One of two independent flip sequences reaches two heads simultaneously. What is the distribution of others tosses.

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Two people start flipping fair coins. We wait for one of them to reach two heads in a row. When that happens, we look at the last two tosses of the other guy. The other guy might have HH, HT, TH and TT. Naively, we might think all of these four are equally likely. However, that would mean there is a 25% chance the two sequences would reach two heads simultaneously. Now that seems high and as a matter of fact is. If you look at the code here (method named get_loser_state_prob at the very end) -

https://github.com/ryu577/stochproc/blob/master/stochproc/competitivecointoss/simulation.py

you will see the distribution actually looks like this -

HH: 0.12
HT: 0.24
TH: 0.32
TT: 0.32

It's completely unintuitive to me why HT and TH should have different probabilities and the distribution in general is something I can't wrap my mind around. Can someone explain why we should expect -

1) HH should be the lowest

2) HT should be lower than TH

3) HT and TT should be roughly the same (since it's simulation, can't tell if they are exactly the same but certainly very close).

asked 4 hours ago

Rohit Pandey

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4
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favorite

https://github.com/ryu577/stochproc/blob/master/stochproc/competitivecointoss/simulation.py

you will see the distribution actually looks like this -

HH: 0.12
HT: 0.24
TH: 0.32
TT: 0.32

1) HH should be the lowest

2) HT should be lower than TH

3) HT and TT should be roughly the same (since it's simulation, can't tell if they are exactly the same but certainly very close).

asked 4 hours ago

Rohit Pandey

851818

add a commentÂ |Â

up vote
4
down vote

favorite

https://github.com/ryu577/stochproc/blob/master/stochproc/competitivecointoss/simulation.py

you will see the distribution actually looks like this -

HH: 0.12
HT: 0.24
TH: 0.32
TT: 0.32

1) HH should be the lowest

2) HT should be lower than TH

3) HT and TT should be roughly the same (since it's simulation, can't tell if they are exactly the same but certainly very close).

asked 4 hours ago

Rohit Pandey

851818

https://github.com/ryu577/stochproc/blob/master/stochproc/competitivecointoss/simulation.py

you will see the distribution actually looks like this -

HH: 0.12
HT: 0.24
TH: 0.32
TT: 0.32

1) HH should be the lowest

2) HT should be lower than TH

3) HT and TT should be roughly the same (since it's simulation, can't tell if they are exactly the same but certainly very close).

probability probability-theory

asked 4 hours ago

Rohit Pandey

851818

asked 4 hours ago

Rohit Pandey

851818

asked 4 hours ago

Rohit Pandey

851818

asked 4 hours ago

Rohit Pandey

851818

asked 4 hours ago

Rohit Pandey

851818

add a commentÂ |Â

2 Answers
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up vote
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So I actually double checked this because it seemed strange to me as well, but it is indeed true.

 result <- c()

for(iter in 1:10000)
 temp <- TRUE
 x1 <- rep(0,2)
 x2 <- rep(0,2)
 while(temp)
 x1[2] <- x1[1]; x2[2] <- x2[1]
 x1[1] <- rbinom(1,1,0.5); x2[1] <- rbinom(1,1,0.5) 

 if(sum(x1)==2)
 result <- c(result,sum(x2))
 temp <- FALSE
 else if(sum(x2)==2)
 result <- c(result,sum(x1))
 temp <- FALSE
 
 


 > sum(result==2)/length(result)
 [1] 0.1164 ## Probs of HH
 > sum(result==1)/length(result)
 [1] 0.565 ## Probs of HT or TH
 > sum(result==0)/length(result)
 [1] 0.3186 ## Probs of TT

I think the basic reason for the asymmetry is because the fact the game ends on this round (as opposed to the previous round) means that both players had gotten at least one tails in the previous two rounds prior to the final round (otherwise the game would have ended the previous round).

This means that each player has a higher probability of getting two tails in the final round, since his last two throws prior to the final round must be one of $HT, TH, TT$ and cannot be $HH$.

Let $X$ be the last two throws of the loser before the final round, $Y$ be his final throw, and $Z$ his last two throws after the final round.

