odds of picking the correct door is 1/3, no matter the door you pick. Each door, therefore, has a value of 1/3. If you pick door 1, you have a 1/3 chance of being correct, and a 2/3 chance in being wrong. Mathematically expressed:

D1 + D2 + D3 = 1

D1 + (D2 + D3) = 1

1/3 + (2/3) = 1

So, Monty opens door 2, revealing a goat. What does that mean? How does it change the expression? Well, what it tells us is that (D2 + D3) isn't (1/3 + 1/3) as we would expect it to be. It still has to equal 2/3, so the quantity is actually (0 + 2/3). So, going against how we normally think probability works, the numbers show that 2/3 times, the prize will be in the door you didn't choose.

The best part of this problem is that it caused so much consternation on the Internet that math teachers actually had their students sit down and test the theory. Time and time again, experimentation showed that switching was the best strategy to win. Yay for fun ways to teach kids math and science!

Vod Canockers |

Vod Canockers wrote:If each person reveals their draw at once, then the first person has 1 in 5, the second has 1 in 4 or 0 in 4...The second person has a chance of (0.8)(0.25) + (0.2)(0) = 0.2 = 1 in 5. (That's the chance the 1st person didn't get it x the chance the second does, plus the chance the first person did get it x the chance the second does (0 if already accounted for). It works the same for the third person, and so on.

Everyone has a 1 in 5 chance.

Overall, but since you know the outcome of the previous draw your instant odds keep changing.

BigDTBone |

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Kirth Gersen wrote:Overall, but since you know the outcome of the previous draw your instant odds keep changing.Vod Canockers wrote:If each person reveals their draw at once, then the first person has 1 in 5, the second has 1 in 4 or 0 in 4...The second person has a chance of (0.8)(0.25) + (0.2)(0) = 0.2 = 1 in 5. (That's the chance the 1st person didn't get it x the chance the second does, plus the chance the first person did get it x the chance the second does (0 if already accounted for). It works the same for the third person, and so on.

Everyone has a 1 in 5 chance.

Your odds at the time of pulling may change but those odds are dependent upon a previous random trial. The sum of your odds for the entire exercise remain constant no matter when you draw.

therealthom |

...

In the end, simple probability ended up solving it for me. At the outset, you have three doors, and you have no knowledge of what's behind any of them. Your odds of picking the correct door is 1/3, no matter the door you pick. Each door, therefore, has a value of 1/3. If you pick door 1, you have a 1/3 chance of being correct, and a 2/3 chance in being wrong. Mathematically expressed:

D1 + D2 + D3 = 1

D1 + (D2 + D3) = 1

1/3 + (2/3) = 1So, Monty opens door 2, revealing a goat. What does that mean? How does it change the expression? Well, what it tells us is that (D2 + D3) isn't (1/3 + 1/3) as we would expect it to be. It still has to equal 2/3, so the quantity is actually (0 + 2/3). So, going against how we normally think probability works, the numbers show that 2/3 times, the prize will be in the door you didn't choose.

The best part of this problem is that it caused so much consternation on the Internet that math teachers actually had their students sit down and test the theory. Time and time again, experimentation showed that switching was the best strategy to win. Yay for fun ways to teach kids math and science!

I'm not getting this. It seems to me that whether or not switching your choice improves your odds depends on Monty's behavior. Does he always show you what's behind one of the doors? Or is he more likely to show you what's behind one of the doors if you have chosen poorly.

Assuming Monty will always show you what's behind one of the doors, you can just as easily pose rewrite your equations thusly:

D1 + D2 + D3 = 1

(D1 + D2) + D3 = 1

2/3 + 1/3 = 1

And darned if you shouldn't always stick to your choice.

If Monty always shows you an empty door your odds were 1/2 when you started and they stay that way.

thejeff |

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Misroi wrote:...

In the end, simple probability ended up solving it for me. At the outset, you have three doors, and you have no knowledge of what's behind any of them. Your odds of picking the correct door is 1/3, no matter the door you pick. Each door, therefore, has a value of 1/3. If you pick door 1, you have a 1/3 chance of being correct, and a 2/3 chance in being wrong. Mathematically expressed:

D1 + D2 + D3 = 1

D1 + (D2 + D3) = 1

1/3 + (2/3) = 1So, Monty opens door 2, revealing a goat. What does that mean? How does it change the expression? Well, what it tells us is that (D2 + D3) isn't (1/3 + 1/3) as we would expect it to be. It still has to equal 2/3, so the quantity is actually (0 + 2/3). So, going against how we normally think probability works, the numbers show that 2/3 times, the prize will be in the door you didn't choose.