Then

$P(Z = HH) = P(X = TH, Y=H)$

Since no other combination gives him two heads at the end. But now consider the probability his last two throws are tails

$P(Z = TT) = P(X=TT, Y=T) + P(X = HT,Y=T)$

Similarly

$P(Z = HT) = P(X=TH, Y=T)$

$P(Z = TH) = P(X=HT, Y=H) + P(X=TT,Y=H)$

Hence there's a clear asymmetry between the probabilities of the final throw caused by the fact he must have at least one tails prior to the final throw.

Notice how the simulation show $P(TH) > P(HT)$, and the above has two summands for $TH$ but only one for $HT$!

answered 2 hours ago

Xiaomi

1555

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up vote
1
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In this answer a way to find the probabilities.

Let $p$ denote the probability at start position, let $q$ denote the probability under condition that in the first round a tail and a head are thrown and let $r$ denote the probability under condition that in the first round two heads are thrown.

If it concerns the probability on HH then:

$p=frac14p+frac12q+frac14r$

$q=frac14p+frac14q$

$r=frac14+frac14p$

Solving this we find $p_HH=p=0.12$.

If it concerns the probability on TH then:

$p=frac14p+frac12q+frac14r$

$q=frac14+frac14p+frac14q$

$r=frac14p$

Solving this we find $p_TH=p=0.32$.

If it concerns the probability on HT then:

$p=frac14p+frac12q+frac14r$

$q=frac14p+frac14q$

$r=frac12+frac14p$

Solving this we find $p_HT=p=0.24$.

If it concerns the probability on TT then:

$p=frac14p+frac12q+frac14r$

$q=frac14+frac14p+frac14q$

$r=frac14p$

Solving this we find $p_TT=p=0.32$.

answered 22 mins ago

drhab

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2 Answers
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active

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2 Answers
2

active

oldest

votes

up vote
3
down vote

So I actually double checked this because it seemed strange to me as well, but it is indeed true.

 result <- c()

for(iter in 1:10000)
 temp <- TRUE
 x1 <- rep(0,2)
 x2 <- rep(0,2)
 while(temp)
 x1[2] <- x1[1]; x2[2] <- x2[1]
 x1[1] <- rbinom(1,1,0.5); x2[1] <- rbinom(1,1,0.5) 

 if(sum(x1)==2)
 result <- c(result,sum(x2))
 temp <- FALSE
 else if(sum(x2)==2)
 result <- c(result,sum(x1))
 temp <- FALSE
 
 


 > sum(result==2)/length(result)
 [1] 0.1164 ## Probs of HH
 > sum(result==1)/length(result)
 [1] 0.565 ## Probs of HT or TH
 > sum(result==0)/length(result)
 [1] 0.3186 ## Probs of TT

This means that each player has a higher probability of getting two tails in the final round, since his last two throws prior to the final round must be one of $HT, TH, TT$ and cannot be $HH$.

Let $X$ be the last two throws of the loser before the final round, $Y$ be his final throw, and $Z$ his last two throws after the final round.

Then

$P(Z = HH) = P(X = TH, Y=H)$

Since no other combination gives him two heads at the end. But now consider the probability his last two throws are tails

$P(Z = TT) = P(X=TT, Y=T) + P(X = HT,Y=T)$

Similarly

$P(Z = HT) = P(X=TH, Y=T)$

$P(Z = TH) = P(X=HT, Y=H) + P(X=TT,Y=H)$

Hence there's a clear asymmetry between the probabilities of the final throw caused by the fact he must have at least one tails prior to the final throw.

Notice how the simulation show $P(TH) > P(HT)$, and the above has two summands for $TH$ but only one for $HT$!

answered 2 hours ago

Xiaomi

1555

New contributor

add a commentÂ |Â

up vote
3
down vote

So I actually double checked this because it seemed strange to me as well, but it is indeed true.

 result <- c()

for(iter in 1:10000)
 temp <- TRUE
 x1 <- rep(0,2)
 x2 <- rep(0,2)
 while(temp)
 x1[2] <- x1[1]; x2[2] <- x2[1]
 x1[1] <- rbinom(1,1,0.5); x2[1] <- rbinom(1,1,0.5) 

 if(sum(x1)==2)
 result <- c(result,sum(x2))
 temp <- FALSE
 else if(sum(x2)==2)
 result <- c(result,sum(x1))
 temp <- FALSE
 
 


 > sum(result==2)/length(result)
 [1] 0.1164 ## Probs of HH
 > sum(result==1)/length(result)
 [1] 0.565 ## Probs of HT or TH
 > sum(result==0)/length(result)
 [1] 0.3186 ## Probs of TT

This means that each player has a higher probability of getting two tails in the final round, since his last two throws prior to the final round must be one of $HT, TH, TT$ and cannot be $HH$.

Let $X$ be the last two throws of the loser before the final round, $Y$ be his final throw, and $Z$ his last two throws after the final round.

Then

$P(Z = HH) = P(X = TH, Y=H)$

Since no other combination gives him two heads at the end. But now consider the probability his last two throws are tails

$P(Z = TT) = P(X=TT, Y=T) + P(X = HT,Y=T)$

Similarly

$P(Z = HT) = P(X=TH, Y=T)$

$P(Z = TH) = P(X=HT, Y=H) + P(X=TT,Y=H)$

Hence there's a clear asymmetry between the probabilities of the final throw caused by the fact he must have at least one tails prior to the final throw.

Notice how the simulation show $P(TH) > P(HT)$, and the above has two summands for $TH$ but only one for $HT$!

answered 2 hours ago

Xiaomi

1555

New contributor

add a commentÂ |Â

up vote
3
down vote

So I actually double checked this because it seemed strange to me as well, but it is indeed true.

 result <- c()

for(iter in 1:10000)
 temp <- TRUE
 x1 <- rep(0,2)
 x2 <- rep(0,2)
 while(temp)
 x1[2] <- x1[1]; x2[2] <- x2[1]
 x1[1] <- rbinom(1,1,0.5); x2[1] <- rbinom(1,1,0.5) 

 if(sum(x1)==2)
 result <- c(result,sum(x2))
 temp <- FALSE
 else if(sum(x2)==2)
 result <- c(result,sum(x1))
 temp <- FALSE
 
 


 > sum(result==2)/length(result)
 [1] 0.1164 ## Probs of HH
 > sum(result==1)/length(result)
 [1] 0.565 ## Probs of HT or TH
 > sum(result==0)/length(result)
 [1] 0.3186 ## Probs of TT

This means that each player has a higher probability of getting two tails in the final round, since his last two throws prior to the final round must be one of $HT, TH, TT$ and cannot be $HH$.

Let $X$ be the last two throws of the loser before the final round, $Y$ be his final throw, and $Z$ his last two throws after the final round.

Then

$P(Z = HH) = P(X = TH, Y=H)$

Since no other combination gives him two heads at the end. But now consider the probability his last two throws are tails

$P(Z = TT) = P(X=TT, Y=T) + P(X = HT,Y=T)$

Similarly

$P(Z = HT) = P(X=TH, Y=T)$

$P(Z = TH) = P(X=HT, Y=H) + P(X=TT,Y=H)$

Hence there's a clear asymmetry between the probabilities of the final throw caused by the fact he must have at least one tails prior to the final throw.

Notice how the simulation show $P(TH) > P(HT)$, and the above has two summands for $TH$ but only one for $HT$!

answered 2 hours ago

Xiaomi

1555

New contributor

So I actually double checked this because it seemed strange to me as well, but it is indeed true.

 result <- c()

for(iter in 1:10000)
 temp <- TRUE
 x1 <- rep(0,2)
 x2 <- rep(0,2)
 while(temp)
 x1[2] <- x1[1]; x2[2] <- x2[1]
 x1[1] <- rbinom(1,1,0.5); x2[1] <- rbinom(1,1,0.5) 

 if(sum(x1)==2)
 result <- c(result,sum(x2))
 temp <- FALSE
 else if(sum(x2)==2)
 result <- c(result,sum(x1))
 temp <- FALSE
 
 


 > sum(result==2)/length(result)
 [1] 0.1164 ## Probs of HH
 > sum(result==1)/length(result)
 [1] 0.565 ## Probs of HT or TH
 > sum(result==0)/length(result)
 [1] 0.3186 ## Probs of TT

This means that each player has a higher probability of getting two tails in the final round, since his last two throws prior to the final round must be one of $HT, TH, TT$ and cannot be $HH$.

Let $X$ be the last two throws of the loser before the final round, $Y$ be his final throw, and $Z$ his last two throws after the final round.