The best part of this problem is that it caused so much consternation on the Internet that math teachers actually had their students sit down and test the theory. Time and time again, experimentation showed that switching was the best strategy to win. Yay for fun ways to teach kids math and science!

I'm not getting this. It seems to me that whether or not switching your choice improves your odds depends on Monty's behavior. Does he always show you what's behind one of the doors? Or is he more likely to show you what's behind one of the doors if you have chosen poorly.

Assuming Monty will always show you what's behind one of the doors, you can just as easily pose rewrite your equations thusly:

D1 + D2 + D3 = 1

(D1 + D2) + D3 = 1

2/3 + 1/3 = 1And darned if you shouldn't always stick to your choice.

If Monty always shows you an empty door your odds were 1/2 when you started and they stay that way.

Monty always shows you an empty door, but he's doing it after you've made your initial choice. He's eliminating one of the bad choices. If you could choose up front or wait until he'd shown you an empty door, you would obviously wait, right? Since you'd have a 50/50 chance at that point, instead of one in three. It's the same if he does it after.

Another way to look at it, closer to what Misroi wrote: You have a 1/3 chance to pick the right door the first time, which means there's a 2/3 chance it's one of the other two. But Monty show you one of those two, so if it's one of those, you know it's the other one. It's still a 2/3 chance it's one of those two, but now, if it is, you know which one. So you're left with 1/3 for your door and 2/3 for the other one.

The way you wrote the equation doesn't make sense. He grouped them that way because you picked D1. So there's the one you picked (D1)=1/3 and the ones you didn't pick (D2+D3)=2/3. But then Monty shows you D2, so you know it's (0+D3) = 2/3.

BigNorseWolf |

It depends upon whether or not the straws are shown as drawn or all at once.

For example if there are 5 straws and all revealed together, then the odds are 1 in 5.

If each person reveals their draw at once, then the first person has 1 in 5, the second has 1 in 4 or 0 in 4, the third 1 in 3 or 0 in 3, the fourth 1 in 2 or 0 in 2, the fifth has 1 in 1 or 0 in 1.

You have to factor in the possibility that someone has drawn the straw before you even get to draw.

ShadowcatX |

therealthom wrote:Monty always shows you an empty door, but he's doing it after you've made your initial choice. He's eliminating one of the bad choices. If you could choose up front or wait until he'd shown you an...Misroi wrote:...

D1 + D2 + D3 = 1

D1 + (D2 + D3) = 1

1/3 + (2/3) = 1I'm not getting this. It seems to me that whether or not switching your choice improves your odds depends on Monty's behavior. Does he always show you what's behind one of the doors? Or is he more likely to show you what's behind one of the doors if you have chosen poorly.

Assuming Monty will always show you what's behind one of the doors, you can just as easily pose rewrite your equations thusly:

D1 + D2 + D3 = 1

(D1 + D2) + D3 = 1

2/3 + 1/3 = 1And darned if you shouldn't always stick to your choice.

If Monty always shows you an empty door your odds were 1/2 when you started and they stay that way.

I'm not sure if you are serious or if you're trolling.

yellowdingo |

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BigNorseWolf wrote:The odds are the same either way, 1 in x number of whatevers. Go first and get it over with.I would go last. Because by that time the short straw is already gone.

The short straw is always drawn last. No point in drawing long straws once you know jimbo is the one who gets eaten by the group.

David knott 242 |

Going last is the best choice if you are both clever and unethical. Assuming that each participant reveals the straw as it is drawn, you will know that you are doomed if none of the other three draw the short straw. At that point you don't draw a straw -- you just start running away as fast as you can.

Steve Geddes |

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I am serious. Assuming Monty's behavior is independent of the correctness of my initial choice, I believe that there is no advantage in switching my choice once he shows me an empty door.

I wish I could explain as clearly and elegantly as Steve Geddes explained the straw problem.