Then

$P(Z = HH) = P(X = TH, Y=H)$

Since no other combination gives him two heads at the end. But now consider the probability his last two throws are tails

$P(Z = TT) = P(X=TT, Y=T) + P(X = HT,Y=T)$

Similarly

$P(Z = HT) = P(X=TH, Y=T)$

$P(Z = TH) = P(X=HT, Y=H) + P(X=TT,Y=H)$

Hence there's a clear asymmetry between the probabilities of the final throw caused by the fact he must have at least one tails prior to the final throw.

Notice how the simulation show $P(TH) > P(HT)$, and the above has two summands for $TH$ but only one for $HT$!

answered 2 hours ago

Xiaomi

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New contributor

answered 2 hours ago

Xiaomi

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up vote
1
down vote

In this answer a way to find the probabilities.

If it concerns the probability on HH then:

$p=frac14p+frac12q+frac14r$

$q=frac14p+frac14q$

$r=frac14+frac14p$

Solving this we find $p_HH=p=0.12$.

If it concerns the probability on TH then:

$p=frac14p+frac12q+frac14r$

$q=frac14+frac14p+frac14q$

$r=frac14p$

Solving this we find $p_TH=p=0.32$.

If it concerns the probability on HT then:

$p=frac14p+frac12q+frac14r$

$q=frac14p+frac14q$

$r=frac12+frac14p$

Solving this we find $p_HT=p=0.24$.

If it concerns the probability on TT then:

$p=frac14p+frac12q+frac14r$

$q=frac14+frac14p+frac14q$

$r=frac14p$

Solving this we find $p_TT=p=0.32$.

answered 22 mins ago

drhab

89.8k541123

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up vote
1
down vote

In this answer a way to find the probabilities.

If it concerns the probability on HH then:

$p=frac14p+frac12q+frac14r$

$q=frac14p+frac14q$

$r=frac14+frac14p$

Solving this we find $p_HH=p=0.12$.

If it concerns the probability on TH then:

$p=frac14p+frac12q+frac14r$

$q=frac14+frac14p+frac14q$

$r=frac14p$

Solving this we find $p_TH=p=0.32$.

If it concerns the probability on HT then:

$p=frac14p+frac12q+frac14r$

$q=frac14p+frac14q$

$r=frac12+frac14p$

Solving this we find $p_HT=p=0.24$.

If it concerns the probability on TT then:

$p=frac14p+frac12q+frac14r$

$q=frac14+frac14p+frac14q$

$r=frac14p$

Solving this we find $p_TT=p=0.32$.

answered 22 mins ago

drhab

89.8k541123

add a commentÂ |Â

up vote
1
down vote

In this answer a way to find the probabilities.

If it concerns the probability on HH then:

$p=frac14p+frac12q+frac14r$

$q=frac14p+frac14q$

$r=frac14+frac14p$

Solving this we find $p_HH=p=0.12$.

If it concerns the probability on TH then:

$p=frac14p+frac12q+frac14r$

$q=frac14+frac14p+frac14q$

$r=frac14p$

Solving this we find $p_TH=p=0.32$.

If it concerns the probability on HT then:

$p=frac14p+frac12q+frac14r$

$q=frac14p+frac14q$

$r=frac12+frac14p$

Solving this we find $p_HT=p=0.24$.

If it concerns the probability on TT then:

$p=frac14p+frac12q+frac14r$

$q=frac14+frac14p+frac14q$

$r=frac14p$

Solving this we find $p_TT=p=0.32$.

answered 22 mins ago

drhab

89.8k541123

In this answer a way to find the probabilities.

If it concerns the probability on HH then:

$p=frac14p+frac12q+frac14r$

$q=frac14p+frac14q$

$r=frac14+frac14p$

Solving this we find $p_HH=p=0.12$.

If it concerns the probability on TH then:

$p=frac14p+frac12q+frac14r$

$q=frac14+frac14p+frac14q$

$r=frac14p$

Solving this we find $p_TH=p=0.32$.

If it concerns the probability on HT then:

$p=frac14p+frac12q+frac14r$

$q=frac14p+frac14q$

$r=frac12+frac14p$

Solving this we find $p_HT=p=0.24$.

If it concerns the probability on TT then:

$p=frac14p+frac12q+frac14r$

$q=frac14+frac14p+frac14q$

$r=frac14p$

Solving this we find $p_TT=p=0.32$.

answered 22 mins ago

drhab

89.8k541123

answered 22 mins ago

drhab

89.8k541123

answered 22 mins ago

drhab

89.8k541123

answered 22 mins ago

drhab

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