You should switch. Try this:

there are three equally likely options, presuming you've chosen door one:

D1 has the prize and he opens either D2 or D3 (you shouldn't switch).

D2 has the prize and he shows you D3 (you should switch)

D3 has the prize and he shows you D2 (you should switch).

They're indistinguishable from your perspective and in two of them, you're better off switching whereas in only one are you better off sticking with your original choice.

Steve Geddes |

1 person marked this as a favorite. |

In a sense (though this isn't strictly correct terminology), his behaviour isn't really independent - if you've picked the right door, he has a choice to make. If you've picked one of the wrong doors he has no choice as to which door to show you.

David knott 242 |

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Of course, the Monty Hall problem assumes that Monty Hall always opens a door. If he has the option to not open a door, then the answer changes according to the criteria he uses to decide whether to open a door. To take an extreme case -- if he only opens a door if you made the correct choice, then sticking with your original choice would be the correct choice.

But if Monty always opens a door after you pick one, then changing is by far the best strategy.

Steve Geddes |

It's a fun mental exercise to work through. There are so many different arguments for why you should switch, but one's intuition keeps asserting itself and its hard to remain convinced.

GreyWolfLord |

Oh, I understand the thing, but its still wonky.

Here's something else I think could be equally as wonky.

Once the first door was revealed...yes, your chances are greater of you winning if you pick the other door...

but I believe they are also worse...

You also stand a better chance of losing as well?

Or is that a false assumption?

Berik |

It is a false assumption, I'm afraid. The probability has to total 1, so if your chance to win has gone up your chance to lose has to go up as well.

Personally I sometimes think it's clearer to think back to the difference between perfect knowledge and no knowledge on the part of Monty for understanding the solution. If I start at the beginning with three doors (2 goats, 1 car), then when I pick one it's pretty clear that my chance of being right is 1/3 regardless of whether Monty knows where the car is.

Now if Monty has to get rid of one door, but *doesn't* know where the car is then there's a 1 in 3 chance that the winning door was discarded, a 1 in 3 chance Monty has the winning door, and a 1 in 3 chance that my door is the winning one. Since there's no way to get at the 'discarded' door, then my chances of winning are equal whether I stick with my original door or move to Monty's.

But in the problem Monty has perfect knowledge and the rules say that he *must* discard a door with a goat. Monty's knowledge doesn't do anything to change my chance of picking the right door in the first place, that's still 1 in 3. But there's zero chance for Monty to discard the door with the car, the rules say he always discards a door with a goat. There must be a 2 in 3 chance that Monty has the door with the car, hence if I can switch then I should.

Hitdice |

That's always been my question: Does Monty ever discard the door I've picked? If not, wasn't my choice always 1 out of 2? (Three statisticians are hunting in the woods and they see a deer. The first one shoots and missed by ten feet on the left. The second shoots and misses ten feet on the right. The third one jumps up and screams "I hit it, I hit it!")

thejeff |

That's always been my question: Does Monty ever discard the door I've picked? If not, wasn't my choice always 1 out of 2? (Three statisticians are hunting in the woods and they see a deer. The first one shoots and missed by ten feet on the left. The second shoots and misses ten feet on the right. The third one jumps up and screams "I hit it, I hit it!")

He never opens the door you picked. He opens one of the other two. The key point that adds information and changes the odds, is that he never opens the winning door.

There is a 1/3 chance you picked the right door the first time, since you were picking out of 3 with no further information. Once he shows you that one of the other doors isn't right, you now know the other door is more likely.Hitdice |

Hitdice wrote:That's always been my question: Does Monty ever discard the door I've picked? If not, wasn't my choice always 1 out of 2? (Three statisticians are hunting in the woods and they see a deer. The first one shoots and missed by ten feet on the left. The second shoots and misses ten feet on the right. The third one jumps up and screams "I hit it, I hit it!")He never opens the door you picked. He opens one of the other two. The key point that adds information and changes the odds, is that he never opens the winning door.

There is a 1/3 chance you picked the right door the first time, since you were picking out of 3 with no further information. Once he shows you that one of the other doors isn't right, you now know the other door is more likely.

I'm not trolling or being snarky, I'm really asking: doesn't the door you choose to begin with get just as more likely as the the one you didn't choose? Assuming that the location of the prize is set at the beginning, rather than decided after Monty's discard.

thejeff |

thejeff wrote:I'm not trolling or being snarky, I'm really asking: doesn't the door you choose to begin with get just as more likely as the the one you didn't choose? Assuming that the location of the prize is set at the beginning, rather than decided after Monty's discard.Hitdice wrote:He never opens the door you picked. He opens one of the other two. The key point that adds information and changes the odds, is that he never opens the winning door.

There is a 1/3 chance you picked the right door the first time, since you were picking out of 3 with no further information. Once he shows you that one of the other doors isn't right, you now know the other door is more likely.

No. Each door is equally likely to start with. You chose one, so 1/3 chance of being right.

There were two doors you didn't choose. 2/3 chance it's in one of those.Then, Monty opens one of those 2 doors. He knows where the prize is and doesn't reveal it.

You still only have a 1/3 chance of having been right in your first choice, but if you were wrong, he has now told you which of the other doors it is, meaning that if your first choice was wrong and you switch, you're guaranteed to be right.

Another way of looking at it, that works out exactly the same, since he shows you what's behind one of the doors: You make your first pick and then are always offered the choice to keep your first door or open both of the other two.

BigDTBone |

thejeff wrote:I'm not trolling or being snarky, I'm really asking: doesn't the door you choose to begin with get just as more likely as the the one you didn't choose? Assuming that the location of the prize is set at the beginning, rather than decided after Monty's discard.Hitdice wrote:

You have three choices [1] [2] [3]

For the sake of argument, let's say you choose door 1. So:

[1] is 1/3 likely to be the winner and

[2] [3] together are collectively 2/3 likely to be the winner.

You're choice is now locked odds because it is removed from the game. Under no condition will Monty reveal your curtain before the end of the game. His knowledge of what is behind your door has no impact on the odds. Your original choice will have 1/3 chance of being correct no matter what happens next.

Now you and Monty are looking at [2] [3]. You know that the chances of one of them being the winner is 2/3. Monty now uses his superior knowledge of the circumstances to reveal that [2] has a pile of garbage behind it. Monty never chooses the winning door to reveal. His knowledge of the trial circumstances allows him to be 100% certain that he will always reveal a loser.

So [2] [3] collectively have 2/3 chance to win, you now know that [2] has a 0/3 chance to win because Monty has shown you it is garbage. But the two doors together still have a 2/3 chance to win. So 2/3 - 0/3 = odds of [3] winning.

So you know that [3] has a 2/3 chance and your original door has the other 1/3 chance.

Krensky |

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The odds are highly dependent on "Monty's" actions. The Marilyn vos Savant (not technically hers, but she made it famous in Parade) answer to switch doors for the best chance is predicated on several assumtions.

The host must always open a door that was not picked by the contestant.

The host must always open a door to reveal a goat and never the car.

The host must always offer the chance to switch between the originally chosen door and the remaining closed door.

The location of the car is random.

The contestant's initial choice is random.

The host will randomly choose between the two goats if the contestant chooses the car initially.

Changing any of these changes the probability.

Interestingly, none of those rules applied to the actual Monty Hall who . Not even the setup of the game. Plus the contestant wasn't playing against probability, he was playing against Monty Hall, and Monty wasn't playing by the same rules or even the same goals as the contestant. The contestant wants the big prize. Monty wanted drama.

Monty Hall explains why Marilyn vos Savant is right and wrong.

As for Let's Make a Deal, the Monty Hall problem never really came up because that setup isn't how the Big Deal game worked anyway.

BigDTBone |

1 person marked this as a favorite. |

If I flip a coin and it comes up Heads 5 times in a row, should I next bet on Tails? It's due.

5 heads in a row has a 3.125% chance of happening. Heads 6 times in a row has a 1.5625% chance to happen. You'll notice that the later is exactly 1/2 the former. This is because the 6th flip still has a 50/50 chance. That is the nature of non-dependent statistical tests.

thejeff |

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Electric Wizard wrote:If I flip a coin and it comes up Heads 5 times in a row, should I next bet on Tails? It's due.5 heads in a row has a 3.125% chance of happening. Heads 6 times in a row has a 1.5625% chance to happen. You'll notice that the later is exactly 1/2 the former. This is because the 6th flip still has a 50/50 chance. That is the nature of non-dependent statistical tests.

I'm sticking with my answer.